My Math Forum Gamma Function

 April 21st, 2017, 09:24 AM #1 Newbie   Joined: Apr 2017 From: India Posts: 17 Thanks: 0 Gamma Function While dealing with famous gamma function, I came across a question that requires me to proof: $\gamma$[m+1/2] = $$\frac{(2m-1)(2m-3)......1)\sqrt{\pi}}{2^m}$$ I have been provided the following hints: $\gamma$(1)=1 $\gamma$(1/2)=$\sqrt{\pi}$ $\gamma$($\alpha$+1)=$\alpha$$\gamma$($\alpha$) $\gamma$(k+1)=k!, if k is non-negative integer $\frac{\gamma(m)\gamma(n)}{\gamma(m+n)}$= $\int$ {u^(m-1)}{(1-u)^(n-1)}du How can I use these hints to prove the above given identity?Well I have tried but my attempt failed.Please provide me with the elaborate solution.
 April 21st, 2017, 06:05 PM #2 Global Moderator   Joined: May 2007 Posts: 6,306 Thanks: 525 (2m-1)/2 = m-(1/2) (2m-3)/2 = m-(3/2) etc. Put them all together and the last term is $\displaystyle \gamma(\frac{1}{2})=\sqrt{\pi}$
 April 21st, 2017, 06:39 PM #3 Newbie   Joined: Apr 2017 From: India Posts: 17 Thanks: 0 I really appreciate your help.But you told me the proof in reverse manner.In the text I am referring to says that: $\gamma$[m+1/2] = [1/2+(m-1)][1/2+(m-1)]........[1/2]$\sqrt{\pi}$ and after this they have given the result that I have already mentioned above. How from left hand side, we have derived the right hand side?
 April 22nd, 2017, 03:58 PM #4 Global Moderator   Joined: May 2007 Posts: 6,306 Thanks: 525 $\displaystyle \gamma (m+1/2)=(m-1/2)(m-3/2).....\gamma (1/2)$ by definition. Thanks from topsquark

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