Der Verdoppelungssatz von Legendre.

Es gilt

Γ (a) Γ (a + \frac{1}{2}) = \frac{\sqrt{π}}{2^{2 a - 1}} Γ (2 a), a > 0 .

(2.12.6.1)

Dazu berechnen wir

\begin{array}{rcl} B (a, a) = \int_{0}^{1} x^{a - 1} {(1 - x)}^{a - 1} d x & = & \int_{0}^{1} {(\frac{1}{4} - {(\frac{1}{2} - x)}^{2})}^{a - 1} d x \\ = & 2 \int_{0}^{1 ∕ 2} {(\frac{1}{4} - {(\frac{1}{2} - x)}^{2})}^{a - 1} d x, \end{array}

durch die Substitution $\frac{1}{2} - x = \frac{\sqrt{t}}{2}$

\begin{array}{rcl} B (a, a) & = & 2 \int_{1}^{0} {(\frac{1}{4} - \frac{t}{4})}^{a - 1} d (\frac{1}{2} - \frac{\sqrt{t}}{2}) \\ = & 2 \cdot 4^{1 - a} 2^{- 2} \int_{0}^{1} {(1 - t)}^{a - 1} t^{- 1 ∕ 2} d t = \frac{1}{2^{2 a - 1}} B (\frac{1}{2}, a) . \end{array}

Daraus folgt nach (2.12.4.5)

\frac{Γ (a) Γ (a)}{Γ (2 a)} = B (a, a) = \frac{1}{2^{2 a - 1}} B (\frac{1}{2}, a) = \frac{1}{2^{2 a - 1}} \cdot \frac{Γ (\frac{1}{2}) Γ (a)}{Γ (\frac{1}{2} + a)} .

Wegen $Γ (\frac{1}{2}) = \sqrt{π}$ impliziert dies (2.12.6.1).