Monday Math 20: The Gamma Function (Part 3/?)

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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7 Responses to “Monday Math 20: The Gamma Function (Part 3/?)”

  1. Monday Math 21: The Gamma Function (Part 4/?) « Twisted One 151’s Weblog Says:

    […] The Gamma Function Part 4: Euler Reflection Formula and the Sine Function: (Part 1) (Part 2) (Part 3) Let us now define the function , for non-integer x, as . Now, we remember that , and if we […]

  2. Monday Math 22: The Gamma Function (Part 5/5) « Twisted One 151’s Weblog Says:

    […] Math 22: The Gamma Function (Part 5/5) Part 5: n-ball (Part 1) (Part 2) (Part 3) (Part 4) In mathematics, a ball is the region of three dimensional space bounded by a sphere. The […]

  3. cody Says:

    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

  4. twistedone151 Says:

    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!

  5. Monday Math 27 « Twisted One 151’s Weblog Says:

    […] . Now, we perform the substitution , giving: . We should recoginze this last integral as a beta function: , so we have: . Now, recall the Euler reflection formula . We see that this applies to the above […]

  6. Monday Math 44 « Twisted One 151’s Weblog Says:

    […] how to find the derivatives of the gamma function, we can also find the partial derivatives of the beta function. We know that . Thus . We see from the symmetry between x and y that . Finding the second partial […]

  7. Monday Math 46 « Twisted One 151’s Weblog Says:

    […] Math 46 Find . First, we note via symmetry that . Now, recall that it was shown here, that . Thus . And so we find that . Now, we can use our previous results for the partial […]

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