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Prove that $\Gamma\left(\frac{1}{2}\right)= \sqrt\pi$

Using $$\Gamma(p)\Gamma(1-p) = \frac{\pi}{\sin(\pi p)}$$

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    $\begingroup$ Hint: what is $\sin(\pi/2)$? $\endgroup$ – gammatester Aug 5 '13 at 12:35
  • $\begingroup$ @Sid: Welcome to MSE! It really helps readability to format questions using MathJax (see FAQ). Regards $\endgroup$ – Amzoti Aug 5 '13 at 12:50
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Let $p=\frac12$, we have $$\Gamma\left(\frac12\right)\Gamma\left(\frac12\right)=\frac{\pi}{\sin\frac{\pi}{2}}=\pi$$ therefore $$\Gamma\left(\frac12\right)=\sqrt \pi$$

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    $\begingroup$ (Because $\Gamma>0$. This is trivial, but forgetting that the square is not injective is often cause of problems) $\endgroup$ – Clement C. Aug 5 '13 at 13:50

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