Gamma Function of Negative 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 $P \left({m}\right)$ be the proposition:
 * $\Gamma \left({-m + \dfrac 1 2}\right) = \dfrac {\left({-1}\right)^m 2^{2 m} m!} {\left({2 m}\right)!} \sqrt \pi$

Basis for the Induction
$P \left({1}\right)$ is the case:

and so $P(1)$ holds.

This is our basis for the induction.

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

So this is our induction hypothesis:
 * $\Gamma \left({-k + \dfrac 1 2}\right) = \dfrac {\left({-1}\right)^k 2^{2 k} k!} {\left({2 k}\right)!} \sqrt \pi$

Then we need to show:
 * $\Gamma \left({-\left({k + 1}\right) + \dfrac 1 2}\right) = \dfrac {\left({-1}\right)^{k + 1} 2^{2 \left({k + 1}\right)} \left({k + 1}\right)!} {\left({2 \left({k + 1}\right)}\right)!} \sqrt \pi$

Induction Step
This is our induction step:

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

Finally:

Therefore: