MHB Can You Solve These Two Difficult Integrals?

[sbhatnagar]

Prove that

\[\int_0^1 \frac{\log(1+x)\log(x)}{1-x}dx=\zeta(3)-\frac{\pi^2}{4}\log(2)\]

\[\int_0^1 \frac{\log(1+x^2)}{1+x}dx=\frac{3}{4}\log^2(2) -\frac{\pi^2}{48}\]

 

[sbhatnagar]

I think these kind of problems have gone out of fashion these days.

Anyway, here is my solution to (1).

Problem 1

Step 1 - Reduction to Euler Sum

\[
\begin{aligned}
\int_0^1 \frac{\log(1+x)\log(x)}{1-x}dx &= \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}\int_0^1 \frac{x^n \log(x)}{1-x}dx \\
&= \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n} \left( H_n^{(2)}-\frac{\pi^2}{6}\right)
\\ &= \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}H_n^{(2)}-\frac{\pi^2}{6}\log(2)
\end{aligned}
\]

where \(\displaystyle H_n^{(2)} = \sum_{k=1}^n \frac{1}{n^2}\)

Step 2 - Evaluation of Euler Sum

The evaluation of the Euler Sum is tricky.

Note that

\(\displaystyle \int_0^1 (-r)^{n-1} \ dr = \frac{(-1)^{n+1}}{n}\) and
\(\displaystyle \int_0^1 \int_0^1 (s t)^{k-1} \ ds \ dt = \frac{1}{k^2}\)

Plugging these into the sum, we obtain

\[
\begin{aligned}
\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}H_n^{(2)} &= \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}\sum_{k=1}^n \frac{1}{k^2} \\
&= \sum_{n=1}^\infty \sum_{k=1}^n \int_0^1 (-r)^{n-1} \ dr \int_0^1 \int_0^1 (s t)^{k-1} \ ds \ dt \\
&= \sum_{k=1}^\infty \int_0^1 \left( \sum_{n=k}^\infty (-r)^{n-1} \right)dr \int_0^1 \int_0^1 (s t)^{k-1} \ ds \ dt \\
&= \sum_{k=1}^\infty \int_0^1 \frac{(-r)^{k-1}}{1+r}dr\int_0^1 \int_0^1 (s t)^{k-1} \ ds \ dt \\
&= \int_0^1 \int_0^1 \int_0^1 \frac{1}{(1+r)(1+rst)}dr \ ds \ dt \\
&= \int_0^1 \int_0^1 \frac{\log(1+rs)}{(1+r)(rs)}ds \ dr \\
&=-\int_0^1 \frac{\text{Li}_2(-r)}{(1+r)r}dr \\
&=\int_0^1 \text{Li}_2(-r) \left(\frac{1}{1+r}-\frac{1}{r} \right)dr \\
&= -\log(2)\frac{\zeta(2)}{2}+\int_0^1 \frac{\log^2(1+r)}{r}dr +\frac{\zeta(3)}{4} \\ &= \zeta(3)-\frac{\zeta(2) \log(2)}{2}
\end{aligned}
\]

Step 3 - Now, the combination of these efforts results in

\[\begin{aligned}\int_0^1 \frac{\log(x) \log(1+x)}{1-x}dx &= \left( \zeta(3)-\frac{\zeta(2) \log(2)}{2}\right) - \frac{\pi^2}{6}\log(2) \\ &= \zeta(3)-\frac{\pi^2}{12}\log(2)-\frac{\pi^2}{6}\log(2) \\ &= \zeta(3)-\frac{\pi^2}{4}\log(2)\end{aligned}\]

 

[MarkFL]

According to our suggested guidelines for posting challenge problems posted here:

http://www.mathhelpboards.com/f28/guidelines-posting-answering-challenging-problem-puzzle-3875/

we ask that our members be given at least a week to respond before posting the solution(s). This gives people a good chance to respond if they have a solution. (Sun)

 

[sbhatnagar]

According to our suggested guidelines for posting challenge problems posted here:

http://www.mathhelpboards.com/f28/guidelines-posting-answering-challenging-problem-puzzle-3875/

we ask that our members be given at least a week to respond before posting the solution(s). This gives people a good chance to respond if they have a solution. (Sun)

I am sorry.:D I thought it would be nice to give some ideas on Euler Sums first.

Anyway, if it is against the rules I shall not post a solution so early.

 

[MarkFL]

Well, I wouldn't say the guidelines are as strict as rules, but when you post a challenging problem, it is best to give people a reasonable amount of time in which to respond. :D