Elliptic
- 33
- 0
Integrating with Gamma: cos(theta)^(2k+1)
- Thread starter Elliptic
- Start date
-
- Tags
- Gamma Integrating
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
5 replies · 2K views
- 17,871
- 1,653
You mean:
[tex]\int_0^{\frac{\pi}{2}} \cos^{2k+1}(\theta)d\theta[/tex]... eg: evaluate the definite integral of an arbitrary odd-power of cosine.
The standard approach is to start by integrating by parts.
You'll end up with a reducing formula which you can turn into a ratio of factorials - apply the limits - after which it is a matter of relating that to the factorial form of the gamma function.
eg. http://mathworld.wolfram.com/CosineIntegral.html
[tex]\int_0^{\frac{\pi}{2}} \cos^{2k+1}(\theta)d\theta[/tex]... eg: evaluate the definite integral of an arbitrary odd-power of cosine.
The standard approach is to start by integrating by parts.
You'll end up with a reducing formula which you can turn into a ratio of factorials - apply the limits - after which it is a matter of relating that to the factorial form of the gamma function.
eg. http://mathworld.wolfram.com/CosineIntegral.html
Elliptic
- 33
- 0
- 17,871
- 1,653
If it was easy there'd be no point setting it as a problem.
I'm not going to do it for you ...
Do you know what a gamma function is? You can represent it as a factorial?
Can you identify where you are having trouble seeing what is going on?
Perhaps you should try to do the derivation for yourself?
I'm not going to do it for you ...
Do you know what a gamma function is? You can represent it as a factorial?
Can you identify where you are having trouble seeing what is going on?
Perhaps you should try to do the derivation for yourself?
Last edited:
Elliptic
- 33
- 0
- 17,871
- 1,653
Really? And I thought I was being mean...
The trig-form of the beta function aye - yep, that's a tad more elegant that the path I was suggesting before (the more usual one)... but relies on a hand-wave: do you know how the beta function is derived?
Also - you have [itex]\frac{1}{2}B(\frac{1}{2},k+1)[/itex] but you've spotted that.
If you look at the cosine formula - you have to evaluate the limits ... at first it looks grim because it gives you a sum of terms like [itex]\sin\theta\cos^{2k}\theta[/itex] which is zero at both limits ... unless k=0 ... which is the first term in the sum, which is 1.
After that it is a matter of subbing in the factorial representation of the gamma function.
Which would be a concrete proof.
Yours is shorter and if you have the beta function in class notes then you should be fine using it.
The trig-form of the beta function aye - yep, that's a tad more elegant that the path I was suggesting before (the more usual one)... but relies on a hand-wave: do you know how the beta function is derived?
Also - you have [itex]\frac{1}{2}B(\frac{1}{2},k+1)[/itex] but you've spotted that.
If you look at the cosine formula - you have to evaluate the limits ... at first it looks grim because it gives you a sum of terms like [itex]\sin\theta\cos^{2k}\theta[/itex] which is zero at both limits ... unless k=0 ... which is the first term in the sum, which is 1.
After that it is a matter of subbing in the factorial representation of the gamma function.
Which would be a concrete proof.
Yours is shorter and if you have the beta function in class notes then you should be fine using it.