{{page>:defs}}
{{tag>:exercise fourier}}

<WRAP center round box 80%>
Let $f$ be the $2\pi$-periodic function defined by $f(x)=x^2$ for all 
$x \in  [-\pi,\pi]$. Find its Fourier coefficients and deduce the value of 
$$
\sum_{p=1}^\infty \frac{1}{p^2}
$$

</WRAP>

Write $c_p(f)=\frac{1}{2\pi}\int_{-\pi}^\pi f(x)\rme^{-ipx}\rmd x$ its Fourier coefficients. Then, the function $f$ being symmetric, we have
$$
c_p(f)=\frac{1}{2\pi} \int_{-\pi}^\pi x^2 \cos(px) \rmd x=\frac{1}{\pi} \int_{0}^\pi x^2 \cos(px) \rmd x
$$
Morever, $c_0(f)=\pi^2/3$. Using twice the integration by parts, we get for all $p\neq 0$, 
\begin{align*}
c_p(f)&=\frac{1}{\pi}\int_0^\pi \lrb{\underbrace{\lrb{x^2 \frac{\sin(px)}{p}}_0^\pi}_{=0}-2\int_0^\pi
x \frac{\sin(px)}{p}}\\
&=\frac{-2}{\pi
p} \lrcb{\lrb{-x\frac{\cos(px)}{p}}_0^\pi+\underbrace{\int_0^\pi \frac{\cos(px)}{p}\rmd x}_{=0}}
 =\frac{2(-1)^p}{
p^2}
\end{align*}
The function $f$ being $C_1$ on subintervals and being locally integrable, we get for all $x\in [-\pi,\pi]$, 
\begin{equation}
x^2=\frac{f(x^+)+f(x^-)}{2}=\frac{\pi^2}{3}+\sum_{p \neq 0} \frac{2(-1)^p}{p^2} \rme^{ipx}
\end{equation}
and taking $x=\pi$, it comes
$$
\sum_{p=1}^\infty \frac{1}{p^2}=\frac{\pi^2}{6}
$$