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Question Number 124827 by mnjuly1970 last updated on 06/Dec/20

$$....nicecalculus...$$$$provethat::$$$${\int }_0^{\frac{\pi }{2}}\frac{log(1+tan\left(x\right))}{tan\left(x\right)}dx=\frac{5{\pi }^2}{48}✓$$$$$$

Answered by mindispower last updated on 06/Dec/20

$$f\left(t\right)={\int }_0^{\mathrm{\infty }}\frac{ln(1+t.tan\left(x\right))}{tan\left(x\right)}dx$$$$q=f\left(1\right),q$$$$f^{\prime }\left(t\right)={\int }_0^{\frac{\pi }{2}}\frac{dx}{1+t.tan\left(x\right)}={\int }_0^{\frac{\pi }{2}}\frac{cos\left(x\right)}{cos\left(x\right)+tsin\left(x\right)}dx$$$$cos\left(x\right)=\alpha (cos\left(x\right)+tsin\left(x\right))+\beta (-sin\left(x\right)+tcos\left(x\right)$$$$\Rightarrow \alpha +\beta t=1$$$$\alpha t-\beta =0$$$$\alpha =\frac{1}{1+t^2},\beta =\frac{t}{1+t^2}$$$$f^{\prime }\left(t\right)=\frac{1}{1+t^2}{\int }_0^{\frac{\pi }{2}}dx+\frac{t}{1+t^2}{\int }_0^{\frac{\pi }{2}}\frac{-sin\left(x\right)+tcos\left(x\right)}{cos\left(x\right)+tsin\left(x\right)}dx$$$$=\frac{\pi }{2(1+t^2)}+{\left[\frac{t}{1+t^2}ln(cos\left(x\right)+tsin\left(x\right))\right]}_0^{\frac{\pi }{2}}$$$$=\frac{\pi }{2(1+t^2)}+\frac{tln\left(t\right)}{1+t^2}$$$$f\left(0\right)={\int }_0^{\frac{\pi }{2}}\frac{ln(1+otan\left(x\right))}{tan\left(x\right)}dx=0$$$$q={\int }_0^1\frac{\pi }{2(1+t^2)}+\frac{tln\left(t\right)}{1+t^2}dt$$$$=\frac{\pi }{2}{\tan }^{-1}\left(1\right)+\sum _{k⩾0}{\int }_0^1{(-1)}^k{t}^{2k+1}ln\left(t\right)dt$$$$=\frac{\pi }{2}.\frac{\pi }{4}+\sum _{k⩾0}\frac{{(-1)}^{k+1}}{{(2k+2)}^2}=\frac{\pi }{8}-\frac{1}{4}\sum _{k⩾0}\frac{{(-1)}^k}{{(k+1)}^2}$$$$=\frac{\pi }{8}+\frac{1}{4}(\sum _{k⩾0}\frac{1}{{(2k+2)}^2}-\frac{1}{4}\sum _{k⩾0}\frac{1}{{(2k+1)}^2})$$$$=\frac{{\pi }^2}{8}+\frac{1}{16}\zeta \left(2\right)-\frac{3}{16}\zeta \left(2\right)$$$$=\frac{{\pi }^2}{8}-\frac{1}{8}.\frac{{\pi }^2}{6}=\frac{5{\pi }^2}{48}={\int }_0^{\frac{\pi }{2}}\frac{ln(1+tan\left(x\right))}{tan\left(x\right)}dx$$

Commented by mnjuly1970 last updated on 06/Dec/20

$$againthankyou$$$$sirmindspower$$$$$$

Commented by mindispower last updated on 06/Dec/20

$$alwayspleasursir$$

Answered by mathmax by abdo last updated on 06/Dec/20

$$\mathrm{let}\mathrm{f}\left(\mathrm{a}\right)={\int }_0^{\frac{\pi }{2}}\frac{\log (1+\mathrm{atanx})}{\mathrm{tanx}}\mathrm{dx}\mathrm{with}\mathrm{a}>0\Rightarrow {\mathrm{f}}^{{}^{\prime }}\left(\mathrm{a}\right)={\int }_0^{\frac{\pi }{2}}\frac{\mathrm{tanx}}{(1+\mathrm{atanx})\mathrm{tanx}}\mathrm{dx}$$$$={\int }_0^{\frac{\pi }{2}}\frac{\mathrm{dx}}{1+\mathrm{atanx}}{=}_{\mathrm{tanx}=\mathrm{t}}{\int }_0^{\mathrm{\infty }}\frac{\mathrm{dt}}{(1+{\mathrm{t}}^2)(1+\mathrm{at})}{=}_{\mathrm{at}=\mathrm{z}}{\int }_0^{\mathrm{\infty }}\frac{\mathrm{dz}}{\mathrm{a}(1+\frac{{\mathrm{z}}^2}{{\mathrm{a}}^2})(\mathrm{z}+1)}$$$$=\mathrm{a}{\int }_0^{\mathrm{\infty }}\frac{\mathrm{dz}}{({\mathrm{z}}^2+{\mathrm{a}}^2)(\mathrm{z}+1)}\mathrm{let}\mathrm{decompose}\mathrm{F}\left(\mathrm{z}\right)=\frac{1}{(\mathrm{z}+1)({\mathrm{z}}^2+{\mathrm{a}}^2)}$$$$\mathrm{F}\left(\mathrm{z}\right)=\frac{\alpha }{\mathrm{z}+1}+\frac{\beta \mathrm{z}+\mathrm{n}}{{\mathrm{z}}^2+{\mathrm{a}}^2}$$$$\alpha =\frac{1}{{\mathrm{a}}^2+1},{\lim }_{\mathrm{z}\to +\mathrm{\infty }}\mathrm{zF}\left(\mathrm{z}\right)=0=\alpha +\beta \Rightarrow \beta =-\frac{1}{{\mathrm{a}}^2+1}$$$$\mathrm{F}\left(0\right)=\frac{1}{{\mathrm{a}}^2}=\alpha +\frac{\mathrm{n}}{{\mathrm{a}}^2}\Rightarrow 1={\mathrm{a}}^2\alpha +\mathrm{n}\Rightarrow \mathrm{n}=1-\frac{{\mathrm{a}}^2}{{\mathrm{a}}^2+1}=\frac{1}{{\mathrm{a}}^2+1}\Rightarrow$$$$\mathrm{F}\left(\mathrm{z}\right)=\frac{1}{({\mathrm{a}}^2+1)(\mathrm{z}+1)}+\frac{-\frac{1}{{\mathrm{a}}^2+1}\mathrm{z}+\frac{1}{{\mathrm{a}}^2+1}}{{\mathrm{z}}^2+{\mathrm{a}}^2}\Rightarrow$$$${\int }_0^{\mathrm{\infty }}\mathrm{F}\left(\mathrm{z}\right)\mathrm{dz}=\frac{1}{{\mathrm{a}}^2+1}{\int }_0^{\mathrm{\infty }}\frac{\mathrm{dz}}{\mathrm{z}+1}-\frac{1}{2({\mathrm{a}}^2+1)}{\int }_0^{\mathrm{\infty }}\frac{2\mathrm{z}-2}{{\mathrm{z}}^2+{\mathrm{a}}^2}\mathrm{dz}$$$$=\frac{1}{{\mathrm{a}}^2+1}{\int }_0^{\mathrm{\infty }}(\frac{1}{\mathrm{z}+1}-\frac{1}{2}\times \frac{2\mathrm{z}}{{\mathrm{z}}^2+1})\mathrm{dz}+\frac{1}{{\mathrm{a}}^2+1}{\int }_0^{\mathrm{\infty }}\frac{\mathrm{dz}}{{\mathrm{z}}^2+{\mathrm{a}}^2}(\to \mathrm{z}=\mathrm{au})$$$$=\frac{1}{{\mathrm{a}}^2+1}{[\ln \mid \frac{\mathrm{z}+1}{\sqrt{{\mathrm{z}}^2+1}}\mid ]}_0^{\mathrm{\infty }}+\frac{1}{{\mathrm{a}}^2+1}{\int }_0^{\mathrm{\infty }}\frac{\mathrm{adu}}{{\mathrm{a}}^2({\mathrm{u}}^2+1)}$$$$=\frac{1}{\mathrm{a}({\mathrm{a}}^2+1)}\times \frac{\pi }{2}\Rightarrow {\mathrm{f}}^{{}^{\prime }}\left(\mathrm{a}\right)=\frac{\pi }{2(1+{\mathrm{a}}^2)}\Rightarrow \mathrm{f}\left(\mathrm{a}\right)=\frac{\pi }{2}\arctan \left(\mathrm{a}\right)+\mathrm{C}$$$$\mathrm{f}\left(0\right)=\mathrm{C}\Rightarrow \mathrm{f}\left(\mathrm{a}\right)=\frac{\pi }{2}\arctan \left(\mathrm{a}\right)\mathrm{and}$$$${\int }_0^{\frac{\pi }{2}}\frac{\log (1+\mathrm{tanx})}{\mathrm{tanx}}\mathrm{dx}=\frac{\pi }{2}\arctan \left(1\right)=\frac{\pi }{2}.\frac{\pi }{4}=\frac{{\pi }^2}{8}$$

Commented by mnjuly1970 last updated on 06/Dec/20

$$thankyousomuchsirmax...$$

Commented by mathmax by abdo last updated on 06/Dec/20

$$\mathrm{you}\mathrm{are}\mathrm{welcome}\mathrm{sir}$$

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