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April 21st, 2017, 10:24 AM   #1
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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.
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April 21st, 2017, 07:05 PM   #2
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(2m-1)/2 = m-(1/2)
(2m-3)/2 = m-(3/2)
etc.

Put them all together and the last term is $\displaystyle \gamma(\frac{1}{2})=\sqrt{\pi}$
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April 21st, 2017, 07:39 PM   #3
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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?
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April 22nd, 2017, 04:58 PM   #4
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$\displaystyle \gamma (m+1/2)=(m-1/2)(m-3/2).....\gamma (1/2)$ by definition.
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