Integral of $\sin^2(\pi N x)/(\pi x)^2$ The integral in question is
$$ \int_{-\infty}^{\infty} {\sin^2(\pi N x)\over(\pi x)^2}dx = N. \,$$
I'm pretty certain this is true, but I'd like to know if there's a simpler way to solve this than integration by parts, which achieves something along the lines of 
$$-{\pi x^2\sin(2 \pi N x)\over 4N} +{x\cos(2 \pi Nx)\over4N^2} - \int{\cos(2\pi N x)\over4N^2} \, dx$$
and involves the sine integral Si and is a bit messier than I'd like. Any help is appreciated, thanks much in advance!
 A: Use Parseval's (Plancherel's) Theorem.  First sub $u=\pi x$ and get
$$\frac{1}{\pi} \int_{-\infty}^{\infty} du \frac{\sin^2{N u}}{u^2} $$
Knowing that the Fourier transform of $\sin{N u}/u$ is $\pi$ when $k \in [-N,N]$ and $0$ otherwise, we get, from Parseval's theorem:
$$\frac{1}{\pi} \int_{-\infty}^{\infty} du \frac{\sin^2{N u}}{u^2} = \frac{1}{\pi} \frac{1}{2 \pi} \int_{-N}^N dk \, \pi^2 = N$$
A: also, you can use the fact the integrant is even, let $\pi N x=t$, and reduce by parts from $(\frac{\sin t}{t})^2$ to complete sine integral $\int _0^ \infty \frac{\sin t}{t}$, then you get N.
A: Note that, the integrand is an even function, se we have

$$ \int_{-\infty}^{\infty} {\sin^2(\pi N x)\over(\pi x)^2}dx = 2\int_{0}^{\infty} {\sin^2(\pi N x)\over(\pi x)^2}dx. $$

Now, make the change of variables $u=\pi Nx$ and proceed using the technique used in this answer.
Note: The following Laplace transforms are needed for the evaluation of the integral 

$$ \mathcal{L}\{t\}(s)=\frac{1}{s^2},\quad\mathcal{L}\{\sin^2(t)\}(s)=\frac{2}{s(s^2+4)}.$$

