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Question Number 114959 by mathdave last updated on 22/Sep/20

$$longtimequestionproposedby$$$$mathabdo$$$${\int }_0^{\mathrm{\infty }}\frac{\ln x}{1+x^2+x^4}dx$$$$$$

Answered by mathdave last updated on 22/Sep/20

$$solution$$$$let$$$$I={\int }_0^{\mathrm{\infty }}\frac{\ln x}{1+x^2+x^4}dx={\int }_0^1\frac{\ln x}{1+x^2+x^4}dx+{\int }_1^{\mathrm{\infty }}\frac{\ln x}{1+x^2+x^4}dx=A+B$$$$putttingx=\frac{1}{x}intointegralB$$$$I={\int }_0^1\frac{\ln x}{1+x^2+x^4}dx-{\int }_0^1\frac{x^2\ln x}{1+x^2+x^4}dx=A-B$$$$let$$$$A={\int }_0^1\frac{\ln x}{1+x^2+x^4}dx={\int }_0^1\frac{\ln x}{1-x^6}dx-{\int }_0^1\frac{x^2\ln x}{1-x^6}dx$$$$A=\underset{n=0}{\sum ^{\mathrm{\infty }}}{\int }_0^1{x}^{6n}.\ln xdx-\underset{n=0}{\sum ^{\mathrm{\infty }}}{\int }_0^1{x}^2.x^{6n}.\ln xdx$$$$A=\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{\partial }{\partial a}{\mid }_{a=1}{\int }_0^1{x}^{6n}.x^{a-1}dx-\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{\partial }{\partial a}{\mid }_{a=1}{\int }_0^1{x}^{6n+2}.x^{a-1}dx$$$$A=\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{\partial }{\partial a}{\mid }_{a=1}\left(\frac{1}{6n+a}\right)-\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{\partial }{\partial a}{\mid }_{a=1}\left(\frac{1}{6n+2+a}\right)$$$$A=-\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{1}{{(6n+1)}^2}+\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{1}{{(6n+3)}^2}=-\frac{1}{36}\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{1}{{(n+\frac{1}{6})}^2}+\frac{1}{36}\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{1}{{(n+\frac{1}{2})}^2}$$$$A=-\frac{1}{36}{\psi }^1\left(\frac{1}{6}\right)+\frac{1}{36}{\psi }^1\left(\frac{1}{2}\right)...........\left(1\right)and$$$$B={\int }_0^1\frac{x^2\ln x}{1+x^2+x^4}dx={\int }_0^1\frac{x^2\ln x}{1-x^6}dx-{\int }_0^1\frac{x^4\ln x}{1-x^6}dx$$$$B=\underset{n=0}{\sum ^{\mathrm{\infty }}}{\int }_0^1{x}^2.\ln x.x^{6n}dx-\underset{n=0}{\sum ^{\mathrm{\infty }}}{\int }_0^1{x}^4.\ln x.x^{6n}dx$$$$B=\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{\partial }{\partial a}{\mid }_{a=1}{\int }_0^1{x}^{6n+2}.x^{a-1}dx-\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{\partial }{\partial a}{\mid }_{a=1}{\int }_0^1{x}^{6n+4}.x^{a-1}dx$$$$B=\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{\partial }{\partial a}{\mid }_{a=1}\left(\frac{1}{6n+2+a}\right)-\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{\partial }{\partial a}{\mid }_{a=1}\left(\frac{1}{6n+4+a}\right)$$$$B=-\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{1}{{(6n+3)}^2}+\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{1}{{(6n+5)}^2}=-\frac{1}{36}\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{1}{{(n+\frac{1}{2})}^2}+\frac{1}{36}\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{1}{{(n+\frac{5}{6})}^2}$$$$B=-\frac{1}{36}{\psi }^1\left(\frac{1}{2}\right)+\frac{1}{36}{\psi }^1\left(\frac{5}{6}\right)........\left(2\right)$$$$butI=A-B$$$$I=-\frac{1}{36}{\psi }^1\left(\frac{1}{6}\right)+\frac{1}{36}{\psi }^1\left(\frac{1}{2}\right)-(-\frac{1}{36}{\psi }^1\left(\frac{1}{2}\right)+\frac{1}{36}{\psi }^1\left(\frac{5}{6}\right))$$$$I=\frac{1}{18}{\psi }^1\left(\frac{1}{2}\right)-\frac{1}{36}[{\psi }^1\left(\frac{1}{6}\right)+{\psi }^1(1-\frac{1}{6})]$$$$using{\psi }^1\left(z\right)+{\psi }^1(1-z)=\frac{{\pi }^2}{{\sin }^2\left(\pi z\right)}$$$${\psi }^1\left(\frac{1}{6}\right)+{\psi }^1(1-\frac{1}{6})=\frac{{\pi }^2}{{\left[\sin \left(\frac{\pi }{6}\right)\right]}^2}=4{\pi }^2$$$$I=\frac{1}{18}{\psi }^1\left(\frac{1}{2}\right)-\frac{4{\pi }^2}{36}=\frac{1}{18}\left(\frac{{\pi }^2}{2}\right)-\frac{4{\pi }^2}{36}=-\frac{3{\pi }^2}{36}=-\frac{{\pi }^2}{12}$$$$\because {\int }_0^{\mathrm{\infty }}\frac{\ln x}{1+x^2+x^4}dx=-\frac{{\pi }^2}{12}$$$$b8ynathdave\left(22/09/2020\right)$$

Commented by I want to learn more last updated on 23/Sep/20

$$\mathrm{And}\mathrm{sir},\mathrm{does}\mathrm{the}\mathrm{identity}\mathrm{has}\mathrm{proof}?$$

Commented by I want to learn more last updated on 23/Sep/20

$$\mathrm{Weldone}\mathrm{sir}.$$$$\mathrm{Please}\mathrm{help}\mathrm{me}\mathrm{with}\mathrm{this}{\int }_0^{\mathrm{\infty }}\frac{10\sin \left(\mathrm{x}\right)}{\mathrm{x}}\mathrm{dx}$$$$\mathrm{Also}\mathrm{without}\mathrm{the}\mathrm{limit}.\int \frac{\sin \left(\mathrm{x}\right)}{\mathrm{x}}\mathrm{dx}.$$$$\mathrm{Thanks}\mathrm{sir}.$$

Commented by mathdave last updated on 23/Sep/20

$$note{\int }_0^{\mathrm{\infty }}\frac{\sin \left(ax\right)}{x^n}dx=\frac{\pi a^{n-1}}{2\mathrm{\Gamma }\left(n\right)\sin \left(\frac{\pi n}{2}\right)}$$$${\int }_0^{\mathrm{\infty }}\frac{\sin \left(x\right)}{x}dx=\frac{\pi {\left(1\right)}^{1-1}}{2\mathrm{\Gamma }\left(1\right)\sin \left(\frac{\pi }{2}\right)}=\frac{\pi }{2}$$$$where\mathrm{\Gamma }\left(1\right)=1,a=1andn=1$$$$\because {\int }_0^{\mathrm{\infty }}\frac{\sin \left(x\right)}{x}dx=\frac{\pi }{2}$$

Commented by mathdave last updated on 23/Sep/20

$$ucanusingIBPfordonewithoutd$$$$symmentricbondaries$$

Commented by I want to learn more last updated on 23/Sep/20

$$\mathrm{Thanks}\mathrm{sir},\mathrm{i}\mathrm{appreciate}.$$

Commented by I want to learn more last updated on 23/Sep/20

$$\mathrm{I}\mathrm{will}\mathrm{love}\mathrm{to}\mathrm{see}\mathrm{your}\mathrm{work}\mathrm{sir}$$

Commented by mathdave last updated on 23/Sep/20

$$yahithasproof$$

Commented by I want to learn more last updated on 23/Sep/20

$$\mathrm{Please}\mathrm{sir}\mathrm{help}\mathrm{me}\mathrm{to}\mathrm{prove}\mathrm{it}\mathrm{when}\mathrm{you}\mathrm{are}\mathrm{less}\mathrm{busy}.\mathrm{Thanks}\mathrm{sir}.\mathrm{I}\mathrm{appreciate}.$$

Commented by Tawa11 last updated on 06/Sep/21

$$\mathrm{great}\mathrm{sir}$$

Answered by Bird last updated on 24/Sep/20

$$letA={\int }_0^{\mathrm{\infty }}\frac{lnx}{x^4+x^2+1}dx$$$$weuse{\int }_0^{\mathrm{\infty }}q\left(x\right)ln\left(x\right)dx$$$$=-\frac{1}{2}Re\left(\mathrm{\Sigma }Res(q\left(z\right){ln}^{2}z,a_k)\right)$$$$w\left(z\right)=\frac{{ln}^{2}z}{z^4+z^2+1}poleofw?$$$$u^2+u+1=0(u=z^2)$$$$\mathrm{\Delta }=-3\Rightarrow u_1=\frac{-1+i\sqrt{3}}{2}=e^{i\frac{2\pi }{3}}and$$$$u_2=\frac{-1-i\sqrt{3}}{2}=e^{-\frac{i2\pi }{3}}$$$$w\left(z\right)=\frac{{ln}^{2}z}{(z^2-e^{\frac{i2\pi }{3}})(z^2-e^{-\frac{i2\pi }{3}})}$$$$=\frac{{ln}^{2}z}{(z-e^{\frac{i\pi }{3}})(z+e^{\frac{i\pi }{3}})(z-e^{-\frac{i\pi }{3}})(z+e^{-\frac{i\pi }{3}})}$$$$Res(w,e^{\frac{i\pi }{3}})=\frac{{\left(\frac{i\pi }{3}\right)}^2}{2e^{\frac{i\pi }{3}}\left(2isin\left(\frac{2\pi }{3}\right)\right)}$$$$=\frac{1}{4i}\times \frac{-{\pi }^2}{9}\times \frac{e^{-\frac{i\pi }{3}}}{\frac{\sqrt{3}}{2}}=-\frac{{\pi }^2}{18i\sqrt{3}}.e^{-\frac{i\pi }{3}}$$$$$$$$Res(f,-e^{\frac{i\pi }{3}})=\frac{{(i\pi +\frac{i\pi }{3})}^2}{-2e^{-\frac{i\pi }{3}}\left(2isin\left(\frac{2\pi }{3}\right)\right)}$$$$=\frac{-{\left(\frac{4\pi }{3}\right)}^2}{-4i}\times \frac{e^{\frac{i\pi }{3}}}{\frac{\sqrt{3}}{2}}=\frac{16}{9}{\pi }^2\times \frac{e^{\frac{i\pi }{3}}}{2i\sqrt{3}}$$$$Res(f,e^{-\frac{i\pi }{3}})=\frac{{(-\frac{i\pi }{3})}^2}{(-2isin\left(\frac{2\pi }{3}\right))2e^{-\frac{i\pi }{3}}}$$$$=-\frac{{\pi }^2}{9}\times \frac{e^{i\frac{\pi }{3}}}{-4i\times \frac{\sqrt{3}}{2}}=\frac{{\pi }^2}{18i\sqrt{3}}.e^{\frac{i\pi }{3}}$$$$Res(f,-e^{-\frac{i\pi }{3}})$$$$=\frac{{(i\pi -\frac{i\pi }{3})}^2}{(-2e^{-\frac{i\pi }{3}})(-2isin\left(\frac{2\pi }{3}\right)}$$$$=\frac{-{\left(\frac{2\pi }{3}\right)}^2}{4i\times \frac{\sqrt{3}}{2}}.e^{\frac{i\pi }{3}}=-\frac{4{\pi }^2}{9}\times \frac{e^{\frac{i\pi }{3}}}{2i\sqrt{3}}$$$$\Rightarrow \mathrm{\Sigma }Res\left(f\right)=-\frac{{\pi }^2}{18i\sqrt{3}}{e}^{-\frac{i\pi }{3}}$$$$+\frac{16}{9}{\pi }^2\frac{e^{\frac{i\pi }{3}}}{2i\sqrt{3}}+\frac{{\pi }^2}{18i\sqrt{3}}{e}^{\frac{i\pi }{3}}$$$$-\frac{4{\pi }^2}{18i\sqrt{3}}\times e^{\frac{i\pi }{3}}$$$$=\frac{i{\pi }^2}{18\sqrt{3}}\{\frac{1}{2}-\frac{i\sqrt{3}}{2}\}$$$$-\frac{16i{\pi }^2}{18\sqrt{3}}\{\frac{1}{2}+\frac{i\sqrt{3}}{2}\}-\frac{i{\pi }^2}{18\sqrt{3}}\{\frac{1}{2}+i\frac{\sqrt{3}}{2}\}$$$$+\frac{2{\pi }^{2}i}{9\sqrt{3}}\{\frac{1}{2}+i\frac{\sqrt{3}}{2}\}...becontinued...$$

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