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Question Number 136210 by Ar Brandon last updated on 19/Mar/21

$$\mathrm{Show}\mathrm{that}{\int }_0^{\mathrm{\infty }}\frac{\mathrm{lnx}}{{({\mathrm{x}}^2+1)}^2}\mathrm{dx}=-\frac{\pi }{4}$$

Commented by Ar Brandon last updated on 19/Mar/21

$$\mathrm{Oh}!\mathrm{Thank}\mathrm{You}\mathrm{Sir}$$😃

Commented by Ar Brandon last updated on 19/Mar/21

$$\mathrm{\Omega }={\int }_0^{\mathrm{\infty }}\frac{\mathrm{lnx}}{{({\mathrm{x}}^2+1)}^2}\mathrm{dx}$$$$=\frac{\partial }{\partial \mathrm{s}}{\mid }_{\mathrm{s}=0}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{x}}^{\mathrm{s}}}{{({\mathrm{x}}^2+1)}^2}\mathrm{dx}\stackrel{\mathrm{u}={\mathrm{x}}^2}{=}\frac{\partial }{\partial \mathrm{s}}{\mid }_{\mathrm{s}=0}\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{u}}^{\frac{\mathrm{s}}{2}-\frac{1}{2}}}{{(\mathrm{u}+1)}^2}\mathrm{du}$$$$=\frac{\partial }{\partial \mathrm{s}}{\mid }_{\mathrm{s}=0}\frac{1}{2}\beta (\frac{\mathrm{s}}{2}+\frac{1}{2},\frac{3}{2}-\frac{\mathrm{s}}{2})=\frac{\partial }{\partial \mathrm{s}}{\mid }_{\mathrm{s}=0}\frac{1}{2}\mathrm{\Gamma }(\frac{\mathrm{s}}{2}+\frac{1}{2})\mathrm{\Gamma }(\frac{3}{2}-\frac{\mathrm{s}}{2})$$$$=\frac{1}{2}{\mid }_{\mathrm{s}=0}\mathrm{\Gamma }(\frac{\mathrm{s}}{2}+\frac{1}{2}){\mathrm{\Gamma }}^{\prime }(\frac{3}{2}-\frac{\mathrm{s}}{2})+\mathrm{\Gamma }(\frac{3}{2}-\frac{\mathrm{s}}{2}){\mathrm{\Gamma }}^{\prime }(\frac{\mathrm{s}}{2}+\frac{1}{2})$$$$=\frac{1}{2}\cdot \frac{1}{2}\mathrm{\Gamma }\left(\frac{1}{2}\right)\mathrm{\Gamma }\left(\frac{3}{2}\right)[\psi \left(\frac{1}{2}\right)-\psi \left(\frac{3}{2}\right)]=\frac{\pi }{8}[\psi \left(\frac{1}{2}\right)-\psi \left(\frac{1}{2}\right)-2]=-\frac{\pi }{4}$$

Commented by mnjuly1970 last updated on 19/Mar/21

$$errorinsignofdrivative$$$$of\psi (\frac{3}{2}-s).......{}^{\prime }=-{\psi }^{\prime }(\frac{3}{2}-s)$$$$...goodluckmybrothermrbrandon...$$

Answered by mnjuly1970 last updated on 19/Mar/21

$$\mathit{\varphi }\stackrel{x^2=y}{=}\frac{1}{4}{\int }_0^{\mathrm{\infty }}\frac{y^{-\frac{1}{2}}ln\left(y\right)}{{(y+1)}^2}dy$$$$f\left(a\right)={\int }_0^{\mathrm{\infty }}\frac{y^{a-\frac{1}{2}}}{{(y+1)}^2}dy$$$$\mathit{\varphi }=\frac{f{}^{\prime }\left(0\right)}{4}...✓$$$$f\left(a\right)={\int }_0^{\mathrm{\infty }}\frac{y^{a+\frac{1}{2}-1}}{{(y+1)}^2}dy=\beta (a+\frac{1}{2},\frac{3}{2}-a)$$$$f\left(a\right)=\mathrm{\Gamma }(a+\frac{1}{2}).\mathrm{\Gamma }(\frac{3}{2}-a)$$$$f{}^{\prime }\left(a\right)={\mathrm{\Gamma }}^{\prime }(a+\frac{1}{2})\mathrm{\Gamma }(\frac{3}{2}-a)-{\mathrm{\Gamma }}^{\prime }(\frac{3}{2}-a)\mathrm{\Gamma }(\frac{1}{2}+a)$$$$\mathit{\varphi }=\frac{1}{4}f{}^{\prime }\left(0\right)=\frac{1}{4}(\frac{1}{2}\psi \left(\frac{1}{2}\right){\mathrm{\Gamma }}^2\left(\frac{1}{2}\right)-\frac{1}{2}\psi \left(\frac{3}{2}\right){\mathrm{\Gamma }}^2\left(\frac{1}{2}\right))$$$$=\frac{\pi }{8}(\psi \left(\frac{1}{2}\right)-\psi \left(\frac{3}{2}\right))$$$$=\frac{\pi }{8}(\psi \left(\frac{1}{2}\right)-\psi \left(\frac{1}{2}\right)-2)=\frac{-\pi }{4}..✓✓$$$$$$

Answered by Dwaipayan Shikari last updated on 19/Mar/21

$$\frac{1}{4}{\int }_0^{\mathrm{\infty }}\frac{u^{-\frac{1}{2}}log\left(u\right)}{{(1+u)}^2}du=\frac{1}{4}{\tau }^{\prime }\left(\frac{1}{2}\right)={\int }_0^{\mathrm{\infty }}\frac{log\left(x\right)}{{(x^2+1)}^2}dx$$$$\tau \left(\alpha \right)={\int }_0^{\mathrm{\infty }}\frac{u^{\alpha -1}}{{(1+u)}^2}du=\frac{\mathrm{\Gamma }\left(\alpha \right)\mathrm{\Gamma }(2-\alpha )}{\mathrm{\Gamma }\left(2\right)}=\frac{\pi (1-\alpha )}{sin\left(\pi \alpha \right)}$$$$\Rightarrow {\tau }^{\prime }\left(\alpha \right)={\int }_0^{\mathrm{\infty }}\frac{u^{\alpha -1}log\left(u\right)}{{(1+u)}^2}du=-{\pi }^{2}cosec\left(\pi \alpha \right)cot\left(\pi \alpha \right)-\frac{\pi }{sin\left(\pi \alpha \right)}$$$$\frac{1}{4}{\tau }^{\prime }\left(\frac{1}{2}\right)=-0-\frac{\pi }{4}=-\frac{\pi }{4}$$

Commented by Ar Brandon last updated on 19/Mar/21

$$\mathrm{Thanks}\mathrm{bro}$$

Commented by Dwaipayan Shikari last updated on 19/Mar/21

$$$$😃

Answered by Ajetunmobi last updated on 19/Mar/21

$$$$$$\mathbf{i}\mathbf{have}\mathbf{drop}\mathbf{the}\mathbf{solution}\mathbf{before}\mathbf{here}$$

Commented by Ar Brandon last updated on 19/Mar/21

$$\mathrm{Alright}!$$$$\mathrm{Before}\mathrm{posting}\mathrm{I}\mathrm{used}\mathrm{the}\mathrm{search}\mathrm{option}$$$$\mathrm{to}\mathrm{check}\mathrm{if}\mathrm{it}\mathrm{was}\mathrm{previously}\mathrm{posted}\mathrm{but}$$$${\mathrm{couldn}}^{\prime }\mathrm{t}\mathrm{find}\mathrm{it}.\mathrm{Thank}\mathrm{You}\mathrm{Sir}$$

Commented by Ajetunmobi last updated on 19/Mar/21

$$$$$$\mathbf{ok}$$

Answered by Ajetunmobi last updated on 19/Mar/21

Answered by mathmax by abdo last updated on 19/Mar/21

$$\mathrm{let}\mathrm{f}\left(\mathrm{a}\right)={\int }_0^{\mathrm{\infty }}\frac{\mathrm{lnx}}{{\mathrm{x}}^2+{\mathrm{a}}^2}\mathrm{dx}\mathrm{with}\mathrm{a}>0\Rightarrow {\mathrm{f}}^{{}^{\prime }}\left(\mathrm{a}\right)=-2\mathrm{a}{\int }_0^{\mathrm{\infty }}\frac{\mathrm{lnx}}{{({\mathrm{x}}^2+{\mathrm{a}}^2)}^2}\mathrm{dx}\Rightarrow$$$${\mathrm{f}}^{{}^{\prime }}\left(1\right)=-2{\int }_0^{\mathrm{\infty }}\frac{\mathrm{lnx}}{{({\mathrm{x}}^2+1)}^2}\Rightarrow {\int }_0^{\mathrm{\infty }}\frac{\mathrm{lnx}}{{({\mathrm{x}}^2+1)}^2}=-\frac{1}{2}{\mathrm{f}}^{{}^{\prime }}\left(1\right)$$$$\mathrm{f}\left(\mathrm{a}\right){=}_{\mathrm{x}=\mathrm{at}}{\int }_0^{\mathrm{\infty }}\frac{\mathrm{lna}+\mathrm{lnt}}{{\mathrm{a}}^2(1+{\mathrm{t}}^2)}\mathrm{adt}=\frac{1}{\mathrm{a}}{\int }_0^{\mathrm{\infty }}\frac{\mathrm{lna}+\mathrm{lnt}}{(1+{\mathrm{t}}^2)}\mathrm{dt}$$$$=\frac{\mathrm{lna}}{\mathrm{a}}{\int }_0^{\mathrm{\infty }}\frac{\mathrm{dt}}{1+{\mathrm{t}}^2}+\frac{1}{\mathrm{a}}{\int }_0^{\mathrm{\infty }}\frac{\mathrm{lnt}}{1+{\mathrm{t}}^2}\mathrm{dt}\mathrm{but}{\int }_0^{\mathrm{\infty }}\frac{\mathrm{lnt}}{1+{\mathrm{t}}^2}\mathrm{dt}=0\left(\mathrm{proved}\right)\Rightarrow$$$$\mathrm{f}\left(\mathrm{a}\right)=\frac{\mathrm{lna}}{\mathrm{a}}.\frac{\pi }{2}=\frac{\pi \mathrm{lna}}{2\mathrm{a}}\Rightarrow {\mathrm{f}}^{{}^{\prime }}\left(\mathrm{a}\right)=\frac{\frac{\pi }{\mathrm{a}}\left(2\mathrm{a}\right)-2\pi \mathrm{lna}}{{4\mathrm{a}}^2}\Rightarrow {\mathrm{f}}^{{}^{\prime }}\left(1\right)=\frac{2\pi }{4}=\frac{\pi }{2}\Rightarrow$$$${\int }_0^{\mathrm{\infty }}\frac{\mathrm{lnx}}{{({\mathrm{x}}^2+1)}^2}\mathrm{dx}=-\frac{1}{2}.\frac{\pi }{2}=-\frac{\pi }{4}$$

Commented by mathmax by abdo last updated on 19/Mar/21

$$\mathrm{folow}\mathrm{my}\mathrm{facebook}\left(\mathrm{abdo}\mathrm{imad}\right)$$

Commented by Ajetunmobi last updated on 19/Mar/21

$$$$$$\mathbf{sir}\mathbf{mathmax}\mathbf{by}\mathbf{abdo}$$$$\mathbf{how}\mathbf{can}\mathbf{we}\mathbf{chat}\mathbf{privately}\mathbf{sir}\mathbf{i}\mathbf{need}\mathbf{to}$$$$\mathbf{talk}\mathbf{about}\mathbf{something}\mathbf{with}\mathbf{you}$$$$\mathbf{you}\mathbf{can}\mathbf{send}\mathbf{me}\mathbf{your}\mathbf{whatsapp}\mathbf{number}$$$$\mathbf{or}\mathbf{your}\mathbf{facebook}\mathbf{name}$$$$\mathbf{Thanks}$$

Commented by Ajetunmobi last updated on 20/Mar/21

$$$$$$\mathbf{there}\mathbf{are}\mathbf{many}\mathbf{abdo}\mathbf{imad}\mathbf{there}$$$$\mathbf{how}\mathbf{will}\mathbf{i}\mathbf{recognize}\mathbf{you}\mathbf{sir}?$$

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