# evaluating the gamma function at 1/2

In the entry on the gamma function it is mentioned that $\Gamma(1/2)=\sqrt{\pi}$. In this entry we reduce the proof of this claim to the problem of computing the area under the bell curve. First note that by definition of the gamma function,

 $\displaystyle\Gamma(1/2)$ $\displaystyle=\int_{0}^{\infty}e^{-x}x^{-1/2}\,dx$ $\displaystyle=2\int_{0}^{\infty}e^{-x}\frac{1}{2\sqrt{x}}\,dx.$

Performing the substitution $u=\sqrt{x}$, we find that $du=\frac{1}{2\sqrt{x}}\,dx$, so

 $\Gamma(1/2)=2\int_{0}^{\infty}e^{-u^{2}}\,du=\int_{\mathbb{R}}e^{-u^{2}}\,du,$

where the last equality holds because $e^{-u^{2}}$ is an even function. Since the area under the bell curve is $\sqrt{\pi}$, it follows that $\Gamma(1/2)=\sqrt{\pi}$.

Title evaluating the gamma function at 1/2 EvaluatingTheGammaFunctionAt12 2013-03-22 16:57:13 2013-03-22 16:57:13 CWoo (3771) CWoo (3771) 4 CWoo (3771) Derivation msc 30D30 msc 33B15 AreaUnderGaussianCurve LaplaceTransformOfPowerFunction