Gamma Function of One Half

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Theorem

Let $\Gamma$ denote the Gamma function.

Then:

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

Its decimal expansion starts:

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

This sequence is A002161 in the On-Line Encyclopedia of Integer Sequences (N. J. A. Sloane (Ed.), 2008).


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

From Euler's Reflection Formula:

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

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

$\blacksquare$


Sources

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