Legendre's Duplication Formula

Theorem
Let $\Gamma$ denote the gamma function.

Then:
 * $\forall z \notin \left\{{-\dfrac n 2: n \in \N}\right\}: \Gamma \left({z}\right) \Gamma \left (z + \dfrac 1 2 \right) = 2^{1 - 2 z} \sqrt \pi \Gamma \left({2 z}\right)$

where $\N$ denotes the natural numbers.

Proof 1
From the Beta function:

Letting $z_1 = z_2 = z$ gives:

Now substituting $u = x^2$ into the Beta function:


 * $\displaystyle \Beta \left({z_1, z_2}\right) = \int_0^1 x^{2z_1 - 2} \left({1 - x^2}\right)^{z_2 - 1} 2x \ \mathrm d x$

Letting $z_1 = \dfrac 1 2$ and $z_2 = z$ gives:


 * $(2): \quad \displaystyle \Beta \left({\frac 1 2, z}\right) = 2 \int_0^1 \left({1 - x^2}\right)^{z - 1} \ \mathrm d x$

Combining results $(1)$ and $(2)$:

From Gamma Function of One Half:
 * $\Gamma \left({\dfrac 1 2}\right) = \sqrt \pi$

It follows that:


 * $\Gamma \left({z}\right) \Gamma \left({z + \dfrac 1 2}\right) = 2^{1 - 2z} \sqrt \pi \Gamma \left({2z}\right)$

Proof 2
From Gauss Multiplication Formula we have:


 * $\displaystyle \prod_{k \mathop = 0}^{n - 1} \Gamma \left({z + \frac k n}\right) = \left({2 \pi}\right)^{\left({n - 1}\right) / 2} n^{1/2 - n z} \Gamma \left({n z}\right)$

Substituting $n=2$ yields: