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Question Number 133791 by mnjuly1970 last updated on 24/Feb/21

$$......advancedintegral....$$$$provethat::$$$$\mathit{\varphi }={\int }_0^{\mathrm{\infty }}\left(\frac{1-e^{-\phi x}}{1+e^{\phi x}}\right)\frac{dx}{x}=??$$$$\phi :=Goldenratio...$$$$$$

Answered by Dwaipayan Shikari last updated on 24/Feb/21

$$I\left(\alpha \right)={\int }_0^{\mathrm{\infty }}\frac{\frac{1}{1+e^{\phi x}}-\frac{e^{-\alpha \phi x}}{1+e^{\phi x}}}{x}dx$$$$I^{\prime }\left(\alpha \right)=\phi {\int }_0^{\mathrm{\infty }}\frac{e^{-\alpha \phi x}}{1+e^{\phi x}}dx=\underset{n=1}{\sum ^{\mathrm{\infty }}}{(-1)}^{n+1}{\int }_0^{\mathrm{\infty }}{e}^{-(\alpha \phi +n\phi )x}dx$$$$=\phi \underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{n+1}}{\phi (\alpha +n)}=(\frac{1}{1+\alpha }-\frac{1}{2+\alpha }+\frac{1}{3+\alpha }-\frac{1}{4+\alpha }+...)$$$$=(\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{1}{n+1}-\frac{1}{2n+2+\alpha }-\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{1}{n+1}-\frac{1}{2n+1+\alpha })$$$$I^{\prime }\left(\alpha \right)=\frac{1}{2}(\psi (1+\frac{\alpha }{2})-\psi (\frac{1}{2}+\frac{\alpha }{2}))$$$$I\left(\alpha \right)=.log\left(\mathrm{\Gamma }(1+\frac{\alpha }{2})\right)-log\left(\mathrm{\Gamma }\left(\frac{1+\alpha }{2}\right)\right)+C$$$$\alpha =0\Rightarrow I\left(0\right)=-log\left(\sqrt{\pi }\right)+C\Rightarrow C=log\left(\sqrt{\pi }\right)$$$$I\left(1\right)=log\left(\frac{\sqrt{\pi }}{2}\right)+log\left(\sqrt{\pi }\right)=log\left(\frac{\pi }{2}\right)$$

Commented by Dwaipayan Shikari last updated on 24/Feb/21

$$Thankingyouforshowingmymistake.Ihaveedited$$

Commented by mnjuly1970 last updated on 24/Feb/21

$$grateful...$$

Commented by mathDivergent last updated on 24/Feb/21

Commented by mathDivergent last updated on 24/Feb/21

$$\mathrm{you}\mathrm{have}\mathrm{mistake}\mathrm{in}2\mathrm{nd}\mathrm{line}$$$$\mathrm{Here}:I^{\prime }\left(\alpha \right)={\int }_0^{\mathrm{\infty }}\frac{e^{-\alpha \phi x}}{1+e^{\phi x}}dx=\underset{n=1}{\sum ^{\mathrm{\infty }}}{(-1)}^{n+1}{\int }_0^{\mathrm{\infty }}{e}^{-(\alpha \phi +n\phi )x}dx$$$$\mathrm{Since}:\frac{\partial }{\partial \alpha }\left(\frac{\frac{1}{1+e^{\phi x}}-\frac{e^{-\alpha \phi x}}{1+e^{\phi x}}}{x}\right)=\phi \frac{e^{-\alpha \phi x}}{1+e^{\phi x}}$$

Commented by mathDivergent last updated on 24/Feb/21

$$good,butIseeonemoremistakepleasecheckyoursolution$$$$(p.s.ansiscorrect)$$

Answered by mathDivergent last updated on 24/Feb/21

$$\ln \frac{\pi }{2}$$

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