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Find a formula for $\Gamma(\frac{n}{2})$ for positive integer n.


I know the following relations; $\Gamma (z+1)=z\Gamma (z)$ and $\Gamma(n+1)=n!$

Please give me a way how to show this. Thank you.

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    $\begingroup$ If you know $\Gamma(\frac12)$, this is trivial. If you don't, you should. $\endgroup$ – Did Dec 29 '13 at 0:06
  • $\begingroup$ Yes $\Gamma(1/2)=\sqrt{\pi}$ @Did $\endgroup$ – user315 Dec 29 '13 at 0:07
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    $\begingroup$ See en.wikipedia.org/wiki/…. $\endgroup$ – lhf Dec 29 '13 at 0:07
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$$ \Gamma(n)=(n-1)! $$ and $$ \begin{align} \Gamma\left(n+\frac12\right) &=\Gamma\left(\frac12\right)\frac12\frac32\frac52\cdots\frac{2n-1}{2}\\ &=\sqrt\pi\frac{1\cdot\color{#A0A0A0}{2}\cdot3\cdot\color{#A0A0A0}{4}\cdot5\cdot\color{#A0A0A0}{6}\cdots(2n-1)\cdot\color{#A0A0A0}{2n}}{2^n(\color{#A0A0A0}{2}\cdot\color{#A0A0A0}{4}\cdot\color{#A0A0A0}{6}\cdots\color{#A0A0A0}{2n})}\\ &=\sqrt\pi\frac{(2n)!}{4^nn!} \end{align} $$


To match the form in the question:

For even $n$, $$ \Gamma\left(\frac n2\right)=\left(\frac n2-1\right)! $$ For odd $n$, $$ \Gamma\left(\frac n2\right)=\sqrt\pi\frac{(n-1)!}{2^{n-1}\frac{n-1}{2}!} $$

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    $\begingroup$ would the downvoter care to comment? $\endgroup$ – robjohn Dec 29 '13 at 0:30
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If $n$ is even, use that second identity. Otherwise, use that first identity to relate $\Gamma\left(\frac{n}{2}\right)$ to $\Gamma\left(\frac{1}{2}\right)=\sqrt\pi$.

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