How do you derive this alternate form of the gamma function?

Click For Summary
The discussion focuses on deriving an alternate form of the gamma function, specifically transforming the integral representation \Gamma(n) = int(0 to infinity)[(x^(n-1))*e^-x]dx into \Gamma(n) = 2int(0 to infinity)[(x^(2n-1))*e^(-x^2)]dx. Participants suggest that a simple change of variables can facilitate this transformation, emphasizing the importance of recognizing that the variable in each integral can differ. One user expresses initial confusion regarding the relationship between the exponential terms, which is clarified through the proposed substitution. The conversation highlights the utility of variable substitution in integral calculus for deriving equivalent expressions. Understanding this technique is essential for successfully manipulating gamma function representations.
Zatman
Messages
85
Reaction score
0
\Gamma(n) = int(0 to infinity)[(x^(n-1))*e^-x]dx

Show that it can also be written as:

\Gamma(n) = 2int(0 to infinity)[(x^(2n-1))*e^(-x^2)]dxI have no idea how to go about this. I have tried integration by parts of each to see if anything relates, but how can you get from an exp(-x) to and exp(-x^2) term?

Any help would be appreciated.

(PS apologies, I am unfamiliar with writing formulae using latex)
 
Physics news on Phys.org
A simple change of variables will do the trick. Take another look at the integrals and see if you can figure out what substitution you should make.

(Maybe it will help you see it if you use the variable 't' instead of 'x' in the second integral)
 
Right, I was getting confused because I thought the x in each was the same.

But that makes sense, thank you. :)
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 9 ·
Replies
9
Views
4K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 17 ·
Replies
17
Views
7K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 200 ·
7
Replies
200
Views
29K