# Gamma Function of One Half

## Theorem

Let $\Gamma$ denote the Gamma function.

Then:

$\map \Gamma {\dfrac 1 2} = \sqrt \pi$

Its decimal expansion starts:

$\map \Gamma {\dfrac 1 2} = 1 \cdotp 77245 \, 38509 \, 05516 \, 02729 \, 81674 \, 83341 \, 14518 \, 27975 \ldots$

## Proof 1

From the definition of the Beta function:

$\Beta \left({x, y}\right) := \dfrac {\Gamma \left({x}\right) \Gamma \left({y}\right)} {\Gamma \left({x + y}\right)}$

Setting $x = y = \dfrac 1 2$:

 $\ds \Beta \left({\dfrac 1 2, \dfrac 1 2}\right)$ $=$ $\ds \frac {\Gamma \left({\dfrac 1 2}\right) \Gamma \left({\dfrac 1 2}\right)} {\Gamma \left({\dfrac 1 2 + \dfrac 1 2}\right)}$ $\ds$ $=$ $\ds \left({\Gamma \left({\dfrac 1 2}\right)}\right)^2$

Then from Beta Function of Half with Half:

$\Beta \left({\dfrac 1 2, \dfrac 1 2}\right) = \pi$

Hence the result.

$\blacksquare$

## Proof 2

$\forall z \notin \Z: \Gamma \left({z}\right) \Gamma \left({1 - z}\right) = \dfrac \pi {\sin \left({\pi z}\right)}$

Setting $z = \dfrac 1 2$:

 $\ds \Gamma \left({\frac 1 2}\right) \Gamma \left({\frac 1 2}\right)$ $=$ $\ds \frac \pi {\sin \left({\frac \pi 2}\right)}$ $\ds$ $=$ $\ds \frac \pi 1$ Sine of Right Angle $\ds \implies \ \$ $\ds \Gamma \left({\frac 1 2}\right)$ $=$ $\ds \pm \sqrt \pi$

By definition of the $\Gamma$ function:

$\forall z \in \R_{\ge 0}: \Gamma \left({z}\right) > 0$

and so the negative square root can be discarded.

Hence:

$\Gamma \left({\dfrac 1 2}\right) = \sqrt \pi$

as required.

$\blacksquare$

## Proof 3

 $\ds \Gamma \left({\dfrac 1 2}\right)$ $=$ $\ds \int_0^{\to \infty} t^{-\frac 1 2} e^{-t} \ \mathrm d t$ Definition of Gamma Function $\ds$ $=$ $\ds \int_0^{\to \infty} u^{-1} e^{-u^2} 2 u \ \mathrm d u$ Integration by Substitution, $\phi \left({u}\right) = u^2$ $\ds$ $=$ $\ds 2 \int_0^{\to \infty} e^{-u^2} \ \mathrm d u$ Linear Combination of Integrals $\ds$ $=$ $\ds \int_{\to -\infty}^{\to \infty} e^{-u^2} \ \mathrm d u$ Definite Integral of Even Function $\ds$ $=$ $\ds \sqrt {\pi}$ Gaussian Integral

$\blacksquare$

## Proof 4

 $\ds \Gamma \left({\frac 1 2}\right)$ $=$ $\ds \frac {0!} {2^0 0!} \sqrt \pi$ Gamma Function of Positive Half-Integer $\ds$ $=$ $\ds \sqrt \pi$ Factorial of Zero

$\blacksquare$

## Proof 5

 $\ds \map \Gamma 1 \, \map \Gamma {\frac 1 2}$ $=$ $\ds 2^0 \sqrt \pi \ \map \Gamma 1$ Legendre's Duplication Formula $\ds \leadsto \ \$ $\ds 0! \, \map \Gamma {\frac 1 2}$ $=$ $\ds 0! \, \sqrt \pi$ Definition of Gamma Function $\ds \leadsto \ \$ $\ds \map \Gamma {\frac 1 2}$ $=$ $\ds \sqrt \pi$ Factorial of Zero

$\blacksquare$