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Question Number 135851 by Ñï= last updated on 16/Mar/21

$${\int }_0^{\pi /2}\left(\cos {}^{2}x\right)ln(1+\tan {}^{4}x)dx=?$$

Answered by mathmax by abdo last updated on 16/Mar/21

$$\mathrm{let}\mathrm{\Phi }={\int }_0^{\frac{\pi }{2}}{\cos }^2\mathrm{x}.\log (1+{\tan }^4\mathrm{x})\mathrm{dx}\mathrm{we}\mathrm{have}{\cos }^2\mathrm{x}=\frac{1}{1+{\tan }^2\mathrm{x}}$$$$\mathrm{we}\mathrm{do}\mathrm{the}\mathrm{channgement}\mathrm{tanx}=\mathrm{t}\Rightarrow$$$$\mathrm{\Phi }={\int }_0^{\mathrm{\infty }}\frac{1}{1+{\mathrm{t}}^2}\log (1+{\mathrm{t}}^4)\frac{\mathrm{dt}}{1+{\mathrm{t}}^2}={\int }_0^{\mathrm{\infty }}\frac{\log (1+{\mathrm{t}}^4)}{{({\mathrm{t}}^2+1)}^2}\mathrm{dt}\mathrm{let}$$$$\mathrm{f}\left(\mathrm{a}\right)={\int }_0^{\mathrm{\infty }}\frac{\ln (1+{\mathrm{at}}^4)}{{({\mathrm{t}}^2+1)}^2}\mathrm{dt}(\mathrm{a}⩾0)\Rightarrow {\mathrm{f}}^{{}^{\prime }}\left(\mathrm{a}\right)={\int }_0^{\mathrm{\infty }}\frac{{\mathrm{t}}^4}{({\mathrm{at}}^4+1){({\mathrm{t}}^2+1)}^2}\mathrm{dt}$$$${\mathrm{f}}^{{}^{\prime }}\left(\mathrm{a}\right)=\frac{1}{2}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{{\mathrm{t}}^4}{\mathrm{a}({\mathrm{t}}^4+{\left({\mathrm{a}}^{\frac{1}{4}}\right)}^4){({\mathrm{t}}^2+1)}^2}\mathrm{dt}({\mathrm{x}}_0={\mathrm{a}}^{\frac{1}{4}})$$$$=\frac{1}{2\mathrm{a}}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{{\mathrm{t}}^4}{({\mathrm{t}}^2-{\mathrm{ix}}_0^2({\mathrm{t}}^2+{\mathrm{ix}}_0^2){(\mathrm{t}-\mathrm{i})}^2{(\mathrm{t}+\mathrm{i})}^2}\mathrm{dt}\Rightarrow$$$${2\mathrm{af}}^{{}^{\prime }}\left(\mathrm{a}\right)={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{{\mathrm{t}}^4}{(\mathrm{t}-{\mathrm{x}}_0{\mathrm{e}}^{\frac{\mathrm{i}\pi }{4}})(\mathrm{t}+{\mathrm{x}}_0{\mathrm{e}}^{\frac{\mathrm{i}\pi }{4}})(\mathrm{t}-{\mathrm{x}}_0{\mathrm{e}}^{-\frac{\mathrm{i}\pi }{4}})(\mathrm{t}+{\mathrm{x}}_0{\mathrm{e}}^{-\frac{\mathrm{i}\pi }{4}}){(\mathrm{t}-\mathrm{i})}^2{(\mathrm{t}+\mathrm{i})}^2}$$$$=2\mathrm{i}\pi \{\mathrm{Res}(\mathrm{w},{\mathrm{x}}_0{\mathrm{e}}^{\frac{\mathrm{i}\pi }{4}})+\mathrm{Res}(\mathrm{w},-{\mathrm{x}}_0{\mathrm{e}}^{-\frac{\mathrm{i}\pi }{4}})+\mathrm{Res}(\mathrm{w},\mathrm{i})\}$$$$\mathrm{Res}(\mathrm{w},{\mathrm{x}}_0{\mathrm{e}}^{\frac{\mathrm{i}\pi }{4}})=\frac{-{\mathrm{x}}_0^4}{{2\mathrm{x}}_0{\mathrm{e}}^{\frac{\mathrm{i}\pi }{4}}{({\mathrm{x}}_0^2\mathrm{i}+1)}^2}$$$$\mathrm{Res}(\mathrm{w},-{\mathrm{x}}_0{\mathrm{e}}^{-\frac{\mathrm{i}\pi }{4}})=\frac{-{\mathrm{x}}_0^4}{-{2\mathrm{x}}_0{\mathrm{e}}^{-\frac{\mathrm{i}\pi }{4}}{(-{\mathrm{ix}}_0^2+1)}^2}$$$$\mathrm{Res}(\mathrm{w},\mathrm{i})={\lim }_{\mathrm{z}\to \mathrm{i}}{\left\{\frac{{\mathrm{t}}^4}{({\mathrm{t}}^4+{\mathrm{x}}_0^4){(\mathrm{t}+\mathrm{i})}^2}\right\}}^{\left(1\right)}=....$$$$...\mathrm{be}\mathrm{continued}....$$

Commented by mathmax by abdo last updated on 16/Mar/21

$$\mathrm{let}\mathrm{try}\mathrm{another}\mathrm{way}\mathrm{\Phi }={\int }_0^{\mathrm{\infty }}\frac{\log (1+{\mathrm{t}}^4)}{{({\mathrm{t}}^2+1)}^2}\mathrm{dt}\Rightarrow$$$$\mathrm{\Phi }{=}_{\mathrm{t}=\frac{1}{\mathrm{x}}}-{\int }_0^{\mathrm{\infty }}\frac{\log (1+\frac{1}{{\mathrm{x}}^2})}{{(\frac{1}{{\mathrm{x}}^2}+1)}^2}(-\frac{\mathrm{dx}}{{\mathrm{x}}^2})={\int }_0^{\mathrm{\infty }}{\mathrm{x}}^4\frac{\log (1+{\mathrm{x}}^2)-2\mathrm{logx}}{{\mathrm{x}}^2{(1+{\mathrm{x}}^2)}^2}$$$$={\int }_0^{\mathrm{\infty }}\frac{{\mathrm{x}}^2}{{(1+{\mathrm{x}}^2)}^2}(\log (1+{\mathrm{x}}^2)-2\mathrm{logx})\mathrm{dx}$$$$={\int }_0^{\mathrm{\infty }}\frac{{\mathrm{x}}^2\log (1+{\mathrm{x}}^2)}{{(1+{\mathrm{x}}^2)}^2}\mathrm{dx}-2{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{x}}^2\mathrm{logx}}{{(1+{\mathrm{x}}^2)}^2}\mathrm{dx}=\mathrm{H}-2\mathrm{K}\mathrm{we}\mathrm{have}$$$$\mathrm{H}={\int }_0^{\mathrm{\infty }}\frac{{\mathrm{x}}^2\log (1+{\mathrm{x}}^2)}{{(1+{\mathrm{x}}^2)}^2}\mathrm{dx}={\int }_0^{\mathrm{\infty }}\frac{\mathrm{x}}{{(1+{\mathrm{x}}^2)}^2}\left(\mathrm{xlog}(1+{\mathrm{x}}^2)\right)\mathrm{dx}$$$$\mathrm{by}\mathrm{parts}{\mathrm{u}}^{{}^{\prime }}=\frac{\mathrm{x}}{{(1+{\mathrm{x}}^2)}^2}\Rightarrow \mathrm{u}=-\frac{1}{2(1+{\mathrm{x}}^2)}\mathrm{and}\mathrm{v}=\mathrm{xlog}(1+{\mathrm{x}}^2)\Rightarrow$$$$\mathrm{H}={[-\frac{1}{2(1+{\mathrm{x}}^2)}\mathrm{xlog}(1+{\mathrm{x}}^2)]}_0^{\mathrm{\infty }}+{\int }_0^{\mathrm{\infty }}\frac{1}{2(1+{\mathrm{x}}^2)}(\log (1+{\mathrm{x}}^2)+\frac{{2\mathrm{x}}^2}{1+{\mathrm{x}}^2})\mathrm{dx}$$$$=\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{\log (1+{\mathrm{x}}^2)}{1+{\mathrm{x}}^2}\mathrm{dx}+{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{x}}^2}{2{(1+{\mathrm{x}}^2)}^2}\mathrm{dx}$$$${\int }_0^{\mathrm{\infty }}\frac{\log (1+{\mathrm{x}}^2)}{1+{\mathrm{x}}^2}\mathrm{dx}{=}_{\mathrm{x}=\tan \theta }{\int }_0^{\frac{\pi }{2}}\frac{\log \left(\frac{1}{{\cos }^2\theta }\right)}{1+{\tan }^2\theta }(1+{\tan }^2\theta )\mathrm{d}\theta$$$$=-2{\int }_0^{\frac{\pi }{2}}\log \left(\cos \theta \right)\mathrm{d}\theta =-2(-\frac{\pi }{2}\log 2)=\pi \log 2$$$${\int }_0^{\mathrm{\infty }}\frac{{\mathrm{x}}^2}{2{(1+{\mathrm{x}}^2)}^2}\mathrm{dx}={\int }_0^{\mathrm{\infty }}\frac{1+{\mathrm{x}}^2-1}{2{(1+{\mathrm{x}}^2)}^2}\mathrm{dx}=\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{\mathrm{dx}}{1+{\mathrm{x}}^2}-\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{\mathrm{dx}}{{(1+{\mathrm{x}}^2)}^2}(\mathrm{x}=\tan \theta )$$$$=\frac{\pi }{4}-\frac{1}{2}{\int }_0^{\frac{\pi }{2}}\frac{1+{\tan }^2\theta }{{(1+{\tan }^2\theta )}^2}\mathrm{d}\theta =\frac{\pi }{4}-\frac{1}{2}{\int }_0^{\frac{\pi }{2}}{\cos }^2\theta \mathrm{d}\theta$$$$=\frac{\pi }{4}-\frac{1}{4}{\int }_0^{\frac{\pi }{2}}(1+\cos \left(2\theta \right))\mathrm{d}\theta =\frac{\pi }{4}-\frac{\pi }{8}-\frac{1}{8}{\left[\sin \left(2\theta \right)\right]}_0^{\frac{\pi }{2}}$$$$=\frac{\pi }{8}\Rightarrow \mathrm{H}=\frac{\pi }{2}\log \left(2\right)+\frac{\pi }{8}$$$$\mathrm{K}={\int }_0^{\mathrm{\infty }}\frac{{\mathrm{x}}^2\mathrm{logx}}{{(1+{\mathrm{x}}^2)}^2}\mathrm{dx}={\int }_0^{\mathrm{\infty }}\frac{({\mathrm{x}}^2+1-1)\mathrm{logx}}{{({\mathrm{x}}^2+1)}^2}\mathrm{dx}$$$$={\int }_0^{\mathrm{\infty }}\frac{\mathrm{logx}}{1+{\mathrm{x}}^2}\mathrm{dx}(=0)-{\int }_0^{\mathrm{\infty }}\frac{\mathrm{logx}}{{({\mathrm{x}}^2+1)}^2}\mathrm{dx}\mathrm{we}\mathrm{have}$$$${\int }_0^{\mathrm{\infty }}\frac{\mathrm{logx}}{{(1+{\mathrm{x}}^2)}^2}\mathrm{dx}={\int }_0^1\frac{\mathrm{logx}}{{(1+{\mathrm{x}}^2)}^2}\mathrm{dx}+{\int }_1^{\mathrm{\infty }}\frac{\mathrm{logx}}{{(1+{\mathrm{x}}^2)}^2}\mathrm{dx}(\mathrm{x}=\frac{1}{\mathrm{t}})$$$$={\int }_0^1\frac{\mathrm{logx}}{{(1+{\mathrm{x}}^2)}^2}\mathrm{dx}-{\int }_0^1\frac{-\mathrm{logt}}{{(1+\frac{1}{{\mathrm{t}}^2})}^2}(-\frac{\mathrm{dt}}{{\mathrm{t}}^2}$$$$={\int }_0^1\frac{\mathrm{logx}}{{(1+{\mathrm{x}}^2)}^2}\mathrm{dx}-{\int }_0^1\frac{{\mathrm{t}}^2\mathrm{logt}}{{({\mathrm{t}}^2+1)}^2}\mathrm{dt}$$$$\mathrm{we}\mathrm{have}\frac{1}{1+{\mathrm{x}}^2}=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{(-1)}^{\mathrm{n}}{\mathrm{x}}^{2\mathrm{n}}\Rightarrow$$$$\frac{-2\mathrm{x}}{{(1+{\mathrm{x}}^2)}^2}=\sum _{\mathrm{n}=1}^{\mathrm{\infty }}2\mathrm{n}{(-1)}^{\mathrm{n}}{\mathrm{x}}^{2\mathrm{n}-1}\Rightarrow \frac{1}{{(1+{\mathrm{x}}^2)}^2}$$$$=\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\mathrm{n}{(-1)}^{\mathrm{n}-1}{\mathrm{x}}^{2\mathrm{n}-2}\Rightarrow$$$${\int }_0^1\frac{\mathrm{logx}}{{(1+{\mathrm{x}}^2)}^2}\mathrm{dx}=\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\mathrm{n}{(-1)}^{\mathrm{n}}\frac{1}{2\mathrm{n}-1}...\mathrm{be}\mathrm{continued}....$$

Commented by Ñï= last updated on 16/Mar/21

$${\int }_0^{\pi /2}\left(\cos {}^{2}x\right)ln(1+\tan {}^{4}x)dx$$$$\stackrel{x\to \frac{\pi }{2}-x}{=}{\int }_0^{\pi /2}\left(\sin {}^{2}x\right)ln(1+\tan {}^{4}x)dx-4{\int }_0^{\pi /2}\left(\sin {}^{2}x\right)ln\left(\tan x\right)dx$$$${\int }_0^{\pi /2}\left(cos{}^{2}x\right)ln(1+\tan {}^{4}x)dx-{\int }_0^{\pi /2}\left(\sin {}^{2}x\right)ln(1+\tan {}^{4}x)dx=-4{\int }_0^{\pi /2}\left(\sin {}^{2}x\right)ln\left(\tan x\right)dx=-\pi$$$${\int }_0^{\pi /2}\left(cos{}^{2}x\right)ln(1+\tan {}^{4}x)dx+{\int }_0^{\pi /2}\left(\sin {}^{2}x\right)ln(1+\tan {}^{4}x)dx={\int }_0^{\pi /2}ln(1+\tan {}^{4}x)dx=\frac{\pi }{2}ln(6+4\sqrt{2})$$$${\int }_0^{\pi /2}\left(cos{}^{2}x\right)ln(1+\tan {}^{4}x)dx=\frac{1}{2}[-\pi +\frac{\pi }{2}ln(6+4\sqrt{2})]$$$${\int }_0^{\pi /2}\left(\sin {}^{2}x\right)ln(1+\tan {}^{4}x)dx=\frac{1}{2}[\pi +\frac{\pi }{2}ln(6+4\sqrt{2})]$$

Commented by mathmax by abdo last updated on 16/Mar/21

$$\mathrm{from}\mathrm{where}\mathrm{come}-\pi \mathrm{and}\mathrm{sign}+\mathrm{sir}?$$

Commented by Ajetunmobi last updated on 17/Mar/21

$$$$$$\mathbf{that}\mathbf{is}\mathbf{the}\mathbf{proof}\mathbf{for}{\int }_0^{\mathrm{\infty }}\frac{\mathbf{ln}\left(\mathbf{x}\right)}{{(1+{\mathbf{x}}^2)}^2}$$

Commented by Ajetunmobi last updated on 16/Mar/21

Commented by Ajetunmobi last updated on 17/Mar/21

$$$$$$\mathbf{the}\mathbf{proposal}$$$$\mathbf{can}\mathbf{u}\mathbf{show}\mathbf{the}\mathbf{full}\mathbf{solution}?$$$$\mathbf{because}\mathbf{the}\mathbf{integral}\mathbf{is}\mathbf{somehow}\mathbf{elegant}\mathbf{and}$$$$\mathbf{also}\mathbf{lenghty}\mathbf{in}\mathbf{my}\mathbf{view}$$

Commented by mathmax by abdo last updated on 17/Mar/21

$$\mathrm{thank}\mathrm{you}\mathrm{sir}$$

Commented by mathmax by abdo last updated on 17/Mar/21

$$\mathrm{let}\mathrm{use}{\int }_0^{\mathrm{\infty }}\mathrm{p}\left(\mathrm{x}\right)\ln \left(\mathrm{x}\right)\mathrm{dx}=-\frac{1}{2}\mathrm{Re}(\mathrm{\Sigma }\mathrm{Res}(\mathrm{p}\left(\mathrm{z}\right){\ln }^2\left(\mathrm{z}\right),{\mathrm{a}}_{\mathrm{i}})\mathrm{we}\mathrm{have}$$$$\mathrm{I}={\int }_0^{\mathrm{\infty }}\frac{\mathrm{lnx}}{{(1+{\mathrm{x}}^2)}^2}\mathrm{dx}\Rightarrow \phi \left(\mathrm{z}\right)=\frac{{\ln }^2\mathrm{z}}{{({\mathrm{z}}^2+1)}^2}=\frac{{\ln }^2\mathrm{z}}{{(\mathrm{z}-\mathrm{i})}^2{(\mathrm{z}+\mathrm{i})}^2}$$$$\mathrm{\Sigma }\mathrm{Res}(\phi ,{\mathrm{z}}_{\mathrm{i}})=\mathrm{Res}(\phi ,\mathrm{i})+\mathrm{Res}(\phi ,-\mathrm{i})$$$$\mathrm{Res}(\phi ,\mathrm{i})={\lim }_{\mathrm{z}\to \mathrm{i}}\frac{1}{(2-1)!}{\left\{\frac{{\ln }^2\mathrm{z}}{{(\mathrm{z}+\mathrm{i})}^2}\right\}}^{\left(1\right)}$$$$={\lim }_{\mathrm{z}\to \mathrm{i}}\frac{\frac{2\mathrm{lnz}}{\mathrm{z}}{(\mathrm{z}+\mathrm{i})}^2-2(\mathrm{z}+\mathrm{i}){\ln }^2\mathrm{z}}{{(\mathrm{z}+\mathrm{i})}^4}$$$$={\lim }_{\mathrm{z}\to \mathrm{i}}\frac{\frac{2\mathrm{lnz}}{\mathrm{z}}(\mathrm{z}+\mathrm{i})-{2\ln }^2\mathrm{z}}{{(\mathrm{z}+\mathrm{i})}^3}=\frac{4\mathrm{lni}-{2\ln }^2\mathrm{i}}{{\left(2\mathrm{i}\right)}^3}$$$$=\frac{4\ln \left({\mathrm{e}}^{\frac{\mathrm{i}\pi }{2}}\right)-2{\left(\ln \left({\mathrm{e}}^{\frac{\mathrm{i}\pi }{2}}\right)\right)}^2}{-8\mathrm{i}}=\frac{2\mathrm{i}\pi -2{\left(\frac{\mathrm{i}\pi }{2}\right)}^2}{-8\mathrm{i}}=-\frac{2\mathrm{i}\pi -\frac{{\pi }^2}{2}}{8\mathrm{i}}$$$$=\frac{1}{8}(-2\pi -\frac{\mathrm{i}{\pi }^2}{2})=-\frac{\pi }{4}-\frac{\mathrm{i}{\pi }^2}{16}$$$$\mathrm{Res}(\phi ,-\mathrm{i})={\lim }_{\mathrm{z}\to -\mathrm{i}}\frac{1}{(2-1)!}{\left\{\frac{{\ln }^2\mathrm{z}}{{(\mathrm{z}-\mathrm{i})}^2}\right\}}^{\left(1\right)}$$$$={\lim }_{\mathrm{z}\to -\mathrm{i}}\frac{\frac{2\mathrm{lnz}}{\mathrm{z}}{(\mathrm{z}-\mathrm{i})}^2-2(\mathrm{z}-\mathrm{i}){\ln }^2\mathrm{z}}{{(\mathrm{z}-\mathrm{i})}^4}$$$$={\lim }_{\mathrm{z}\to -\mathrm{i}}\frac{\frac{2\mathrm{lnz}}{\mathrm{z}}(\mathrm{z}-\mathrm{i})-{2\ln }^2\mathrm{z}}{{(\mathrm{z}-\mathrm{i})}^3}=\frac{4\ln \left({\mathrm{e}}^{-\frac{\mathrm{i}\pi }{2}}\right)-2{\left(\ln \left({\mathrm{e}}^{-\frac{\mathrm{i}\pi }{2}}\right)\right)}^2}{{(-2\mathrm{i})}^3}$$$$=\frac{-2\mathrm{i}\pi -2{(-\frac{\mathrm{i}\pi }{2})}^2}{8\mathrm{i}}=\frac{-2\mathrm{i}\pi -\frac{{\pi }^2}{2}}{8\mathrm{i}}=-\frac{\pi }{4}+\frac{\mathrm{i}{\pi }^2}{16}$$$$\Rightarrow \mathrm{\Sigma }\mathrm{Res}(\phi ,{\mathrm{z}}_{\mathrm{i}})=-\frac{\pi }{2}\Rightarrow {\int }_0^{\mathrm{\infty }}\frac{\mathrm{lnx}}{{(1+{\mathrm{x}}^2)}^2}=-\frac{1}{2}(-\frac{\pi }{2})=\frac{\pi }{4}$$

Commented by Ajetunmobi last updated on 17/Mar/21

$$\mathbf{welcome}\mathbf{sir}$$

Commented by Ajetunmobi last updated on 17/Mar/21

$$\mathbf{I}\mathbf{dont}\mathbf{know}\mathbf{residue}\mathbf{theorem}\mathbf{though}$$$$\mathbf{but}\mathbf{the}\mathbf{answer}\mathbf{is}-\frac{\mathit{\pi }}{4}$$

Commented by Ajetunmobi last updated on 17/Mar/21

$$\mathbf{the}\mathbf{answer}\mathbf{by}\mathbf{the}\mathbf{proposal}\mathbf{is}\mathbf{correct}$$$$\mathbf{but}\mathbf{i}\mathbf{dont}\mathbf{really}\mathbf{understand}\mathbf{is}\mathbf{solution}$$$$\mathbf{because}\mathbf{the}\mathbf{proposal}\mathbf{never}\mathbf{show}\mathbf{the}\mathbf{full}$$$$\mathbf{solution}$$

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