Prove: $\int_0^{\infty}\left(\frac{\sin x}{x}\right)^2dx=\pi/2$ I am dealing exercise 12 in Chapter 8 of Rudin's Principles of Mathematical Analysis. Given the function $f$: 
  $$f(x) =
\begin{cases}
1,  & \text{if $|x|\le\delta$} \\
0, & \text{if $\delta<|x|\lt \pi$}  \\
\end{cases}
,$$
where $0<\delta<\pi$. 
I have done the first three part: 
(a)find the Fourier coefficients of $f$; 
(b)Conclude that $$\sum_{n=1}^{\infty}{{\sin^2(n\delta)}\over n}={{\pi-\delta}\over 2};$$
(c)Deduce from Parseval's Theorem that $$\sum_{n=1}^{\infty}{{\sin^2(n\delta)}\over \delta n^2}={{\pi-\delta}\over 2}$$
However, I got trouble with: 
(d) Let $\delta\to 0$ and prove that $$\int_0^{\infty}\Big({{\sin x}\over{x}}\Big)^2dx={\pi\over2}.$$
I got confused about how to convert the sum in (c) to integral in (d). Thanks. 
 A: Given $\epsilon\gt0$, select $M > 1/\epsilon$, such that: 
$$
\left|\int_0^\infty \left(\frac{\sin x}{x}\right)^2 dx - \int_0^M \left(\frac{\sin x}{x}\right)^2 dx\right|\lt \epsilon \tag{1}
$$
Then, select $N_0$ so that $N\gt N_0$ implies:
$$
\left|\int_0^M \left(\frac{\sin x}{x}\right)^2 dx - \sum_{i=1}^N{\frac{\sin^2\left(n\frac{M}{N}\right)}{n^2\left(\frac{M}{N}\right)}}\right|\lt \epsilon \tag{2}
$$
This is possible, since:
$$\sum_{i=1}^N{\frac{\sin^2\left(n\frac{M}{N}\right)}{n^2\left(\frac{M}{N}\right)}} = \sum_{i=1}^N{\left(\frac{\sin\left(n\frac{M}{N}\right)}{n\frac{M}{N}}\right)^2\cdot\frac{M}{N}} \longrightarrow \int_0^M \left(\frac{\sin x}{x}\right)^2 dx$$
as $N\rightarrow\infty$, as the expression on the left is a riemann sum of the integral on the right.
Finally, we note that:
$$
\left|\sum_{i=1}^N{\frac{\sin^2\left(n\frac{M}{N}\right)}{n^2\left(\frac{M}{N}\right)}} - \frac{\pi-\frac{M}{N}}{2}\right|=\sum_{i=N+1}^\infty{\frac{\sin^2\left(n\frac{M}{N}\right)}{n^2\left(\frac{M}{N}\right)}}\le\frac{N}{M}\sum_{i=N+1}^\infty{\frac{1}{n^2}}\le \frac{1}{M}\lt\epsilon \tag{3}
$$
Combining $(1)$, $(2)$, and $(3)$, we get:
$$
\left|\int_0^\infty \left(\frac{\sin x}{x}\right)^2 dx - \frac{\pi-\frac{M}{N}}{2}\right|\lt3\epsilon
$$
when $N>N_0$. In particular, taking $N\rightarrow\infty$, we get:
$$
\left|\int_0^\infty \left(\frac{\sin x}{x}\right)^2 dx - \frac{\pi}{2}\right|\le3\epsilon
$$
