1
$\begingroup$

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?

$\endgroup$

1 Answer 1

1
$\begingroup$

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.)

It won't always just be half the residue for higher-order poles, but you can still do the calculation with an indented contour to get the correct answer.

$\endgroup$
5
  • $\begingroup$ It seems like the first singularity always contributes half, is this always the case? $\endgroup$ Commented Jun 2, 2018 at 15:23
  • $\begingroup$ What do you mean by "first"? $\endgroup$
    – Chappers
    Commented Jun 2, 2018 at 15:29
  • $\begingroup$ 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. $\endgroup$ Commented Jun 2, 2018 at 15:39
  • $\begingroup$ Ah, right. The point is that writing down half the residue won't always work. The indentation and take the limit technique will work. $\endgroup$
    – Chappers
    Commented Jun 2, 2018 at 15:45
  • $\begingroup$ Could you explain that technique please in your answer if possible? $\endgroup$ Commented Jun 2, 2018 at 15:57

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .