- #1
coomast
- 279
- 1
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
\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