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Need Help with Gamma Functions

  1. May 20, 2008 #1
    I'am not sure how to solve this gamma function [tex]\Gamma[/tex](5/4). any help ?

    sorry if this is in the wrong section.
  2. jcsd
  3. May 20, 2008 #2


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    The Gamma function is the generalization of factoral. The simplest representation is an integral:

    Gamma(x)=int(0,oo)tx-1e-tdt, for x>0.

    It can be extended by analytic continuation.
  4. May 20, 2008 #3
    yeah the original integral was [tex]\int[/tex][0,4]Y[tex]^{3/2}[/tex](16-Y[tex]^{2}[/tex])[tex]^{1/2}[/tex] dy

    which i simplified to

    64[tex]\int[/tex](0,1) t[tex]^{(5/4)-1}[/tex](1-t)[tex]^{(3/2)-1}[/tex] dt

    using Gamma(x)=int(0,oo)tx-1e-tdt, =>> [tex]\beta[/tex](5/4, 3/2)

    hence i am trying to solve [tex]\beta[/tex](5/4, 3/2)
    = [tex]\Gamma[/tex](5/4)[tex]\Gamma[/tex](3/2) / [tex]\Gamma[/tex]((5/2)+(3/2))

    i am trying to solve this to get an answer in terms of a number.

    i dont know how to solve [tex]\Gamma[/tex](5/4)
    Last edited: May 20, 2008
  5. May 20, 2008 #4
    [tex]\Gamma\left(\frac{5}{4}\right) = \Gamma\left(1+\frac{1}{4}\right) = \frac{1}{4}\Gamma\left(\frac{1}{4}\right)[/tex]

    [tex]\Gamma\left(\frac{1}{4}\right)[/tex] has no known basic expression, but is known to be transcendental.
  6. May 20, 2008 #5
    thanks for your help
  7. May 20, 2008 #6
    I read that (1/q)! has no representation for integer q>2, except decimal form. But, if you are concerned with gamma(1/4), you can see it to 1,000,000 decimals at:


    Working with Euler's reflection formula, I get

    (1/4)!(3/4)! = [tex]\frac{3\pi\sqrt2}{16}=.833041[/tex]
    Last edited by a moderator: Apr 23, 2017
  8. May 20, 2008 #7
    how about [tex]\Gamma[/tex](0) I read somewhere that its a complex infinity but dont understand what that means
  9. May 20, 2008 #8
    mjk1: how about (0)

    Well, when you have a form like[tex]\int_{0}^{\infty} \frac{e^-x}{x} dx[/tex], you have trouble.

    If you check out the Euler's reflection formula, we have

    [tex]\Gamma(Z)\Gamma(1-Z) = \frac{\pi}{sin\pi(x)} for Z=0.[/tex]
    Last edited: May 21, 2008
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