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I am trying to evaluate: $$I = \int_{-\infty}^{\infty} \frac{\sin^2(x)}{x^2} \, dx.$$ Using a contour semi-circle (upper plane), I can get: $$ \oint_{C} f(z) \,dz = \oint_{C} \frac{1 - e^{2iz}}{z^2} \, dz.$$ The whole issue is the $z^2$. I cannot use the residue theory, because it lies on the contour.

I don’t want a full solution. I really want to try on my own, I just need some guidance!

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  • $\begingroup$ Maybe you make a smaaall half-circle around the origin? $\endgroup$ – Karl Feb 8 '15 at 18:13
  • $\begingroup$ The integrand only contains removable singularities. $\endgroup$ – robjohn♦ Feb 8 '15 at 18:36
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The integrand contains a removable singularity at $z=0$. That is, the integrand is an entire function.

This answer shows how to compute $$ \int_{-\infty}^\infty\left(\frac{\sin(x)}x\right)^n\mathrm{d}x $$ by first showing that it equals $$ \int_{-\infty-i}^{\infty-i}\left(\frac{\sin(x)}x\right)^n\mathrm{d}x $$ Using $\sin(x)=\dfrac{e^{ix}-e^{-ix}}{2i}$ and the binomial theorem, we apply contour integration to get $$ \int_{-\infty}^\infty\left(\frac{\sin(x)}x\right)^n\mathrm{d}x=\frac{2\pi}{2^n(n-1)!}\sum_{k=0}^{\lfloor n/2\rfloor}(-1)^k\binom{n}{k}(n-2k)^{n-1} $$ For $n=2$, this gives $$ \int_0^\infty\frac{\sin^2(x)}{x^2}\mathrm{d}x=\frac\pi2 $$

Another approach, using Riemann Sums, is shown in this answer.

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Avoid $z=0$ with a semicircle $c_\epsilon$ of radius $\epsilon>0$ small (clockwise sense). You will have to compute $$ \lim_{\epsilon\to0}\int_{c_\epsilon}\frac{1-e^{2iz}}{z^2}\,dz. $$

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  • $\begingroup$ I tried that too. I got: $$\frac{1 - e^{2i\epsilon e^{i\theta}}}{(\epsilon e^{i\theta}}$$ But when you let $\epsilon \to 0$ you get division by $0$. $\endgroup$ – Lebes Feb 8 '15 at 18:17
  • $\begingroup$ You get an indeterminate form $\frac{0}{0}$, which can be dealt with using L'Hopital's rule. $\endgroup$ – user170231 Feb 8 '15 at 18:19
  • $\begingroup$ But it becomes 0/0 forever. $\endgroup$ – Amad27 Feb 8 '15 at 18:37
  • $\begingroup$ Have you done the computation? $(1-e^{2iz})/z^2\sim -2\,i/z$, and when you integrate over the semicircle the limit of the integral is half the residue. $\endgroup$ – Julián Aguirre Feb 8 '15 at 19:13
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How about good old integration by parts? $$ \int_{-\infty}^{+\infty} \frac{\sin^2(x)}{x^2} dx = \Big|_{-\infty}^{+\infty} -\frac{\sin^2(x)}{x} + \int_{-\infty}^{+\infty} \frac {2 \sin(x)\cos(x)}{x} dx = 0+ \int_{-\infty}^{+\infty} \frac{\sin(2x)}{x} dx = \int_{-\infty}^{+\infty} \frac{\sin(x)}{x} dx = \pi $$ where at the end I use the substitution $u=2x$ and the final integral is well-known to equal $\pi$ ( if you are not familiar with this result, tell me, I can prove that too)

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Another way to think about this is to note that $$\displaystyle \int\frac{\sin^2(x)}{x^2} dx = \int \frac{\sin(x)}{x} dx$$ Then you can use the well know solution to the Dirichlet integral. If you don't see how the two integrals are related I can prove it two. Hint: use integration by parts.

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