Complex Analysis Integral Question

[RJLiberator]

Homework Statement

Computer the integral:

Integral from 0 to infinity of (d(theta)/(5+4sin(theta))

Homework Equations

integral 0 to 2pi (d(theta)/1+asin(theta)) = 2pi/(sqrt(1-a^2)) (-1<a<1)

The Attempt at a Solution

I've seen this integral be computed from 0 to 2pi, where the answer is 2pi/3 using the integral from 0 to 2pi (d(theta)/1+asin(theta)) = 2pi/(sqrt(1-a^2)) (-1<a<1)

Is there a typo in the bounds? Should the bounds be from 0 to 2pi instead of 0 to infinity? :/ Or can this problem be solved with another trick/the same trick?

 

[RUber]

It looks like the bounds should be 0 to 2pi. Otherwise you have:
$\int_0^\infty \frac{1}{5+4\sin\theta} \, d\theta = \infty \int_0^{2\pi} \frac{1}{5+4\sin\theta} \, d\theta$
since there are an infinite number of periods between 0 and 2pi.
The function is always positive >1/9, so there is no practical application to integrating over infinity, since the integral will clearly be greater than $\infty \frac19$.

 

[RJLiberator]

Beautiful! As I suspected.
Thank you for this.