1. integral

$\zeta (2) = \sum^{\infty}_{n=1} \frac{1}{n^{2}} = Li_{2} (1) = -\int^{1}_{0} \frac{\ln(1-x)}{x} \ dx$

So how do I show that $-\int^{1}_{0} \frac{\ln(1-x)}{x} \ dx = \frac{\pi^{2}}{6}$ ?

The only solution I've found is to write it as $\int^{1}_{0} \int^{1}_{0} \frac{1}{1+xy} \ dy \ dx$ and then make a crazy change of variables that rotates the coordinate system by $\frac{\pi}{4}$. But that gets really messy.

2. Originally Posted by Random Variable
$\zeta (2) = \sum^{\infty}_{n=1} \frac{1}{n^{2}} = Li_{2} (1) = -\int^{1}_{0} \frac{\ln(1-t)}{t} \ dt$

So how do I show that $-\int^{1}_{0} \frac{\ln(1-x)}{x} \ dx = \frac{\pi^{2}}{6}$ ?

The only solution I've found is to write it as $\int^{1}_{0} \int^{1}_{0} \frac{1}{1+xy} \ dy \ dx$ and then make a crazy change of variables that rotates the coordinate system by $\frac{\pi}{4}$. But that gets really messy.

Hmmm...I believe you would have to use a power series representation here.

For $0 \le x \le 1$

$\frac{ ln(1-x) }{ x} = \frac{ -(x + \frac{x^2}{2} + \frac{x^3}{3}+...) }{x} = -(1+ \frac{x}{2} + \frac{x^2}{3}+...)$

If we take the integral of this,

$-\int^{1}_{0} \frac{\ln(1-x)}{x} \ dx = \int_0^1 (1+ \frac{x}{2} + \frac{x^2}{3}+...) = (1 + \frac{1}{2^2} + \frac{1}{3^2}+...)$

Note how this is the series,

$\sum \frac{1}{n^2} = \frac{ ( \pi)^2 }{6}$

So $-\int^{1}_{0} \frac{\ln(1-x)}{x} \ dx = \frac{ ( \pi)^2 }{6}$

Edit- Can you explain your way? I don't quite comprehend what you were trying to do but it seems like a cool exercise (so if you could, could you post your solution?)

3. Originally Posted by AllanCuz
Hmmm...I believe you would have to use a power series representation here.

For $0 \le x \le 1$

$\frac{ ln(1-x) }{ x} = \frac{ -(x + \frac{x^2}{2} + \frac{x^3}{3}+...) }{x} = -(1+ \frac{x}{2} + \frac{x^2}{3}+...)$

If we take the integral of this,

$-\int^{1}_{0} \frac{\ln(1-x)}{x} \ dx = \int_0^1 (1+ \frac{x}{2} + \frac{x^2}{3}+...) = (1 + \frac{1}{2^2} + \frac{1}{3^2}+...)$

Note how this is the series,

$\sum \frac{1}{n^2} = \frac{ ( \pi)^2 }{6}$

So $-\int^{1}_{0} \frac{\ln(1-x)}{x} \ dx = \frac{ ( \pi)^2 }{6}$

Edit- Can you explain your way? I don't quite comprehend what you were trying to do but it seems like a cool exercise (so if you could, could you post your solution?)
Yes, that's definitely true. But I'm not trying to show that the integral is equal to $\sum^{\infty}_{n=1} \frac{1}{n^{2}}$. I'm trying to evaluate the integral to show that $\sum^{\infty}_{n=1}\frac{1}{n^{2}} = \frac{\pi^{2}}{6}$

The rotation equations are $x'= x \cos \theta - y \sin \theta$ and $y' = x \sin \theta + y \cos \theta$

so just let $\theta = \frac{\pi}{4}$

4. Originally Posted by Random Variable
Yes, that's definitely true. But I'm not trying to show that the integral is equal to $\sum^{\infty}_{n=1} \frac{1}{n^{2}}$. I'm trying to evaluate the integral to show that $\sum^{\infty}_{n=1}\frac{1}{n^{2}} = \frac{\pi^{2}}{6}$

The rotation equations are $x'= x \cos \theta - y \sin \theta$ and $y' = x \sin \theta + y \cos \theta$

so just let $\theta = \frac{\pi}{4}$
Yikes! I kind of jumped the gun on that one eh?

This should help: Math Forum - Ask Dr. Math

It would appear that you have to represent the integral in a completely differn't form.

5. Originally Posted by Random Variable
$\zeta (2) = \sum^{\infty}_{n=1} \frac{1}{n^{2}} = Li_{2} (1) = -\int^{1}_{0} \frac{\ln(1-x)}{x} \ dx$

So how do I show that $-\int^{1}_{0} \frac{\ln(1-x)}{x} \ dx = \frac{\pi^{2}}{6}$ ?

The only solution I've found is to write it as $\int^{1}_{0} \int^{1}_{0} \frac{1}{1+xy} \ dy \ dx$ and then make a crazy change of variables that rotates the coordinate system by $\frac{\pi}{4}$. But that gets really messy.
$J=-\int_0^1\frac{\ln(1-x)}{x}dx$. Let $-\ln(1-x)=z\implies 1-e^{-z}=x\implies e^{-z}dz=dx$

So, our integral becomes $J=\int_0^{\infty}\frac{z e^{-z}}{1-e^{-z}}dz$ multiplying top and bottom gives $J=\int_0^{\infty}\frac{z}{e^z-1}dz=\cdots$ ah crap, you need the zeta function.

Why not just do Fourier series or use contour integration?