Integral of Square of Reciprocal of √(a + cos t)

[e(ho0n3]

[SOLVED] Integral of Square of Reciprocal of √(a + cos t)

Homework Statement
Demonstrate that

$$\int_0^\pi \frac{dt}{(a + \cos t)^2} = \frac{a\pi}{(\sqrt{a^2 - 1})^3}$$

where a > 0.

The attempt at a solution
I want to compute this using the Residue Theorem. The first thing I did was make the substitution $\theta / 2 = t$ so that

$$\int_0^\pi \frac{dt}{(a + \cos t)^2} = \int_0^{2\pi} \frac{d\theta}{2(a + \cos \theta/2)^2}$$

I then made the substitution $z = e^{i\theta/2}$ so that

$$\int_0^{2\pi} \frac{d\theta}{2(a + \cos \theta/2)^2} = \int_{|z|=1} \frac{4z}{i(z^2 + 2az + 1)^2} \, dz = \int_{|z|=1} \frac{4z}{i(z - z_0)^2(z - z_1)^2} \, dz$$

where

$$\begin{align*}z_0 &= \frac{-a + \sqrt{a^2 - 1}}{2}\\<br /> z_1 &= \frac{-a - \sqrt{a^2 - 1}}{2}\end{align*}$$

Call the integrand of the last integral f(z). By the Residue Theorem

$$\int_{C} f(z) \, dz = 2\pi i \sum_{u \in D} \text{Res}(f; u)$$

where u is a pole of f, C is the unit circle and D is its interior. The only possible poles are $z_0, z_1$. I know $z_0$ is in D for all a > 0. However, $z_1$ is not in D if a > 5/4. What do I in this case? I would condition on the value of a but then my answer won't match up with that of the problem statement.

 

[Dick]

First off I think the condition that you want on a is a>1. The right hand side of what you want to prove is imaginary and the left side doesn't exist if 0<a<=1. Second, check the roots of your quadratic. You have an extra factor of 1/2 in there. I think you will find z0 is always in D and z1 never is.

 

[Dick]

Also z=exp(i*theta/2) for 0 to 2pi doesn't go all of the way around |z|=1. I think the substitution you want is just z=exp(i*theta). Start by just changing the limits on your original integral to 0 to 2pi. That will give you twice the answer you want, so you can just divide the result by 2. cos(theta) is symmetric around theta=pi.

 

[e(ho0n3]

I followed your suggestions and obtained the answer. Thanks a lot.

 

[tron_2.0]

ahh this is a good problem