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

$$\mathbf{Simplify}$$$$\frac{\mathbf{\Gamma }\left(\frac{\mathbf{p}}{2}\right)\mathbf{\Gamma }\left(\frac{1}{2}\right)}{\mathbf{\Gamma }(\frac{\mathbf{p}}{2}+\frac{1}{2})}$$

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

$$\mathrm{\Gamma }\left(\mathrm{m}\right)\mathrm{\Gamma }(\mathrm{m}+\frac{1}{2})=\frac{\sqrt{\pi }}{2^{2\mathrm{m}-1}}\mathrm{\Gamma }\left(2\mathrm{m}\right)$$$$\mathrm{\Gamma }(\frac{\mathrm{p}}{2}+\frac{1}{2})=\frac{\sqrt{\pi }}{2^{\mathrm{p}-1}}\cdot \frac{\mathrm{\Gamma }\left(\mathrm{p}\right)}{\mathrm{\Gamma }\left(\frac{\mathrm{p}}{2}\right)}$$$$\frac{\mathbf{\Gamma }\left(\frac{\mathbf{p}}{2}\right)\mathbf{\Gamma }\left(\frac{1}{2}\right)}{\mathbf{\Gamma }(\frac{\mathbf{p}}{2}+\frac{1}{2})}=\frac{\sqrt{\pi }\mathrm{\Gamma }\left(\frac{\mathrm{p}}{2}\right)}{\sqrt{\pi }\mathrm{\Gamma }\left(\mathrm{p}\right)}\cdot \mathrm{\Gamma }\left(\frac{\mathrm{p}}{2}\right)2^{2\mathrm{p}-1}$$$$=\frac{{\mathrm{\Gamma }}^2\left(\frac{\mathrm{p}}{2}\right)}{\mathrm{\Gamma }\left(\mathrm{p}\right)}{2}^{\mathrm{p}-1}={\mathrm{\Gamma }}^2\left(\frac{\mathrm{p}}{2}\right)\frac{2^{\mathrm{p}-1}}{(\mathrm{p}-1)!}$$

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