Exercise. Evaluate the improper integral $$\int_{-\infty}^{\infty}\frac{\sin^3(x)}{x^3}\mathrm{d}x$$ using the complex epsilon method.

First we add the epsilon term.


Using the fact that

$$\sin^3(x)=\frac{3\sin x-\sin(3x)}{4},$$

we are left to evaluate

$$\lim\limits_{\varepsilon\to0^+}\int_{-\infty}^{\infty}\frac{3\sin x-\sin(3x)}{4(x^3+\varepsilon^3)}\mathrm{d}x=\lim\limits_{\varepsilon\to0^+}\int_{-\infty}^{\infty}\Im\left[\frac{3\operatorname{e}^{\mathrm{i}z}-\operatorname{e}^{3\mathrm{i}z}}{4(z^3+\varepsilon^3)}\right]\mathrm{d}z.$$

There are a couple of singularities, namely

$$z_k=(-\varepsilon^3)^{1/3}=(\operatorname{e}^{\mathrm{i}\pi}\varepsilon^3)^{1/3}=\varepsilon\operatorname{exp}\left(\mathrm{i}\frac{\pi+2\pi k}{3}\right),\ \ \ k\in\{0,1,2\}.$$

Considering the singularities on the upper complex plane (including the real number line) suffices. As such,

$$\lim\limits_{\varepsilon\to0^+}\int_{-\infty}^{\infty}\Im\overbrace{\left[\frac{3\operatorname{e}^{\mathrm{i}z}-\operatorname{e}^{3\mathrm{i}z}}{4(z^3+\varepsilon^3)}\right]}^{f(z)}\mathrm{d}z=\lim\limits_{\varepsilon\to0^+}\Im\left[2\pi \mathrm{i}\sum_{k=0}^{1}\operatorname{\underset{z=z_k}{Res}}f(z) + \lim\limits_{R\to\infty}\int_{C_R}f(z)\right]\tag1$$

where $C_R$ is the semi-circle between $-R$ and $R$ big enough to encompass the aformementioned singularities.

The answer key instead says this should be

$$\lim\limits_{\varepsilon\to0^+}\Im\left[2\pi \mathrm{i}\sum_{k=0}^{0}\operatorname{\underset{z=z_k}{Res}}f(z) + \mathrm{i}\pi\operatorname{\underset{z=-\varepsilon}{Res}}f(z) + \lim\limits_{R\to\infty}\int_{C_R}f(z)\right]\tag{1'}$$

which I have checked will give the correct answer for the integral, that is $3\pi/4$.

  • Question: Where to does the $2$ disappear in $(1')$? In other words, why does the residue at $-\varepsilon$ have the coefficient $\mathrm{i}\pi$ instead of $2\mathrm{i}\pi$ as the rest?
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    $\begingroup$ The singularity at $-\epsilon$ is not inside the contour, it is on it -- so strictly speaking the residue theorem doesn't apply (you can use a tiny semicircular indent which will give the unexplained factor of 1/2). $\endgroup$ – Matthew Towers Apr 18 '17 at 13:08
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    $\begingroup$ You have to add a little semi-circle to $C_R$ to go around the singularity at $-\epsilon$. See here: math.stackexchange.com/questions/319959/… $\endgroup$ – Hans Lundmark Apr 18 '17 at 13:09
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    $\begingroup$ Could someone recommend some resources to learn about this complex epsilon method? Or perhaps some way to look it up, because I couldn't arrive at any results. Thank you in advance! $\endgroup$ – Fede Poncio Apr 18 '17 at 13:18
  • $\begingroup$ Thank you very much for your links, and especially for expanding my mathematical vocabulary with indent... That'll make my life easier. So it is basically this theorem, only a more general version. @HansLundmark $\endgroup$ – Linear Christmas Apr 18 '17 at 13:23
  • $\begingroup$ @m_t_ And if I increase the power by one, to $$\sin^4(x)/x^4,$$ we are back to the usual theorem because the indent integral (around zero now, probably) is zero? $\endgroup$ – Linear Christmas Apr 18 '17 at 13:25

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