**The Gamma Function Part 3: Beta Function and Legendre duplication:**

(Part 1)

(Part 2)

Consider the function defined by

. The integral converges for *x*,*y*>0; and the substitution in the integral shows that .

Now, consider the product of two gamma functions :

Now, let us make a change in variables from (*u*,*v*) to (*r*,*t*) via , . Then and we have:

Now, for the first integral, we make the transformation to find:

And thus we see that:

by our definition of the beta function; solving for gives us:

Note that if we take our integral definition for beta and make the change of variables given by , then we find:

And thus:

Now, we can use the beta function to derive a gamma function relation called the Legendre duplication formula. First, we consider :

Now, make the substitution :

Reexamining the beta function integral, we make the change to find:

Comparing this integral to the resulting integral we had for , we see:

.

Solving for , we find:

This relation is also sometimes written as:

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Tags: Beta Function, Gamma Funtion, Legendre Duplication, Math, Monday Math

This entry was posted on May 19, 2008 at 12:01 am and is filed under Math/Science. You can follow any responses to this entry through the RSS 2.0 feed.
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July 5, 2008 at 1:58 pm |

I like your Beta function tutorial. You know a good thing to add would be to show how B(n,1/2)=(2^(2n)(n!)^2)/(n(2n)!)

Hey, are you the same Twisted One from MHF?. If so, this is galactus. I just stumbled onto this by googling Beta. Cool

July 5, 2008 at 9:24 pm |

Yes, I’m the same TwistedOne151. The bit you suggest isn’t really needed for the proof in question, but it could be done using the above work, particularly where we see that B(1/2,n)=2^(2n-1)((Γ(n))^2)/(Γ(2n)).

From this, we get:

B(1/2,n)=2^(2n-1)(((n-1)!)^2)/((2n-1)!).

Multiply both numerator and denominator by 2n^2; then use 2*2^(2n-1)=2^(2n),

2n*(2n-1)!=(2n)!, and n(n-1)!=n!

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