Legendre duplication formula

Remarks

One proof applies the Bohr–Mollerup theorem to $f(x) = 2^{x-1}\Gamma(\tfrac{x}{2})\Gamma(\tfrac{x+1}{2})/\sqrt{\pi}$ for $x > 0$: it satisfies $f(1) = 1$, $f(x+1) = x f(x)$, and $\ln f$ is convex, so $f = \Gamma$; then put $x = 2z$. Another proof uses the beta function: the substitution $t \mapsto 4t(1-t)$ gives $B(z, z) = 2^{1-2z} B(z, \tfrac12)$. The identity theorem extends the formula from $(0, \infty)$ to all allowed complex $z$. Check at $z = 1$: the left side is $\Gamma(1)\Gamma(\tfrac32) = \tfrac{\sqrt{\pi}}{2}$, and the right side is $2^{-1}\sqrt{\pi}\,\Gamma(2) = \tfrac{\sqrt{\pi}}{2}$. It is the case $m = 2$ of the Gauss multiplication formula. In Mathlib the statement is Complex.Gamma_mul_Gamma_add_half.

Going deeper

For a student

The formula is how one computes $\Gamma$ at half-integers, giving $\Gamma(n + \tfrac12) = \frac{(2n)!}{4^n n!}\sqrt{\pi}$. It appears in the functional equation of the Riemann zeta function, where it converts between the symmetric form with $\Gamma(s/2)$ and the form with $\Gamma(s)$ and a cosine. It also gives the volume of the unit ball in odd dimensions in closed form.