# 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,

$\mathrm{\Gamma}(1/2)$ | $={\displaystyle {\int}_{0}^{\mathrm{\infty}}}{e}^{-x}{x}^{-1/2}\mathit{d}x$ | ||

$=2{\displaystyle {\int}_{0}^{\mathrm{\infty}}}{e}^{-x}{\displaystyle \frac{1}{2\sqrt{x}}}\mathit{d}x.$ |

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

$$\mathrm{\Gamma}(1/2)=2{\int}_{0}^{\mathrm{\infty}}{e}^{-{u}^{2}}\mathit{d}u={\int}_{\mathbb{R}}{e}^{-{u}^{2}}\mathit{d}u,$$ |

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 |