Gamma Function of One Half/Proof 4

Theorem

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

Proof

Due to the organization of pages at $\mathsf{Pr} \infty \mathsf{fWiki}$, this argument is circular. In particular: This uses Gamma Function of Positive Half-Integer in its proof, which itself uses Gamma Function of One Half, so it's circular. You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by resolving this issue. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{CircularStructure}} from the code. If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page.

\(\ds \map \Gamma {\frac 1 2}\)

\(=\)

\(\ds \frac {0!} {2^0 0!} \sqrt \pi\)

Gamma Function of Positive Half-Integer

\(\ds \)

\(=\)

\(\ds \sqrt \pi\)

Factorial of Zero

$\blacksquare$