Integral of $\sin^2(\pi N x)/(\pi x)^2$

Question

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!

This is a minor variant of 13344 and is likely to attract essentially the same answers. — Mårten W, Sep 13 '13 at 06:42

score 1 · Accepted Answer · answered Sep 13 '13 at 01:05

1

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:

$$\frac{1}{\pi} \int_{-\infty}^{\infty} du \frac{\sin^2{N u}}{u^2} = \frac{1}{\pi} \frac{1}{2 \pi} \int_{-N}^N dk \, \pi^2 = N$$

answered Sep 13 '13 at 01:05

Ron Gordon

138,521

You mean that the Fourier transform of $\chi_{[-N,N]}$ is $\sin(Nu)/u$, right? – detnvvp Sep 13 '13 at 01:09
No, I mean that the FT of sinc is chi. As even functions, the FT and the inverse for each look the same. – Ron Gordon Sep 13 '13 at 01:17
But, $\sin(Nu)/u$ is not in $L^1$. – detnvvp Sep 13 '13 at 01:18
@detnvvp: that's why we have Plancherel's Theorem. http://en.wikipedia.org/wiki/Plancherel_theorem – Ron Gordon Sep 13 '13 at 01:19
Oh, you are taking the Fourier transform in the $L^2$ sense; right. – detnvvp Sep 13 '13 at 01:20

score 1 · Answer 2 · answered Sep 13 '13 at 01:12

1

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.

answered Sep 13 '13 at 01:12

Stefanoski

11

score 0 · Answer 3 · edited Apr 13 '17 at 12:21

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

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

Integral of $\sin^2(\pi N x)/(\pi x)^2$

3 Answers3