TITLE: Establishing zeta(3) as a definite integral and its computation.
QUESTION [17 upvotes]: I am a 19 yr old student new to all these ideas. I made the transformation $X(z)=\sum_{n=1}^\infty z^n/n^2$. Therefore $X(1)=\pi^2/6$ as we all know (it is $\zeta(2)$). To calculate $X(1)$, I integrated
$$\frac{Y(z)}{z}=\sum_{n=1}^\infty \frac{z^{n-1}}{n} $$
between 0 to 1;
We all know $Y(z)=-\log(1-z)$; so doing integral of $Y(z)/z$ from 0 to 1, I get
$$X(1)=\int_0^1\ln\frac{z}{z-1}dz$$
that is $\pi^2/6$; Now integrating $X(z)/z$ between 0 to 1, I get $\sum_{n=1}^\infty 1/n^3$;
therefore performing that integral I get $\sum1/n^3$ as 
$$\int_0^1 \frac{(\log x)^2 dx}{1-x};$$
which Iam unable to do.
So can anyone give an idea of how to find 
$$\sum_{n=1}^\infty\frac1{n^3}=\zeta(3)=\int_0^1 \frac{(\log x)^2 dx}{1-x};$$
given
$$\zeta(2)=\int_0^1 \frac{\log x\ dx}{1-x}=\frac{\pi^2}{6}.$$
Please take time to read all these things and help me out by giving some suggestion.

REPLY [2 votes]: Following your way:
Let $F(x) = \displaystyle\int_0^x{-\frac{\ln(1-t)}{t}}dt$, where $x  \in [0,1]$. Easy to see that $F(1)$ yields $\zeta(2)$. 
$F(x)$ is continuous and derivable on $(0,1)$ [The integrand being continuous on $(0,1)$] and $F'(x)= -\displaystyle\frac{\ln(1-x)}{x}$. Again, $F(x)=x+\frac{x^2}{2^2}+\frac{x^3}{3^2}+... $.
It can be readily seen that: $\displaystyle\int_0^x{\frac{F(x)}{x}}dx=\mathrm{Li}_3(x).$
Integrating by parts:$\displaystyle\int_0^x{\frac{F(x)}{x}}dx=$ $\displaystyle{[{F(x){\ln(x)}-\int{F'(x)\ln(x)dx}}]_0^x=[F(x)\ln(x)+\frac{\ln(1-x)\ln(x)}{x}]_0^x}$
$F(0)=0$ [follows from the continuity of integral functions] and $\lim_{x \to 0}\ln(1-x)\ln(x)=0$. 
Hence, we finally arrive at:  
$\zeta(3)= \displaystyle\sum_{n=1}^\infty{\frac{1}{n^3}}=\displaystyle\int_0^1{\frac{\ln(1-x)\ln(x)}{x}dx}$