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 $x = y = \dfrac 1 2$:

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


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:

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

Setting $z = \dfrac 1 2$:

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

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

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

$\blacksquare$


Proof 4

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

$\blacksquare$


Proof 5

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

$\blacksquare$


Sources