# Calculating integrals using the residue theorem

Calculate the following definite integral

$$\int_{- \infty}^{\infty}\frac{\sin 3x}{x(x^2+1)}dx$$

using the residue theorem

This was my attempted solution:

Solution: $$\int_{- \infty}^{\infty}\frac{\sin 3x}{x(x^2+1)}dx = \int_{-\infty}^{\infty} \operatorname{Im}\left[\frac{e^{i3z}}{z(z^2+1)}\right]dz = \operatorname{Im}\left[\int_{- \infty}^{\infty} \frac{e^{i3z}}{z(z^2+1)} dz\right]$$

Now note that $f(z) = \frac{e^{i3z}}{z(z^2+1)}$ has singularities at $z_0 = 0$ and $z_1 = i$ and $z_2 = -i$ then using the residue theorem we have $$\int_{- \infty}^{\infty} \frac{e^{i3z}}{z(z^2+1)} dz = 2\pi i \sum_{k=0}^2 \operatorname{Res}_{z = z_k} f(z)$$

Then calculating the residues at $z_0$, $z_1$ (we don't need to calcualte $z_2$ since it's not in the upper half-plane) we get $\operatorname{Res}_{z = z_0} f(z) = 1$ and $\operatorname{Res}_{z = z_1} f(z) = \frac{1}{-2e^3}$

Hence $$\sum_{k=0}^2 \operatorname{Res}_{z = z_k} f(z) = 1- \frac{1}{2e^3}$$ and we have that $$\operatorname{Im}\left[\int_{- \infty}^{\infty} \frac{e^{i3z}}{z(z^2+1)} dz\right] = \pi\left(2 -\frac{1}{e^3} \right)$$ and thus $$\int_{- \infty}^{\infty}\frac{\sin 3x}{x(x^2+1)}dx = \pi\left(2 -\frac{1}{e^3} \right)$$

However the solution that my lecturer gave was $\pi\left(1 -\frac{1}{e^3} \right)$, is his solution incorrect or is mine incorrect and if so why?

The singularity at $z=0$ is a simple pole on the contour, so it contributes half its residue.
To avoid the singularity, use an contour indented with a small semicircle of radius $\varepsilon$, expand the integrand about the pole using the Laurent expansion, and evaluate the integral around the small semicircle using this, and take $\varepsilon \downarrow 0$. (And this gives the half in the simple pole case.)
• Sorry I explained that poorly, disregard that last comment of mine. What I meant was this, take this integral $$\int_{-\infty}^{\infty}\frac{\sin x -x}{x^3} dx$$. To calculate that I'll take the imaginary part of the integral $$\int_{-\infty}^{\infty} \frac{e^{iz}-z}{z^3}dz$$ and apply the residue theorem. Then note that $z_0 = 0$ is a pole (the only pole) of order $n =3$, however it also contributes half it's residue. (This is judging by the answer I get which is 2x the actual answer). I don't see how your line of reasoning extends to this case. – Perturbative Jun 2 '18 at 15:39