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world:exofourier


2017/10/07 23:39 ·

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 $$\sum_{p=1}^\infty \frac{1}{p^2}$$

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 $x\in [-\pi,\pi]$, $$x^2=\frac{f(x^+)+f(x^-)}{2}=\frac{\pi^2}{3}+\sum_{p \neq 0} \frac{2(-1)^p}{p^2} \rme^{ipx}$$ and taking $x=\pi$, it comes $$\sum_{p=1}^\infty \frac{1}{p^2}=\frac{\pi^2}{6}$$