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Question Number 142347 by mathmax by abdo last updated on 30/May/21

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

Commented by mathmax by abdo last updated on 31/May/21

$$\mathrm{\Phi }={\int }_0^{\mathrm{\infty }}\frac{{\mathrm{x}}^2\mathrm{logx}}{1+{\mathrm{x}}^6}\mathrm{dx}\mathrm{changement}\mathrm{x}={\mathrm{t}}^{\frac{1}{6}}\mathrm{give}$$$$\mathrm{\Phi }=\frac{1}{6}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{t}}^{\frac{1}{3}}\mathrm{logt}}{1+\mathrm{t}}\frac{1}{6}{\mathrm{t}}^{\frac{1}{6}-1}\mathrm{dt}$$$$=\frac{1}{36}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{t}}^{\frac{1}{3}+\frac{1}{6}-1}}{1+\mathrm{t}}\mathrm{logt}\mathrm{dt}=\frac{1}{36}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{t}}^{-\frac{1}{2}}}{1+\mathrm{t}}\mathrm{logt}\mathrm{dt}$$$$\mathrm{let}\mathrm{f}\left(\mathrm{a}\right)={\int }_0^{\mathrm{\infty }}\frac{{\mathrm{t}}^{\mathrm{a}-1}}{1+\mathrm{t}}\mathrm{dt}\mathrm{we}\mathrm{know}\mathrm{f}\left(\mathrm{a}\right)=\frac{\pi }{\sin \left(\pi \mathrm{a}\right)}(0<\mathrm{a}<1)$$$$\mathrm{and}{\mathrm{f}}^{{}^{\prime }}\left(\mathrm{a}\right)={\int }_0^{\mathrm{\infty }}\frac{\partial }{\partial \mathrm{a}}\left(\frac{{\mathrm{e}}^{(\mathrm{a}-1)\mathrm{logt}}}{1+\mathrm{t}}\right)\mathrm{dt}={\int }_0^{\mathrm{\infty }}\frac{{\mathrm{t}}^{\mathrm{a}-1}}{1+\mathrm{t}}\mathrm{logt}\mathrm{dt}$$$$\mathrm{but}{\mathrm{f}}^{{}^{\prime }}\left(\mathrm{a}\right)=-{\pi }^2\frac{\cos \left(\pi \mathrm{a}\right)}{{\sin }^2\left(\pi \mathrm{a}\right)}$$$$\Rightarrow \mathrm{\Phi }=\frac{1}{36}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{t}}^{\frac{1}{2}-1}}{1+\mathrm{t}}\mathrm{logt}\mathrm{dt}=\frac{1}{36}{\mathrm{f}}^{{}^{\prime }}\left(\frac{1}{2}\right)=\frac{1}{36}(-{\pi }^2).\frac{\cos \left(\frac{\pi }{2}\right)}{{\sin }^2\left(\frac{\pi }{2}\right)}$$$$=0$$

Answered by Dwaipayan Shikari last updated on 31/May/21

$$I\left(a\right)={\int }_0^{\mathrm{\infty }}\frac{x^a}{x^6+1}dx$$$$I\left(a\right)=\frac{1}{6}{\int }_0^{\mathrm{\infty }}\frac{u^{\frac{a}{6}-\frac{5}{6}}}{u+1}du\Rightarrow I\left(a\right)=\frac{\pi }{6sin\left(\frac{a+1}{6}\pi \right)}$$$$I^{\prime }\left(a\right)=-\frac{\pi }{36}cosec\left(\frac{a+1}{6}\pi \right)cot\left(\frac{a+1}{6}\pi \right)={\int }_0^{\mathrm{\infty }}\frac{x^{a}log\left(x\right)}{x^6+1}dx$$$$I^{\prime }\left(2\right)={\int }_0^{\mathrm{\infty }}\frac{x^{2}log\left(x\right)}{x^6+1}dx=0$$

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