Evaluation of a Subclass \( K_{p,r} =\int x^p \left ( 1+ x^2\right )^r dx \) of the Binomial Integral

1. Preliminary

The traditional analytical approach for evaluating integrals relies on a toolkit of substitution and reduction formulas through integration by parts. The results of such manual derivations, accumulated and compiled over centuries, resulted in a large number of formulas for various integrals and are now available in encyclopedic handbooks [1,8], which serve as the authoritative reference for practitioners seeking a specific solution.

In the meantime, the modern algorithmic approach is embodied by symbolic computation packages such as Maple [9] and Mathematica [13], which implement sophisticated procedures, for instance the Risch-Norman algorithm, to find antiderivatives automatically [7]. While immensely powerful, these algorithms operate as black boxes, but offer limited insight into the analytical relationships between solutions, mathematical structure of the solutions, or a methodology for manual derivation.

Among a large number of integrals, the evaluation of binomial integrals of the form $\int x^p{\left(a+b{x}^n\right)}^{r}dx$ is a classical problem in integral calculus with importance in physics and engineering [3,4]. The theoretical limits of this problem were established by Chebyshev, whose theorem delineates the specific conditions under which these integrals can be expressed in terms of elementary functions [5]. The class of integrals

$$\begin{array}{ccc}\hfill K_{p,r}& =& \int x^p{\left(1+x^2\right)}^{r}dx\hfill \end{array}$$

(1)

is equivalent to a subclass of the binomial integrals $\int x^p{\left(a+b{x}^n\right)}^{r}dx$ for $ab>0$ and constitutes a classical yet nontrivial problem in integral calculus, particularly when the parameters p and r extend beyond simple non-negative integers. Such cases arise naturally in various applications: for instance, the integral $K_{2,{\displaystyle \frac{1}{2}}}$ appears in the computation of surface areas of solids of revolution [17]. From a more general perspective, integrals of this type can be expressed in terms of Gauss hypergeometric functions [2,12]. In modern mathematical physics, related integrals with integer and half-integer parameters are frequently analyzed using harmonic polylogarithms [10] and special-function expansions, particularly in the study of Bessel-type functions and Feynman integrals [11,14,15]. Some of the formulas in [8, 2.27, pp. 99-101]] can be interpreted as special cases of $K_{p,r}$.

Despite the breadth of existing results [1,6,8], several limitations remain apparent in both pedagogical and research-oriented treatments of the integral $K_{p,r}$. For instance, standard substitution techniques frequently lead to case-by-case calculations, which tend to obscure the underlying analytical structure of the solutions [17]. In addition, reduction formulas obtained through integration by parts often encounter ‘walls’ that interrupt recursive evaluation, requiring problem-dependent initial conditions. Furthermore, although closed-form expressions are available for many individual cases, these results are typically scattered across tables or derived in isolation [1,8] with little emphasis on a unified analytical organization on the parameter space $(p,r)$.

In this letter, we develop a unified analytical and recursive framework for the systematic evaluation of $K_{p,r}$ The approach in this letter is based on a detailed analysis of the parameter plane $(p,r)$. We identify five fundamental lines $p=1$, $p=-1$, $r=0$, $r=-1$, and $p+2r+3=0$ along which the integral admits direct closed-form evaluation. These cases provide a complete and natural set of initial conditions that anchor the recursive structure of the problem. Building upon these initial conditions, we derive two general recursion formulas that enable the evaluation of $K_{p,r}$ at any admissible point in the parameter plane while systematically bypassing the recursion barriers encountered, mostly on the line $p+2r+1=0$, in conventional approaches.

The main contributions of this work can be summarized as follows:

• The parameter plane of $(p,r)$ is systematically analyzed to provide a unified analytical framework for the categorization and derivation of $K_{p,r}$. Five fundamental lines are identified upon which the integral $K_{p,r}$ can be evaluated directly to serve as the foundational initial conditions.
• Two general recursion formulas are derived that enable the methodical generation of solutions at any admissible point $(p,r)$.
• By identifying a complete and sufficient set of initial conditions, traditional recursion walls are systematically bypassed, thus ensuring recursive stability.
• This work culminates in a compendium of explicit, closed-form formulas, serving as a valuable reference.

2. Evaluation of ${\mathit{K}}_{\mathit{p},\mathit{r}}$ with Basic Techniques

In this section we first apply basic techniques of integral for the evaluation of the integral $K_{p,r}$. Throughout the letter, except when necessary, integration constants are omitted for brevity.

It is simple to have $K_{p,r}=\int {\displaystyle \sum _{q=0}^r}\left({\displaystyle \genfrac{}{}{0pt}{}{r}{q}}\right)x^{p+2q}dx$ or

$$\begin{array}{ccc}\hfill K_{p,r}& =& \left({\displaystyle \genfrac{}{}{0pt}{}{r}{r_1}}\right)I_{A_{0r}}\left(r_1\right)\ln \left|x\right|+{\displaystyle \sum _{q=0,q\ne r_1}^r}\left({\displaystyle \genfrac{}{}{0pt}{}{r}{q}}\right){\displaystyle \frac{x^{p+2q+1}}{p+2q+1}}\hfill \end{array}$$

(2)

when p is an integer and r is a non-negative integer, where $A_{0r}=\{0,1,\cdots ,r\}$,

$$\begin{array}{ccc}\hfill r_1& =& -{\displaystyle \frac{p+1}{2}},\hfill \end{array}$$

(3)

and $I_A\left(x\right)$ is the indicator function of x defined by $I_A\left(x\right)=1$ if $x\in A$ and $I_A\left(x\right)=0$ if $x\notin A$.

2.1. Evaluation via Substitution

We start with the most basic method for the evaluation of the integral $K_{p,r}$.

2.1.1. Substitution 1

With $x=tan\theta$, we have

$$\begin{array}{ccc}\hfill K_{p,r}& =& \int {\displaystyle \frac{{sin}^p\theta }{{cos}^{p+2r+2}\theta }}d\theta .\hfill \end{array}$$

(4)

When ${\displaystyle \frac{p-1}{2}}$ is an integer, (4) can be rewritten as

$$\begin{array}{ccc}\hfill K_{p,r}& =& -\int {\displaystyle \frac{{\left(1-v^2\right)}^{\frac{p-1}{2}}}{v^{p+2r+2}}}dv\hfill \end{array}$$

(5)

with $v=cos\theta ={\displaystyle \frac{1}{\sqrt{1+x^2}}}$. When ${\displaystyle \frac{p-1}{2}}$ is a non-positive integer, (5) can be expressed as

$$\begin{array}{ccc}\hfill K_{p,r}& =& -\int {\displaystyle \frac{v^{-p-2r-2}}{{\left(1-v\right)}^{-\frac{p-1}{2}}{\left(1+v\right)}^{-\frac{p-1}{2}}}}dv,\hfill \end{array}$$

(6)

for which we can use partial fraction expansion if $2r$ is an integer. On the other hand, (5) will produce the formula $K_{p,r}=-\int {\displaystyle \sum _{q=0}^m}\left({\displaystyle \genfrac{}{}{0pt}{}{m}{q}}\right){\left(-v^2\right)}^q{v}^{-(2m+2r+3)}dv$ or

$$\begin{array}{ccc}\hfill K_{p,r}& =& -{\displaystyle \sum _{q=0,q\ne C_{30}}^m}{\displaystyle \frac{{(-1)}^q\left(\genfrac{}{}{0pt}{}{m}{q}\right)}{2\left(q-C_{30}\right)}}{\left(1+x^2\right)}^{C_{30}-q}+{\displaystyle \frac{1}{2}}{C}_{31}\ln \left(1+x^2\right)\hfill \end{array}$$

(7)

when $m={\displaystyle \frac{p-1}{2}}$ is a non-negative integer, where $C_{30}=m+r+1={\displaystyle \frac{1}{2}}(p+2r+1)$ and $C_{31}=I_{A_{0m}}\left(C_{30}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{m}{C_{30}}}\right){(-1)}^{C_{30}}$ with $A_{0m}=\{0,1,\cdots ,m\}$.

When $p=0$, we get $K_{0,r}=\int {\displaystyle \frac{d\theta }{{cos}^{2r+2}\theta }}$ from (4). Thus, $K_{0,-{\displaystyle \frac{1}{2}}}=\int {\displaystyle \frac{d\theta }{cos\theta }}=\ln |tan\theta +sec\theta |$ or

$$\begin{array}{ccc}\hfill K_{0,-{\displaystyle \frac{1}{2}}}& =& \ln \left|x+\sqrt{1+x^2}\right|,\hfill \end{array}$$

(8)

$K_{0,-1}=\int d\theta =\theta$ or

$$\begin{array}{ccc}\hfill K_{0,-1}& =& {tan}^{-1}x,\hfill \end{array}$$

(9)

and $K_{0,-{\displaystyle \frac{3}{2}}}=\int cos\theta d\theta =sin\theta$ or

$$\begin{array}{ccc}\hfill K_{0,-{\displaystyle \frac{3}{2}}}& =& {\displaystyle \frac{x}{\sqrt{1+x^2}}}\hfill \end{array}$$

(10)

for $r=-{\displaystyle \frac{1}{2}}$, $-1$, and $-{\displaystyle \frac{3}{2}}$, respectively.

and $C_{31}=I_{\left\{0\right\}}\left(0\right)\left({\displaystyle \genfrac{}{}{0pt}{}{0}{0}}\right){(-1)}^0=1$. Then we obtain $K_{1,-1}=-{\displaystyle \sum _{q=0,q\ne 0}^0}{\displaystyle \frac{{(-1)}^q\left(\genfrac{}{}{0pt}{}{0}{q}\right)}{2\left(q-0\right)}}{\left(1+x^2\right)}^{0-q}+{\displaystyle \frac{1}{2}}\ln \left(1+x^2\right)$ or

$$\begin{array}{ccc}\hfill K_{1,-1}& =& {\displaystyle \frac{1}{2}}\ln \left(1+x^2\right)\hfill \end{array}$$

(11)

from (7). Similarly, we get $K_{1,1}=-{\displaystyle \sum _{q=0,q\ne 2}^0}{\displaystyle \frac{{(-1)}^q\left(\genfrac{}{}{0pt}{}{0}{q}\right)}{2\left(q-2\right)}}{\left(1+x^2\right)}^{2-q}$ or

$$\begin{array}{ccc}\hfill K_{1,1}& =& {\displaystyle \frac{1}{4}}{\left(1+x^2\right)}^2\hfill \end{array}$$

(12)

from (7) for $(p,r)=(1,1)$ with $m={\displaystyle \frac{p-1}{2}}=0$, $C_{30}=m+r+1=2$, and $C_{31}=0$ from $I_{\left\{0\right\}}\left(2\right)=0$.

When $(p,r)=(-1,-1)$, we get $2m+2r+3=-1$, and thus $K_{-1,-1}=-\int {\displaystyle \frac{vdv}{1-v^2}}=\int \left({\displaystyle \frac{-\frac{1}{2}}{1-v}}+{\displaystyle \frac{\frac{1}{2}}{1+v}}\right)dv={\displaystyle \frac{1}{2}}\ln |1-v|+{\displaystyle \frac{1}{2}}\ln |1+v|$ or

$$\begin{array}{ccc}\hfill K_{-1,-1}& =& {\displaystyle \frac{1}{2}}\ln {\displaystyle \frac{x^2}{1+x^2}}\hfill \end{array}$$

(13)

using (6) by noting that $v=cos\theta ={\displaystyle \frac{1}{\sqrt{1+x^2}}}$.

2.1.2. Substitution 2

With $1+x^2=t$, we get

$$\begin{array}{ccc}\hfill K_{p,r}& =& {\displaystyle \frac{1}{2}}\int {(t-1)}^{{\displaystyle \frac{p-1}{2}}}{t}^{r}dt,\hfill \end{array}$$

(14)

which will produce the formula $K_{p,r}={\displaystyle \frac{1}{2}}\int {\displaystyle \sum _{q=0}^m}\left({\displaystyle \genfrac{}{}{0pt}{}{m}{q}}\right){(-1)}^{m-q}{t}^{q+r}dt$ or

$$\begin{array}{ccc}\hfill K_{p,r}& =& C_{22}\ln \left(1+x^2\right)+{\displaystyle \frac{1}{2}}{\displaystyle \sum _{q=0,q\ne C_{20}}^m}{\displaystyle \frac{{(-1)}^{m-q}\left(\genfrac{}{}{0pt}{}{m}{q}\right)}{q-C_{20}}}{\left(1+x^2\right)}^{q-C_{20}}\hfill \end{array}$$

(15)

when $m={\displaystyle \frac{p-1}{2}}$ is a non-negative integer, where $C_{22}={\displaystyle \frac{1}{2}}{I}_{A_{0m}}\left(C_{20}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{m}{C_{20}}}\right){(-1)}^{m-C_{20}}$ and $C_{20}=-r-1$.

We get $K_{3,-2}={\displaystyle \frac{1}{2}}\int {\displaystyle \frac{t-1}{t^2}}dt={\displaystyle \frac{1}{2}}\int \left({\displaystyle \frac{1}{t}}-{\displaystyle \frac{1}{t^2}}\right)dt$: that is,

$$\begin{array}{ccc}\hfill K_{3,-2}& =& {\displaystyle \frac{1}{2}}\left\{\ln \left(1+x^2\right)+{\displaystyle \frac{1}{1+x^2}}\right\}\hfill \end{array}$$

(16)

from (14) or (15).

When $(p,r)=(1,-1)$, we get $m={\displaystyle \frac{p-1}{2}}=0$, $C_{20}=-r-1=0$, $C_{22}={\displaystyle \frac{1}{2}}{I}_{\left\{0\right\}}\left(0\right)\left({\displaystyle \genfrac{}{}{0pt}{}{0}{0}}\right){(-1)}^0={\displaystyle \frac{1}{2}}$, and thus, as obtained already in (11) also, $K_{1,-1}={\displaystyle \frac{1}{2}}\ln \left(1+x^2\right)$ from (15).

When $(p,r)=(1,1)$, we get $m={\displaystyle \frac{p-1}{2}}=0$, $C_{20}=-r-1=-2$, and $C_{22}=0$ because $I_{\left\{0\right\}}(-2)=0$. Then, from (15), we have $K_{1,1}={\displaystyle \frac{1}{2}}{\displaystyle \sum _{q=0}^0}{\displaystyle \frac{{(-1)}^{0-q}\left(\genfrac{}{}{0pt}{}{0}{q}\right)}{q-(-2)}}{\left(1+x^2\right)}^{q-(-2)}={\displaystyle \frac{1}{4}}{\left(1+x^2\right)}^2$ as in (12).

2.1.3. Substitution 3

With $\sqrt{1+x^2}=t$, we get

$$\begin{array}{ccc}\hfill K_{p,r}& =& \int {\left(t^2-1\right)}^{{\displaystyle \frac{p-1}{2}}}{t}^{2r+1}dt,\hfill \end{array}$$

(17)

similar to the second method but more useful especially when r is an integral multiple of ${\displaystyle \frac{1}{2}}$. When $m={\displaystyle \frac{p-1}{2}}$ is a non-negative integer, (17) will produce (15) again. On the other hand, when ${\displaystyle \frac{p-1}{2}}$ is a non-positive integer, (17) can be expressed as

$$\begin{array}{ccc}\hfill K_{p,r}& =& \int {\displaystyle \frac{t^{2r+1}}{{\left(t^2-1\right)}^{-\frac{p-1}{2}}}}dt,\hfill \end{array}$$

(18)

and subsequently evaluated via partial fraction expansion if $2r+1$ is an integer. Note that (17) is almost the same as (14), yet more useful because it can be evaluated via partial fraction expansion when $2r+1$ is an integer thus including the case where r is an integer.

From (17) with $(p,r)=\left(1,-{\displaystyle \frac{1}{2}}\right)$, we get $K_{1,-{\displaystyle \frac{1}{2}}}=\int dt=t$ or

$$\begin{array}{ccc}\hfill K_{1,-{\displaystyle \frac{1}{2}}}& =& \sqrt{1+x^2}.\hfill \end{array}$$

(19)

We get $K_{-1,-{\displaystyle \frac{1}{2}}}=\int {\displaystyle \frac{dt}{t^2-1}}=\int \left({\displaystyle \frac{\frac{1}{2}}{t-1}}+{\displaystyle \frac{-\frac{1}{2}}{t+1}}\right)dt$ or

$$\begin{array}{ccc}\hfill K_{-1,-{\displaystyle \frac{1}{2}}}& =& {\displaystyle \frac{1}{2}}\ln {\displaystyle \frac{\sqrt{1+x^2}-1}{\sqrt{1+x^2}+1}}\hfill \end{array}$$

(20)

from (18).

2.2. Evaluation via Integration by Parts

We address two ways of integration by parts, which will eventually lead us to useful recursion formulas for the evaluation of $K_{p.r}$.

2.2.1. Integration by Parts 1

When $p\ne -1$, with $u^{\prime }=x^p$, $v={\left(1+x^2\right)}^r$, $u={\displaystyle \frac{x^{p+1}}{p+1}}$, and $v^{\prime }=2rx{\left(1+x^2\right)}^{r-1}$, we get

$$\begin{array}{ccc}& & K_{p,r}={\displaystyle \frac{x^{p+1}{\left(1+x^2\right)}^r}{p+1}}-{\displaystyle \frac{2r}{p+1}}\int x^{p+2}{\left(1+x^2\right)}^{r-1}dx.\hfill \end{array}$$

(21)

Letting $r=0$ in (21), we get the result $K_{p,0}=\int x^{p}dx$ or

$$\begin{array}{ccc}\hfill K_{p,0}& =& {\displaystyle \frac{x^{p+1}}{p+1}}\hfill \end{array}$$

(22)

for $p\ne -1$. Obviously, $K_{-1,0}=\int {\displaystyle \frac{dx}{x}}$ or

$$\begin{array}{ccc}\hfill K_{-1,0}& =& \ln \left|x\right|.\hfill \end{array}$$

(23)

Next, (21) can be rewritten as

$$\begin{array}{ccc}& & K_{p,r}={\displaystyle \frac{x^{p+1}}{p+1}}{\left(1+x^2\right)}^r-{\displaystyle \frac{2r}{p+1}}\int \left(1+x^2-1\right)x^p{\left(1+x^2\right)}^{r-1}dx\hfill \\ & & ={\displaystyle \frac{x^{p+1}}{p+1}}{\left(1+x^2\right)}^r-{\displaystyle \frac{2r}{p+1}}{K}_{p,r}+{\displaystyle \frac{2r}{p+1}}{K}_{p,r-1},\hfill \end{array}$$

(24)

from which we get the following two results.

First, when $-{\displaystyle \frac{2r}{p+1}}\ne 1$, we get $\left(1+{\displaystyle \frac{2r}{p+1}}\right)K_{p,r}={\displaystyle \frac{x^{p+1}{\left(1+x^2\right)}^r}{p+1}}+{\displaystyle \frac{2r}{p+1}}{K}_{p,r-1}$, which can be expressed as

$$\begin{array}{ccc}\hfill K_{p,r}& =& {\displaystyle \frac{x^{p+1}{\left(1+x^2\right)}^r}{p+2r+1}}+{\displaystyle \frac{2r}{p+2r+1}}{K}_{p,r-1}\hfill \end{array}$$

(25)

for $p+2r+1\ne 0$ and $p\ne -1$. When $r=0$, (25) is the same as (22). Although we assumed $p\ne -1$ from the starting point of (21), note that (25) holds even for $p=-1$, producing

$$\begin{array}{ccc}\hfill K_{-1,r}& =& {\displaystyle \frac{1}{2r}}{\left(1+x^2\right)}^r+K_{-1,r-1}\hfill \end{array}$$

(26)

for $r\ne 0$. The recursion (26) can of course be obtained alternatively as $K_{-1,r}=\int {\displaystyle \frac{1}{x}}{\left(1+x^2\right)}^{r}dx=\int \left(1+x^2\right){\displaystyle \frac{1}{x}}{\left(1+x^2\right)}^{r-1}dx=\int \left\{x{\left(1+x^2\right)}^{r-1}+{\displaystyle \frac{1}{x}}{\left(1+x^2\right)}^{r-1}\right\}dx=K_{1,r-1}+K_{-1,r-1}={\displaystyle \frac{1}{2r}}{\left(1+x^2\right)}^r+K_{-1,r-1}$, where $K_{1,r-1}=\int x{\left(1+x^2\right)}^{r-1}dx={\displaystyle \frac{1}{2r}}{\left(1+x^2\right)}^r$ is used.

On the other hand, when $-{\displaystyle \frac{2r}{p+1}}=1$ in (24), we get $0={\displaystyle \frac{x^{p+1}{\left(1+x^2\right)}^r}{p+1}}+{\displaystyle \frac{2r}{p+1}}{K}_{p,r-1}$ or

$$\begin{array}{ccc}\hfill K_{p,-{\displaystyle \frac{p+3}{2}}}& =& {\displaystyle \frac{x^{p+1}}{p+1}}{\left(1+x^2\right)}^{-{\displaystyle \frac{p+1}{2}}}\hfill \end{array}$$

(27)

for $p\ne -1$.

When $p=0$ and $r=-{\displaystyle \frac{1}{2}}$, we get $K_{0,-{\displaystyle \frac{1}{2}}}={\displaystyle \frac{x}{\sqrt{1+x^2}}}+K_{2,-{\displaystyle \frac{3}{2}}}$ from (21). Thus, $K_{2,-{\displaystyle \frac{3}{2}}}=K_{0,-{\displaystyle \frac{1}{2}}}-{\displaystyle \frac{x}{\sqrt{1+x^2}}}$ or

$$\begin{array}{ccc}\hfill K_{2,-{\displaystyle \frac{3}{2}}}& =& \ln \left|x+\sqrt{1+x^2}\right|-{\displaystyle \frac{x}{\sqrt{1+x^2}}}\hfill \end{array}$$

(28)

using the results $K_{0,-{\displaystyle \frac{1}{2}}}=\ln \left|x+\sqrt{1+x^2}\right|$ shown in (8).

From (25), we have $K_{2,-{\displaystyle \frac{1}{2}}}={\displaystyle \frac{x^3}{2\sqrt{1+x^2}}}-{\displaystyle \frac{1}{2}}{K}_{2,-{\displaystyle \frac{3}{2}}}$ or

$$\begin{array}{ccc}\hfill K_{2,-{\displaystyle \frac{1}{2}}}& =& {\displaystyle \frac{x^3+x}{2\sqrt{1+x^2}}}-{\displaystyle \frac{1}{2}}\ln \left|x+\sqrt{1+x^2}\right|\hfill \end{array}$$

(29)

with (28) and subsequently $K_{2,{\displaystyle \frac{1}{2}}}={\displaystyle \frac{x^3}{4}}\sqrt{1+x^2}+{\displaystyle \frac{1}{4}}{K}_{2,-{\displaystyle \frac{1}{2}}}$ or

$$\begin{array}{ccc}\hfill K_{2,{\displaystyle \frac{1}{2}}}& =& {\displaystyle \frac{2x^5+3x^3+x}{8\sqrt{1+x^2}}}-{\displaystyle \frac{1}{8}}\ln \left|x+\sqrt{1+x^2}\right|\hfill \end{array}$$

(30)

with (29).

When $r=1$, we easily have $K_{-1,1}=\int \left(x^{-1}+x\right)dx=\ln \left|x\right|+{\displaystyle \frac{x^2}{2}}$ and $K_{-3,1}=\int \left(x^{-1}+x^{-3}\right)dx=\ln \left|x\right|-{\displaystyle \frac{1}{2x^2}}$ for $p=-1$ and $p=-3$, respectively. Then, collecting the results (22) and $K_{p,1}={\displaystyle \frac{x^{p+1}\left(1+x^2\right)}{p+3}}+{\displaystyle \frac{2x^{p+1}}{(p+1)(p+3)}}={\displaystyle \frac{x^{p+1}}{p+1}}+{\displaystyle \frac{x^{p+3}}{p+3}}$ from (25), we have

$$\begin{array}{ccc}\hfill K_{p,1}& =& \left\{\begin{array}{cc}\ln \left|x\right|+{\displaystyle \frac{x^{2(p+2)}}{2(p+2)}},\hfill & p=-1,-3,\hfill \\ {\displaystyle \frac{x^{p+1}}{p+1}}+{\displaystyle \frac{x^{p+3}}{p+3}},\hfill & p\ne -1,-3,\hfill \end{array}\right.\hfill \end{array}$$

(31)

which is the same as that we would obtain from the direct evaluation of $\int x^p\left(1+x^2\right)dx$ apparently.

For $p=\cdots ,-3,-2,-1,0,1,\cdots$, we get

$$\begin{array}{ccc}& ⋮& \hfill \end{array}$$

$$\begin{array}{ccc}\hfill K_{-3,0}& =& -{\displaystyle \frac{x^{-2}}{2}}\left(1+x^2\right),\hfill \end{array}$$

(32)

$$\begin{array}{ccc}\hfill K_{-2,-{\displaystyle \frac{1}{2}}}& =& -x^{-1}{\left(1+x^2\right)}^{{\displaystyle \frac{1}{2}}},\hfill \end{array}$$

(33)

$$\begin{array}{ccc}\hfill \{K_{-1,-1}& =& {\displaystyle \frac{1}{2}}\ln {\displaystyle \frac{x^2}{1+x^2}}\}\hfill \end{array}$$

(34)

$$\begin{array}{ccc}\hfill K_{0,-{\displaystyle \frac{3}{2}}}& =& x{\left(1+x^2\right)}^{-{\displaystyle \frac{1}{2}}},\hfill \end{array}$$

(35)

$$\begin{array}{ccc}\hfill K_{1,-2}& =& {\displaystyle \frac{x^2}{2}}{\left(1+x^2\right)}^{-1},\hfill \\ & ⋮& \hfill \end{array}$$

(36)

from (27): note that (34), not obtained from (27) but shown in (13), is also included for easy reference, comparison, and convenience.

2.2.2. Integration by Parts 2

When $r\ne -1$, with $u=x^{p-1}$, $v^{\prime }=x{\left(1+x^2\right)}^r$, $u^{\prime }=(p-1)x^{p-2}$, and $v={\displaystyle \frac{{\left(1+x^2\right)}^{r+1}}{2(r+1)}}$, we get

$$\begin{array}{ccc}& & K_{p,r}={\displaystyle \frac{x^{p-1}}{2(r+1)}}{\left(1+x^2\right)}^{r+1}-{\displaystyle \frac{p-1}{2(r+1)}}\int x^{p-2}{\left(1+x^2\right)}^{r+1}dx.\hfill \end{array}$$

(37)

If $p=1$, we easily get

$$\begin{array}{c}\hfill K_{1,r}={\displaystyle \frac{{\left(1+x^2\right)}^{r+1}}{2(r+1)}}\end{array}$$

(38)

for $r\ne -1$ from (37): for $r=-1$, recollect $K_{1,-1}={\displaystyle \frac{1}{2}}\ln \left(1+x^2\right)$ shown in (11).

Subsequently, (37) can be expressed as

$$\begin{array}{ccc}\hfill K_{p,r}& =& {\displaystyle \frac{x^{p-1}}{2(r+1)}}{\left(1+x^2\right)}^{r+1}-{\displaystyle \frac{p-1}{2(r+1)}}\int \left(1+x^2\right)x^{p-2}{\left(1+x^2\right)}^{r}dx\hfill \\ & =& {\displaystyle \frac{x^{p-1}}{2(r+1)}}{\left(1+x^2\right)}^{r+1}-{\displaystyle \frac{p-1}{2(r+1)}}{K}_{p,r}-{\displaystyle \frac{p-1}{2(r+1)}}{K}_{p-2,r},\hfill \end{array}$$

(39)

which can be used to obtain the following two results.

First, when $-{\displaystyle \frac{p-1}{2(r+1)}}\ne 1$ and $r\ne -1$, we get $\left\{1+{\displaystyle \frac{p-1}{2(r+1)}}\right\}{K}_{p,r}={\displaystyle \frac{x^{p-1}}{2(r+1)}}{\left(1+x^2\right)}^{r+1}-{\displaystyle \frac{p-1}{2(r+1)}}{K}_{p-2,r}$, or

$$\begin{array}{ccc}\hfill K_{p,r}& =& {\displaystyle \frac{x^{p-1}{\left(1+x^2\right)}^{r+1}}{p+2r+1}}-{\displaystyle \frac{p-1}{p+2r+1}}{K}_{p-2,r}\hfill \end{array}$$

(40)

for $p+2r+1\ne 0$ and $r\ne -1$. When $p=1$, (40) is the same as (38). In (37), we have assumed $r\ne -1$: yet, producing

$$\begin{array}{ccc}\hfill K_{p,-1}& =& {\displaystyle \frac{x^{p-1}}{p-1}}-K_{p-2,-1},\hfill \end{array}$$

(41)

(40) holds true even for $r=-1$. The result (41) can also be shown in an alternative way as $K_{p,-1}=\int {\displaystyle \frac{x^p}{1+x^2}}dx=\int {\displaystyle \frac{x^{p-2}+x^p-x^{p-2}}{1+x^2}}dx=\int \left\{{\displaystyle \frac{x^{p-2}\left(1+x^2\right)}{1+x^2}}-{\displaystyle \frac{x^{p-2}}{1+x^2}}\right\}dx=K_{p-2,0}-K_{p-2,-1}={\displaystyle \frac{x^{p-1}}{p-1}}-K_{p-2,-1}$ with $K_{p-2,0}={\displaystyle \frac{x^{p-1}}{p-1}}$ from (22).

Secondly, when $-{\displaystyle \frac{p-1}{2(r+1)}}=1$ and $r\ne -1$, we get $0={\displaystyle \frac{x^{p-1}}{2(r+1)}}{\left(1+x^2\right)}^{r+1}-{\displaystyle \frac{p-1}{2(r+1)}}{K}_{p-2,r}$ or equivalently $K_{p-2,-{\displaystyle \frac{p+1}{2}}}={\displaystyle \frac{x^{p-1}}{p-1}}{\left(1+x^2\right)}^{-{\displaystyle \frac{p-1}{2}}}$ for $p\ne 1$: this result is the same as (27) when p is replaced with $p+2$.

We have $K_{0,-{\displaystyle \frac{3}{2}}}=-{\displaystyle \frac{1}{2x\sqrt{1+x^2}}}-{\displaystyle \frac{1}{2}}{K}_{-2,-{\displaystyle \frac{3}{2}}}$ from (40), and thus $K_{-2,-{\displaystyle \frac{3}{2}}}=-{\displaystyle \frac{1}{x\sqrt{1+x^2}}}-2K_{0,-{\displaystyle \frac{3}{2}}}$ or

$$\begin{array}{ccc}\hfill K_{-2,-{\displaystyle \frac{3}{2}}}& =& -{\displaystyle \frac{2x^2+1}{x\sqrt{1+x^2}}}\hfill \end{array}$$

(42)

with $K_{0,-{\displaystyle \frac{3}{2}}}={\displaystyle \frac{x}{\sqrt{1+x^2}}}$ shown in (10).

3. Explicit Formulas from Direct Evaluation

Let us now consider the main problem of obtaining the explicit formulas of the integral $K_{p,r}$ for p an integer and r an integral multiple of ${\displaystyle \frac{1}{2}}$ in this and following sections. We first consider the cases for which direct evaluation is possible.

In the following developments, ${\displaystyle \prod _{a=b}^c}$ and ${\displaystyle \sum _{a=b}^c}$ denote the product and sum, respectively, for $a=b,b+1,b+2,\cdots ,c$ even when b is not an integer. We also let ${\displaystyle \sum _{a=b}^c}g\left(a\right)=0$ and ${\displaystyle \prod _{a=b}^c}h\left(a\right)=1$ if $c<b$. Here, $(p,r)$ denotes a point on the p–r plane for p integer and r an integral multiple of ${\displaystyle \frac{1}{2}}$, and often represents $K_{p,r}$ also.

3.1. The Cases $p=\pm 1$, $r=0$, $r=-1$, and $p+2r+3=0$

For $r=0$, from (22) and (23), we easily get

$$\begin{array}{ccc}\hfill K_{p,0}& =& \left\{\begin{array}{cc}\ln \left|x\right|,\hfill & p=-1,\hfill \\ {\displaystyle \frac{x^{p+1}}{p+1}},\hfill & p\ne -1.\hfill \end{array}\right.\hfill \end{array}$$

(43)

Next, by combining the results shown in (11) and (38), we have

$$\begin{array}{ccc}\hfill K_{1,r}& =& \left\{\begin{array}{cc}{\displaystyle \frac{1}{2}}\ln \left(1+x^2\right),\hfill & r=-1,\hfill \\ {\displaystyle \frac{{\left(1+x^2\right)}^{r+1}}{2(r+1)}},\hfill & r\ne -1\hfill \end{array}\right.\hfill \end{array}$$

(44)

for $p=1$. Similarly, collecting the results shown in (13) and (27), we have

$$\begin{array}{ccc}\hfill K_{p,-{\displaystyle \frac{p+3}{2}}}& =& \left\{\begin{array}{cc}{\displaystyle \frac{1}{2}}\ln {\displaystyle \frac{x^2}{1+x^2}},\hfill & p=-1,\hfill \\ {\displaystyle \frac{x^{p+1}}{p+1}}{\left(1+x^2\right)}^{-{\displaystyle \frac{p+1}{2}}},\hfill & p\ne -1\hfill \end{array}\right.\hfill \end{array}$$

(45)

for $p+2r+3=0$.

Next, for $p=-1$, the result (A8) shown in Appendix A.1 can be expressed as

$$\begin{array}{ccc}\hfill K_{-1,r}& =& \left\{\begin{array}{c}{\displaystyle \frac{1}{2}}\left\{\ln \left|v_r\left(x\right)\right|+{\displaystyle \sum _{k=1}^{h_{2r+1}}}\left({\displaystyle \genfrac{}{}{0pt}{}{h_{2r+1}}{k}}\right){\displaystyle \frac{v_r^k\left(x\right)}{k}}\right\},\hfill \\ r=0,\pm 1,\cdots ,\hfill \\ {\displaystyle \frac{1}{2}}\left\{\ln {\displaystyle \frac{\sqrt{1+x^2}-1}{\sqrt{1+x^2}+1}}+{\displaystyle \sum _{k=0}^{h_{2r+1}-{\displaystyle \frac{1}{2}}}}{\displaystyle \frac{{\left(1+x^2\right)}^{k_{2r+1}{s}_r}}{k_{2r+1}}}\right\},\hfill \\ r=\pm {\displaystyle \frac{1}{2}},\pm {\displaystyle \frac{3}{2}},\cdots \hfill \end{array}\right.\hfill \end{array}$$

(46)

after some steps. Here,

$$\begin{array}{ccc}\hfill h_y& =& {\displaystyle \frac{1}{2}}\left|y\right|-{\displaystyle \frac{1}{2}},\hfill \end{array}$$

(47)

$$\begin{array}{ccc}\hfill k_y& =& h_y-k,\hfill \end{array}$$

(48)

and

$$\begin{array}{ccc}\hfill v_y\left(x\right)& =& \left\{\begin{array}{cc}{x}^2,\hfill & y=0,1,\cdots ,\hfill \\ {\displaystyle \frac{-x^2}{1+x^2}},\hfill & y=-1,-2,\cdots \hfill \end{array}\right.\hfill \end{array}$$

(49)

with

$$\begin{array}{ccc}\hfill s_y& =& \left\{\begin{array}{cc}1,\hfill & y\ge 0,\hfill \\ -1,\hfill & y<0\hfill \end{array}\right.\hfill \end{array}$$

(50)

the sign of y: note that we have chosen $s_0=1$, not $s_0=0$. From (46), we get $K_{-1,-2}={\displaystyle \frac{1}{2}}\ln \left|v_{-2}\left(x\right)\right|+{\displaystyle \frac{1}{2}}{v}_{-2}\left(x\right)$ or

$$\begin{array}{ccc}\hfill K_{-1,-2}& =& {\displaystyle \frac{1}{2}}\left(\ln {\displaystyle \frac{x^2}{1+x^2}}-{\displaystyle \frac{x^2}{1+x^2}}\right),\hfill \end{array}$$

(51)

for instance.

Similarly, when $r=-1$, the result (A17) for $K_{p,-1}$ shown in Appendix A.2 can be equivalently expressed as

$$\begin{array}{c}\hfill K_{p,-1}=\left\{\begin{array}{c}{s}_p{\displaystyle \frac{{(-1)}^{h_p+\frac{1}{2}}}{2}}\left\{2{tan}^{-1}{x}^{s_p}+{\displaystyle \sum _{k=1}^{h_p+{\displaystyle \frac{1}{2}}}}{\displaystyle \frac{{(-1)}^{k_p-\frac{1}{2}}}{k_p+1}}{x}^{2(k_p+1)s_p}\right\},\hfill \\ p=0,\pm 2,\pm 4\cdots ,\hfill \\ s_p{\displaystyle \frac{{(-1)}^{h_p}}{2}}\left\{\ln \left|1+x^{2s_p}\right|+{\displaystyle \sum _{k=1}^{h_p}}{\displaystyle \frac{{(-1)}^k}{k}}\left({\displaystyle \genfrac{}{}{0pt}{}{h_p}{k}}\right){\left(1+x^{2s_p}\right)}^k\right\},\hfill \\ p=\pm 1,\pm 3,\cdots .\hfill \end{array}\right.\end{array}$$

(52)

3.2. Discussion

For every point on the five lines $r=0$, $p=1$, $p=-1$, $r=-1$, and $p+2r+3=0$, the integral $K_{p,r}$ now has readily available formula. The formulas can be used also as the initial conditions (IC’s) in obtaining the formulas $K_{p,r}$ for other points via recursions as discussed in Section 4. Figure 1 illustrates the five lines on which every point now has an explicit formula for the integral $K_{p,r}$.

4. Evaluation via Recursions

For the points not on the five lines $r=0$, $p=1$, $p=-1$, $r=-1$, and $p+2r+3=0$, and also for some points on the five lines, the integral can be evaluated via recursions as described in this section.

4.1. Evaluation of $K_{p,r}$ When $p+2r+1=0$

If $\{p+2r+1=0\}\cap \{p=-1\}$, we easily get $K_{-1,0}=\int {\displaystyle \frac{1}{x}}dx=\ln \left|x\right|$ as discussed in (23) also. For $\{p+2r+1=0\}\cap \{p\ne -1\}$, we can obtain $K_{p,-{\displaystyle \frac{p+1}{2}}}$ from $K_{p+2,-{\displaystyle \frac{p+3}{2}}}$ or vise versa for $p=0,\pm 1,\pm 2,\cdots$, with the recursion

$$\begin{array}{ccc}& & K_{p,-{\displaystyle \frac{p+1}{2}}}={\displaystyle \frac{x^{p+1}{\left(1+x^2\right)}^{-\frac{p+1}{2}}}{p+1}}+K_{p+2,-{\displaystyle \frac{p+3}{2}}},\hfill \end{array}$$

(53)

which can be derived from (21) or (37).

In using the recursion (53), two IC’s are required because the recursion describes a relationship between two $K_{p,r}$’s with a difference of 2 in the subscript p and we assume integers for p. In addition, due to the ‘wall’ effect (that is, neither $K_{-1,0}$ from $K_{1,-1}$ nor $K_{1,-1}$ from $K_{-1,0}$ can be obtained when $p=-1$) caused by the factors $p+1$ in the denominator on the right-hand side of (53), one additional initial condition is required. In short, for the recursion (53), a complete set of three IC’s would be $\left\{K_{1,-1},K_{0,-{\displaystyle \frac{1}{2}}},K_{-1,0}\right\}$, of which the first, second, and third elements can be replaced by any element from $\left\{K_{2m-1,-m}:m=1,2,\cdots \right\}$, $\left\{K_{2m,-m-{\displaystyle \frac{1}{2}}}:m=0,\pm 1,\pm 2,\cdots \right\}$, and $\left\{K_{2m-1,-m}:m=0,-1,-2,\cdots \right\}$, respectively.

Now, by adding

$$\begin{array}{ccc}\hfill K_{p,-{\displaystyle \frac{p+1}{2}}}& =& {\displaystyle \frac{x^{p+1}{\left(1+x^2\right)}^{-\frac{p+1}{2}}}{p+1}}+K_{p+2,-{\displaystyle \frac{p+3}{2}}},\hfill \end{array}$$

(54)

$$\begin{array}{ccc}\hfill K_{p+2,-{\displaystyle \frac{p+3}{2}}}& =& {\displaystyle \frac{x^{p+3}{\left(1+x^2\right)}^{-\frac{p+3}{2}}}{p+3}}+K_{p+4,-{\displaystyle \frac{p+5}{2}}},\hfill \\ & ⋮& \hfill \end{array}$$

(55)

$$\begin{array}{ccc}\hfill K_{s-2,-{\displaystyle \frac{s-1}{2}}}& =& {\displaystyle \frac{x^{s-1}{\left(1+x^2\right)}^{-\frac{s-1}{2}}}{s-1}}+K_{s,-{\displaystyle \frac{s+1}{2}}},\hfill \end{array}$$

(56)

obtained from (53), we get

$$\begin{array}{ccc}\hfill K_{p,-{\displaystyle \frac{p+1}{2}}}& =& K_{s,-{\displaystyle \frac{s+1}{2}}}+{\displaystyle \sum _{k=p:2}^{s-2}}{\displaystyle \frac{x^{k+1}{\left(1+x^2\right)}^{-\frac{k+1}{2}}}{k+1}}\hfill \end{array}$$

(57)

for $s=p+2m$ with m a positive integer, where ‘$k=p:2$’ denotes ‘$k=p,p+2,\cdots$’. It is easy to see that the result (57) can also be used for $s=p+2m$ with m a negative integer after an interchange of the subscripts p and s. Subsequently, the IC’s $K_{0,-{\displaystyle \frac{1}{2}}}=\ln \left|x+\sqrt{1+x^2}\right|$, $K_{1,-1}={\displaystyle \frac{1}{2}}\ln \left(1+x^2\right)$, and $K_{-1,0}=\ln \left|x\right|$ shown in (8), (11), and (23), respectively, can be used to obtain $K_{p,-{\displaystyle \frac{p+1}{2}}}$ recursively for $p=\pm 2,\pm 4,\cdots$, for $p=3,5,\cdots$, and for $p=-3,-5,\cdots$, respectively. More specifically, we will eventuslly get

$$\begin{array}{ccc}\hfill K_{p,r_1}& =& v_{p,r_1}\left(x\right)+{\displaystyle \sum _{k=L_{p,r_1}:2}^{U_{p,r_1}}}{w}_{p,r_1,k}\left(x\right),\hfill \end{array}$$

(58)

where $r_1=-{\displaystyle \frac{p+1}{2}}$ as defined in (3),

$$\begin{array}{ccc}\hfill L_{p,r_1}& =& \left\{\begin{array}{cc}0,\hfill & p=0,2,\cdots ,\hfill \\ 1,\hfill & p=1,3,\cdots ,\hfill \\ p,\hfill & p=-1,-2,\cdots \hfill \end{array}\right.\hfill \end{array}$$

(59)

denotes the lower bound of the summation on the right-hand side of (58),

$$\begin{array}{ccc}\hfill U_{p,r_1}& =& \left\{\begin{array}{cc}p-2,\hfill & p=1,2,\cdots ,\hfill \\ -3,\hfill & p=-1,-3,\cdots ,\hfill \\ -2,\hfill & p=0,-2,-4,\cdots \hfill \end{array}\right.\hfill \end{array}$$

(60)

denotes the upper bound of the summation on the right-hand side of (58),

$$\begin{array}{ccc}\hfill v_{p,r_1}\left(x\right)& =& \left\{\begin{array}{cc}\ln \sqrt{1+x^2},\hfill & p=1,3,\cdots ,\hfill \\ \ln \left|x\right|,\hfill & p=-1,-3,\cdots ,\hfill \\ \ln \left|x+\sqrt{1+x^2}\right|,\hfill & p=0,\pm 2,\pm 4,\cdots ,\hfill \end{array}\right.\hfill \end{array}$$

(61)

and

$$\begin{array}{ccc}\hfill w_{p,r_1,k}\left(x\right)& =& -{\displaystyle \frac{s_p}{k+1}}{x}^{k+1}{\left(1+x^2\right)}^{-{\displaystyle \frac{k+1}{2}}}.\hfill \end{array}$$

(62)

We get $K_{3,-2}={\displaystyle \frac{1}{2}}\ln \left(1+x^2\right)-{\displaystyle \frac{x^2}{2\left(1+x^2\right)}}$, $K_{5,-3}={\displaystyle \frac{1}{2}}\ln \left(1+x^2\right)-{\displaystyle \frac{x^2}{2\left(1+x^2\right)}}-{\displaystyle \frac{x^4}{4{\left(1+x^2\right)}^2}}$, $K_{2,-{\displaystyle \frac{3}{2}}}=\ln \left|x+\sqrt{1+x^2}\right|-{\displaystyle \frac{x}{\sqrt{1+x^2}}}$, and $K_{-2,{\displaystyle \frac{1}{2}}}=\ln |x+\sqrt{1+x^2}|-{\displaystyle \frac{\sqrt{1+x^2}}{x}}$ from (58). Similarly, we get $K_{-3,1}=\ln \left|x\right|-{\displaystyle \frac{1+x^2}{2x^2}}$ and $K_{-5,3}=\ln \left|x\right|-{\displaystyle \frac{1+x^2}{2x^2}}-{\displaystyle \frac{{\left(1+x^2\right)}^2}{4x^4}}$.

4.2. Other Points of $(p,r)$

When p is fixed, to obtain $K_{p,r}$ from $K_{p,r-1}$ or vice versa for $r=0,\pm {\displaystyle \frac{1}{2}},\pm 1,...$, we can use

$$\begin{array}{ccc}\hfill K_{p,r}& =& {\displaystyle \frac{x^{p+1}{\left(1+x^2\right)}^r}{p+2r+1}}+{\displaystyle \frac{2r}{p+2r+1}}{K}_{p,r-1}\hfill \end{array}$$

(63)

shown in (25) under the condition $p+2r+1\ne 0$. Next, when r is fixed, to obtain $K_{p,r}$ from $K_{p-2,r}$ or vise versa for $p=0,\pm 1,\pm 2,...$, we can use

$$\begin{array}{ccc}\hfill K_{p,r}& =& {\displaystyle \frac{x^{p-1}{\left(1+x^2\right)}^{r+1}}{p+2r+1}}-{\displaystyle \frac{p-1}{p+2r+1}}{K}_{p-2,r}\hfill \end{array}$$

(64)

shown in (40) under the condition $p+2r+1\ne 0$.

Clearly, when using the two recursions (63) and (64), it is natural to employ the known formulas of the points on the six lines $r=0$, $p=1$, $p=-1$, $r=-1$, $p+2r+3=0$, and $p+2r+1=0$ as the IC’s. From Figure 1, it is observed that there now exist four (if $p=\cdots ,-5,-4,-2,0,2,3,\cdots$) and three (if $p=-3$) formulas of $K_{p,r}$ readily available as the candidates of the IC’s when using (63) for p fixed. Similarly, when using (64) for r fixed, four (if $r=\cdots ,-3,-{\displaystyle \frac{5}{2}},-{\displaystyle \frac{3}{2}},-{\displaystyle \frac{1}{2}},{\displaystyle \frac{1}{2}},1,{\displaystyle \frac{3}{2}},\cdots$) and three (if $r=-2$) formulas of $K_{p,r}$ are readily available as the candidates of the IC’s.

4.2.1. Initial Conditions for the Recursion ( 63)

To obtain $K_{p,r}$ with a fixed value of $p\ne -1$ via the recursion (63), let $r_1=-{\displaystyle \frac{p+1}{2}}$ as in (3), $r_L=min\left(r_1,0\right)$, and $r_U=max\left(r_1,0\right)$. The recursion (63) implies that $K_{p,r}$ can be obtained from $K_{p,r-1}$ and vice versa in general: however, when $r=0$ and when $r=r_1$, neither $K_{p,r}$ from $K_{p,r-1}$ nor $K_{p,r-1}$ from $K_{p,r}$ can be obtained. Based on this observation, we divide the points $(p,r)$ into three groups such that $r\in G_{p,L}$, $r\in G_{p,B}$, and $r\in G_{p,U}$, where $G_{p,L}=\left(-\infty ,r_L-{\displaystyle \frac{1}{2}}\right]$, $G_{p,B}=\left[r_L,r_U-{\displaystyle \frac{1}{2}}\right]$, and $G_{p,U}=\left[r_U,\infty \right)$ are called the group sets of r. For each of the group sets, we need two IC’s because we have assumed integral multiples of ${\displaystyle \frac{1}{2}}$ for r and the recursion (63) describes a relationships between two $K_{p,r}$’s with a difference of 1 in the subscript r. In the meantime, an IC $K_{p,r}$ with r an odd multiple of ${\displaystyle \frac{1}{2}}$ for one of the three group sets can be used also in another set across a border $r=r_L$ or $r=r_U$ when the border is an integer. Likewise, an IC $K_{p,r}$ with r an integer can be used across a border when the border is an odd multiple of ${\displaystyle \frac{1}{2}}$: that is, two of the six IC’s are redundant. In other words, the number of elements for a complete set of IC’s is at most four: refer to Section 4.2.3 for more detail.

Specifically, we would first choose $K_{p,r_U}$ and $K_{p,r_L-1}$. Next, if $r_1$ is an odd multiple of ${\displaystyle \frac{1}{2}}$ (equivalently, if p is an even integer), it suffices to choose $K_{p,r_U-1}$ and $K_{p,r_L}$ because one of $r_L$ and $r_U$ is 0 and the other is an odd multiple of ${\displaystyle \frac{1}{2}}$. If $r_1$ is an integer, on the other hand, both $r_L$ and $r_U$ are integers: then, we could choose $\left\{K_{p,r_U-1},K_{p,\hslash }\right\}$ or $\left\{K_{p,r_L},K_{p,\hslash }\right\}$ as the remaining two IC’s, where ℏ denotes any odd multiple of ${\displaystyle \frac{1}{2}}$. In short, a complete set of IC’s for the recursion (63) is

$$\begin{array}{ccc}\hfill I_{1,p}& =& \left\{\begin{array}{c}\left\{K_{p,r_U},K_{p,r_L-1},K_{p,r_U-1},K_{p,r_L}\right\},\hfill \\ \mathrm{if} p \mathrm{is} \mathrm{an} \mathrm{even} \mathrm{integer},\hfill \\ \left\{K_{p,r_U},K_{p,r_L-1},K_{p,\hslash },K_{p,r_L}\right\}or\hfill \\ \left\{K_{p,r_U},K_{p,r_L-1},K_{p,r_U-1},K_{p,\hslash }\right\},\hfill \\ \mathrm{if} p \mathrm{is} \mathrm{an} \mathrm{odd} \mathrm{integer},\hfill \end{array}\right.\hfill \end{array}$$

(65)

In the set $I_{1,p}$, the IC’s $K_{p,r_U}$ and $K_{p,r_L-1}$ can be replaced by any element of the sets ${\left\{K_{p,r_U+m}\right\}}_{m=0}^{\infty }$ and ${\left\{K_{p,r_L-m}\right\}}_{m=1}^{\infty }$, respectively. In addition, when $r_1$ is an integer and an odd multiple of ${\displaystyle \frac{1}{2}}$, the IC $K_{p,r_U-1}$ can be replaced by any element of the sets ${\left\{K_{p,r_U-m}\right\}}_{m=1}^{r_U-r_L}$ and ${\left\{K_{p,r_U-m}\right\}}_{m=1}^{\infty }$, respectively. Similarly, the IC $K_{p,r_L}$ can be replaced by any element of the sets ${\left\{K_{p,r_L+m}\right\}}_{m=0}^{r_U-r_L-1}$ and ${\left\{K_{p,r_L+m}\right\}}_{m=0}^{\infty }$ when $r_1$ is an integer and an odd multiple of ${\displaystyle \frac{1}{2}}$, respectively.

4.2.2. Initial Conditions for the Recursion ( 64)

To obtain $K_{p,r}$ with a fixed value of $r\ne -1$ via the recursion (64), let $p_1=-2r-1$, $p_L=min\left(p_1,1\right)$, and $p_U=max\left(p_1,1\right)$. Denote by $G_{L,r}=\left(-\infty ,p_L-1\right]$, $G_{B,r}=\left[p_L,p_U-1\right]$, and $G_{U,r}=\left[p_U,\infty \right)$ the group sets of p. Following steps similar to those in Section 4.2.1, a complete set of IC’s for the recursion (64) is

$$\begin{array}{ccc}\hfill I_{2,r}& =& \left\{\begin{array}{c}\left\{K_{p_U,r},K_{p_L-2,r},K_{p_U-2,r},K_{p_L,r}\right\},\hfill \\ \mathrm{if} r \mathrm{is} \mathrm{an} \mathrm{odd} \mathrm{multiple} \mathrm{of} {\displaystyle \frac{1}{2}},\hfill \\ \left\{K_{p_U,r},K_{p_L-2,r},K_{\odot ,r},K_{p_L,r}\right\}or\hfill \\ \left\{K_{p_U,r},K_{p_L-2,r},K_{p_U-2,r},K_{\odot ,r}\right\},\hfill \\ \mathrm{if} r \mathrm{is} \mathrm{an} \mathrm{integer},\hfill \end{array}\right.\hfill \end{array}$$

(66)

where ⊙ denotes any even integer.

In the set $I_{2,r}$, the IC’s $K_{p_U,r}$ and $K_{p_L-2,r}$ can be replaced by any element of the sets ${\left\{K_{p_U+2m,r}\right\}}_{m=0}^{\infty }$ and ${\left\{K_{p_L-2m,r}\right\}}_{m=1}^{\infty }$, respectively. Similarly, the IC $K_{p_U-2,r}$ can be replaced by any element of the sets ${\left\{K_{p_U-2m,r}\right\}}_{m=1}^{\infty }$ and ${\left\{K_{p_U-2m,r}\right\}}_{m=1}^{{\displaystyle \frac{1}{2}}\left(p_U-p_L\right)}$ when $p_1$ is an even and an odd integer, respectively. Likewise, the IC $K_{p_L,r}$ can be replaced by any element of the sets ${\left\{K_{p_L+2m,r}\right\}}_{m=0}^{{\displaystyle \frac{1}{2}}\left(p_U-p_L\right)-1}$ and ${\left\{K_{p_L+2m,r}\right\}}_{m=0}^{\infty }$ when $p_1$ is an odd and an even integers, respectively. Note that ${\displaystyle \frac{1}{2}}\left(p_U-p_L\right)={\displaystyle \frac{1}{2}}\left|p_1-1\right|=|r+1|$.

4.2.3. Special Cases

Let us note that, although $I_{1,-3}$ in effect contains only three elements because $r_L=r_U-1$ when $p=-3$, it will still be a set of a sufficient number of IC’s for solving the recursion (63). Similarly, when $r=-2$, the set $I_{2,-2}$ will still be a complete set for solving the recursion (64) although it contains only three elements because $p_L=p_U-2$.

Each of the sets $I_{1,-1}$ for $p=-1$ and $I_{2,-1}$ for $r=-1$ contains only two elements because $r_L=r_U=0$ and $p_L=p_U=1$, respectively: the number two of IC’s will be insufficient for solving the recursions. Yet, when $p=-1$ and when $r=-1$, the integral $K_{p,r}$ does not need to be calculated via recursions but can easily be obtained via direct evaluations as we have discussed already in Section 3.1 with details delineated in Appendices Appendix A.1 and Appendix A.2.

4.3. Evaluation of $K_{p,r}$ via the Recursion ( 63)

Now, with $a_{p,r}={\displaystyle \frac{x^{p+1}{\left(1+x^2\right)}^r}{p+2r+1}}$ and $b_{p,r}={\displaystyle \frac{2r}{p+2r+1}}$, if we add

$$\begin{array}{ccc}\hfill K_{p,r}& =& a_{p,r}+b_{p,r}{K}_{p,r-1}\hfill \end{array}$$

(67)

$$\begin{array}{ccc}\hfill b_{p,r}{K}_{p,r-1}& =& b_{p,r}{a}_{p,r-1}+b_{p,r}{b}_{p,r-1}{K}_{p,r-2}\hfill \end{array}$$

(68)

$$\begin{array}{ccc}\hfill b_{p,r}{b}_{p,r-1}{K}_{p,r-2}& =& b_{p,r}{b}_{p,r-1}{a}_{p,r-2}+b_{p,r}{b}_{p,r-1}{b}_{p,r-2}{K}_{p,r-3}\hfill \\ & ⋮& \hfill \end{array}$$

(69)

$$\begin{array}{ccc}\hfill K_{p,s+1}{\displaystyle \prod _{m=s+2}^r}{b}_{p,m}& =& a_{p,s+1}{\displaystyle \prod _{m=s+2}^r}{b}_{p,m}+K_{p,s}{\displaystyle \prod _{m=s+1}^r}{b}_{p,m}\hfill \end{array}$$

(70)

we get

$$\begin{array}{ccc}\hfill K_{p,r}& =& {\displaystyle \sum _{k=s+1}^r}{a}_{p,k}{\displaystyle \prod _{m=k+1}^r}{b}_{p,m}+K_{p,s}{\displaystyle \prod _{m=s+1}^r}{b}_{p,m}\hfill \end{array}$$

(71)

for $r>s$.

Now, note that ${\displaystyle \prod _{m=k+1}^r}{b}_{p,m}={\displaystyle \frac{2k+2}{p+2k+3}}{\displaystyle \frac{2k+4}{p+2k+5}}\cdots {\displaystyle \frac{2r}{p+2r+1}}={\displaystyle \frac{\left(2r\right)!!(p+2k+1)!!}{(p+2r+1)!!\left(2k\right)!!}}={\displaystyle \frac{S(p,r)}{S(p,k)}}$, where

$$\begin{array}{c}\hfill S(a,b)={\displaystyle \frac{\left(2b\right)!!}{(a+2b+1)!!}}\end{array}$$

(72)

with the double factorial defined as

$$\begin{array}{ccc}\hfill k!!& =& \left\{\begin{array}{cc}-1,\hfill & ifk=-2,\hfill \\ 1,\hfill & ifk=-1,0,\hfill \\ k(k-2)!!,\hfill & ifk=1,2,\cdots ,\hfill \\ {\displaystyle \frac{(k+2)!!}{k+2}},\hfill & ifk=-3,-4,\cdots .\hfill \end{array}\right.\hfill \end{array}$$

(73)

Some values of $k!!$ are shown in Table 1 for easy reference.

Then, for $p\ne -1$, we finally get from (71)

$$\begin{array}{ccc}\hfill K_{p,r}& =& S(p,r)\left\{{\displaystyle \frac{K_{p,s}}{S(p,s)}}+{\displaystyle \sum _{k=s+1}^r}{\displaystyle \frac{a_{p,k}}{S(p,k)}}\right\}\hfill \end{array}$$

(74)

useful when calculating $K_{p,r}$ from the initial condition $K_{p,s}$ for $s<r$, and

$$\begin{array}{ccc}\hfill K_{p,r}& =& S(p,r)\left\{{\displaystyle \frac{K_{p,s}}{S(p,s)}}-{\displaystyle \sum _{k=r+1}^s}{\displaystyle \frac{a_{p,k}}{S(p,k)}}\right\}\hfill \end{array}$$

(75)

useful when calculating $K_{p,r}$ from the initial condition $K_{p,s}$ for $s>r$. The two formulas (74) and (75) can be used when $\{r,s\}\subset G_{p,L}$, $\{r,s\}\subset G_{p,B}$, or $\{r,s\}\subset G_{p,U}$, but not when r and s are in different group sets.

Let us now describe in detail how to obtain the integral $K_{p,r}$ for a fixed r based on (74) and (75). The procedures differ slightly depending on the value of $r_1$ or, equivalently, of p. First, when the set of IC’s is $\{K_{p,r_U}$, $K_{p,r_U-1}$, $K_{p,r_L}$, $K_{p,r_L-1}\}$, that is, when $r_1$ is an odd multiple of ${\displaystyle \frac{1}{2}}$ or equivalently when p is an even integer, use the initial condition $K_{p,r_U}$ in (74) for obtaining $K_{p,r}$ for $r=r_U+1,r_U+2,\cdots$; use the initial condition $K_{p,r_L-1}$ in (75) for obtaining $K_{p,r}$ for $r=r_L-2,r_L-3,\cdots$; use the initial condition $K_{p,r_L}$ in (74) for obtaining $K_{p,r}$ for $r=r_L+1,r_L+2,\cdots$; and use the initial condition $K_{p,r_U-1}$ in (75) for obtaining $K_{p,r}$ for $r=r_U-2,r_U-3,\cdots$. In effect, as mentioned before indirectly, among the two sets $\left\{K_{p,r_U},K_{p,r_U-1}\right\}$ and $\left\{K_{p,r_L},K_{p,r_L-1}\right\}$, one set is used when obtaining $K_{p,r}$ for r odd multiple of ${\displaystyle \frac{1}{2}}$ and the other set for r even multiple of ${\displaystyle \frac{1}{2}}$.

On the other hand, when the set of IC’s is $\{K_{p,r_U}$, $K_{p,r_U-1}$, $K_{p,r_L}$, $K_{p,\hslash }\}$ or $\{K_{p,r_U}$, $K_{p,r_U-1}$, $K_{p,r_L-1}$, $K_{p,\hslash }\}$, that is, when $r_1$ is an integer or equivalently when p is an odd integer, use the initial condition $K_{p,r_U}$ in (74) for obtaining $K_{p,r}$ for $r=r_U+1,r_U+2,\cdots$ and the initial condition $K_{p,r_L-1}$ in (75) for obtaining $K_{p,r}$ for $r=r_L-2,r_L-3,\cdots$: these two steps are the same as those in the first case. The initial condition $K_{p,K_{p,\hslash }}$ can next be used in (74) for obtaining $K_{p,r}$ for $r=\hslash +1,\hslash +2,\cdots$; and also in (75) for obtaining $K_{p,r}$ for $r=\hslash -1,\hslash -2,\cdots$. In addition, we need to obtain $K_{p,r}$ for $r=r_L,r_L+1,\cdots ,r_U-1$: for that goal, we may use the initial condition $K_{p,r_L}$ in (74) for obtaining $K_{p,r}$ for $r=r_L+1,r_L+2,\cdots ,r_U-1$ or use the initial condition $K_{p,r_U-1}$ in (75) for obtaining $K_{p,r}$ for $r=r_U-2,r_U-3,\cdots ,r_L$. Table 2 summarizes how $K_{p,r}$ can be obtained when the value of p is fixed.

When $p=1$, a result that is the same as (44) subject to an integral constant can be obtained based on (74) and (75) as shown in Appendix A.3.

Example 1.

When $p=0$, we can obtain $\{K_{0,r}:$$r=1$, 2, $\cdots$}, $\{K_{0,r}:$$r={\displaystyle \frac{1}{2}}$, ${\displaystyle \frac{3}{2}}$, $\cdots \}$, $\left\{K_{0,r}:r=-2,-3,\cdots \right\}$, and $\left\{K_{0,r}:r=-{\displaystyle \frac{5}{2}},-{\displaystyle \frac{7}{2}},\cdots \right\}$ with the IC’s $K_{0,0}=\int dx=x$, $K_{0,-{\displaystyle \frac{1}{2}}}=\ln \left|x+\sqrt{1+x^2}\right|$ from (8), $K_{0,-1}={tan}^{-1}x$ from (9), and $K_{0,-{\displaystyle \frac{3}{2}}}={\displaystyle \frac{x}{\sqrt{1+x^2}}}$ from (10), respectively. Specifically, we can get

$$\begin{array}{ccc}\hfill K_{0,r}& =& {\displaystyle \frac{\left(2r\right)!!}{(2r+1)!!}}\left\{v_{0,r}\left(x\right)+{\displaystyle \sum _{k=L_{0,r}}^{U_{0,r}}}{w}_{0,r,k}\left(x\right)\right\},\hfill \end{array}$$

(76)

where

$$\begin{array}{ccc}\hfill L_{0,r}& =& \left\{\begin{array}{cc}1,\hfill & r=1,2,\cdots ,\hfill \\ {\displaystyle \frac{1}{2}},\hfill & r={\displaystyle \frac{1}{2}},{\displaystyle \frac{3}{2}},\cdots ,\hfill \\ r+1,\hfill & r=0,-{\displaystyle \frac{1}{2}},-1,-{\displaystyle \frac{3}{2}},\cdots ,\hfill \end{array}\right.\hfill \end{array}$$

(77)

$$\begin{array}{ccc}\hfill U_{0,r}& =& \left\{\begin{array}{cc}-1,\hfill & r=-1,-2\cdots ,\hfill \\ -{\displaystyle \frac{3}{2}},\hfill & r=-{\displaystyle \frac{1}{2}},-{\displaystyle \frac{3}{2}},\cdots ,\hfill \\ r,\hfill & r=0,{\displaystyle \frac{1}{2}},1,{\displaystyle \frac{3}{2}},\cdots ,\hfill \end{array}\right.\hfill \end{array}$$

(78)

$$\begin{array}{ccc}\hfill v_{0,r}\left(x\right)& =& \left\{\begin{array}{cc}x,\hfill & r=0,1,\cdots ,\hfill \\ -{tan}^{-1}x,\hfill & r=-1,-2,\cdots ,\hfill \\ \ln \left|x+\sqrt{1+x^2}\right|,\hfill & r=-{\displaystyle \frac{1}{2}},{\displaystyle \frac{1}{2}},\cdots ,\hfill \\ {\displaystyle \frac{x}{\sqrt{1+x^2}}},\hfill & r=-{\displaystyle \frac{3}{2}},-{\displaystyle \frac{5}{2}},\cdots ,\hfill \end{array}\right.\hfill \end{array}$$

(79)

and

$$\begin{array}{ccc}\hfill w_{0,r,k}\left(x\right)& =& s_r{\displaystyle \frac{(2k-1)!!}{\left(2k\right)!!}}x{\left(1+x^2\right)}^k\hfill \end{array}$$

(80)

after some steps by noting that $S(0,b)={\displaystyle \frac{\left(2b\right)!!}{(2b+1)!!}}$.

Example 2.

It is easy to see that $K_{0,1}={\displaystyle \frac{2!!}{3!!}}\left\{1+{\displaystyle \frac{1!!}{2!!}}\left(1+x^2\right)\right\}x$ or

$$\begin{array}{ccc}\hfill K_{0,1}& =& {\displaystyle \frac{x}{3}}\left(3+x^2\right),\hfill \end{array}$$

(81)

$K_{0,{\displaystyle \frac{1}{2}}}={\displaystyle \frac{1!!}{2!!}}\left\{\ln \left|x+\sqrt{1+x^2}\right|+{\displaystyle \frac{0!!}{1!!}}x{\left(1+x^2\right)}^{{\displaystyle \frac{1}{2}}}\right\}$ or

$$\begin{array}{ccc}\hfill K_{0,{\displaystyle \frac{1}{2}}}& =& {\displaystyle \frac{1}{2}}\left(x\sqrt{1+x^2}+\ln \left|x+\sqrt{1+x^2}\right|\right),\hfill \end{array}$$

(82)

$K_{0,-2}={\displaystyle \frac{(-4)!!}{(-3)!!}}\left\{-{tan}^{-1}x-{\displaystyle \frac{(-3)!!}{(-2)!!}}x{\left(1+x^2\right)}^{-1}\right\}$ or

$$\begin{array}{ccc}\hfill K_{0,-2}& =& {\displaystyle \frac{1}{2}}\left({tan}^{-1}x+{\displaystyle \frac{x}{1+x^2}}\right),\hfill \end{array}$$

(83)

$$\begin{array}{ccc}\hfill K_{0,{\displaystyle \frac{3}{2}}}& =& {\displaystyle \frac{3!!}{4!!}}\left\{\ln \left|x+\sqrt{1+x^2}\right|+{\displaystyle \frac{0!!}{1!!}}x{\left(1+x^2\right)}^{{\displaystyle \frac{1}{2}}}+{\displaystyle \frac{2!!}{3!!}}x{\left(1+x^2\right)}^{{\displaystyle \frac{3}{2}}}\right\}\hfill \\ & =& {\displaystyle \frac{1}{8}}\left\{\left(2x^3+5x\right)\sqrt{1+x^2}+3\ln \left|x+\sqrt{1+x^2}\right|\right\},\hfill \end{array}$$

(84)

and $K_{0,-{\displaystyle \frac{5}{2}}}={\displaystyle \frac{(-5)!!}{(-4)!!}}{\displaystyle \frac{(-2)!!}{(-3)!!}}x\left\{{\displaystyle \frac{1}{\sqrt{1+x^2}}}+{\displaystyle \frac{1}{2{\left(1+x^2\right)}^{\frac{3}{2}}}}\right\}$ or

$$\begin{array}{ccc}\hfill K_{0,-{\displaystyle \frac{5}{2}}}& =& {\displaystyle \frac{\left(2x^2+3\right)x}{3{\left(1+x^2\right)}^{\frac{3}{2}}}}\hfill \end{array}$$

(85)

based on (76).

4.4. Evaluation of $K_{p,r}$ via the Recursion ( 64)

Following steps similar to those leading to (74) and (75), with details provided in Appendix A.4, we can get

$$\begin{array}{ccc}\hfill K_{p,r}& =& T(p,r)\left\{{(-1)}^{{\displaystyle \frac{p-s}{2}}}{\displaystyle \frac{K_{s,r}}{T(s,r)}}+{\displaystyle \sum _{k=s+2:2}^p}{(-1)}^{{\displaystyle \frac{p-k}{2}}}{\displaystyle \frac{c_{k,r}}{T(k,r)}}\right\}\hfill \end{array}$$

(86)

and

$$\begin{array}{ccc}\hfill K_{p,r}& =& T(p,r)\left\{{(-1)}^{{\displaystyle \frac{p-s}{2}}}{\displaystyle \frac{K_{s,r}}{T(s,r)}}-{\displaystyle \sum _{k=p+2:2}^s}{(-1)}^{{\displaystyle \frac{p-k}{2}}}{\displaystyle \frac{c_{k,r}}{T(k,r)}}\right\}\hfill \end{array}$$

(87)

useful in the calculation of ${\left\{K_{p,r}\right\}}_{p=0,\pm 1,\pm 2,\cdots }$ for $r\ne -1$ with the initial condition $K_{s,r}$ when $s<p$ and $s>p$, respectively, where

$$\begin{array}{ccc}\hfill c_{p,r}& =& {\displaystyle \frac{x^{p-1}{\left(1+x^2\right)}^{r+1}}{p+2r+1}}\hfill \end{array}$$

(88)

and

$$\begin{array}{ccc}\hfill T(a,b)& =& {\displaystyle \frac{(a-1)!!}{(a+2b+1)!!}}.\hfill \end{array}$$

(89)

The two formulas (86) and (87) can be used when $\{p,s\}\subset G_{L,r}$, $\{p,s\}\subset G_{B,r}$, or $\{p,s\}\subset G_{U,r}$, but not when p and s are in different group sets. After a discussion similar to that in Section 4.3, we will get Table 3 as the details on evaluating $K_{p,r}$ for a fixed value of r.

Example 3.

The result (A35) shown in Appendix A.5 for $r=-{\displaystyle \frac{1}{2}}$ can be written as

$$\begin{array}{ccc}\hfill K_{p,-{\displaystyle \frac{1}{2}}}& =& s_p{\displaystyle \frac{(p-1)!!}{p!!}}\left\{v_{p,-{\displaystyle \frac{1}{2}}}\left(x\right)+{\displaystyle \sum _{k=L_{p,-{\displaystyle \frac{1}{2}}}:2}^{U_{p,-{\displaystyle \frac{1}{2}}}}}{w}_{p,-{\displaystyle \frac{1}{2}},k}\left(x\right)\right\},\hfill \end{array}$$

(90)

where

$$\begin{array}{ccc}\hfill L_{p,-{\displaystyle \frac{1}{2}}}& =& \left\{\begin{array}{cc}2,\hfill & p=0,2,\cdots ,\hfill \\ 3,\hfill & p=1,3,\cdots ,\hfill \\ p+2,\hfill & p=-1,-2,\cdots ,\hfill \end{array}\right.\hfill \end{array}$$

(91)

$$\begin{array}{ccc}\hfill U_{p,-{\displaystyle \frac{1}{2}}}& =& \left\{\begin{array}{cc}p,\hfill & p=0,1,\cdots ,\hfill \\ -1,\hfill & p=-1,-3,\cdots ,\hfill \\ -2,\hfill & p=-2,-4,\cdots ,\hfill \end{array}\right.\hfill \end{array}$$

(92)

$$\begin{array}{ccc}\hfill v_{p,-{\displaystyle \frac{1}{2}}}\left(x\right)& =& \left\{\begin{array}{cc}{(-1)}^{{\displaystyle \frac{p}{2}}}\ln \left|x+\sqrt{1+x^2}\right|,\hfill & p=0,2,\cdots ,\hfill \\ {(-1)}^{{\displaystyle \frac{p-1}{2}}}\sqrt{1+x^2},\hfill & p=1,3,\cdots ,\hfill \\ {(-1)}^{{\displaystyle \frac{p}{2}}}\ln {\displaystyle \frac{\sqrt{1+x^2}-1}{\left|x\right|}},\hfill & p=-1,-3,\cdots ,\hfill \\ {(-1)}^{{\displaystyle \frac{p+2}{2}}}{\displaystyle \frac{\sqrt{1+x^2}}{x}},\hfill & p=-2,-4,\cdots ,\hfill \end{array}\right.\hfill \end{array}$$

(93)

and

$$\begin{array}{ccc}\hfill w_{p,-{\displaystyle \frac{1}{2}},k}\left(x\right)& =& {\displaystyle \frac{(k-2)!!{(-1)}^{\frac{p-k}{2}}}{(k-1)!!}}{x}^{k-1}\sqrt{1+x^2}.\hfill \end{array}$$

(94)

Note that we have used ${\displaystyle \frac{1}{2}}\ln {\displaystyle \frac{\sqrt{1+x^2}-1}{\sqrt{1+x^2}+1}}=\ln {\displaystyle \frac{\sqrt{1+x^2}-1}{\left|x\right|}}$ in (93).

Example 4.

With (90), we can obtain $K_{3,-{\displaystyle \frac{1}{2}}}={\displaystyle \frac{1}{3}}\left(x^2-2\right)\sqrt{1+x^2}$, $K_{5,-{\displaystyle \frac{1}{2}}}={\displaystyle \frac{1}{15}}\left(3x^4-4x^2+8\right)\sqrt{1+x^2}$, $K_{2,-{\displaystyle \frac{1}{2}}}=-{\displaystyle \frac{1}{2}}\ln \left|x+\sqrt{1+x^2}\right|+{\displaystyle \frac{x}{2}}\sqrt{1+x^2}$, $K_{4,-{\displaystyle \frac{1}{2}}}={\displaystyle \frac{3}{8}}\ln \left|x+\sqrt{1+x^2}\right|+{\displaystyle \frac{x}{8}}\left(2x^2-3\right)\sqrt{1+x^2}$, $K_{-4,-{\displaystyle \frac{1}{2}}}=-{\displaystyle \frac{\sqrt{1+x^2}}{x}}{\displaystyle \frac{(-5)!!}{(-4)!!{(-1)}^{\frac{-2}{2}}}}\left\{1+{(-1)}^{{\displaystyle \frac{-2-2}{2}}}{\displaystyle \frac{(-4)!!}{(-3)!!}}{x}^{-2}\right\}={\displaystyle \frac{\sqrt{1+x^2}}{3x^3}}\left(2x^2-1\right)$, and $K_{-6,-{\displaystyle \frac{1}{2}}}=-{\displaystyle \frac{\sqrt{1+x^2}}{15x^5}}\left(8x^4-4x^2+3\right)$, for instance.

We can also obtain $K_{-3,-{\displaystyle \frac{1}{2}}}={\displaystyle \frac{(-4)!!}{(-3)!!}}{\displaystyle \frac{1}{{(-1)}^{\frac{-3+1}{2}}}}\left\{-{\displaystyle \frac{1}{2}}\ln {\displaystyle \frac{\sqrt{1+x^2}-1}{\sqrt{1+x^2}+1}}-{(-1)}^{{\displaystyle \frac{-1+1}{2}}}{\displaystyle \frac{(-3)!!}{(-2)!!}}{\displaystyle \frac{\sqrt{1+x^2}}{x^2}}\right\}=-{\displaystyle \frac{1}{4}}\ln {\displaystyle \frac{\sqrt{1+x^2}-1}{\sqrt{1+x^2}+1}}-{\displaystyle \frac{\sqrt{1+x^2}}{2x^2}}$ and $K_{-5,-{\displaystyle \frac{1}{2}}}={\displaystyle \frac{-(-6)!!}{(-5)!!{(-1)}^{\frac{-5+1}{2}}}}\left[{\displaystyle \frac{1}{2}}\ln {\displaystyle \frac{\sqrt{1+x^2}-1}{\sqrt{1+x^2}+1}}+\sqrt{1+x^2}\left\{{(-1)}^{{\displaystyle \frac{-1+1}{2}}}{\displaystyle \frac{(-3)!!}{(-2)!!}}{x}^{-2}+{(-1)}^{{\displaystyle \frac{-1+3}{2}}}{\displaystyle \frac{(-5)!!}{(-4)!!}}{x}^{-4}\right\}\right]={\displaystyle \frac{3}{16}}\ln {\displaystyle \frac{\sqrt{1+x^2}-1}{\sqrt{1+x^2}+1}}+{\displaystyle \frac{\sqrt{1+x^2}}{8x^4}}\left(3x^2-2\right)$.

Example 5.

The result (A40) shown in Appendix A.6 for $r=-2$ can be expressed as

$$\begin{array}{ccc}\hfill K_{p,-2}& =& s_p{v}_{p,-2}\left(x\right)+{\displaystyle \sum _{k=L_{p,-2}:2}^{U_{p,-2}}}{w}_{p,-2,k}\left(x\right),\hfill \end{array}$$

(95)

where

$$\begin{array}{ccc}\hfill L_{p,-2}& =& \left\{\begin{array}{cc}2,\hfill & p=0,2,\cdots ,\hfill \\ 5,\hfill & p=1,3,\cdots ,\hfill \\ p+2,\hfill & p=-1,-2,\cdots ,\hfill \end{array}\right.\hfill \end{array}$$

(96)

$$\begin{array}{ccc}\hfill U_{p,-2}& =& \left\{\begin{array}{cc}0,\hfill & p=0,-2,-4,\cdots ,\hfill \\ -1,\hfill & p=-1,-3,\cdots ,\hfill \\ p,\hfill & p=1,2,\cdots ,\hfill \end{array}\right.\hfill \end{array}$$

(97)

$$\begin{array}{ccc}\hfill v_{p,-2}\left(x\right)& =& \left\{\begin{array}{cc}-{\displaystyle \frac{1}{2\left(1+x^2\right)}},\hfill & p=1,\hfill \\ {(-1)}^{{\displaystyle \frac{p+2}{2}}}{\displaystyle \frac{p-1}{2}}\left({tan}^{-1}x+{\displaystyle \frac{x}{1+x^2}}\right),\hfill & p=0,\pm 2,\cdots ,\hfill \\ s_p{(-1)}^{{\displaystyle \frac{p+1}{2}}}{\displaystyle \frac{p-1}{4}}\left(\ln {\displaystyle \frac{{\left|1+x^2\right|}^{s_p}}{{\left|x\right|}^{s_p-1}}}-{\displaystyle \frac{x^2}{1+x^2}}\right),\hfill & p=-1,\pm 3,\pm 5,\cdots ,\hfill \end{array}\right.\hfill \end{array}$$

(98)

and

$$\begin{array}{ccc}\hfill w_{p,-2,k}\left(x\right)& =& {\displaystyle \frac{{(-1)}^{\frac{p-k}{2}}{s}_p}{(k-1)(k-3)}}{\displaystyle \frac{x^{k-1}}{1+x^2}}.\hfill \end{array}$$

(99)

Thus, for instance, we have

$$\begin{array}{ccc}\hfill K_{5,-2}& =& -\ln \left|1+x^2\right|+{\displaystyle \frac{x^2}{1+x^2}}+{\displaystyle \frac{x^4}{2\left(1+x^2\right)}}\hfill \end{array}$$

(100)

when $p=5$ from (95) or (A36),

$$\begin{array}{ccc}\hfill K_{-3,-2}& =& -\ln {\displaystyle \frac{x^2}{1+x^2}}+{\displaystyle \frac{x^2}{1+x^2}}-{\displaystyle \frac{1}{2x^2\left(1+x^2\right)}}\hfill \end{array}$$

(101)

when $p=-3$ from (95) or (A37),

$$\begin{array}{ccc}\hfill K_{2,-2}& =& {\displaystyle \frac{1}{2}}\left({tan}^{-1}x-{\displaystyle \frac{x}{1+x^2}}\right)\hfill \end{array}$$

(102)

when $p=2$ from (95) or (A38), and

$$\begin{array}{ccc}\hfill K_{-2,-2}& =& -{\displaystyle \frac{3}{2}}{tan}^{-1}x-{\displaystyle \frac{3x^2+2}{2x\left(1+x^2\right)}}\hfill \end{array}$$

(103)

when $p=-2$ from (95) or (A39).

Table 4 shows some explicit formulas of $K_{p,r}=\int x^p{\left(1+x^2\right)}^{r}dx$ that have been obtained by the steps described in Section 2, Section 3 and Section 4.

5. Conclusions

In this letter, by employing the well-known techniques, i.e., substitution (change of variables) and integration by parts, for integration, we have addressed the integral $K_{p,r}=\int x^p{\left(1+x^2\right)}^{r}dx$, where we have focused on the cases of integers for p and integral multiples of ${\displaystyle \frac{1}{2}}$ for r. The cases $p=\pm 1$, $r=0$, $r=-1$, and $p+2r+3=0$ of the integral have first been obtained by direct calculations. For the remaining cases of the integral including the special case of $p+2r+1=0$, explicit formulas for the integral have been obtained via recursive formulas derived based on integration by parts: in these cases, the formulas obtained by direct calculations are used as the initial conditions. Detailed discussions on the issue of choosing the initial conditions have also been provided. Some of the explicit formulas of the integral are tabulated, and examples and applications of the results are also included.

Appendix A.1. the Case p = −1

First, the case $(p,r)=(-1,-1)$ has already been addressed in (45). When $\{p=-1\}\cap \{r\ne -1\}$, we can use partial fraction expansion based on $K_{-1,r}=-\int {\displaystyle \frac{v^{-2r-1}dv}{\left(1-v\right)\left(1+v\right)}}$ from (6) with $v={\displaystyle \frac{1}{\sqrt{1+x^2}}}$ or

$$\begin{array}{ccc}\hfill K_{-1,r}& =& \int {\displaystyle \frac{t^{2r+1}}{t^2-1}}dt\hfill \end{array}$$

(A1)

from (18) with $t=\sqrt{1+x^2}$. In addition, the formula

$$\begin{array}{ccc}\hfill K_{-1,r}& =& {\displaystyle \frac{1}{2}}\int {\displaystyle \frac{t^r}{t-1}}dt\hfill \end{array}$$

(A2)

from (14) with $t=1+x^2$ is more and less convenient than (A1) when r is even and odd, respectively, multiple of ${\displaystyle \frac{1}{2}}$.

First, for $r=0,1,\cdots$, we have $K_{-1,r}={\displaystyle \frac{1}{2}}\int {\displaystyle \frac{t^r}{t-1}}dt={\displaystyle \frac{1}{2}}\int {(v+1)}^r{\displaystyle \frac{dv}{v}}={\displaystyle \frac{1}{2}}\int {\displaystyle \sum _{k=0}^r}\left({\displaystyle \genfrac{}{}{0pt}{}{r}{k}}\right)v^{k-1}dv$ with $v=t-1=x^2$ from (A2). Thus,

$$\begin{array}{ccc}\hfill K_{-1,r}& =& \ln \left|x\right|+{\displaystyle \sum _{k=1}^r}{\displaystyle \frac{\left(\genfrac{}{}{0pt}{}{r}{k}\right)}{2k}}{x}^{2k}.\hfill \end{array}$$

(A3)

For example, we get $K_{-1,0}=\ln \left|x\right|$ and $K_{-1,1}=\ln \left|x\right|+{\displaystyle \frac{x^2}{2}}$.

Subsequently, for $r=-1,-2\cdots$, we have $K_{-1,r}={\displaystyle \frac{1}{2}}\int {\displaystyle \frac{dt}{t^q(t-1)}}={\displaystyle \frac{1}{2}}\int {\displaystyle \frac{y^{q-1}}{y-1}}dy$ with $q=-r$, $y={\displaystyle \frac{1}{t}}$, and $dt=-{\displaystyle \frac{dy}{y^2}}$ from (A2). Thus, letting $v=y-1$, we get $K_{-1,r}={\displaystyle \frac{1}{2}}\int {\displaystyle \frac{y^{q-1}}{y-1}}dy={\displaystyle \frac{1}{2}}\left\{\ln \left|v\right|+{\displaystyle \sum _{k=1}^{q-1}}{\displaystyle \frac{\left(\genfrac{}{}{0pt}{}{q-1}{k}\right)}{k}}{v}^k\right\}$ or

$$\begin{array}{ccc}\hfill K_{-1,r}& =& {\displaystyle \frac{1}{2}}\ln {\displaystyle \frac{x^2}{1+x^2}}+{\displaystyle \sum _{k=1}^{-r-1}}{\displaystyle \frac{\left(\genfrac{}{}{0pt}{}{-r-1}{k}\right)}{2k}}{\left({\displaystyle \frac{-x^2}{1+x^2}}\right)}^k.\hfill \end{array}$$

(A4)

For example, we get $K_{-1,-1}={\displaystyle \frac{1}{2}}\ln {\displaystyle \frac{x^2}{1+x^2}}$ and $K_{-1,-2}={\displaystyle \frac{1}{2}}\ln {\displaystyle \frac{x^2}{1+x^2}}-{\displaystyle \frac{x^2}{2\left(1+x^2\right)}}$: the latter result can alternatively be obtained as $K_{-1,-2}=\int {\displaystyle \frac{dt}{\left(t^2-1\right)t^3}}={\displaystyle \frac{1}{2}}\int {\displaystyle \frac{dq}{{\left(q+1\right)}^{2}q}}={\displaystyle \frac{1}{2}}\int \left\{{\displaystyle \frac{1}{q}}+{\displaystyle \frac{-1}{q+1}}+{\displaystyle \frac{-1}{{(q+1)}^2}}\right\}dq={\displaystyle \frac{1}{2}}\ln \left|{\displaystyle \frac{q}{q+1}}\right|+{\displaystyle \frac{1}{2(q+1)}}$ or

$$\begin{array}{ccc}\hfill K_{-1,-2}& =& {\displaystyle \frac{1}{2}}\ln {\displaystyle \frac{x^2}{1+x^2}}+{\displaystyle \frac{1}{2\left(1+x^2\right)}}\hfill \end{array}$$

(A5)

with $q=t^2-1=x^2$ from (A1): the last term ${\displaystyle \frac{1}{2\left(1+x^2\right)}}$ of (A5) is equivalent to the last term $-{\displaystyle \frac{x^2}{2\left(1+x^2\right)}}$ of $K_{-1,-2}$ shown in (51) because $-{\displaystyle \frac{x^2}{2\left(1+x^2\right)}}=-{\displaystyle \frac{1+x^2-1}{2\left(1+x^2\right)}}=-{\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{2\left(1+x^2\right)}}$.

When $r=m+{\displaystyle \frac{1}{2}}$ for $m=0,1,\cdots$, using that ${\displaystyle \frac{t^{2m+2}}{t^2-1}}=t^{2m}+t^{2m-2}+\cdots +1+{\displaystyle \frac{1}{t^2-1}}={\displaystyle \sum _{k=0}^m}{t}^{2(m-k)}+{\displaystyle \frac{\frac{1}{2}}{t-1}}+{\displaystyle \frac{-\frac{1}{2}}{t+1}}$, we get $K_{-1,m+{\displaystyle \frac{1}{2}}}=\int {\displaystyle \frac{t^{2m+2}}{t^2-1}}dt$ or

$$\begin{array}{ccc}\hfill K_{-1,m+{\displaystyle \frac{1}{2}}}& =& {\displaystyle \frac{1}{2}}\ln {\displaystyle \frac{\sqrt{1+x^2}-1}{\sqrt{1+x^2}+1}}+{\displaystyle \sum _{k=0}^m}{\displaystyle \frac{{\left(1+x^2\right)}^{m-k+\frac{1}{2}}}{2(m-k)+1}}\hfill \end{array}$$

(A6)

from (A1). For example, $K_{-1,{\displaystyle \frac{1}{2}}}={\displaystyle \frac{1}{2}}\ln {\displaystyle \frac{\sqrt{1+x^2}-1}{\sqrt{1+x^2}+1}}+\sqrt{1+x^2}$ when $m=0$ and $K_{-1,{\displaystyle \frac{3}{2}}}={\displaystyle \frac{1}{2}}\ln {\displaystyle \frac{\sqrt{1+x^2}-1}{\sqrt{1+x^2}+1}}+{\displaystyle \frac{1}{3}}{\left(1+x^2\right)}^{{\displaystyle \frac{3}{2}}}+\sqrt{1+x^2}$ when $m=1$.

Subsequently, when $r=q+{\displaystyle \frac{1}{2}}$ for $q=-1,-2\cdots$, using (A1) we get $K_{-1,q+{\displaystyle \frac{1}{2}}}=\int {\displaystyle \frac{t^{2q+2}}{t^2-1}}dt=-\int {\displaystyle \frac{v^{-2q-2}}{\frac{1}{v^2}-1}}{\displaystyle \frac{dv}{v^2}}=\int {\displaystyle \frac{v^{-2q-2}}{v^2-1}}dv={\displaystyle \frac{1}{2}}\ln {\displaystyle \frac{\sqrt{1+x^2}-1}{\sqrt{1+x^2}+1}}+{\displaystyle \sum _{k=0}^{-q-2}}{\displaystyle \frac{{\left(1+x^2\right)}^{q+k+2-\frac{1}{2}}}{2(-q-2-k)+1}}$ or

$$\begin{array}{ccc}\hfill K_{-1,q+{\displaystyle \frac{1}{2}}}& =& {\displaystyle \frac{1}{2}}\ln {\displaystyle \frac{\sqrt{1+x^2}-1}{\sqrt{1+x^2}+1}}-{\displaystyle \sum _{k=0}^{-q-2}}{\displaystyle \frac{{\left(1+x^2\right)}^{q+k+\frac{3}{2}}}{2q+2k+3}}\hfill \end{array}$$

(A7)

by replacing m and $\sqrt{1+x^2}$ in (A6) with $-q-2$ (from $2m+2=-2q-2$ we get $m=-q-2$) and ${\displaystyle \frac{1}{\sqrt{1+x^2}}}$, respectively. For example, we get $K_{-1,-{\displaystyle \frac{1}{2}}}={\displaystyle \frac{1}{2}}\ln {\displaystyle \frac{\sqrt{1+x^2}-1}{\sqrt{1+x^2}+1}}$ when $q=-1$ and $K_{-1,-{\displaystyle \frac{3}{2}}}={\displaystyle \frac{1}{2}}\ln {\displaystyle \frac{\sqrt{1+x^2}-1}{\sqrt{1+x^2}+1}}+{\displaystyle \frac{1}{\sqrt{1+x^2}}}$ when $q=-2$.

Collecting the results (A3), (A4), (A6), and (A7), we have

$$\begin{array}{ccc}\hfill K_{-1,r}& =& \left\{\begin{array}{cc}\ln \left|x\right|+{\displaystyle \sum _{k=1}^r}{\displaystyle \frac{\left(\genfrac{}{}{0pt}{}{r}{k}\right)}{2k}}{x}^{2k},\hfill & r=0,1,\cdots ,\hfill \\ {\displaystyle \frac{1}{2}}\ln {\displaystyle \frac{x^2}{1+x^2}}+{\displaystyle \sum _{k=1}^{-r-1}}{\displaystyle \frac{\left(\genfrac{}{}{0pt}{}{-r-1}{k}\right)}{2k}}{\left({\displaystyle \frac{-x^2}{1+x^2}}\right)}^k,\hfill & r=-1,-2,\cdots ,\hfill \\ {\displaystyle \frac{1}{2}}\ln {\displaystyle \frac{\sqrt{1+x^2}-1}{\sqrt{1+x^2}+1}}+{\displaystyle \sum _{k=0}^{r-{\displaystyle \frac{1}{2}}}}{\displaystyle \frac{{\left(1+x^2\right)}^{r-k}}{2(r-k)}},\hfill & r={\displaystyle \frac{1}{2}},{\displaystyle \frac{3}{2}},\cdots ,\hfill \\ {\displaystyle \frac{1}{2}}\ln {\displaystyle \frac{\sqrt{1+x^2}-1}{\sqrt{1+x^2}+1}}-{\displaystyle \sum _{k=0}^{-r-{\displaystyle \frac{3}{2}}}}{\displaystyle \frac{{\left(1+x^2\right)}^{r+k+1}}{2(r+k+1)}},\hfill & r=-{\displaystyle \frac{1}{2}},-{\displaystyle \frac{3}{2}},\cdots .\hfill \end{array}\right.\hfill \end{array}$$

(A8)

In passing, let us note that a result equivalent to (A8) can be obtained via (26) alternatively. For example, by using (26) recursively and recollecting (23), we have $K_{-1,r}=K_{-1,0}+{\displaystyle \frac{1}{2}}{\displaystyle \sum _{k=1}^r}{\displaystyle \frac{1}{k}}{\left(1+x^2\right)}^k$ or

$$\begin{array}{ccc}\hfill K_{-1,r}& =& \ln \left|x\right|+{\displaystyle \frac{1}{2}}{\displaystyle \sum _{k=1}^r}{\displaystyle \frac{1}{k}}{\left(1+x^2\right)}^k\hfill \end{array}$$

(A9)

for $r=0,1,\cdots$. The term ${\displaystyle \sum _{k=1}^r}{\displaystyle \frac{1}{k}}{\left(1+x^2\right)}^k={\displaystyle \sum _{k=1}^r}{\displaystyle \frac{1}{k}}{\displaystyle \sum _{j=0}^k}\left({\displaystyle \genfrac{}{}{0pt}{}{k}{j}}\right)x^{2j}$ in (A9) can be expressed as ${\displaystyle \sum _{k=1}^r}{\displaystyle \frac{1}{k}}{\displaystyle \sum _{j=0}^k}\left({\displaystyle \genfrac{}{}{0pt}{}{k}{j}}\right)x^{2j}={\displaystyle \sum _{k=1}^r}{\displaystyle \frac{1}{k}}{\displaystyle \sum _{j=0}^r}\left({\displaystyle \genfrac{}{}{0pt}{}{k}{j}}\right)x^{2j}={\displaystyle \sum _{j=0}^r}{x}^{2j}{\displaystyle \sum _{k=1}^r}{\displaystyle \frac{1}{k}}\left({\displaystyle \genfrac{}{}{0pt}{}{k}{j}}\right)={\displaystyle \sum _{j=1}^r}{x}^{2j}{\displaystyle \sum _{k=1}^r}{\displaystyle \frac{1}{j}}\left({\displaystyle \genfrac{}{}{0pt}{}{k-1}{j-1}}\right)+x^0{\displaystyle \sum _{k=1}^r}{\displaystyle \frac{1}{k}}={\displaystyle \sum _{j=1}^r}{x}^{2j}{\displaystyle \frac{1}{j}}{\displaystyle \sum _{t=0}^{r-1}}\left({\displaystyle \genfrac{}{}{0pt}{}{t}{j-1}}\right)+C_r={\displaystyle \sum _{j=1}^r}{x}^{2j}{\displaystyle \frac{1}{j}}\left({\displaystyle \genfrac{}{}{0pt}{}{r}{j}}\right)+C_r$, where $C_r={\displaystyle \sum _{k=1}^r}{\displaystyle \frac{1}{k}}$: we have used that $\left({\displaystyle \genfrac{}{}{0pt}{}{k}{j}}\right)=0$ when the integers j and k satisfy $j>k\ge 0$ [[16] Table 1.4] and that ${\displaystyle \sum _{t=0}^r}\left({\displaystyle \genfrac{}{}{0pt}{}{t}{s}}\right)={\displaystyle \sum _{t=s}^r}\left({\displaystyle \genfrac{}{}{0pt}{}{t}{s}}\right)=\left({\displaystyle \genfrac{}{}{0pt}{}{r+1}{s+1}}\right)$ [[16] (1.E.22)]. In essence, (A9) is equivalent to the first line on the right-hand side (A8).

Appendix A.2. the Case r = −1

The cases $(0,-1)$, $(p,r)=(1,-1)$, and $(-1,-1)$ have been addressed in (9), (44), and (45), respectively. Let us next consider the remaining cases $\{p\ne \pm 1,0\}\cap \{r=-1\}$. First, recollect that ${\displaystyle \sum _{k=1}^m}{(-1)}^{k+1}{x}^{2(m-k)}={\displaystyle \frac{{(-1)}^{m+1}\left\{1-{\left(-x^2\right)}^m\right\}}{1-\left(-x^2\right)}}$ or

$$\begin{array}{ccc}\hfill {\displaystyle \sum _{k=1}^m}{(-1)}^{k+1}{x}^{2(m-k)}& =& {\displaystyle \frac{{(-1)}^{m+1}+x^{2m}}{1+x^2}}\hfill \end{array}$$

(A10)

and ${(-1)}^{m+1}+{(-1)}^m=0$ for $m=1,2,\cdots$. Then, for $p=2m$ with $m=1,2,\cdots$, we have $K_{2m,-1}=\int {\displaystyle \frac{x^{2m}}{1+x^2}}dx=\int \left\{{\displaystyle \sum _{k=1}^m}{(-1)}^{k+1}{x}^{2(m-k)}+{\displaystyle \frac{{(-1)}^m}{1+x^2}}\right\}dx$ or

$$\begin{array}{ccc}\hfill K_{2m,-1}& =& {\displaystyle \sum _{k=1}^m}{\displaystyle \frac{{(-1)}^{k+1}}{2(m-k)+1}}{x}^{2(m-k)+1}+{(-1)}^m{tan}^{-1}x\hfill \end{array}$$

(A11)

using (A10). Similarly, when $p=2m$ with $m=-1,-2,\cdots$, we get $K_{2m,-1}=\int {\displaystyle \frac{x^{2m}}{1+x^2}}dx=-\int {\displaystyle \frac{y^{-2m}}{1+y^2}}dy$ or

$$\begin{array}{ccc}\hfill K_{2m,-1}& =& -{(-1)}^{-m}{tan}^{-1}{\displaystyle \frac{1}{x}}-{\displaystyle \sum _{k=1}^{-m}}{\displaystyle \frac{{(-1)}^{k+1}{x}^{2m+2k-1}}{2(-m-k)+1}}\hfill \end{array}$$

(A12)

with $y={\displaystyle \frac{1}{x}}$ and $dx=-{\displaystyle \frac{dy}{y^2}}$. Note that, when $x=tan\theta$ or $\theta ={tan}^{-1}x$, we have ${tan}^{-1}{\displaystyle \frac{1}{x}}={\displaystyle \frac{\pi }{2}}-\theta =-{tan}^{-1}x+{\displaystyle \frac{\pi }{2}}$. Thus, for example, $K_{-2,-1}=-{\displaystyle \frac{1}{x}}+{tan}^{-1}{\displaystyle \frac{1}{x}}+c_1$ from (A12) while it can also be expressed as $K_{-2,-1}=-{\displaystyle \frac{1}{x}}-{tan}^{-1}x+c_2$, which can be derived as $K_{-2,-1}=\int {\left({\displaystyle \frac{1-t^2}{2t}}\right)}^2{\displaystyle \frac{2}{1+t^2}}dt={\displaystyle \frac{1}{2}}\int {\displaystyle \frac{1-2t^2+t^4}{t^2\left(1+t^2\right)}}dt={\displaystyle \frac{1}{2}}\int \left\{1+{\displaystyle \frac{-3t^2+1}{t^2\left(1+t^2\right)}}\right\}dt={\displaystyle \frac{1}{2}}\int \left(1+{\displaystyle \frac{1}{t^2}}+{\displaystyle \frac{-4}{1+t^2}}\right)dt={\displaystyle \frac{1}{2}}\left(t-{\displaystyle \frac{1}{t}}-4{tan}^{-1}t\right)={\displaystyle \frac{1}{2}}\left(-{\displaystyle \frac{1-t^2}{t}}-2\theta \right)={\displaystyle \frac{1}{2}}\left(-{\displaystyle \frac{2}{tan\theta }}-2{tan}^{-1}x\right)$ or

$$\begin{array}{ccc}\hfill K_{-2,-1}& =& -{\displaystyle \frac{1}{x}}-{tan}^{-1}x.\hfill \end{array}$$

(A13)

We also easily get $K_{2,-1}=\int \left(1-{\displaystyle \frac{1}{1+x^2}}\right)dx$ or

$$\begin{array}{ccc}\hfill K_{2,-1}& =& x-{tan}^{-1}x.\hfill \end{array}$$

(A14)

Secondly, for $p=2m+1$ with $m=0,1,\cdots$, we have $K_{2m+1,-1}=\int {\displaystyle \frac{x^{2m+1}}{1+x^2}}dx={\displaystyle \frac{1}{2}}\int {\displaystyle \frac{{(t-1)}^m}{t}}dt={\displaystyle \frac{1}{2}}\int {\displaystyle \sum _{k=0}^m}\left({\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right){(-1)}^{m-k}{t}^{k-1}dt$ or

$$\begin{array}{ccc}\hfill K_{2m+1,-1}& =& {\displaystyle \frac{{(-1)}^m}{2}}\ln \left|1+x^2\right|+{\displaystyle \sum _{k=1}^m}{\displaystyle \frac{\left(\genfrac{}{}{0pt}{}{m}{k}\right){(-1)}^{m-k}}{2k}}{\left(1+x^2\right)}^k\hfill \end{array}$$

(A15)

from (14) with $t=1+x^2$. Similarly, when $p=-2m-1$ for $m=0,1,\cdots$, with $y={\displaystyle \frac{1}{x}}$ and $dx=-{\displaystyle \frac{dy}{y^2}}$, we get $K_{-2m-1,-1}=\int {\displaystyle \frac{dx}{x^{2m+1}\left(1+x^2\right)}}=-\int {\displaystyle \frac{y^{2m+1}}{1+y^2}}dy$, which is basically the same as (A15) with x replaced by y. Thus,

$$\begin{array}{ccc}\hfill K_{-2m-1,-1}& =& {\displaystyle \frac{1}{2}}{(-1)}^{m+1}\ln {\displaystyle \frac{1+x^2}{x^2}}\hfill \\ & & +{\displaystyle \sum _{k=1}^m}{\displaystyle \frac{\left(\genfrac{}{}{0pt}{}{m}{k}\right){(-1)}^{m-k+1}}{2k}}{\left({\displaystyle \frac{1+x^2}{x^2}}\right)}^k.\hfill \end{array}$$

(A16)

For example, we get $K_{1,-1}={\displaystyle \frac{1}{2}}\ln \left|1+x^2\right|$ when $m=0$ and $K_{3,-1}={\displaystyle \frac{1}{2}}\left(1+x^2-\ln \left|1+x^2\right|\right)$ when $m=1$ from (A15), and $K_{-1,-1}=-{\displaystyle \frac{1}{2}}\ln {\displaystyle \frac{1+x^2}{x^2}}={\displaystyle \frac{1}{2}}\ln {\displaystyle \frac{x^2}{1+x^2}}$ when $m=0$ and $K_{-3,-1}={\displaystyle \frac{1}{2}}\ln {\displaystyle \frac{1+x^2}{x^2}}-{\displaystyle \frac{1+x^2}{2x^2}}$ when $m=1$ from (A16).

Collecting the results (A11), (A12), (A15), and (A16), we get

$$\begin{array}{ccc}\hfill K_{p,-1}& =& \left\{\begin{array}{cc}{(-1)}^{{\displaystyle \frac{p}{2}}}{tan}^{-1}x+{\displaystyle \sum _{k=1}^{{\displaystyle \frac{p}{2}}}}{\displaystyle \frac{{(-1)}^{k+1}}{p-2k+1}}{x}^{p-2k+1},\hfill & p=0,2,\cdots ,\hfill \\ -{(-1)}^{-{\displaystyle \frac{p}{2}}}{tan}^{-1}{\displaystyle \frac{1}{x}}\hfill \\ -{\displaystyle \sum _{k=1}^{-{\displaystyle \frac{p}{2}}}}{\displaystyle \frac{{(-1)}^{k+1}}{-p-2k+1}}{\left({\displaystyle \frac{1}{x}}\right)}^{-p-2k+1},\hfill & p=-2,-4,\cdots ,\hfill \\ {\displaystyle \frac{1}{2}}{(-1)}^{{\displaystyle \frac{p-1}{2}}}\ln \left|1+x^2\right|\hfill \\ +{\displaystyle \sum _{k=1}^{{\displaystyle \frac{p-1}{2}}}}{\displaystyle \frac{{(-1)}^{\frac{p-1}{2}-k}}{2k}}\left({\displaystyle \genfrac{}{}{0pt}{}{\frac{p-1}{2}}{k}}\right){\left(1+x^2\right)}^k,\hfill & p=1,3,\cdots ,\hfill \\ -{\displaystyle \frac{1}{2}}{(-1)}^{{\displaystyle \frac{\left|p\right|-1}{2}}}\ln {\displaystyle \frac{1+x^2}{x^2}}\hfill \\ +{\displaystyle \sum _{k=1}^{{\displaystyle \frac{\left|p\right|-1}{2}}}}{\displaystyle \frac{{(-1)}^{\frac{\left|p\right|-1}{2}-k+1}}{2k}}\left({\displaystyle \genfrac{}{}{0pt}{}{\frac{\left|p\right|-1}{2}}{k}}\right){\left({\displaystyle \frac{1+x^2}{x^2}}\right)}^k,\hfill & p=-1,-3,\cdots .\hfill \end{array}\right.\hfill \end{array}$$

(A17)

Similarly to the case of $p=-1$ shown in Appendix A.1, the recursion (41) may alternatively be employed to obtain (A17). For example, by adding

$$\begin{array}{ccc}\hfill K_{p,-1}& =& {\displaystyle \frac{x^{p-1}}{p-1}}-K_{p-2,-1}\hfill \\ \hfill (-1)K_{p-2,-1}& =& (-1){\displaystyle \frac{x^{p-3}}{p-3}}-(-1)K_{p-4,-1}\hfill \\ \hfill {(-1)}^2{K}_{p-4,-1}& =& {(-1)}^2{\displaystyle \frac{x^{p-5}}{p-5}}-{(-1)}^2{K}_{p-6,-1}\hfill \\ & ⋮& \hfill \\ \hfill {(-1)}^{{\displaystyle \frac{p-s}{2}}}{K}_{s,-1}& =& {(-1)}^{{\displaystyle \frac{p-s}{2}}}{\displaystyle \frac{x^{s-1}}{s-1}}-{(-1)}^{{\displaystyle \frac{p-s}{2}}}{K}_{s-2,-1}\hfill \end{array}$$

we will get

$$\begin{array}{ccc}\hfill K_{p,-1}& =& -{(-1)}^{{\displaystyle \frac{p-s}{2}}}{K}_{s-2,-1}+{\displaystyle \sum _{k=2:2}^{p-s+2}}{(-1)}^{{\displaystyle \frac{k-2}{2}}}{\displaystyle \frac{x^{p-k+1}}{p-k+1}}.\hfill \end{array}$$

(A18)

It is easy to see that (A18), with $s=2$ and $K_{0,-1}={tan}^{-1}x$ from (9), is the same as the first line on the right-hand of (A17).

Appendix A.3. the Case p = 1

When $p=1$, with $S(1,k)={\displaystyle \frac{\left(2k\right)!!}{(2k+2)!!}}={\displaystyle \frac{1}{2k+2}}$ and $a_{1,k}={\displaystyle \frac{x^2{\left(1+x^2\right)}^k}{2k+2}}$, (74) and (75) can be expressed as $K_{1,r}=S(1,r)\left\{{\displaystyle \frac{K_{1,s}}{S(1,s)}}+{\displaystyle \sum _{k=s+1}^r}{\displaystyle \frac{a_{1,k}}{S(1,k)}}\right\}$ or

$$\begin{array}{ccc}\hfill K_{1,r}& =& {\displaystyle \frac{1}{2r+2}}\left\{(2s+2)K_{1,s}+x^2{\displaystyle \sum _{k=s+1}^r}{\left(1+x^2\right)}^k\right\}\hfill \end{array}$$

(A19)

for $s<r$, and $K_{1,r}=S(1,r)\left\{{\displaystyle \frac{K_{1,s}}{S(1,s)}}-{\displaystyle \sum _{k=r+1}^s}{\displaystyle \frac{a_{1,k}}{S(1,k)}}\right\}$ or

$$\begin{array}{ccc}\hfill K_{1,r}& =& {\displaystyle \frac{1}{2r+2}}\left\{(2s+2)K_{1,s}-x^2{\displaystyle \sum _{k=r+1}^s}{\left(1+x^2\right)}^k\right\}\hfill \end{array}$$

(A20)

for $s>r$, respectively. In addition, from $r_1=-{\displaystyle \frac{1+1}{2}}=-1$ and $r_2=0$, we can choose the set $\{K_{1,0}$, $K_{1,-1}$, $K_{1,-{\displaystyle \frac{1}{2}}}$, $K_{1,-2}\}$ of IC’s.

Now, with $s=0$, $S(1,0)={\displaystyle \frac{1}{2}}$, and the initial condition $K_{1,0}={\displaystyle \frac{x^2}{2}}$ from (43), we get $K_{1,r}={\displaystyle \frac{1}{2r+2}}\left\{x^2+x^2{\displaystyle \sum _{k=1}^r}{\left(1+x^2\right)}^k\right\}={\displaystyle \frac{x^2}{2r+2}}\left\{1+\left(1+x^2\right){\displaystyle \frac{1-{\left(1+x^2\right)}^r}{1-\left(1+x^2\right)}}\right\}$ or

$$\begin{array}{ccc}\hfill K_{1,r}& =& {\displaystyle \frac{{\left(1+x^2\right)}^{r+1}-1}{2(r+1)}}\hfill \end{array}$$

(A21)

for $r=1,2,\cdots$ from (A19). With $s=-2$, $S(1,-2)=-{\displaystyle \frac{1}{2}}$, and the initial condition $K_{1,-2}={\displaystyle \frac{x^2}{2\left(1+x^2\right)}}$ shown in (36), we get $K_{1,r}={\displaystyle \frac{1}{2r+2}}\left\{-2K_{1,-2}-x^2{\displaystyle \sum _{k=r+1}^{-2}}{\left(1+x^2\right)}^k\right\}={\displaystyle \frac{1}{2r+2}}\left\{-{\displaystyle \frac{x^2}{1+x^2}}-x^2{\left(1+x^2\right)}^{r+1}{\displaystyle \frac{1-{\left(1+x^2\right)}^{-r-2}}{1-\left(1+x^2\right)}}\right\}$ or

$$\begin{array}{ccc}\hfill K_{1,r}& =& {\displaystyle \frac{{\left(1+x^2\right)}^{r+1}-1}{2(r+1)}}\hfill \end{array}$$

(A22)

for $r=-3,-4,\cdots$ from (A20).

Similarly, with $s=-{\displaystyle \frac{1}{2}}$, $S\left(1,-{\displaystyle \frac{1}{2}}\right)=1$, and the initial condition $K_{1,-{\displaystyle \frac{1}{2}}}=\sqrt{1+x^2}$ shown in (19), we get $K_{1,r}={\displaystyle \frac{1}{2r+2}}\left\{\sqrt{1+x^2}+x^2{\displaystyle \sum _{k={\displaystyle \frac{1}{2}}}^r}{\left(1+x^2\right)}^k\right\}={\displaystyle \frac{1}{2r+2}}\left\{\sqrt{1+x^2}+x^2\sqrt{1+x^2}{\displaystyle \frac{1-{\left(1+x^2\right)}^{r+\frac{1}{2}}}{1-\left(1+x^2\right)}}\right\}$ or

$$\begin{array}{ccc}\hfill K_{1,r}& =& {\displaystyle \frac{{\left(1+x^2\right)}^{r+1}}{2(r+1)}}\hfill \end{array}$$

(A23)

for $r={\displaystyle \frac{1}{2}},{\displaystyle \frac{3}{2}},\cdots$ from (A19). We can also obtain $K_{1,r}={\displaystyle \frac{1}{2r+2}}\left\{\sqrt{1+x^2}-{\displaystyle \sum _{k=r+1}^{-{\displaystyle \frac{1}{2}}}}{x}^2{\left(1+x^2\right)}^k\right\}={\displaystyle \frac{1}{2r+2}}\left\{\sqrt{1+x^2}\right.\left.-x^2{\left(1+x^2\right)}^{r+1}{\displaystyle \frac{1-{\left(1+x^2\right)}^{-r-\frac{1}{2}}}{1-\left(1+x^2\right)}}\right\}$ or

$$\begin{array}{ccc}\hfill K_{1,r}& =& {\displaystyle \frac{{\left(1+x^2\right)}^{r+1}}{2(r+1)}}\hfill \end{array}$$

(A24)

for $r=-{\displaystyle \frac{3}{2}},-{\displaystyle \frac{5}{2}},\cdots$ from (A20).

Combining the results (A21)–(A24), and noting (11), we easily see that the result is the same as the result (44) subject to an integral constant.

Appendix A.4. Derivation of (86) and (87)

With $c_{p,r}={\displaystyle \frac{x^{p-1}{\left(1+x^2\right)}^{r+1}}{p+2r+1}}$ and $d_{p,r}=-{\displaystyle \frac{p-1}{p+2r+1}}$, if we add

$$\begin{array}{ccc}\hfill K_{p,r}& =& c_{p,r}+d_{p,r}{K}_{p-2,r},\hfill \end{array}$$

(A25)

$$\begin{array}{ccc}\hfill d_{p,r}{K}_{p-2,r}& =& d_{p,r}{c}_{p-2,r}+d_{p-2,r}{d}_{p,r}{K}_{p-4,r},\hfill \end{array}$$

(A26)

$$\begin{array}{ccc}\hfill d_{p,r}{d}_{p-2,r}{K}_{p-4,r}& =& d_{p-2,r}{d}_{p,r}{c}_{p-4,r}+d_{p-4,r}{d}_{p-2,r}{d}_{p,r}{K}_{p-6,r},\hfill \\ & ⋮& \hfill \end{array}$$

(A27)

$$\begin{array}{ccc}\hfill K_{s-2,r}{\displaystyle \prod _{m=s:2}^p}{d}_{m,r}& =& c_{s+2,r}{\displaystyle \prod _{m=s+2:2}^p}{d}_{m,r}+K_{s,r}{\displaystyle \prod _{m=s+2:2}^p}{d}_{m,r},\hfill \end{array}$$

(A28)

we get

$$\begin{array}{ccc}\hfill K_{p,r}& =& {\displaystyle \sum _{k=s+2:2}^p}{c}_{k,r}{\displaystyle \prod _{m=k+2:2}^p}{d}_{m,r}+K_{s,r}{\displaystyle \prod _{m=s+2:2}^p}{d}_{m,r}\hfill \end{array}$$

(A29)

for $p>s$. Noting that ${\displaystyle \prod _{m=k+2:2}^p}{d}_{m,r}={\displaystyle \frac{-(k+1)}{k+2r+3}}{\displaystyle \frac{-(k+3)}{k+2r+5}}\cdots {\displaystyle \frac{-(p-1)}{p+2r+1}}={\displaystyle \frac{{(-1)}^{\frac{p-k}{2}}(p-1)!!}{(p+2r+1)!!}}{\displaystyle \frac{(k+2r+1)!!}{(k-1)!!}}$ can be rewritten as

$$\begin{array}{ccc}\hfill {\displaystyle \prod _{m=k+2:2}^p}{d}_{m,r}& =& {(-1)}^{{\displaystyle \frac{p-k}{2}}}{\displaystyle \frac{T(p,r)}{T(k,r)}},\hfill \end{array}$$

(A30)

we get (86) and (87) from (A29).

Appendix A.5. the Case $r=-{\displaystyle \frac{1}{2}}$

When $r=-{\displaystyle \frac{1}{2}}$, we get $p_1=-2\left(-{\displaystyle \frac{1}{2}}\right)-1=0$, $p_L=min(0,1)=0$, $p_U=max(0,1)=1$, $c_{k,-{\displaystyle \frac{1}{2}}}={\displaystyle \frac{1}{k}}{x}^{k-1}{\left(1+x^2\right)}^{{\displaystyle \frac{1}{2}}}$, and $T\left(k,-{\displaystyle \frac{1}{2}}\right)={\displaystyle \frac{(k-1)!!}{k!!}}$. Let us choose the set $\{K_{1,-{\displaystyle \frac{1}{2}}}$, $K_{-2,-{\displaystyle \frac{1}{2}}}$, $K_{0,-{\displaystyle \frac{1}{2}}}$, $K_{-1,-{\displaystyle \frac{1}{2}}}\}$ of IC’s.

First, from (86), with $s=1$, $T\left(1,-{\displaystyle \frac{1}{2}}\right)={\displaystyle \frac{0!!}{1!!}}=1$, and $K_{1,-{\displaystyle \frac{1}{2}}}=\sqrt{1+x^2}$ shown in (19), we get $K_{p,-{\displaystyle \frac{1}{2}}}={\displaystyle \frac{(p-1)!!}{p!!}}\left\{{(-1)}^{{\displaystyle \frac{p-1}{2}}}\sqrt{1+x^2}+{\displaystyle \sum _{k=3:2}^p}{(-1)}^{{\displaystyle \frac{p-k}{2}}}{c}_{k,r}{\displaystyle \frac{k!!}{(k-1)!!}}\right\}$ or

$$\begin{array}{ccc}\hfill K_{p,-{\displaystyle \frac{1}{2}}}& =& {\displaystyle \frac{(p-1)!!}{p!!}}\{{(-1)}^{{\displaystyle \frac{p-1}{2}}}\sqrt{1+x^2}\hfill \\ & & +\sqrt{1+x^2}{\displaystyle \sum _{k=3:2}^p}{(-1)}^{{\displaystyle \frac{p-k}{2}}}{\displaystyle \frac{(k-2)!!}{(k-1)!!}}{x}^{k-1}\}\hfill \end{array}$$

(A31)

for $p=3,5,\cdots$; and with $s=0$, $T\left(0,-{\displaystyle \frac{1}{2}}\right)={\displaystyle \frac{(-1)!!}{0!!}}=1$, and $K_{0,-{\displaystyle \frac{1}{2}}}=\ln \left|x+\sqrt{1+x^2}\right|$ shown in (8), we get

$$\begin{array}{ccc}\hfill K_{p,-{\displaystyle \frac{1}{2}}}& =& {\displaystyle \frac{(p-1)!!}{p!!}}\{{(-1)}^{{\displaystyle \frac{p}{2}}}\ln \left|x+\sqrt{1+x^2}\right|\hfill \\ & & +\sqrt{1+x^2}{\displaystyle \sum _{k=2:2}^p}{(-1)}^{{\displaystyle \frac{p-k}{2}}}{\displaystyle \frac{(k-2)!!}{(k-1)!!}}{x}^{k-1}\}\hfill \end{array}$$

(A32)

for $p=2,4,\cdots$.

Next, from (87), with $s=-2$, $T\left(-2,-{\displaystyle \frac{1}{2}}\right)={\displaystyle \frac{(-3)!!}{(-2)!!}}=1$, and $K_{-2,-{\displaystyle \frac{1}{2}}}=-{\displaystyle \frac{\sqrt{1+x^2}}{x}}$ shown in (29), we get

$$\begin{array}{ccc}\hfill K_{p,-{\displaystyle \frac{1}{2}}}& =& -{\displaystyle \frac{(p-1)!!}{p!!}}\{{(-1)}^{{\displaystyle \frac{p+2}{2}}}{\displaystyle \frac{\sqrt{1+x^2}}{x}}\hfill \\ & & +\sqrt{1+x^2}{\displaystyle \sum _{k=p+2:2}^{-2}}{(-1)}^{{\displaystyle \frac{p-k}{2}}}{\displaystyle \frac{(k-2)!!}{(k-1)!!}}{x}^{k-1}\}\hfill \end{array}$$

(A33)

for $p=-4,-6,\cdots$; and with $s=-1$, $T\left(-1,-{\displaystyle \frac{1}{2}}\right)={\displaystyle \frac{(-2)!!}{(-1)!!}}=-1$, and $K_{-1,-{\displaystyle \frac{1}{2}}}={\displaystyle \frac{1}{2}}\ln {\displaystyle \frac{\sqrt{1+x^2}-1}{\sqrt{1+x^2}+1}}$ shown in (20), we get

$$\begin{array}{ccc}\hfill K_{p,-{\displaystyle \frac{1}{2}}}& =& -{\displaystyle \frac{(p-1)!!}{p!!}}\{{\displaystyle \frac{{(-1)}^{\frac{p+1}{2}}}{2}}\ln {\displaystyle \frac{\sqrt{1+x^2}-1}{\sqrt{1+x^2}+1}}\hfill \\ & & +\sqrt{1+x^2}{\displaystyle \sum _{k=p+2:2}^{-1}}{(-1)}^{{\displaystyle \frac{p-k}{2}}}{\displaystyle \frac{(k-2)!!}{(k-1)!!}}{x}^{k-1}\}\hfill \end{array}$$

(A34)

for $p=-3,-5,\cdots$.

Collecting the results (A31)–(A34), we have

$$\begin{array}{ccc}\hfill K_{p,-{\displaystyle \frac{1}{2}}}& =& \left\{\begin{array}{c}{\displaystyle \frac{(p-1)!!}{p!!}}\sqrt{1+x^2}\left\{{(-1)}^{{\displaystyle \frac{p-1}{2}}}+{\displaystyle \sum _{k=3:2}^p}{(-1)}^{{\displaystyle \frac{p-k}{2}}}{\displaystyle \frac{(k-2)!!}{(k-1)!!}}{x}^{k-1}\right\},\hfill \\ p=1,3,\cdots ,\hfill \\ {\displaystyle \frac{(p-1)!!}{p!!}}\{{(-1)}^{{\displaystyle \frac{p}{2}}}\ln \left|x+\sqrt{1+x^2}\right|\hfill \\ +\sqrt{1+x^2}{\displaystyle \sum _{k=2:2}^p}{(-1)}^{{\displaystyle \frac{p-k}{2}}}{\displaystyle \frac{(k-2)!!}{(k-1)!!}}{x}^{k-1}\},\hfill \\ p=0,2,\cdots ,\hfill \\ -{\displaystyle \frac{(p-1)!!}{p!!}}\sqrt{1+x^2}\{{(-1)}^{{\displaystyle \frac{p+2}{2}}}{\displaystyle \frac{1}{x}}\hfill \\ +{\displaystyle \sum _{k=p+2:2}^{-2}}{(-1)}^{{\displaystyle \frac{p-k}{2}}}{\displaystyle \frac{(k-2)!!}{(k-1)!!}}{x}^{k-1}\},\hfill \\ p=-2,-4,\cdots ,\hfill \\ -{\displaystyle \frac{(p-1)!!}{p!!}}\{{\displaystyle \frac{1}{2}}{(-1)}^{{\displaystyle \frac{p+1}{2}}}\ln {\displaystyle \frac{\sqrt{1+x^2}-1}{\sqrt{1+x^2}+1}}\hfill \\ +\sqrt{1+x^2}{\displaystyle \sum _{k=p+2:2}^{-1}}{(-1)}^{{\displaystyle \frac{p-k}{2}}}{\displaystyle \frac{(k-2)!!}{(k-1)!!}}{x}^{k-1}\},\hfill \\ p=-1,-3,\cdots .\hfill \end{array}\right.\hfill \end{array}$$

(A35)

Appendix A.6. the Case r = −2

When $r=-2$, we get $p_1=3$, $p_L=1$, $p_U=3$, $c_{p,-2}={\displaystyle \frac{x^{p-1}}{(p-3)\left(1+x^2\right)}}$, and $T\left(k,-2\right)={\displaystyle \frac{(k-1)!!}{(k-3)!!}}=k-1$. We can choose the IC’s $K_{3,-2}={\displaystyle \frac{1}{2}}\ln \left|1+x^2\right|-{\displaystyle \frac{x^2}{2\left(1+x^2\right)}}$ from1 (16), $K_{-1,-2}={\displaystyle \frac{1}{2}}\left(\ln {\displaystyle \frac{x^2}{1+x^2}}-{\displaystyle \frac{x^2}{1+x^2}}\right)$ from (103), $K_{1,-2}=-{\displaystyle \frac{1}{2\left(1+x^2\right)}}$ from (44), and $K_{0,-2}={\displaystyle \frac{1}{2}}\left({tan}^{-1}x+{\displaystyle \frac{x}{1+x^2}}\right)$ from (83), where we have chosen 0 as ⊙.

First, from (86) with $s=3$, $T(3,-2)=2$, and $K_{3,-2}={\displaystyle \frac{1}{2}}\ln \left|1+x^2\right|-{\displaystyle \frac{x^2}{2\left(1+x^2\right)}}$, we get

$$\begin{array}{ccc}\hfill K_{p,-2}& =& (p-1)\{{\displaystyle \frac{{(-1)}^{\frac{p-3}{2}}}{4}}\left(\ln \left|1+x^2\right|-{\displaystyle \frac{x^2}{1+x^2}}\right)\hfill \\ & & +{\displaystyle \frac{1}{1+x^2}}{\displaystyle \sum _{k=5:2}^p}{\displaystyle \frac{{(-1)}^{\frac{p-k}{2}}{x}^{k-1}}{(k-1)(k-3)}}\}\hfill \end{array}$$

(A36)

for $p=5,7,\cdots$. Second, from (87) with $s=-1$, $T(-1,-2)=-2$, and $K_{-1,-2}={\displaystyle \frac{1}{2}}\ln {\displaystyle \frac{x^2}{1+x^2}}-{\displaystyle \frac{x^2}{2\left(1+x^2\right)}}$, we get

$$\begin{array}{ccc}\hfill K_{p,-2}& =& (p-1)\{-{\displaystyle \frac{{(-1)}^{\frac{p+1}{2}}}{4}}\left(\ln {\displaystyle \frac{x^2}{1+x^2}}-{\displaystyle \frac{x^2}{1+x^2}}\right)\hfill \\ & & -{\displaystyle \frac{1}{1+x^2}}{\displaystyle \sum _{k=p+2:2}^{-1}}{\displaystyle \frac{{(-1)}^{\frac{p-k}{2}}{x}^{k-1}}{(k-1)(k-3)}}\}\hfill \end{array}$$

(A37)

for $p=-3,-5,\cdots$.

Third, with the IC $K_{1,-2}=-{\displaystyle \frac{1}{2\left(1+x^2\right)}}$, we do not have any other formula to obtain: this is a consequence of $p_L=p_U-2$. Fourth, with the IC $K_{0,-2}={\displaystyle \frac{1}{2}}\left({tan}^{-1}x+{\displaystyle \frac{x}{1+x^2}}\right)$, we can obtain every formula for p even. Specifically, from (86) with $s=0$ and $T(0,-2)=-1$, we get

$$\begin{array}{ccc}\hfill K_{p,-2}& =& (p-1)\{-{\displaystyle \frac{{(-1)}^{\frac{p}{2}}}{2}}\left({tan}^{-1}x+{\displaystyle \frac{x}{1+x^2}}\right)\hfill \\ & & +{\displaystyle \frac{1}{1+x^2}}{\displaystyle \sum _{k=2:2}^p}{\displaystyle \frac{{(-1)}^{\frac{p-k}{2}}{x}^{k-1}}{(k-1)(k-3)}}\}\hfill \end{array}$$

(A38)

for $p=2,4,\cdots$. Similarly, from (87) with $s=0$ and $T(0,-2)=-1$, we get

$$\begin{array}{ccc}\hfill K_{p,-2}& =& (p-1)\{-{\displaystyle \frac{{(-1)}^{\frac{p}{2}}}{2}}\left({tan}^{-1}x+{\displaystyle \frac{x}{1+x^2}}\right)\hfill \\ & & -{\displaystyle \frac{1}{1+x^2}}{\displaystyle \sum _{k=p+2:2}^0}{\displaystyle \frac{{(-1)}^{\frac{p-k}{2}}{x}^{k-1}}{(k-1)(k-3)}}\}\hfill \end{array}$$

(A39)

for $p=-2,-4,\cdots$.

Combining the results (A36)–(A39) above, we have

$$\begin{array}{ccc}\hfill K_{p,-2}& =& \left\{\begin{array}{c}-{\displaystyle \frac{1}{2\left(1+x^2\right)}},\hfill \\ p=1,\hfill \\ (p-1)\{{\displaystyle \frac{{(-1)}^{\frac{p-3}{2}}}{4}}\left(\ln \left|1+x^2\right|-{\displaystyle \frac{x^2}{1+x^2}}\right)\hfill \\ +{\displaystyle \frac{1}{1+x^2}}{\displaystyle \sum _{k=5:2}^p}{\displaystyle \frac{{(-1)}^{\frac{p-k}{2}}{x}^{k-1}}{(k-1)(k-3)}}\},\hfill \\ p=3,5,\cdots ,\hfill \\ (p-1)\{-{\displaystyle \frac{{(-1)}^{\frac{p+1}{2}}}{4}}\left(\ln {\displaystyle \frac{x^2}{1+x^2}}-{\displaystyle \frac{x^2}{1+x^2}}\right)\hfill \\ -{\displaystyle \frac{1}{1+x^2}}{\displaystyle \sum _{k=p+2:2}^{-1}}{\displaystyle \frac{{(-1)}^{\frac{p-k}{2}}{x}^{k-1}}{(k-1)(k-3)}}\},\hfill \\ p=-1,-3,\cdots ,\hfill \\ (p-1)\{-{\displaystyle \frac{{(-1)}^{\frac{p}{2}}}{2}}\left({tan}^{-1}x+{\displaystyle \frac{x}{1+x^2}}\right)\hfill \\ +{\displaystyle \frac{1}{1+x^2}}{\displaystyle \sum _{k=2:2}^p}{\displaystyle \frac{{(-1)}^{\frac{p-k}{2}}{x}^{k-1}}{(k-1)(k-3)}}\}\hfill \\ p=0,2,\cdots ,\hfill \\ (p-1)\{-{\displaystyle \frac{{(-1)}^{\frac{p}{2}}}{2}}\left({tan}^{-1}x+{\displaystyle \frac{x}{1+x^2}}\right)\hfill \\ -{\displaystyle \frac{1}{1+x^2}}{\displaystyle \sum _{k=p+2:2}^0}{\displaystyle \frac{{(-1)}^{\frac{p-k}{2}}{x}^{k-1}}{(k-1)(k-3)}}\},\hfill \\ p=-2,-4,\cdots .\hfill \end{array}\right.\hfill \end{array}$$

(A40)