# Thread: Challenging Integral?.

1. ## Challenging Integral?.

This may not be that bad, but give it a go for those who wish.

Let's see what interesting methods are come up with.

$\int_{0}^{1}ln(x)ln(1-x)dx$

2. Originally Posted by galactus
This may not be that bad, but give it a go for those who wish.

Let's see what interesting methods are come up with.

$\int_{0}^{1}ln(x)ln(1-x)dx$
We know that $-\sum_{n=0}^{\infty}\frac{x^{n+1}}{n+1}=\ln(1-x)$

so then I think what you can do is this

$-\int{\ln(x)}\sum_{n=0}^{\infty}\frac{x^{n+1}}{n+1} dx=-\sum_{n=0}^{\infty}\frac{1}{n+1}\int\ln(x)x^{n+1}$

$\int_0^{1}\ln(x)\cdot{x^{n+1}}=\bigg[x^{n}\bigg[\frac{x^2\ln(x)}{n+2}-\frac{x^2}{(n+2)^2}\bigg]\bigg]\bigg|_0^{1}=\frac{-1}{(n+2)^2}$

so we have

$\sum_{n=0}^{\infty}\frac{1}{(n+1)(n+2)^2}=2-\frac{\pi^2}{6}$

by PFD and use of a riemann zeta table =D

NOTE: what I mean by PFD was that after I decomposed it it gave a telescoping series and a riemann zeta function, which came out to be $1-\bigg(\frac{\pi^2}{6}-1\bigg)$, the latter riemann zeta was gotten using the CRC table of values

3. Very good. I knew you fellas would chew it up and spit it out.

Here is what I done. Which is similar to you fellas line of thought.

$\int_{0}^{1}ln(x)ln(1-x)dx$

$-\int_{0}^{1}ln(x)\sum_{k=1}^{\infty}\frac{x^{k}}{k }dx$

$-\sum_{k=1}^{\infty}\frac{1}{k}\int_{0}^{1}x^{k}ln( x)dx$

$\sum_{k=1}^{\infty}\frac{1}{k(k+1)^{2}}$

$\sum_{k=1}^{\infty}\frac{1}{k(k+1)}-\sum_{k=1}^{\infty}\frac{1}{(k+1)^{2}}$

$=1-\left(\frac{{\pi}^{2}}{6}-1\right)$

$=\boxed{2-\frac{{\pi}^{2}}{6}}$