Gamma Function of Positive Half-Integer

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Theorem

\(\ds \map \Gamma {m + \frac 1 2}\) \(=\) \(\ds \frac {\paren {2 m}!} {2^{2 m} m!} \sqrt \pi\)
\(\ds \) \(=\) \(\ds \frac {1 \times 3 \times 5 \times \cdots \times \paren {2 m - 1} } {2^m} \sqrt \pi\)

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:

\(\ds \map \Gamma {1 + \frac 1 2}\) \(=\) \(\ds \frac 1 2 \map \Gamma {\frac 1 2}\) Gamma Difference Equation
\(\ds \) \(=\) \(\ds \frac {\sqrt \pi} 2\) Gamma Function of One Half
\(\ds \) \(=\) \(\ds \frac 2 4 \sqrt \pi\)
\(\ds \) \(=\) \(\ds \frac {2!} {2^{2 \times 1} 1!} \sqrt \pi\) Definition of Factorial
\(\ds \) \(=\) \(\ds \frac {\paren {2 m}!} {2^{2 m} m!} \sqrt \pi\) where $m = 1$

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:

\(\ds \map \Gamma {k + 1 + \frac 1 2}\) \(=\) \(\ds \paren {k + \frac 1 2} \map \Gamma {k + \frac 1 2}\) Gamma Difference Equation
\(\ds \) \(=\) \(\ds \paren {k + \frac 1 2} \dfrac {\paren {2 k}!} {2^{2 k} k!} \sqrt \pi\) Induction Hypothesis
\(\ds \) \(=\) \(\ds \frac {\paren {2 k + 1} \paren {2 k}!} {2 \times 2^{2 k} k!} \sqrt \pi\) simplifying
\(\ds \) \(=\) \(\ds \frac {\paren {2 k}! \paren {2 k + 1} \paren {2 k + 2} } {2 \paren {2 k + 2} 2^{2 k} k!} \sqrt \pi\) multiplying top and bottom by $2 k + 2$
\(\ds \) \(=\) \(\ds \frac {\paren {2 k + 2}!} {2^{2 k + 1} \paren {2 \paren {k + 1} } k!} \sqrt \pi\)
\(\ds \) \(=\) \(\ds \frac {\paren {2 \paren {k + 1} }!} {2^{2 \paren {k + 1} } \paren {k + 1}!} \sqrt \pi\)


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


Finally:

\(\ds \frac {\paren {2 m}!} {2^{2 m} m!}\) \(=\) \(\ds \frac {1 \times 2 \times 3 \times \cdots \times 2 m} {2^{2 m} \ 1 \times 2 \times 3 \times \cdots \times m}\)
\(\ds \) \(=\) \(\ds \frac {1 \times 2 \times 3 \times \cdots \times \paren {2 m - 1} \times 2 m} {2^m \ 2^m \ \paren {1 \times 2 \times 3 \times \cdots \times m} }\)
\(\ds \) \(=\) \(\ds \frac {1 \times 2 \times 3 \times \cdots \times \paren {2 m - 1} \times 2 m} {2^m \paren {\paren {2 \times 1} \times \paren {2 \times 2} \times \paren {2 \times 3} \times \cdots \times 2 m} }\)
\(\ds \) \(=\) \(\ds \frac {1 \times 2 \times 3 \times \cdots \times \paren {2 m - 1} \times 2 m} {2^m \paren {2 \times 4 \times 6 \times \cdots \times 2 m} }\)
\(\ds \) \(=\) \(\ds \frac {1 \times 3 \times 5 \times \cdots \times \paren {2 m - 1} } {2^m}\)


Therefore:

\(\ds \forall m \in \Z_{>0}: \ \ \) \(\ds \map \Gamma {m + \frac 1 2}\) \(=\) \(\ds \frac {\paren {2 m}!} {2^{2 m} m!} \sqrt \pi\)
\(\ds \) \(=\) \(\ds \frac {1 \times 3 \times 5 \times \cdots \times \paren {2 m - 1} } {2^m} \sqrt \pi\)

$\blacksquare$


Sources