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Question Number 162535 by mnjuly1970 last updated on 30/Dec/21

$$$$$$\mathrm{prove}\mathrm{that}$$$$\mathrm{\Omega }={\int }_0^{\mathrm{\infty }}\frac{\ln \left(\frac{1}{x}\right)}{x^4+17x^2+16}dx\stackrel{?}{=}\frac{\pi }{60}\ln \left(2\right)$$$$$$

Commented by Ar Brandon last updated on 30/Dec/21

$$\mathrm{Synthax}\mathrm{error},\mathrm{Sir}.\mathrm{Your}dx\mathrm{is}\mathrm{missing}.$$Haha 😅😁

Commented by amin96 last updated on 30/Dec/21

$$$$very big mistake😂😂

Commented by mnjuly1970 last updated on 30/Dec/21

$$yesyouarerightsir...$$

Answered by Ar Brandon last updated on 24/Mar/22

$$\mathrm{\Omega }={\int }_0^{\mathrm{\infty }}\frac{\ln \left(\frac{1}{x}\right)}{x^4+17x^2+16}dx=-{\int }_0^{\mathrm{\infty }}\frac{\ln x}{x^4+17x^2+16}dx$$$$\phi \left(z\right)=\frac{{\left(\ln z\right)}^2}{z^4+17z^2+16},\mathrm{poles}:(z^2+16)(z^2+1)=0\Rightarrow z_1=4e^{\frac{\pi }{2}i},z_2=4e^{-\frac{\pi }{2}i},z_3=e^{\frac{\pi }{2}i},z_4=e^{-\frac{\pi }{2}i}$$$$\mathrm{\Omega }=-\frac{1}{2}Re(Res(\phi ,z_1)+Res(\phi ,z_2)+Res(\phi ,z_3)+Res(\phi ,z_4))$$$$Res(\phi ,z_1)=\underset{z\to z_1}{\lim }\left(\frac{{\left(\ln z\right)}^2}{(z-z_2)(z-z_3)(z-z_4)}\right)=\frac{{(\ln 4+\frac{\pi }{2}i)}^2}{\left(8i\right)\left(3i\right)\left(5i\right)}=\frac{i}{120}({\ln }^{2}4-\frac{{\pi }^2}{4}+i\pi \ln 4)$$$$Res(\phi ,z_2)=\underset{z\to z_2}{\lim }\left(\frac{{\left(\ln z\right)}^2}{(z-z_1)(z-z_3)(z-z_4)}\right)=\frac{{(\ln 4-\frac{\pi }{2}i)}^2}{(-8i)(-5i)(-3i)}=-\frac{i}{120}({\ln }^{2}4+\frac{{\pi }^2}{4}-i\pi \ln 4)$$$$Res(\phi ,z_3)=\underset{z\to z_3}{\lim }\left(\frac{{\left(\ln z\right)}^2}{(z-z_1)(z-z_2)(z-z_4)}\right)=\frac{{\left(\frac{\pi }{2}i\right)}^2}{(-3i)\left(5i\right)\left(2i\right)}=\frac{i}{30}\left(\frac{{\pi }^2}{4}\right)=i\frac{{\pi }^2}{120}$$$$Res(\phi ,z_4)=\underset{z\to z_4}{\lim }\left(\frac{{\left(\ln z\right)}^2}{(z-z_1)(z-z_2)(z-z_3)}\right)=\frac{{(-\frac{\pi }{2}i)}^2}{(-5i)\left(3i\right)(-2i)}=-\frac{i}{30}\left(\frac{{\pi }^2}{4}\right)=-i\frac{{\pi }^2}{120}$$$$\mathrm{\Omega }=-\frac{1}{2}(-\frac{\pi \ln 4}{120}-\frac{\pi \ln 4}{120})=\frac{\pi \ln 4}{120}=\frac{\pi }{60}\ln \left(2\right)$$

Commented by mnjuly1970 last updated on 30/Dec/21

$$verynicesolution..residuestheory..$$$$thanksalotsirbrandon..$$

Answered by mindispower last updated on 30/Dec/21

$$=-{\int }_0^{\mathrm{\infty }}-\frac{ln\left(x\right)}{(x^2+16)(x^2+1)}dx=-\frac{1}{15}{\int }_0^{\mathrm{\infty }}\frac{ln\left(x\right)}{x^2+1}+\frac{ln\left(x\right)}{x^2+16}dx$$$$=\frac{1}{15}(-{\int }_0^{\mathrm{\infty }}\frac{ln\left(x\right)}{x^2+1}dx+\int \frac{ln\left(4y\right)}{16(y^2+1)}4y$$$$=-\frac{3}{60}{\int }_0^{\mathrm{\infty }}\frac{ln\left(x\right)}{1+x^2}+\frac{ln\left(4\right)}{60}{\int }_0^{\mathrm{\infty }}\frac{dx}{1+x^2}$$$${\int }_0^{\mathrm{\infty }}\frac{ln\left(x\right)}{1+x^2}dx,x\to \frac{1}{x}\Rightarrow =-{\int }_0^{\mathrm{\infty }}\frac{ln\left(x\right)}{1+x^2}dx$$$$\Rightarrow {\int }_0^{\mathrm{\infty }}\frac{ln\left(x\right)}{1+x^2}dx=0$$$$\mathrm{\Omega }=\frac{\ln \left(4\right)}{60}{\int }_0^{\mathrm{\infty }}\frac{dx}{1+x^2}=\frac{ln\left(4\right)}{60}\underset{x\to \mathrm{\infty }}{\lim }{\left[{\tan }^{-1}\left(z\right)\right]}_0^x$$$$=\frac{ln\left(4\right)}{60}.\frac{\pi }{2}=\frac{\pi ln\left(2\right)}{60}$$

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