# Integral ∫10ln(1−x)ln(1+x)ln2xdx{\large\int}_0^1\ln(1-x)\ln(1+x)\ln^2x\,dx

This problem was posted at I&S a week ago, and no attempts to solve it have been posted there yet. It looks very alluring, so I decided to repost it here:

Prove:

I found a paper where some similar integrals are evaluated: J. A. M. Vermaseren, Harmonic sums, Mellin transforms and Integrals, Int. J. Mod. Phys. A, 14 (1999), 2037-2076, DOI: 10.1142/S0217751X99001032
, but it’s not quite easy to read for me. Maybe it could be of some help for this problem.

This answer is split into 3 main steps.

Step 1: Expressing the integral as a sum

Step 2a: Value of $\displaystyle \sum^\infty_{n=1}\frac{(-1)^nH_n}{n}$

See here for the details.

Step 2b: Value of $\displaystyle \sum^\infty_{n=1}\frac{(-1)^nH_n}{n^2}$

See here for the details.

Step 2c: Value of $\displaystyle \sum^\infty_{n=1}\frac{(-1)^nH_n}{n^3}$

Tunk-Fey did a calculation of this type here.

Step 2d: Value of $\displaystyle \sum^\infty_{n=1}\frac{(-1)^nH_n^{(2)}}{n}$

Therefore

You can use polylogarithm identities to simplify the last equation. I took the easy way out and used Wolfram Alpha. Note that contour integration is a slightly more efficient method to solve this sum, however this method is required if I want to solve $\displaystyle \sum^\infty_{n=1}\frac{(-1)^nH_n^{(2)}}{n^2}$ as well.

Step 2e: Value of $\displaystyle \sum^\infty_{n=1}\frac{(-1)^nH_n^{(3)}}{n}$

Therefore

Step 2f: Value of $\displaystyle \sum^\infty_{n=1}\frac{(-1)^nH_n^{(2)}}{n^2}$

This part is rather similar to Tunk-Fey’s answer, so he certainly deserves credit.

The blue integral is

The orange integral is

So

Therefore

The grey terms miraculously cancel.

Step 3a: Evaluating $\displaystyle 2\sum^\infty_{j=1}\frac{(-1)^jH_{j+1}}{j(j+1)^3}$

Step 3b: Evaluating $\displaystyle -\frac{\pi^2}{3}\sum^\infty_{j=1}\frac{(-1)^j}{j(j+1)^2}$

Step 3c: Evaluating $\displaystyle 2\sum^\infty_{j=1}\frac{(-1)^jH_{j+1}^{(2)}}{j(j+1)^2}$

Step 3d: Evaluating $\displaystyle -2\zeta(3)\sum^\infty_{j=1}\frac{(-1)^j}{j(j+1)}$

Step 3e: Evaluating $\displaystyle 2\sum^\infty_{j=1}\frac{(-1)^jH_{j+1}^{(3)}}{j(j+1)}$

Step 4: Obtaining the final result

Summing the results from steps 3a, 3b, 3c, 3d and 3e gives

hence completing the proof.