Lobachevsky's Formula for Integrals If the function $f:\mathbb{R}\to \mathbb{R}$ satisfies $f(x+\pi)=f(\pi-x)=f(x), \forall x \in \mathbb{R}$ then 
$$\int_0^\infty f(x)\frac{\sin x }{x} \mathrm{d}x = \int _0^\frac\pi2 f(x) \mathrm{d}x$$
How can I prove this equality?
 A: Indeed, let
$$I=\int_0^\infty f(x)\frac{\sin x}{x}dx$$
By asumption $f$ is even and $\pi$-periodic. Moreover,
$$\eqalign{
2I&=\lim_{n\to\infty}\int_{-n\pi}^{(n+1)\pi}\frac{\sin(x)}{x}f(x)dx
=\lim_{n\to\infty}\sum_{k=-n}^n\int_{k\pi}^{(k+1)\pi}\frac{\sin(x)}{x}f(x)dx\cr
&=\lim_{n\to\infty}\sum_{k=-n}^n\int_{0}^{ \pi}\frac{(-1)^k}{x+k\pi}\sin(x)f(x)dx\cr
&=\lim_{n\to\infty}\int_{0}^{ \pi}\left(\sum_{k=-n}^n\frac{(-1)^k}{x+k\pi}\right)\sin(x)f(x)dx\cr
&=\lim_{n\to\infty}\int_{0}^{ \pi}U_n(x)f(x)dx\cr
}
$$
with
$$
U_n(x) =\sin(x)\sum_{k=-n}^n\frac{(-1)^k}{x+k\pi}
$$
But it is well-known that $\lim_{n\to+\infty}U_n(x)=1$ for every $x\in(0,\pi)$ and that the sequence $\{U_n\}$ is bounded on $[0,\pi]$, so, by the Dominated Convergence Theorem, we conclude that
$$
2I=\int_0^\pi f(x)dx=2\int_0^{\pi/2}f(x)dx.
$$
and the desired conclusion follows.
Remark. Concerning the statements about the sequence $\{U_n\}$ these are well-known, and a simple proof can be found here. Indeed,
consider $\alpha\in(0,1)$, and let $f_\alpha$ be the $2\pi$-periodic function that coincides  with $x\mapsto e^{i\alpha x}$ on the interval $(-\pi,\pi)$. It is easy to check that the exponential Fourier coefficients of $f_\alpha$  are
given by
$$
C_n(f_\alpha)=\frac{1}{2\pi}\int_{-\pi}^{\pi}f_\alpha(x)e^{-inx}dx=\sin(\alpha\pi)\frac{(-1)^n}{\alpha \pi-n\pi}
$$
and since $f_\alpha$ is continuously differentiable on $(-\pi,\pi)$ the Fourier series of $f_\alpha$ converges to $f_\alpha$ on this interval:
$$
\forall\,x\in(-\pi,\pi),\quad e^{i\alpha x}=\lim_{n\to\infty}\sum_{k=-n}^{n}C_k(f_\alpha)e^{ikx}=\lim_{n\to\infty}\sum_{k=-n}^{n}\frac{(-1)^k\sin(\alpha\pi)}{\alpha\pi- k\pi}e^{ikx}
$$
Taking $x=0$, we get
$$
1=\lim_{n\to\infty}U_n(\alpha\pi)
$$
