# Integral of Sinc Function Squared Over The Real Line [duplicate]

I am trying to evaluate $$\int_{-\infty}^{\infty} \frac{\sin(x)^2}{x^2} dx$$ Would a contour work? I have tried using a contour but had no success. Thanks.

Edit: About 5 minutes after posting this question I suddenly realised how to solve it. Therefore, sorry about that. But thanks for all the answers anyways.

• Very good link. Isn't the integrand an even function? Aug 19, 2014 at 15:51
• Check this technique which is not included in the above link. Aug 19, 2014 at 16:14

Why not to try integration by parts? This just gives:

$$\int_{\mathbb{R}}\frac{\sin^2 x}{x^2}\,dx=\int_{\mathbb{R}}\frac{\sin(2x)}{x}\,dx=\pi.$$

With the same approach you can also find the values of $$I_m = \int_{\mathbb{R}}\frac{\sin^m(x)}{x^m}\,dx.$$

• This is precisely the method I found. :) Aug 19, 2014 at 20:56
• @Assaultous2 I know this is not an obligation on M.SE but it would be nice if you upvote the one who gives an answer to your OP to show your appreciation to him. +1 Jack! Aug 20, 2014 at 10:59

This is an approach using contour integration. \begin{align} \int_{\mathbb{R}}\frac{\sin^2{x}}{x^2}{\rm d}x &=-\frac{1}{4}\lim_{\epsilon \to 0}\int_{\mathbb{R}}\frac{e^{2ix}-2+e^{-2ix}}{(x-i\epsilon)^2}{\rm d}x\tag1\\ &=-\frac{1}{4}\lim_{\epsilon \to 0}2\pi i\lim_{z \to i\epsilon}\frac{{\rm d}}{{\rm d}z}(e^{2iz}-2)\tag2\\ &=\frac{1}{4}\lim_{\epsilon \to 0}4\pi e^{-2\epsilon}\\ &=\pi \end{align} Explanation:
$(1)$: Expand $\sin^2{x}$ in terms of complex exponentials and shift the pole upwards.
$(2)$: Split the integral into $2$. Integrate the first along a semicircle in the uhp, and the second along a semicircle in the lhp. The second integral $=0$ since it encloses no poles.

Alternatively, one may use the fact that $$\int^\infty_0t^{n-1}e^{-xt}{\rm d}t=\frac{\Gamma(n)}{x^n}$$ It follows that \begin{align} 2\int^\infty_0\frac{\sin^2{x}}{x^2}{\rm d}x &=2\int^\infty_0t\int^\infty_0e^{-xt}\sin^2{x} \ {\rm d}x \ {\rm d}t\\ &=2\int^\infty_0\int^\infty_0e^{-xt}\sin{2x} \ {\rm d}x \ {\rm d}t \tag{Integrated by parts}\\ &=2\int^\infty_0\frac{2}{t^2+4}{\rm d}t\\ &=\pi \end{align}

• There isn't really a "pole" because the discontinuity is removable. $f(z) = \begin{cases} \frac{\sin(z)}{z} & \text{ if } z \neq 0 \\ 1 & \text{ if } z=0 \\ \end{cases}$ defines an entire function. Jun 6, 2015 at 18:33
• Oh hmmm maybe you mean there are poles after you split $\frac{\sin^2(x)}{x^2}$ as a sum of three functions using $\sin(x) = \frac{e^{ix}-e^{-ix}}{2i}$. I guess that makes sense. Jun 6, 2015 at 19:06

Here is another approach. Write the integral as

$$I = {2}\int_{0}^{\infty} \frac{\sin^2(x)}{x^2}.$$

Recalling the Mellin transform of a function $f$

$$\int_{0}^{\infty} x^{s-1} f(x)dx$$

our integral is the Mellin transform of $\sin(x)^2$ with $s=-1$. The Mellin transform is $\sin(x)^2$ given by

$$-\frac{1}{2}\,{\frac {\sqrt {\pi }\,\Gamma \left( 1+s/2 \right) }{s\,\Gamma \left( -s/2 + 1/2 \right) }}.$$

Taking the limit as $s\to -1$ gives the desired answer $\frac{\pi}{2}$. See other approaches.

• I think you should have $2$, not $\frac{1}{2}$, on the outside.
– Ian
Aug 19, 2014 at 16:24
• @Ian: You are right. Thanks. Aug 19, 2014 at 16:25
• I feel like this is circular because you are assuming the result of a more general integral and then deriving your answer from that. I don't think it is any different from using a table to look up the answer. Aug 19, 2014 at 16:28
• @user157227: It is a very powerful techniques which allows you to handle very difficult problems. Aug 19, 2014 at 18:32

What happens if you split $\cos2x$ into $e^{2ix}$ and $e^{-2ix}$? The two pieces need different contours - one above the real line, the other below.