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Question Number 134719 by 0731619177 last updated on 06/Mar/21

Commented by Dwaipayan Shikari last updated on 06/Mar/21

$${\pi }^{\frac{3}{2}}-\frac{3\pi }{2}.\frac{\mathrm{\Gamma }\left(\frac{5}{3}\right)}{\mathrm{\Gamma }\left(\frac{7}{6}\right)}$$

Answered by CutieJanab last updated on 06/Mar/21

$$pleasesolvetbis$$

Answered by Dwaipayan Shikari last updated on 06/Mar/21

$$\frac{1}{\sqrt{\pi }}\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{\mathrm{\Gamma }(n+\frac{1}{2})\mathrm{\Gamma }\left(\frac{1}{2}\right)}{n!(2n+1)(3n+2)}\mathrm{\Gamma }(n+\frac{1}{2})\mathrm{\Gamma }\left(\frac{1}{2}\right)=\pi {\left(\frac{1}{2}\right)}_n$$$$=\frac{2\pi }{\sqrt{\pi }}\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{\left(\frac{1}{2}\right)}_n}{n!(2n+1)}-\frac{3\pi }{\sqrt{\pi }}\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{\left(\frac{1}{2}\right)}_n}{(3n+2)n!}$$$$=2\sqrt{\pi }{\int }_0^1\frac{1}{\sqrt{1-x^2}}dx-3\sqrt{\pi }{\int }_0^{1}x{(1-x^3)}^{-\frac{1}{2}}dx$$$$={\pi }^{\frac{3}{2}}-\sqrt{\pi }{\int }_0^1{u}^{-\frac{1}{3}}{(1-u)}^{-\frac{1}{2}}du$$$$={\pi }^{\frac{3}{2}}-\pi .\frac{\mathrm{\Gamma }\left(\frac{1}{2}\right)\mathrm{\Gamma }\left(\frac{2}{3}\right)}{\mathrm{\Gamma }\left(\frac{7}{6}\right)}={\pi }^{\frac{3}{2}}-\frac{3\pi }{2}.\frac{\mathrm{\Gamma }\left(\frac{5}{3}\right)}{\mathrm{\Gamma }\left(\frac{7}{6}\right)}\sim 0.983$$

Commented by 0731619177 last updated on 06/Mar/21

$$answeriswrong$$

Commented by 0731619177 last updated on 06/Mar/21

Commented by 0731619177 last updated on 06/Mar/21

$$butanswerispositive$$

Commented by Dwaipayan Shikari last updated on 06/Mar/21

$$Sir!Youhaveused(\frac{2}{3}-1)!insteadof(\frac{5}{3}-1)!$$

Commented by Dwaipayan Shikari last updated on 06/Mar/21

$$Hahhahah..SirPoweragain..$$😛

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