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.
$$\lim\limits_{\varepsilon\to0^+}\int_{-\infty}^{\infty}\frac{\sin^3(x)}{x^3+\varepsilon^3}\mathrm{d}x$$
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?
