I'm trying to evaluate the integral:

$$\int_{-\infty}^{\infty}\frac{\sin^2{x}}{x^2}dx$$ with contour integration and am not sure if the basic idea of what I'm doing is correct.

I know that $$\sin{x} = \frac{e^{ix} - e^{-ix}}{2i}$$ and thus $\sin^2{x} = e^{2ix} + e^{-2ix} -2$.

Thus, can I solve

$$\int_{\infty}^{\infty}\frac{e^{2iz}+e^{-2iz} -2}{-4z^2}dz$$ using the indented semicircle contour and take its real part to obtain the solution to my original integral?


Your method almost works. Note that the integrals $\int_{-\infty}^{\infty}\frac{e^{2iz}}{-4z^2}dz$ and $\int_{-\infty}^{\infty}\frac{e^{-2iz}}{-4z^2}dz$ require different contours.

Alternatively, note that $$ \sin^2x = \frac{1 - \cos (2x)}{2} = \Re \left( \frac{1 - e^{2ix}}{2} \right) $$ So that we can calculate $$ \Re \left(\int_{-\infty}^{\infty}\frac{1 - e^{2iz}}{2z^2}dz \right) $$

  • $\begingroup$ I got taken away for a while and did not see your answer before I posted. If you think it is too similar, I will remove it. $\endgroup$
    – robjohn
    Dec 16 '14 at 1:37
  • $\begingroup$ I think that your post is far more useful for posterity. Spiffy hat! $\endgroup$ Dec 16 '14 at 1:39
  • 1
    $\begingroup$ Thanks! I need to find some werewolves, now... $\endgroup$
    – robjohn
    Dec 16 '14 at 1:40

Consider the contour $$ \gamma_1=[-R,R]\cup R-i[0,1]\cup[R,-R]-i\cup-R-i[1,0] $$ Since there are no singularities of $\frac{\sin^2(z)}{z^2}$, we get $$ \int_{\gamma_1}\frac{\sin^2(z)}{z^2}\mathrm{d}z=0 $$ Since the integral over each of the two short segments is bounded by $\dfrac1{R^2}\to0$ we get $$ \int_{-\infty}^\infty\frac{\sin^2(x)}{x^2}\mathrm{d}x =\int_{-\infty-i}^{\infty-i}\frac{\sin^2(x)}{x^2}\mathrm{d}x $$ Consider the contour $$ \gamma_2=[-R,R]-i\cup Re^{i[0,\pi]}-i $$ which encloses the singularity at $z=0$, and the contour $$ \gamma_3=[-R,R]-i\cup Re^{-i[0,\pi]}-i $$ which encloses no singularities. Since the integral over each of the two arcs is bounded by $\dfrac\pi{R}\to0$, we get $$ \begin{align} \int_{-\infty}^\infty\frac{\sin^2(x)}{x^2}\mathrm{d}x &=\int_{-\infty}^\infty\frac{2-e^{2ix}-e^{-2ix}}{4x^2}\mathrm{d}x\\ &=\int_{\gamma_2}\frac{2-e^{2iz}}{4z^2}\mathrm{d}z -\int_{\gamma_3}\frac{e^{-2iz}}{4z^2}\mathrm{d}z\\[6pt] &=\pi-0 \end{align} $$ since the residue of $\dfrac{2-e^{2iz}}{4z^2}$ at $z=0$ is $-\dfrac{i}{2}$ and $\gamma_3$ does not enclose any singularities.

Therefore, we get $$ \int_{-\infty}^\infty\frac{\sin^2(x)}{x^2}\mathrm{d}x=\pi $$


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