Gamma Function of Positive Half-Integer

Theorem
where:
 * $m + \dfrac 1 2$ is a half-integer such that $m > 0$
 * $\Gamma$ denotes the Gamma function.

Proof
Proof by induction:

For all $m \in \Z_{> 0}$, let $\map P m$ be the proposition:
 * $\map \Gamma {m + \dfrac 1 2} = \dfrac {\paren {2 m}!} {2^{2 m} m!} \sqrt \pi$

Basis for the Induction
$\map P 1$ is the case:

and so $\map P 1$ holds.

This is our basis for the induction.

Induction Hypothesis
Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.

So this is our induction hypothesis:
 * $\map \Gamma {k + \dfrac 1 2} = \dfrac {\paren {2 k}!} {2^{2 k} k!} \sqrt \pi$

Then we need to show:
 * $\map \Gamma {k + 1 + \dfrac 1 2} = \dfrac {\paren {2 \paren {k + 1} }!} {2^{2 \paren {k + 1} } \paren {k + 1}!} \sqrt \pi$

Induction Step
This is our induction step:

So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.

Finally:

Therefore: