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Gamma Function identity ?

  1. Feb 18, 2008 #1
    Will some one help me to prove this identity

    G(n)+G(1-n)= pi/ sin npi 0<n<1

    B(m,n) = (m-1)! / n(n+1).....(n+m+1) ,for beta function
     
  2. jcsd
  3. Feb 18, 2008 #2
    You mean
    [tex]\Gamma(n)*\Gamma(1-n)=\frac{\pi}{\sin(n\,\pi)}[/tex]

    First of all use the identity
    [tex]B(x,y)=\frac{\Gamma(x)\,\Gamma(y)}{\Gamma(x+y)}[/tex]
    with [itex]x=n,\,y=1-n[/itex] to arrive to [itex]B(n,1-n)=\Gamma(n)\,\Gamma(1-n)[/itex], i.e.

    [tex]\Gamma(n)\,\Gamma(1-n)=\int_0^\infty\frac{u^{n-1}}{u+1}\,d\,u[/tex]

    which can be calculated with the use of residues.
     
  4. Feb 19, 2008 #3
    by residue it will give limit u tending to -1 [(-1)^n-1] Integral = 2pi i * Residue

    which =2pi i *(-1)^n-1 ,how to proceed further
     
  5. Feb 19, 2008 #4
    I cann't understand that you are saying. In order to calculate the integral choose a keyhole contour like this
    Contour I.jpg
     
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