evaluating the gamma function at 1/2

In the entry on the gamma function it is mentioned that $\mathrm{\Gamma }(1/2)=\sqrt{\pi }$. 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,

$\mathrm{\Gamma }(1/2)$ $={\displaystyle {\int }_0^{\mathrm{\infty }}}{e}^{-x}{x}^{-1/2}𝑑x$

$=2{\displaystyle {\int }_0^{\mathrm{\infty }}}{e}^{-x}{\displaystyle \frac{1}{2\sqrt{x}}}𝑑x.$

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

$$\mathrm{\Gamma }(1/2)=2{\int }_0^{\mathrm{\infty }}{e}^{-u^2}𝑑u={\int }_{ℝ}{e}^{-u^2}𝑑u,$$

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 $\mathrm{\Gamma }(1/2)=\sqrt{\pi }$.

Title	evaluating the gamma function at 1/2
Canonical name	EvaluatingTheGammaFunctionAt12
Date of creation	2013-03-22 16:57:13
Last modified on	2013-03-22 16:57:13
Owner	CWoo (3771)
Last modified by	CWoo (3771)
Numerical id	4
Author	CWoo (3771)
Entry type	Derivation
Classification	msc 30D30
Classification	msc 33B15
Related topic	AreaUnderGaussianCurve
Related topic	LaplaceTransformOfPowerFunction