I am looking for a proof that for a function $F$ that is 1-periodic:
$$\int_{-\infty}^{\infty}F(x)\operatorname{sinc}^2(\pi x)dx=\int_{0}^{1}F(x)dx$$
edit : with the sinus cardinal function : $\operatorname{sinc} : x \to \frac{\sin(x)}{x}$ if $x \neq 0$, else $1$.
Here is what I have: $$ \int^\infty_{-\infty}F(x)\frac{\sin^2(\pi x)}{\pi^2x^2}dx=\sum^\infty_{n=-\infty}\int^{n+1}_nF(x)\frac{\sin^2(\pi x)}{\pi^2 x^2}dx $$
\begin{align} \int^\infty_{-\infty}F(x)\frac{\sin^2(\pi x)}{\pi^2x^2}dx&=\sum_{n\in\mathbb{Z}}\int^{n+1}_nF(x)\frac{\sin^2(\pi x)}{\pi^2 x^2}dx\\ &= \sum_{n\in\mathbb{Z}}\int^1_0 F(x+n)\frac{\sin^2(\pi x+n\pi)}{\pi^2(x+n)^2}dx\\ &=\int^1_0F(x)\sin^2(\pi x)\Big(\sum_{n\in\mathbb{Z}}\frac{1}{\pi^2(n+x)^2}\Big)\,dx \end{align} The term $\sum_{n\in\mathbb{Z}}\frac{1}{\pi^2(n+x)^2}=\csc^2(\pi x)$ (notice that poles and residuals seems fine). This computation can check any standard book in Complex Analysis (Lang's book for instance). The rest should follow through easily.
