How to evaluate $\int_0^1\frac{\tanh ^{-1}(x)\log(x)}{(1-x) x (x+1)} \operatorname d \!x$?

Question

How to evaluate the following integral

$$\int_0^1\frac{\tanh ^{-1}(x)\log(x)}{(1-x) x (x+1)} \operatorname d \!x $$

The numerical result is $= -1.38104$ and when I look at it, I have no idea how to work on it. Could you provide me hits to evaluate the integral above?

Thank you.

Answer 1

Your integral can be rewritten as \begin{align} \int^1_0\frac{{\rm artanh}\ {x}\ln{x}}{x(1-x)(1+x)}{\rm d}x =\int^1_0\frac{{\rm artanh}\ {x}\ln{x}}{x}{\rm d}x+\int^1_0\frac{x{\rm artanh}\ {x}\ln{x}}{1-x^2}{\rm d}x \end{align} The first integral is \begin{align} \int^1_0\frac{{\rm artanh}\ {x}\ln{x}}{x}{\rm d}x =&\chi_2(x)\ln{x}\Bigg{|}^1_0-\int^1_0\frac{\chi_2(x)}{x}{\rm d}x\\ =&-\chi_3(1)=-\frac{7}{8}\zeta(3) \end{align} The second integral is \begin{align} \int^1_0\frac{x{\rm artanh}\ {x}\ln{x}}{1-x^2}{\rm d}x =&\sum^\infty_{n=0}\sum^n_{k=0}\frac{1}{2k+1}\int^1_0 x^{2n+2}\ln{x}\ {\rm d}x\\ =&\sum^\infty_{n=0}\frac{\frac{1}{2}H_n-H_{2n+1}}{(2n+3)^2}\\ =&\frac{1}{2}\sum^\infty_{n=1}\frac{H_n}{(2n+1)^2}-\sum^\infty_{n=1}\frac{H_{2n}}{(2n+1)^2} \end{align} For the first sum, consider $\displaystyle f(z)=\frac{\left(\gamma+\psi_0(-z)\right)^2}{(2z+1)^2}$. At the positive integers, \begin{align} \sum^\infty_{n=1}{\rm Res}(f,n) =&\sum^\infty_{n=1}\operatorname*{Res}_{z=n}\left[\frac{1}{(2z+1)^2(z-n)^2}+\frac{2H_n}{(2z+1)^2(z-n)}\right]\\ =&2\sum^\infty_{n=1}\frac{H_n}{(2n+1)^2}-\frac{7}{2}\zeta(3)+4 \end{align} At $z=0$, \begin{align} {\rm Res}(f,0)=\operatorname*{Res}_{z=0}\frac{1}{z^2(2z+1)^2}=-4 \end{align} At $z=-\frac{1}{2}$, \begin{align} {\rm Res}\left(f,-\tfrac12\right)=\frac{1}{4}\frac{{\rm d}}{{\rm d}z}(\gamma+\psi_0(-z))^2\Bigg{|}_{z=-\frac{1}{2}}=\frac{\pi^2}{2}\ln{2} \end{align} Since the sum of residues is zero, $$\sum^\infty_{n=1}\frac{H_n}{(2n+1)^2}=\frac{7}{4}\zeta(3)-\frac{\pi^2}{4}\ln{2}$$ For the second sum, consider $\displaystyle f(z)=\frac{\pi\cot(\pi z)(\gamma+\psi_0(-2z))}{(2z+1)^2}$. Note that the poles at the half integers are cancelled by the zeroes of $\pi\cot(\pi z)$. At the positive integers, \begin{align} \sum^\infty_{n=1}{\rm Res}(f,n) =&\sum^\infty_{n=1}\operatorname*{Res}_{z=n}\left[\frac{1}{2(2z+1)^2(z-n)^2}+\frac{H_{2n}}{(2z+1)^2(z-n)}\right]\\ =&\sum^\infty_{n=1}\frac{H_{2n}}{(2n+1)^2}-\frac{7}{4}\zeta(3)+2 \end{align} At the negative integers, \begin{align} \sum^\infty_{n=1}{\rm Res}(f,-n) =&\sum^\infty_{n=1}\frac{H_{2n+1}}{(2n+1)^2}+1\\ =&\sum^\infty_{n=1}\frac{H_{2n}}{(2n+1)^2}+\frac{7}{8}\zeta(3) \end{align} At $z=0$, \begin{align} {\rm Res}(f,0)=-2 \end{align} Since the sum of residues is zero, $$\sum^\infty_{n=1}\frac{H_{2n}}{(2n+1)^2}=\frac{7}{16}\zeta(3)$$ Hence, if I had not made any mistakes, the original integral is simply $$\int^1_0\frac{\operatorname{artanh}{x}\ln{x}}{x(1-x)(1+x)}{\rm d}x=-\frac{7}{16}\zeta(3)-\frac{\pi^2}{8}\ln{2}$$

Answer 2

As I said in comments, it seems that there is no explicit antiderivative. Even numerical methods could be problematic because the integrand is undefined at the bounds.

One possible way is to perform a Taylor expansion around $x=0$. We so can get $$\frac{\tanh ^{-1}(x)}{(1-x) x (x+1)}=1+\frac{4 x^2}{3}+\frac{23 x^4}{15}+\frac{176 x^6}{105}+\frac{563 x^8}{315}+\frac{6508 x^{10}}{3465}+\frac{88069 x^{12}}{45045}+O\left(x^{13}\right)$$ and we are left with a series of integrals $$I_n=\int x^{2n}\log(x)dx=\frac{x^{2 n+1} ((2 n+1) \log (x)-1)}{(2 n+1)^2}$$ So $$J_n=\int_0^1 x^{2n}\log(x)dx=-\frac{1}{(2 n+1)^2}$$ Using the above expansion, the result of the integral is $$A_{12}=-\frac{118164390159964}{91398648466125}\approx -1.29285$$ Adding more terms in the expansion, we have $$A_{14}=-\frac{118985678275612}{91398648466125} \approx -1.30183$$ $$A_{16}=-\frac{587809463058901481}{449041559914072125}\approx -1.30903$$ $$A_{18}=-\frac{4049985594549210440279}{3079976059450620705375}\approx -1.31494$$ Now, I am tired with fractions !! This will converge but not very fast (because the coefficients of the Taylor expansion constantly increase).

If we use the previous results and report them on a graph, it looks that we can approximate more or less $$A_{2n}\approx\frac{a+b n}{1+c n}=-\frac{1.01437+0.827943 n}{1+0.604336 n}$$ for which the limit should be $-1.37000$.

Making the model slightly more complex $$A_{2n}\approx\frac{a+b n^d}{1+c n^d}=-\frac{0.98352+0.984458 n^{0.903476}}{1+0.714051 n^{0.903476}}$$ leads to a limit equal to $-1.37869$.