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

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.

• Analytically. Apply Simpson's Rule, or similar. – Graham Kemp Oct 23 '14 at 3:47
• the value on the picture is from wolfram alpha – Simple Oct 23 '14 at 3:49
• There is no closed form for the antiderivative. Numerical methods seem to be the way. – Claude Leibovici Oct 23 '14 at 4:27
• Numerical evaluation to 10000 digits – Alice Ryhl Oct 23 '14 at 13:06
• Not that it matters, but $$I~=~\displaystyle\int_0^1\dfrac{\text{arctanh }x}{1-x^2}\cdot\dfrac{\ln x}xdx~=~\dfrac14\int_0^1\Big(\text{arctanh }^2x\Big)'\cdot\Big(\ln^2x\Big)'dx$$ since $$\quad\text{arctanh }'x=\dfrac1{1-x^2}\quad\text{and}\quad\ln'x=\dfrac1x$$ – Lucian Oct 23 '14 at 14:52

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 $$\int^1_0\frac{\operatorname{artanh}{x}\ln{x}}{x(1-x)(1+x)}{\rm d}x=-\frac{7}{16}\zeta(3)-\frac{\pi^2}{8}\ln{2}$$

• Isn't it $\operatorname{arctanh}$ instead of $\operatorname{artanh}$ – Alice Ryhl Oct 23 '14 at 12:48
• @Darksonn Both notations are accepted if I am not wrong. – M.N.C.E. Oct 23 '14 at 12:52
• I'm learning something new everyday from your posts, +1 of course. One question: is it possible to solve 2nd integral without residue approach? – Venus Oct 23 '14 at 12:57
• Your result seems to be correct, I've confirmed it to 1000 digits – Alice Ryhl Oct 23 '14 at 12:57
• This is a pure beauty ! Thanks for your nice work. Cheers :-) – Claude Leibovici Oct 23 '14 at 14:09

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 $-1.37869$.

• I think such closed-form exist since the denominator of the integrand can be decomposed as $$\frac{1}{(1-x)x(1+x)}=\frac{1}{x}-\frac{1}{2(x+1)}-\frac{1}{2(x-1)}$$and seeing this OP (although, it's not the same), I am quite sure it does exist but it's just my feeling & I could certainly be wrong :-) – Venus Oct 23 '14 at 12:02
• @Venus A closed form indeed exists. In fact, this integral is alot simpler than the one in the link you provided. – M.N.C.E. Oct 23 '14 at 12:04
• @M.N.C.E. Really?? You must have the answer, I presume? – Venus Oct 23 '14 at 12:06
• @Venus. I thought the same as you but I did not find any way to compute any of the integrals. Cheers :-) – Claude Leibovici Oct 23 '14 at 12:17
• I am working on an answer and will upload it shortly. Thanks. – M.N.C.E. Oct 23 '14 at 12:20