Expressing $\Gamma$(n+$\frac{1}{2}$) for n $\in$ $\mathbb{Z}$ in Factorials

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SUMMARY

The expression for the Gamma function at half-integer values, specifically $\Gamma(n+\frac{1}{2})$, is defined for non-negative integers as $\Gamma(n+\frac{1}{2}) = \frac{(2n-1)!}{2^n} \sqrt{\pi}$. For negative integers, the relationship $\Gamma(n) = \frac{\Gamma(n+1)}{n}$ is utilized, leading to the conclusion that $\Gamma(-\frac{1}{2})$ can be expressed in terms of $\Gamma(\frac{1}{2})$. This establishes a clear method for calculating $\Gamma(n+\frac{1}{2})$ across both positive and negative integer domains.

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Homework Statement



Express \Gamma (n+\frac{1}{2}) for n\in\mathbb{Z} in terms of factorials (separately for positive and negative n).

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The Attempt at a Solution



I've got for n\geqslant 0 that \displaystyle \Gamma \left(n+\frac{1}{2} \right) = \frac{(2n-1)!}{2^n} \sqrt{\pi} but what do I do when n<0 ?
 
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The same thing. Use ##n\Gamma(n)=\Gamma(n+1)##, so ##(-1/2)\Gamma(-1/2) = \Gamma(1/2)## and so on.
 

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