# Improper integral using the Residue Theorem

1. Dec 13, 2005

### TD

I'm trying to find

$$\int\limits_0^{ + \infty } {\frac{{\sin x}} {{x^3 + x}}dx}$$

Since the function is even, I can compute it as

$$\frac{1} {2}\int\limits_{ - \infty }^{ + \infty } {\frac{{\sin x}} {{x^3 + x}}dx}$$

To use the residue theorem, I construct a large semi-circle C with center O and radius R. To avoid the real singularity at x = 0, we construct another (small) semi-circle c with radius r. The closed path is then given by C + [-R,-r] + c + [r,R]. We then let R go to infinity and r to 0. So we have:

$$\oint {\frac{{\sin x}} {{x^3 + x}}dx} = \int\limits_C {\frac{{\sin x}} {{x^3 + x}}dx} + \int\limits_c {\frac{{\sin x}} {{x^3 + x}}dx} + \int\limits_{ - R}^{ - r} {\frac{{\sin x}} {{x^3 + x}}} dx + \int\limits_r^R {\frac{{\sin x}} {{x^3 + x}}} dx$$

From the residue theorem, I found the LHS to be $\pi \sinh 1$, by applying the theorem to the only enclosed pole i.
Then, taking the limits or R and r to inf and 0, I found the integral over C to be 0 and over c to be $-\pi$ which results in:

$$\int\limits_{ - \infty }^{ + \infty } {\frac{{\sin x}} {{x^3 + x}}} dx = \pi \sinh 1 + \pi$$

Now that appears to be too large, does anyone know what mistake(s) I made or what I should do? Thanks!

2. Dec 13, 2005

### shmoe

A couple mistakes, the integral over C does not go to zero and there's no pole at x=0, so the integral over c will go to zero.

It's probably easiest to replace sin(x) with $$e^{ix}$$, which is bounded in the upper half plane, making the integral over C much easier to deal with, though you'll now have a pole at z=0.

3. Jun 7, 2010

### TheLifeKaotik

There is indeed a pole at z=0, by Jordan's... but you only need to worry about when doing exactly that kind of problem. If I'm not mistaken you have to set the LHS to Im(2*pi*i*Resi (eiz/(z3+z))