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

$\forall z \notin \Z: \map \Gamma z \map \Gamma {1 - z} = \dfrac \pi {\map \sin {\pi z} }$

Setting $z = \dfrac 1 2$:

 $\ds \map \Gamma {\frac 1 2} \map \Gamma {\frac 1 2}$ $=$ $\ds \frac \pi {\map \sin {\frac \pi 2} }$ $\ds$ $=$ $\ds \frac \pi 1$ Sine of Right Angle $\ds \leadsto \ \$ $\ds \map \Gamma {\frac 1 2}$ $=$ $\ds \pm \sqrt \pi$

By definition of the gamma function:

$\forall z \in \R_{\ge 0}: \map \Gamma z > 0$

and so the negative square root can be discarded.

Hence:

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

as required.

$\blacksquare$

Proof 3

 $\ds \map \Gamma {\dfrac 1 2}$ $=$ $\ds \int_0^{\to \infty} t^{-\frac 1 2} e^{-t} \rd t$ Definition of Gamma Function $\ds$ $=$ $\ds \int_0^{\to \infty} u^{-1} e^{-u^2} 2 u \rd u$ Integration by Substitution, $\map \phi u = u^2$ $\ds$ $=$ $\ds 2 \int_0^{\to \infty} e^{-u^2} \rd u$ Linear Combination of Integrals $\ds$ $=$ $\ds \int_{\to -\infty}^{\to \infty} e^{-u^2} \rd u$ 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

$\blacksquare$