Prove that $ \int_0^\infty (\frac{\sin x}{x})^2 = \frac{\pi}{2}$. I need to prove that 
$ \int_0^\infty (\frac{\sin x}{x})^2 = \frac{\pi}{2}$. 
I have proved that $\sum_1^\infty \frac {\sin^2(n \delta)}{n^2 \delta}=\frac{\pi-\delta}{2}$ for $0<\delta<\pi$ and I'm supposed to use this identity.
 A: Use your identity for $\delta  \to 0$. In this case, by assuming $x = n\delta $, and noting that $dx = (n + 1)\delta  - n\delta  = \delta $ and ${x_{\min }} = 1\delta  = 0$ and ${x_{\max }} = +\infty \delta  = +\infty$, your identity will give $$\mathop {\lim }\limits_{\delta  \to 0} \sum\limits_{n = 1\atop x = n\delta }^\infty  {{{\left( {\frac{{\sin (x)}}{x}} \right)}^2}dx}  = \int\limits_0^{ + \infty } {{{\left( {\frac{{\sin (x)}}{x}} \right)}^2}dx}  = \frac{\pi }{2}$$which is (hopefully) your desired result ;)
A: Here is an alternate proof. It hinges on a lemma, which isn't too difficult to prove (it requires cutting $\Bbb R_+$ in two halves $[0,A]$ and $[A,+\infty)$ for some meaningful $A$, and using Riemann sums on the segment)

Suppose $f:\Bbb R_+\to\Bbb R$ is continuous integrable and eventually decreasing. Then for any $\delta>0$, the sum 
  $$S_\delta=\sum_{n=0}^\infty f(n\delta)\cdot\delta$$
  is (absolutely) convergent, and has a limit when $\delta\to 0$ with
  $$\lim_{\delta\to 0}\;S_\delta=\int_{\Bbb R_+}f$$

This lemma then easily implies that if $\phi:\Bbb R_+\to \Bbb R$ is continuous, integrable and dominated by some non negative, integrable, eventually decreasing function $f$ (in the sense that $|\phi|\leq f$), then, for any $\delta>0$, the series
$$\Sigma_\delta=\sum_{n=0}^\infty\phi(n\delta)\cdot\delta$$
is (absolutely) convergent, and has a limit as $\delta\to 0$, with
$$\lim_{\delta\to 0}\;\Sigma_\delta=\int_{\Bbb R_+}\phi$$

This gives you the desired result, with $\phi(x)=\frac{\sin(x)^2}{x^2}$ and $f(x)$ defined to be equal to $1$ on $[0,1]$, and equal to $\frac1{x^2}$ on $[1,+\infty)$.
