Question Number 127512 by mohammad17 last updated on 30/Dec/20 | ||
Commented by sts last updated on 30/Dec/20 | ||
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Answered by Dwaipayan Shikari last updated on 30/Dec/20 | ||
$$\beta\left({n},{n}\right)=\frac{\Gamma^{\mathrm{2}} \left({n}\right)}{\Gamma\left(\mathrm{2}{n}\right)}=\frac{\Gamma^{\mathrm{2}} \left({n}\right)}{\Gamma\left({n}\right)\Gamma\left({n}+\frac{\mathrm{1}}{\mathrm{2}}\right)}\mathrm{2}^{\mathrm{1}−\mathrm{2}{n}} \sqrt{\pi}=\frac{\Gamma\left({n}\right)\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}{\Gamma\left({n}+\frac{\mathrm{1}}{\mathrm{2}}\right)}\mathrm{2}^{\mathrm{1}−\mathrm{2}{n}} \\ $$$$=\beta\left({n},\frac{\mathrm{1}}{\mathrm{2}}\right)\mathrm{2}^{\mathrm{1}−\mathrm{2}{n}} \\ $$$$\left[{As}\Gamma\left({n}\right)\Gamma\left({n}+\frac{\mathrm{1}}{\mathrm{2}}\right)=\mathrm{2}^{\mathrm{1}−\mathrm{2}{n}} \sqrt{\pi}\:\Gamma\left(\mathrm{2}{n}\right)\right] \\ $$ | ||