# Logarithmic integrals and Euler sums

At various places e.g.

Calculate $$\int_0^1\frac{\log^2(1+x)\log(x)\log(1-x)}{1-x}dx$$

and

How to prove $$\int_0^1x\ln^2(1+x)\ln(\frac{x^2}{1+x})\frac{dx}{1+x^2}$$

logarithmic integrals are connected to Euler-sums. In view of the last link I'm wondering about the following integral $$\int_0^1 \frac{x}{x^2+1} \, \log(x)\log(x+1) \, {\rm d}x \, .$$ I see I can throw it into Wolfram Alpha and get some disgusting anti-derivative with Li's up to $${\rm Li}_3$$. Anyway is there some manually more tractable way to solve this?

I have tried two things of which both don't seem to lead anywhere so far.

For the first one:

I expressed $$\frac{x}{x^2+1}$$ by it's Mellin transform $$\frac{\pi/2}{\cos\left(\frac{\pi s}{2}\right)}$$ and interchanged the integral order $$\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty} {\rm d}s \, \frac{\pi/2}{\cos\left(\frac{\pi s}{2}\right)} \left( -\frac{{\rm d}}{{\rm d}s} \right)\int_0^1 {\rm d}x \, x^{-s} \log(x+1)$$ where the constant $$c>-1$$ is right of the first pole of the cosine at $$s=-1$$ and the contour can be closed in a circle on the left hand side of the plane to use the residue theorem. The $$x$$-integral is equal to $$\int_0^1 {\rm d}x \, x^{-s} \log(x+1) = \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n(n+1-s)} = \sum_{n=1}^\infty \frac{(-1)^{n+1}}{1-s} \left( \frac{1}{n} - \frac{1}{n+1-s} \right)\\ = {\frac {\Psi \left( 1-s/2 \right) - \Psi \left( 3/2-s/2 \right) }{2(1-s)}} + {\frac {\log \left( 2 \right) }{1-s}}$$ where $$\Psi$$ is the Digamma function, related to the harmonic numbers $$H_n$$. Deriving with respect to $$s$$ and picking up the residue $$(-1)^k$$ of the Mellin transform at $$s=-2k-1$$ ($$k=0,1,2,3,...$$) one obtains $$\sum_{k=0}^\infty (-1)^{k+1} \Bigg\{ {\frac {\Psi \left( 3/2+k \right) - \Psi \left( 2+k \right) }{ 8\left( 1+k \right) ^{2}}} - {\frac {\Psi' \left( 3/2+k \right) - \Psi' \left( 2+k \right) }{8(1+k)}} + {\frac {\log \left( 2 \right) }{ 4\left( 1+k \right) ^{2}}} \Bigg\}$$ where $$\Psi'$$ is the derivative of the Digamma function related to $$H_{n,2}$$. The terms with integral argument presumably can be evaluated in closed form, but I'm wondering if the half-integer argument terms can be also evaluated just by algebraic manipulations?

Second:

I tried to find closed form for the integral by partial integration \begin{align} I(a) &=\int_0^1 \frac{\log(x) \log(x+1)}{x+a} \, {\rm d}a \\ &=-\frac{\log(2)}{a+1} - \int_0^1 \frac{x\left(\log(x)-1\right)}{(x+1)(x+a)} \, {\rm d}x + \int_0^1 \frac{x\left(\log(x)-1\right) \log(x+1)}{(x+a)^2} \, {\rm d}x \\ &=-\frac{\log(2)}{a+1} - \int_0^1 \frac{x\left(\log(x)-1\right)}{(x+1)(x+a)} \, {\rm d}x - \int_0^1 \left( \frac{\log(x+1)}{x+a} - \frac{a\log(x+1)}{(x+a)^2} \right) + I(a) + a I'(a) \end{align} and thus $$I(a) = \int_\infty^a \frac{{\rm d}a'}{a'} \Bigg\{ \frac{\log(2)}{a'+1} + \int_0^1 \frac{x\left(\log(x)-1\right)}{(x+1)(x+a')} \, {\rm d}x + \int_0^1 \left( \frac{\log(x+1)}{x+a'} - \frac{a'\log(x+1)}{(x+a')^2} \right) {\rm d}x \Bigg\}$$ of which many terms are easy to integrate, but there is one combination which seems very difficult, namely something like $$\int \frac{{\rm Li}_2(a')}{a'+1} \, {\rm d}a' \, .$$ $$a=\pm i$$ at the end. Any insights?

• I find using Mathematic that $\int_0^1 \frac{x}{x^2+1} \, \log(x)\log(x+1) \, {\rm d}x \, = \frac{1}{64} \left(\pi ^2 \log (4)-15 \zeta (3)\right) \simeq -0.0679481$ – Dr. Wolfgang Hintze Jan 7 at 21:02

## 1 Answer

This is not an answer but it is too long for a comment.

Concerning the disgusting antiderivative, I tried (using another CAS) to undertand how it was computed. If I am not mistaken, they seem to start using $$\frac x{x^2+1}=\frac 12 \left(\frac 1 {x+i}+\frac 1 {x-i} \right)$$ and from here starts the nightmare.

The problem is that we face almost the same kind of problem considering $$I(a)=\int \frac{{\rm Li}_2(a)}{a+1} \, da$$ The result given by a CAS is again awful but simplifies a lot when making $$a=\pm i$$. $$I(+i)=\frac{1}{4} C (\pi +2 i \log (2))-\text{Li}_3(1-i)-\text{Li}_3(1+i)+\frac{1}{192} \pi ^2 (\log (1024)-i \pi )$$ $$I(-i)=\frac{1}{4} C (\pi -2 i \log (2))-\text{Li}_3(1-i)-\text{Li}_3(1+i)+\frac{1}{192} \pi ^2 (\log (1024)+i \pi )$$

• What CAS did you use? – Diger Jan 6 at 16:16
• @Diger. An old tool we developed in my group thirty years ago for our specific needs plus other. For $I(a)$ I took the result from WA and made $a=\pm i$ and simplified (not very funny). I suppose that you know that the result is $\frac{1}{64} \left(\pi ^2 \log (4)-15 \zeta (3)\right)$ – Claude Leibovici Jan 7 at 5:38
• Hey, Thanks and no I did not know because I could not be bothered to copy the result from the WolframAlpha page and simplify it. I don't have Mathematica and Maple can't integrate :-(. Actually I just realized that it is not surprising that the same kind of problem arises with $$\int \frac{{\rm Li_2}(x)}{x+a} \, {\rm d}x$$ as with the original integral in view of $${\rm Li_2}(-x) + {\rm Li_2}(1+x) = \frac{\pi^2}{6} - \log(-x)\log(1+x) \, .$$ – Diger Jan 7 at 11:27
• @Diger. I did copy the result thanks to my wife since I am almost blind. It just took a loooong time to doublecheck everything. But we made it ! Cheers. – Claude Leibovici Jan 7 at 11:31
• You typed it of manually? :-O – Diger Jan 7 at 11:33