# Sharing knowledge

1. Dec 14, 2007

### coomast

Some time ago I had to do a calculation concerning a physical problem. In this calculation some integrals were needed to be solved and the method on how to do this was briefly described. I was able to solve the problem completely and found the method so beautiful that I would like to share it with all of you. Consider the following integrals:

$$\int \frac{1}{1+\epsilon \cdot cos \theta}d\theta$$

$$\int \frac{1}{\left(1+\epsilon \cdot cos \theta\right)^2}d\theta$$

$$\int \frac{1}{\left(1+\epsilon \cdot cos \theta\right)^3}d\theta$$

$$\int \frac{cos \theta}{1+\epsilon \cdot cos \theta}d\theta$$

$$\int \frac{cos \theta}{\left(1+\epsilon \cdot cos \theta\right)^2}d\theta$$

$$\int \frac{cos \theta}{\left(1+\epsilon \cdot cos \theta\right)^3}d\theta$$

$$\int \frac{sin \theta}{1+\epsilon \cdot cos \theta}d\theta$$

$$\int \frac{sin \theta}{\left(1+\epsilon \cdot cos \theta\right)^2}d\theta$$

$$\int \frac{sin \theta}{\left(1+\epsilon \cdot cos \theta\right)^3}d\theta$$

Some of these are straightforward, but some are not. In order to solve the non trivial ones in a systematic way, one can use the following substitution:

$$1+\epsilon \cdot cos \theta = \frac{1-\epsilon^2}{1-\epsilon \cdot cos \gamma}$$

With:

$$0\leq \theta \leq 2\pi$$

$$0\leq \gamma \leq 2\pi$$

The following relations can be obtained:

$$cos \theta = \frac{cos \gamma -\epsilon}{1-\epsilon \cdot cos \gamma}$$

$$sin \theta = \frac{\sqrt{1-\epsilon^2} sin \gamma}{1-\epsilon \cdot cos \gamma}$$

$$cos \gamma = \frac{\epsilon +cos \theta}{1+\epsilon \cdot cos \theta}$$

$$sin \gamma = \frac{\sqrt{1-\epsilon^2} sin \theta}{1+\epsilon \cdot cos \theta}$$

$$d \theta = \frac{\sqrt{1-\epsilon^2}}{1-\epsilon \cdot cos \gamma}d \gamma$$

$$d \gamma = \frac{\sqrt{1-\epsilon^2}}{1+\epsilon \cdot cos \theta}d \theta$$

After rewriting some of the integrals into smaller ones, substituting this and rearranging it is possible to solve them in a fairly easy way. As an example, let's take the one before the last, it was:

$$\int \frac{sin \theta}{\left(1+\epsilon \cdot cos \theta\right)^2}d\theta$$

Using the substitution we get:

$$\int \frac{sin \gamma}{1-\epsilon^2}d\gamma=\frac{-cos \gamma}{1-\epsilon^2}+C$$

The solution is now obtained by substituting either

$$\gamma = arccos \left( \frac{\epsilon +cos \theta}{1+\epsilon \cdot cos \theta} \right)$$

or

$$\gamma = arcsin \left( \frac{\sqrt{1-\epsilon^2} sin \theta}{1+\epsilon \cdot cos \theta} \right)$$

This substitution is called the "Sommerfeld substitution" after the "inventor" and is one of the nicest substitutions I've ever encountered for solving integrals in the real with "classical" functions. I hope this is helpful in solving other types of integrals you might be working on.

best regards,

Coomast

2. Feb 10, 2008

### nancy

your information was very useful thanks

3. Mar 14, 2008

### hamamo

very nice
and very useful
thanx