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Question Number 139455 by mnjuly1970 last updated on 27/Apr/21

$$\text{You can't use 'macro parameter character \#' in math mode}$$$$evaluate:$$$$\mathit{\varphi }:=\underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{k-1}\mathrm{\Gamma }\left(\frac{k}{2}\right)}{k\mathrm{\Gamma }\left(\frac{k+1}{2}\right)}=?$$

Answered by Dwaipayan Shikari last updated on 27/Apr/21

$$\underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{k+1}\mathrm{\Gamma }\left(\frac{k}{2}\right)}{k\mathrm{\Gamma }\left(\frac{k+1}{2}\right)}=\frac{1}{\sqrt{\pi }}\underset{k=1}{\sum ^{\mathrm{\infty }}}{(-1)}^{k+1}\frac{1}{k}{\int }_0^1{x}^{\frac{k}{2}-1}{(1-x)}^{-\frac{1}{2}}dx$$$$=\frac{1}{\sqrt{\pi }}{\int }_0^1\frac{log(1+\sqrt{x})}{x\sqrt{1-x}}dx=\frac{2}{\sqrt{\pi }}{\int }_0^1\frac{log(1+u)}{u\sqrt{1-u^2}}du=\frac{2}{\sqrt{\pi }}{\int }_0^{\frac{\pi }{2}}\frac{log(1+sin\theta )}{sin\theta }d\theta$$$$=\frac{2}{\sqrt{\pi }}\varsigma \left(1\right)$$$$\varsigma \left(g\right)={\int }_0^{\frac{\pi }{2}}\frac{log(1+gsin\theta )}{sin\theta }d\theta \Rightarrow {\varsigma }^{\prime }\left(g\right)={\int }_0^{\frac{\pi }{2}}\frac{1}{1+gsin\theta }d\theta$$$$=2{\int }_0^1\frac{1}{1+\frac{2gt}{1+t^2}}.\frac{1}{1+t^2}dt=2{\int }_0^1\frac{1}{{(t+g)}^2+{\left(\sqrt{1-g^2}\right)}^2}dt=\frac{2}{\sqrt{1-g^2}}({tan}^{-1}\sqrt{\frac{1+g}{1-g}}-{tan}^{-1}\frac{g}{\sqrt{1-g^2}})$$$$\varsigma \left(g\right)=2\int \frac{1}{\sqrt{1-g^2}}{tan}^{-1}\sqrt{\frac{1+g}{1-g}}dg-\frac{2}{\sqrt{1-g^2}}{tan}^{-1}\frac{g}{\sqrt{1-g^2}}$$$$=2\int \frac{1}{\sqrt{1-g^2}}{tan}^{-1}\frac{\frac{1+g-g}{\sqrt{1-g^2}}}{1+\frac{g+g^2}{1-g^2}}dg=2\int \frac{1}{\sqrt{1-g^2}}{tan}^{-1}\frac{\sqrt{1-g^2}}{1+g}dg$$$$g=cos\zeta \Rightarrow 2\int \frac{-sin\zeta }{sin\zeta }{tan}^{-1}\left(tan\frac{\zeta }{2}\right)d\zeta =-\frac{{\zeta }^2}{2}+C=-\frac{{\left({cos}^{-1}\left(g\right)\right)}^2}{2}+C$$$$\varsigma \left(0\right)=-\frac{{\pi }^2}{8}+C=0\Rightarrow C=\frac{{\pi }^2}{8}$$$$\varsigma \left(1\right)=-\frac{0}{2}+\frac{{\pi }^2}{8}$$$$So\frac{2}{\sqrt{\pi }}{\int }_0^{\frac{\pi }{2}}\frac{log(1+sin\theta )}{sin\theta }d\theta =\frac{2}{\sqrt{\pi }}.\frac{{\pi }^2}{8}=\frac{{\pi }^{3/2}}{4}$$

Commented by mnjuly1970 last updated on 27/Apr/21

$$thanksalotmrpayan...$$

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