In Gradshteyn & Ryzhik's Tables of integrals, series, and products (1996) it is stated without proof that
If $f$ is an even, $\pi$-periodic function then $$\int_0^\infty f(x)\frac{\sin{x}}{x}\ dx=\int_0^{\pi/2} f(x)\ dx$$ assuming that the LHS integral exists.
I tried to prove this by breaking up the integral over $[0,\infty[$ into intervals of length $\pi$ and summing to infinity but I could not get any further than finding:
$$\int_0^\infty f(x)\frac{\sin{x}}{x}\ dx=\sum_{n=0}^\infty\int_0^\pi f(x)\frac{(-1)^n\sin{x}}{n\pi+x}\ dx$$
which doesn't seem too fruitful. My gut tells me that Fourier transforms should get involved at some point. Does anyone have a proof or a sketch of a proof?
I think one should work on $(-\pi/2,\pi/2)$ instead. The calculation could go like this (I leave it to you to fill in some details):
$$ \begin{aligned} \int_0^{+\infty}f(x)\frac{\sin x}{x}\,dx &=\frac12\int_{-\infty}^{+\infty}f(x)\frac{\sin x}{x}\,dx\\ &=\frac12\sum_{k=-\infty}^{+\infty}\int_{-\pi/2+k\pi}^{\pi/2+k\pi}f(x)\frac{\sin x}{x}\,dx\\ &=\frac12\sum_{k=-\infty}^{+\infty}\int_{-\pi/2}^{\pi/2}f(u+\pi k)\frac{\sin(u+\pi k)}{u+\pi k}\,du\\ &=\frac12\sum_{k=-\infty}^{+\infty}\int_{-\pi/2}^{\pi/2}f(u)(-1)^k\frac{\sin u}{u+\pi k}\,du\\ &=\frac12\int_{-\pi/2}^{\pi/2}f(u)\frac{\sin u}{u}+f(u)\sin u\sum_{k=1}^{+\infty}(-1)^k\Bigl[\frac{1}{u+\pi k}+\frac{1}{u-\pi k}\Bigr]\,du\\ &=\frac{1}{2}\int_{-\pi/2}^{\pi/2}f(u)\frac{\sin u}{u}\biggl(1+\sum_{k=1}^{+\infty}\frac{(-1)^ku^2}{u^2-\pi^2k^2}\biggr)\,du\\ &=\frac{1}{2}\int_{-\pi/2}^{\pi/2}f(u)\frac{\sin u}{u}\biggl(1+\frac{u}{\sin u}-1\biggr)\,du\\ &=\frac12\int_{-\pi/2}^{\pi/2}f(u)\,du\\ &=\int_0^{\pi/2}f(u)\,du. \end{aligned} $$