Calculating Fourier series of infinite sum in two ways

Question

Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be in the Schwartz class. Show that $$\sum_{n\in\mathbb{Z}}f(2\pi n)=\dfrac{1}{2\pi}\sum_{k\in\mathbb{Z}}\hat{f}(k)$$ by calculating the Fourier series of $$F(x)=\sum_{n\in\mathbb{Z}}f(x-2\pi n)$$ in two ways.

The Fourier coefficients of $F(x)$ are $$\hat{F}(n)=\dfrac{1}{2\pi}\int_{-\pi}^\pi F(x)e^{-inx}dx=\dfrac{1}{2\pi}\int_{-\pi}^\pi \left(\sum_{k\in\mathbb{Z}}f(x-2\pi k)\right)e^{-inx}dx$$ How can we continue from here?

Answer 1

Well, you COULD look it up here: http://www.proofwiki.org/wiki/Poisson_Summation_Formula

Answer 2

Let $$ F(x)=\sum_{n\in\mathbb{Z}}f(x+2n\pi) $$ then $$ \begin{align} \sum_{k\in\mathbb{Z}}\hat{f}(k) &=\sum_{k\in\mathbb{Z}}\frac1{2\pi}\int_{-\infty}^\infty f(x)\,e^{-ikx}\,\mathrm{d}x\\ &=\frac1{2\pi}\sum_{n\in\mathbb{Z}}\sum_{k\in\mathbb{Z}}\int_{(2n-1)\pi}^{(2n+1)\pi} f(x)\,e^{-ikx}\,\mathrm{d}x\\ &=\frac1{2\pi}\sum_{n\in\mathbb{Z}}\sum_{k\in\mathbb{Z}}\int_{-\pi}^\pi f(x+2n\pi)\,e^{-ikx}\,\mathrm{d}x\\ &=\frac1{2\pi}\sum_{k\in\mathbb{Z}}\int_{-\pi}^\pi F(x)\,e^{-ikx}\,\mathrm{d}x\\ &=\sum_{k\in\mathbb{Z}}\hat{F}(k)\\[4pt] &=F(0)\\[8pt] &=\sum_{n\in\mathbb{Z}}f(2n\pi) \end{align} $$