Hint on how to prove $\zeta ( 2) =\pi ^{2}/6$ using the complex Fourier series of $f(x)=x$ I know how to prove $\zeta (2)=\pi ^{2}/6$ by using the trigonometric Fourier series expansion of $x^{2}/4$. How can one prove the same result using the complex Fourier series of $f(x)=x$ for $0\leq x\leq 1$? Any suggestion?
 A: Use the definition:
Say $f$ is defined on $[-\pi, \pi]$.
If $f(z) = \sum_{-\infty}^{\infty} {c_{n} e^{inz}}$
then
$c_{n} = \frac{1}{2\pi}\int_{-\pi}^{\pi}{f(z)e^{-inz}} dz$
If you put $f(z) = z$, can you work out what $c_{n}$ turns out to be?
To integrate, you can try integration by parts. 
A: Extending off from Aryabhatta answer:

For our situation:  We have $f(x)=x ~~~{\text{ for }} 0\leq x\leq 1$
$2L=1,\Rightarrow L=\frac{1}{2}$
So restating we have: 
$f(x) = \displaystyle\sum_{n=-\infty}^{\infty} {c_{n} e^{inx}}, \text{ where }c_{n} = \displaystyle\frac{1}{2\pi}\int_{-\frac{1}{2}}^{\frac{1}{2}}{f(x)e^{-inx}} ~\mathrm{d}x,~~~~~~n=0,~\pm 1,~\pm 2, \cdots~ $
$
\Rightarrow~~ c_{n} = \displaystyle\frac{1}{2\pi}\int_{-\frac{1}{2}}^{\frac{1}{2}}{xe^{-inx}}~\mathrm{d}x
$
After integrating the complex Fourier coefficient we see that we get the following:
$\Rightarrow~~~~\displaystyle c_n=i\left(\frac{\cos(\frac{n}{2})}{2\pi n}-\frac{\sin(\frac{n}{2})}{\pi n^2}\right),~~~\text{for }n \in \mathbb{R}$
Lastly plugging back $c_n$ into $f(x)$ we then get our desired result for $n=0,~\pm 1,~\pm 2, \cdots~$.
Please update if you see any mistakes with any of the work. It has been quite some time since I work with Fourier Series and went off from my head. Feel free to edit mistakes as necessary if willing.
Thanks.
A: This is not related but you would like to see this article: A Short Proof of ζ (2) = π2/6 T.H. Marshall American Math monthly April 2010.
