Question Number 132302 by Lordose last updated on 13/Feb/21
$$\boldsymbol{\mathrm{Simplify}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\boldsymbol{\Gamma}\left(\frac{\boldsymbol{\mathrm{p}}}{\mathrm{2}}\right)\boldsymbol{\Gamma}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}{\boldsymbol{\Gamma}\left(\frac{\boldsymbol{\mathrm{p}}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\right)} \\ $$
Answered by Ar Brandon last updated on 13/Feb/21
$$\Gamma\left(\mathrm{m}\right)\Gamma\left(\mathrm{m}+\frac{\mathrm{1}}{\mathrm{2}}\right)=\frac{\sqrt{\pi}}{\mathrm{2}^{\mathrm{2m}−\mathrm{1}} }\Gamma\left(\mathrm{2m}\right) \\ $$$$\Gamma\left(\frac{\mathrm{p}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\right)=\frac{\sqrt{\pi}}{\mathrm{2}^{\mathrm{p}−\mathrm{1}} }\centerdot\frac{\Gamma\left(\mathrm{p}\right)}{\Gamma\left(\frac{\mathrm{p}}{\mathrm{2}}\right)} \\ $$$$\frac{\boldsymbol{\Gamma}\left(\frac{\boldsymbol{\mathrm{p}}}{\mathrm{2}}\right)\boldsymbol{\Gamma}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}{\boldsymbol{\Gamma}\left(\frac{\boldsymbol{\mathrm{p}}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\right)}=\frac{\sqrt{\pi}\:\Gamma\left(\frac{\mathrm{p}}{\mathrm{2}}\right)}{\:\sqrt{\pi}\Gamma\left(\mathrm{p}\right)}\centerdot\Gamma\left(\frac{\mathrm{p}}{\mathrm{2}}\right)\mathrm{2}^{\mathrm{2p}−\mathrm{1}} \\ $$$$=\frac{\Gamma^{\mathrm{2}} \left(\frac{\mathrm{p}}{\mathrm{2}}\right)}{\Gamma\left(\mathrm{p}\right)}\mathrm{2}^{\mathrm{p}−\mathrm{1}} =\Gamma^{\mathrm{2}} \left(\frac{\mathrm{p}}{\mathrm{2}}\right)\frac{\mathrm{2}^{\mathrm{p}−\mathrm{1}} }{\left(\mathrm{p}−\mathrm{1}\right)!} \\ $$