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

Related:

AreaUnderGaussianCurve, LaplaceTransformOfPowerFunction

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Derivation

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## Mathematics Subject Classification

30D30*no label found*33B15

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