How do I evaluate integrals of the form: $\int_{0}^{1}{\dfrac{\operatorname{arctanh}(x)x^{n}}{1+x^{n+1}} \ dx}$

Question

Few cases have been evaluated:

$n=1$: https://math.stackexchange.com/q/4795632

$n=3:$ https://math.stackexchange.com/q/4801245

I want to know if there is a method to find all related integrals of the form $$\int_{0}^{1}{\dfrac{\operatorname{arctanh}(x)x^{n}}{1+x^{n+1}} \ dx}$$

How do I evaluate these kinds of integrals?

Answer 1

Integrate by parts to rewrite the integral as follows \begin{align} I_n=&\int_0^1 \frac{x^{n-1}\tanh^{-1} x}{1+x^n}dx\\ =& \frac1n \int_0^1 \tanh^{-1} x\ d\left(\ln\frac{1+x^n}2\right) \overset{ibp}=-\frac1n \int_0^1\frac{\ln\frac{1+x^n}2}{1-x^2}dx \end{align} Odd Case: Apply $$1+x^{2m+1}=(1+x)\prod_{k=1}^m (x^2+2x\cos a_k+1), \>\>\>\>\>a_k=\frac{2k\pi}{2m+1} $$ along with $\prod_{k=1}^m 2(1+\cos a_k)=1$ and $ \int_0^1\frac{\ln(1+t)}t dt=\frac{\pi^2}{12}$ \begin{align} I_{2m+1}=& \frac{-1}{2m+1} \int_0^1 \bigg(\ln\frac{1+x}2 + \sum_{k=1}^m \ln\frac{x^2+2x\cos a_k+1}{2+2\cos a_k}\bigg)\overset{x=\frac{1-t}{1+t}}{\frac{dx}{1-x^2}}\\ = & \ \frac{\pi^2}{24} -\frac1{2(2m+1)} \sum_{k=1}^m\int_0^1 \frac{\ln{(t^2 \tan^2\frac{a_k}2+1})}{t}dt \\ =& \ \frac{\pi^2}{24} +\frac1{4(2m+1)} \sum_{k=1}^m \text{Li}_2(-\tan^2\frac{a_k}2) \end{align}

Even Case: Analogous to the odd case $$ I_{2m} = \frac{\pi^2}{24} +\frac1{8m} \sum_{k=1}^{m} \text{Li}_2(-\tan^2\frac{b_k}2),\>\>\>\>\>b_k= \frac{(2k-1)\pi}{2m} $$ which can be further reduced per the symmetry of $b_k$, as well as the inverse formula $\text{Li}_2 (-z)+ \text{Li}_2 (-\frac1z)=-\frac{\pi^2}6-\frac12\ln^2 z $, i.e. \begin{align} I_{2m} = &\ \frac{\pi^2}{24}+\frac1{8m}\bigg[-\frac{m\pi^2}{12} -\frac12\sum_{k=1}^{[\frac m2]} \ln^2(\tan^2\frac{b_k}2)\bigg]\\ =&\ \frac{\pi^2}{32}-\frac1{4m}\sum_{k=1}^{[\frac m2]} \ln^2(\tan\frac{b_k}2) \end{align}

Answer 2

$\newcommand{\artanh}{\operatorname{artanh}}\newcommand{\d}{\,\mathrm{d}}\newcommand{\res}{\operatorname{Res}}$If $n$ is even, good luck.

If $n=2m-1$ is odd then we get: $$I=\frac{1}{2}\int_{-1}^1\frac{\artanh(x)x^{2m-1}}{1+x^{2m}}\d x=\frac{1}{4}\int_{-1}^1\log\left(\frac{1+x}{1-x}\right)\cdot\frac{x^{2m-1}}{1+x^{2m}}\d x$$But in fact we want to consider: $$J=\int_{-1}^1\log\left(\frac{1+x}{1-x}\right)^2\cdot\frac{x^{2m-1}}{1+x^{2m}}\d x$$We may extend the logarithmic term to a holomorphic function on $\Bbb C\setminus[-1,1]$, using the logarithm $\log$ induced by $0\le\arg<2\pi$, cutting on $[0,\infty)$.

Viewing the integrand of $J$ then as a meromorphic function on $\Bbb C\setminus[-1,1]$ we can run a clockwise dogbone contour integral about $[-1,1]$ and pick up a branch-jump of $2\pi i$ in the logarithm as we move from just above $[-1,1]$ to just below. I explain this in more detail in other posts. The important point is that this difference gives us $\log()^2-(\log()^2-4\pi^2+4\pi i\log())=4\pi^2-4\pi i\log()$.

The dogbone integral will pick up residues at every pole outside of the strip $[-1,1]$ including the one at infinity. Although the integrand uniformly vanishes at $\infty$, there is a nonzero residue! There are simple poles at each $2m$th root $\zeta$ of $-1$ and $\frac{(z-\zeta)\cdot z^{2m-1}}{1+z^{2m}}\to\frac{\zeta^{2m-1}}{2m\cdot\zeta^{2m-1}}=\frac{1}{2m}$ as $z\to\zeta$, making the residue evaluation $1/2m$ multiplied with the relevant logarithmic term. Be careful with the arguments; it turns out we get pure imaginary terms, and $\log(it)=\begin{cases}\log|t|+\frac{\pi i}{2}&t>0\\\log|t|+\frac{3\pi i}{2}&t<0\end{cases}$ with the branch that I'm using. To catch the residue at infinity, I explain it in the linked posts but recall if $f$ is meromorphic near $\infty$ then $f(z)=f(\infty)\color{red}{-}\res(f;\infty)z^{-1}+o(z^{-1})+\omega(1)$ for large $z$; in our case as $f$ vanishes at $\infty$ we know all the positive $z$-power terms - the $\omega(1)$ part - are zero and $f(\infty)=0$; thus $f(z)=-\res(f;\infty)z^{-1}+o(z^{-1})$ and $\res(f;\infty)=-\lim_{z\to\infty}zf(z)$. It is useful to recall the logarithmic term is continuous (indeed analytic) on $\Bbb C\setminus[-1,1]$ so it plays nicely with limits.

Combining all the above remarks with some algebra, the residue theorem gives us: $$\begin{align}-16\pi i\cdot I&=4\pi^2\int_{-1}^1\frac{x^{2m-1}}{1+x^{2m}}\d x-16\pi i\cdot I\\&=2\pi i\sum_{k=1}^{2m}\res(e^{\pi i(2k-1)/2m})+2\pi i\res(\infty)\\&=2\pi i\sum_{k=1}^{2m}\frac{1}{2m}\log(i\cot(\pi(2k-1)/4m))^2-2\pi i\log(-1)^2\\&=\frac{\pi i}{m}\left[2\sum_{k=1}^{m}\log^2\cot\frac{\pi(2k-1)}{4m}-2\pi i\underset{=0\text{ as $\ln(1/x)=-\ln(x)$}}{\underbrace{\sum_{k=1}^m\log\cot\frac{\pi(2k-1)}{4m}}}-\frac{5m\pi^2}{2}\right]+2\pi^3i\end{align}$$So: $$\int_0^1\frac{\artanh(x)\cdot x^{2m-1}}{1+x^{2m}}\d x=\frac{\pi^2}{32}-\frac{1}{8m}\sum_{k=1}^{m}\log^2\cot\frac{\pi(2k-1)}{4m},\quad\quad\forall m\in\Bbb N$$

I appreciate that this is rather terse, but I've got to run to a lecture.