Simplify-p-2-1-2-p-2-1-2-

Question Number 132302 by Lordose last updated on 13/Feb/21

Simplify

\frac{Γ (\frac{p}{2}) Γ (\frac{1}{2})}{Γ (\frac{p}{2} + \frac{1}{2})}

Answered by Ar Brandon last updated on 13/Feb/21

Γ (m) Γ (m + \frac{1}{2}) = \frac{\sqrt{π}}{2^{2 m - 1}} Γ (2 m)

Γ (\frac{p}{2} + \frac{1}{2}) = \frac{\sqrt{π}}{2^{p - 1}} \cdot \frac{Γ (p)}{Γ (\frac{p}{2})}

\frac{Γ (\frac{p}{2}) Γ (\frac{1}{2})}{Γ (\frac{p}{2} + \frac{1}{2})} = \frac{\sqrt{π} Γ (\frac{p}{2})}{\sqrt{π} Γ (p)} \cdot Γ (\frac{p}{2}) 2^{2 p - 1}

= \frac{Γ^{2} (\frac{p}{2})}{Γ (p)} 2^{p - 1} = Γ^{2} (\frac{p}{2}) \frac{2^{p - 1}}{(p - 1)!}