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Question Number 143545 by mathmax by abdo last updated on 15/Jun/21

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

Answered by Pagnol last updated on 15/Jun/21

$$\mathrm{I}={\int }_0^{\mathrm{\infty }}\frac{\mathrm{logx}}{{\mathrm{x}}^6+1}\mathrm{dx},\mathrm{u}={\mathrm{x}}^6\Rightarrow \mathrm{x}={\mathrm{u}}^{\frac{1}{6}}\Rightarrow \mathrm{dx}=\frac{1}{6}{\mathrm{u}}^{-\frac{5}{6}}\mathrm{du}$$$$=\frac{1}{36}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{u}}^{-\frac{5}{6}}\mathrm{logu}}{\mathrm{u}+1}\mathrm{du}$$$$\mathrm{I}\left(\alpha \right)={\int }_0^{\mathrm{\infty }}\frac{{\mathrm{u}}^{\alpha }}{\mathrm{u}+1}\mathrm{du}=\beta (\alpha +1,-\alpha )=\frac{\mathrm{\Gamma }(\alpha +1)\mathrm{\Gamma }(-\alpha )}{\mathrm{\Gamma }\left(1\right)}$$$${\mathrm{I}}^{\prime }\left(\alpha \right)={\int }_0^{\mathrm{\infty }}\frac{{\mathrm{u}}^{\alpha }\mathrm{lnu}}{\mathrm{u}+1}\mathrm{du}=-\mathrm{\Gamma }(\alpha +1){\mathrm{\Gamma }}^{\prime }(-\alpha )+{\mathrm{\Gamma }}^{\prime }(\alpha +1)\mathrm{\Gamma }(-\alpha )$$$${\mathrm{I}}^{\prime }(-\frac{5}{6})=-\mathrm{\Gamma }\left(\frac{1}{6}\right){\mathrm{\Gamma }}^{\prime }\left(\frac{5}{6}\right)+{\mathrm{\Gamma }}^{\prime }\left(\frac{1}{6}\right)\mathrm{\Gamma }\left(\frac{5}{6}\right)$$$$=\mathrm{\Gamma }\left(\frac{1}{6}\right)\mathrm{\Gamma }\left(\frac{5}{6}\right)[\psi \left(\frac{1}{6}\right)-\psi \left(\frac{5}{6}\right)]$$$$=\frac{\pi }{\sin \left(\frac{\pi }{6}\right)}[-\pi \cot \left(\frac{5}{6}\pi \right)]=2\pi \left(\pi \sqrt{3}\right)=2\sqrt{3}{\pi }^2$$$$\mathrm{I}=\frac{1}{36}{\mathrm{I}}^{\prime }(-\frac{5}{6})=\frac{\sqrt{3}}{18}{\pi }^2$$

Answered by mathmax by abdo last updated on 15/Jun/21

$$\mathrm{\Phi }={\int }_0^{\mathrm{\infty }}\frac{\mathrm{logx}}{{\mathrm{x}}^6+1}\mathrm{dx}\mathrm{changement}{\mathrm{x}}^6=\mathrm{t}\mathrm{give}$$$$\mathrm{\Phi }=\frac{1}{6}{\int }_0^{\mathrm{\infty }}\frac{\log \left({\mathrm{t}}^{\frac{1}{6}}\right)}{1+\mathrm{t}}{\mathrm{t}}^{\frac{1}{6}-1}\mathrm{dt}=\frac{1}{36}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{t}}^{\frac{1}{6}-1}\mathrm{logt}}{1+\mathrm{t}}\mathrm{dt}$$$$\mathrm{let}\mathrm{f}\left(\mathrm{a}\right)={\int }_0^{\mathrm{\infty }}\frac{{\mathrm{t}}^{\mathrm{a}-1}}{1+\mathrm{t}}\mathrm{dt}\Rightarrow {\mathrm{f}}^{{}^{\prime }}\left(\mathrm{a}\right)={\int }_0^{\mathrm{\infty }}\frac{{\mathrm{t}}^{\mathrm{a}-1}\mathrm{logt}}{1+\mathrm{t}}\mathrm{dt}\Rightarrow {\mathrm{f}}^{{}^{\prime }}\left(\frac{1}{6}\right)=36\mathrm{\Phi }\Rightarrow$$$$\mathrm{\Phi }=\frac{1}{36}{\mathrm{f}}^{{}^{\prime }}\left(\frac{1}{6}\right)\mathrm{we}\mathrm{know}\mathrm{f}\left(\mathrm{a}\right)=\frac{\pi }{\sin \left(\pi \mathrm{a}\right)}\Rightarrow {\mathrm{f}}^{{}^{\prime }}\left(\mathrm{a}\right)=-\frac{{\pi }^2\cos \left(\pi \mathrm{a}\right)}{{\sin }^2\left(\pi \mathrm{a}\right)}\Rightarrow$$$${\mathrm{f}}^{{}^{\prime }}\left(\frac{1}{6}\right)=-\frac{{\pi }^2\times \frac{\sqrt{3}}{2}}{{\left(\frac{1}{2}\right)}^2}=-4{\pi }^2.\frac{\sqrt{3}}{2}=-2{\pi }^2\sqrt{3}\Rightarrow$$$$\mathrm{\Phi }=\frac{1}{36}(-2{\pi }^2\sqrt{3})=-\frac{{\pi }^2}{18}\sqrt{3}$$

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