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Question Number 130136 by mathmax by abdo last updated on 22/Jan/21

$$\mathrm{calculate}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{x}}^2\mathrm{lnx}}{1+{\mathrm{x}}^6}\mathrm{dx}$$

Answered by Dwaipayan Shikari last updated on 22/Jan/21

$$I^{\prime }\left(2\right)=\frac{-{\pi }^2}{6^2}.\frac{cos\left(\frac{2+1}{6}\pi \right)}{{sin}^2\left(\frac{2+1}{6}\pi \right)}=0\left(ProvedEarlier\right)$$

Answered by Ar Brandon last updated on 22/Jan/21

$$\mathcal{I}={\int }_0^1\frac{{\mathrm{x}}^2\mathrm{lnx}}{1+{\mathrm{x}}^6}\mathrm{dx}+{\int }_1^{\mathrm{\infty }}\frac{{\mathrm{x}}^2\mathrm{lnx}}{1+{\mathrm{x}}^6}\mathrm{dx}$$$$={\int }_0^1\frac{{\mathrm{x}}^2\mathrm{lnx}}{1+{\mathrm{x}}^6}\mathrm{dx}-{\int }_0^1\frac{{\mathrm{t}}^2\mathrm{lnt}}{1+{\mathrm{t}}^2}\mathrm{dt}=0$$

Answered by mathmax by abdo last updated on 23/Jan/21

$$\mathrm{I}={\int }_0^{\mathrm{\infty }}\frac{{\mathrm{x}}^2\mathrm{lnx}}{1+{\mathrm{x}}^6}\mathrm{dx}\mathrm{we}\mathrm{do}\mathrm{the}\mathrm{changement}{\mathrm{x}}^6=\mathrm{t}\Rightarrow$$$$\mathrm{I}=\frac{1}{6}{\int }_0^{\mathrm{\infty }}\frac{{\left({\mathrm{t}}^{\frac{1}{6}}\right)}^2\ln \left({\mathrm{t}}^{\frac{1}{6}}\right)}{1+\mathrm{t}}{\mathrm{t}}^{\frac{1}{6}-1}\mathrm{dt}=\frac{1}{36}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{t}}^{\frac{1}{3}+\frac{1}{6}-1}\ln \left(\mathrm{t}\right)}{1+\mathrm{t}}\mathrm{dt}$$$$=\frac{1}{36}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{t}}^{-\frac{1}{2}}\ln \left(\mathrm{t}\right)}{1+\mathrm{t}}\mathrm{dt}\mathrm{but}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{t}}^{\mathrm{a}-1}\mathrm{lnt}}{1+\mathrm{t}}\mathrm{dt}=-\frac{{\pi }^2\cos \left(\pi \mathrm{a}\right)}{{\sin }^2\left(\pi \mathrm{a}\right)}\Rightarrow$$$${\int }_0^{\mathrm{\infty }}\frac{{\mathrm{t}}^{-\frac{1}{2}}\ln \left(\mathrm{t}\right)}{1+\mathrm{t}}\mathrm{dt}=-\frac{{\pi }^2\cos \left(\frac{\pi }{2}\right)}{{\sin }^2\left(\frac{\pi }{2}\right)}=0\Rightarrow \mathrm{I}=0$$

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