Gamma Function of One Half/Proof 1

Theorem

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

From the definition of the Beta function:

$\map \Beta {x, y} := \dfrac {\map \Gamma x \map \Gamma y} {\map \Gamma {x + y} }$

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

$\ds \map \Beta {\dfrac 1 2, \dfrac 1 2}$

$=$

$\ds \frac {\map \Gamma {\dfrac 1 2} \map \Gamma {\dfrac 1 2} } {\map \Gamma {\dfrac 1 2 + \dfrac 1 2} }$

$\ds $

$=$

$\ds \paren {\map \Gamma {\dfrac 1 2} }^2$

$\map \Beta {\dfrac 1 2, \dfrac 1 2} = \pi$

Hence the result.

$\blacksquare$

1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): $\S 17.7 \ (6)$
1997: Donald E. Knuth: The Art of Computer Programming: Volume 1: Fundamental Algorithms (3rd ed.) ... (previous) ... (next): $\S 1.2.6$: Binomial Coefficients: Exercise $43$