# Two Challenging Integrals

#### sbhatnagar

##### Active member
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

##### Active member
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}

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#### sbhatnagar

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