# Challenge integral

• Aug 3rd 2009, 06:04 PM
Bruno J.
Challenge integral
This one was given to me as a challenge. I was happy I could solve it!

Show that
$\displaystyle \int_0^1\frac{\log(x)\log(1-x)}{x}\: dx = \zeta(3) = \sum_{n=1}^\infty\frac{1}{n^3}$
• Aug 3rd 2009, 08:01 PM
Krizalid
It's quite easy.

\displaystyle \begin{aligned} \int_{0}^{1}{x^{j-1}\ln (x)\,dx}&=-\int_{0}^{1}{\int_{x}^{1}{\frac{x^{j-1}}{t}\,dt}\,dx} \\ & =-\int_{0}^{1}{\int_{0}^{t}{\frac{x^{j-1}}{t}\,dx}\,dt} \\ & =-\frac{1}{j}\int_{0}^{1}{t^{j-1}\,dt} \\ & =-\frac{1}{j^{2}}. \end{aligned}

Hence we get $\displaystyle \int_{0}^{1}{\frac{\ln (x)\ln (1-x)}{x}\,dx}=-\sum\limits_{j=1}^{\infty }{\left( \frac{1}{j}\int_{0}^{1}{x^{j-1}\ln (x)\,dx} \right)}=\sum\limits_{j=1}^{\infty }{\frac{1}{j^{3}}},$ as required. $\displaystyle \quad\blacksquare$
• Aug 3rd 2009, 08:02 PM
TwistedOne151
First, use integration by parts with $\displaystyle u=\ln(x)\ln(1-x)$, $\displaystyle dv=\frac{dx}{x}$, so that $\displaystyle du=\frac{\ln(1-x)}{x}-\frac{\ln(x)}{1-x}$, $\displaystyle v=\ln{x}$, and
$\displaystyle \int_0^1\frac{\ln(x)\ln(1-x)}{x}\,dx=\left[\ln^2(x)\ln(1-x)\right]_0^1-\int_0^1\left(\frac{\ln(1-x)}{x}-\frac{\ln(x)}{1-x}\right)\ln(x)\,dx$
$\displaystyle \int_0^1\frac{\ln(x)\ln(1-x)}{x}\,dx=0-\int_0^1\frac{\ln(x)\ln(1-x)}{x}\,dx+\int_0^1\frac{\ln^2(x)}{1-x}\,dx$
$\displaystyle 2\int_0^1\frac{\ln(x)\ln(1-x)}{x}\,dx=\int_0^1\frac{\ln^2(x)}{1-x}\,dx$
$\displaystyle \int_0^1\frac{\ln(x)\ln(1-x)}{x}\,dx=\frac{1}{2}\int_0^1\frac{\ln^2(x)}{1-x}\,dx$
Now, let $\displaystyle u=-\ln(x)$: x=0 becomes $\displaystyle u=\infty$, x=1 becomes $\displaystyle u=0$, and $\displaystyle x=e^{-u}$, and $\displaystyle dx=-e^{-u}\,du$, so our right-hand integral becomes
$\displaystyle \frac{1}{2}\int_0^1\frac{\ln^2(x)}{1-x}\,dx=\frac{1}{2}\int_{\infty}^0\frac{(-u)^2}{1-e^{-u}}\cdot-e^{-u}\,du$
$\displaystyle =\frac{1}{2}\int_0^{\infty}\frac{u^2e^{-u}}{1-e^{-u}}\,du$
$\displaystyle =\frac{1}{2}\int_0^{\infty}\frac{u^2}{e^u-1}\,du$.
Now,
$\displaystyle \int_0^{\infty}\frac{x^{s-1}}{e^x-1}\,dx=\Gamma(s)\zeta(s)$ (see here).
So
$\displaystyle \frac{1}{2}\int_0^{\infty}\frac{u^2}{e^u-1}\,du=\frac{1}{2}\Gamma(3)\zeta(3)$
$\displaystyle =\frac{1}{2}(2!)\zeta(3)$
$\displaystyle =\zeta(3)$.

-Kevin C.
• Aug 3rd 2009, 08:31 PM
Random Variable
Quote:

$\displaystyle \int_{0}^{1}{\frac{\ln (x)\ln (1-x)}{x}\,dx}=-\sum\limits_{j=1}^{\infty }{\left( \frac{1}{j}\int_{0}^{1}{x^{j-1}\ln (x)\,dx} \right)}$

I don't understand what you did here.
• Aug 3rd 2009, 08:32 PM
Krizalid
$\displaystyle \ln (1-x)=-\sum\limits_{j=1}^{\infty }{\frac{x^{j}}{j}},$ and then switch sum and integral.
• Aug 4th 2009, 10:28 AM
Moo
Quote:

Originally Posted by Krizalid
$\displaystyle \ln (1-x)=-\sum\limits_{j=1}^{\infty }{\frac{x^{j}}{j}},$ and then switch sum and integral.

This kind of stuff has to be justified precisely huh...
(It has been brought to my attention just a while ago lol)
• Aug 4th 2009, 12:52 PM
Krizalid
you know the justification, and it's quite known and I won't do it.