Prove that $\sum_{k=1}^{\infty}\frac{1}{k^2}=\frac{\pi^2}{6}$

Let $f=\frac{1}{2}-x$ on the interval $[0,1)$, and extend $f$ to be periodic on $\R$. 

a. Show that $\hat{f}(0)=0$, and $\hat{f}(k)=(2\pi i k)^{-1}$ if $k\neq 0$.
b. Prove that $\sum_{k=1}^{\infty}\frac{1}{k^2}=\frac{\pi^2}{6}$. (Use Parseval identity.)