Prove: $\int_0^{\infty}\left(\frac{\sin x}{x}\right)^2dx=\pi/2$

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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 $$\sum_{n=1}^{\infty}{{\sin(n\delta)}\over n}={{\pi-\delta}\over 2};$$

(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.