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Question Number 131547 by liberty last updated on 06/Feb/21

$$\mathrm{J}={\int }_0^{\mathrm{\infty }}\frac{{\mathrm{x}}^8-1}{{\mathrm{x}}^{10}+1}\mathrm{dx}?$$

Answered by rs4089 last updated on 06/Feb/21

$$0$$

Answered by liberty last updated on 06/Feb/21

$$\mathrm{consider}\mathrm{I}={\int }_0^{\mathrm{\infty }}\frac{{\mathrm{x}}^8}{{\mathrm{x}}^{10}+1}\mathrm{dx}$$$$\mathrm{replace}\mathrm{x}\mathrm{by}\frac{1}{\mathrm{x}}$$$$\mathrm{I}={\int }_{\mathrm{\infty }}^0\frac{{\left(\frac{1}{\mathrm{x}}\right)}^8}{{\left(\frac{1}{\mathrm{x}}\right)}^{10}+1}.(-\frac{1}{{\mathrm{x}}^2})\mathrm{dx}$$$$\mathrm{I}=-{\int }_{\mathrm{\infty }}^0\frac{{\mathrm{x}}^2}{1+{\mathrm{x}}^{10}}.\left(\frac{\mathrm{dx}}{{\mathrm{x}}^2}\right)={\int }_0^{\mathrm{\infty }}\frac{\mathrm{dx}}{{\mathrm{x}}^{10}+1}$$$$\mathrm{so}\mathrm{we}\mathrm{have}\mathrm{J}={\int }_0^{\mathrm{\infty }}\frac{{\mathrm{x}}^8}{{\mathrm{x}}^{10}+1}\mathrm{dx}-{\int }_0^{\mathrm{\infty }}\frac{\mathrm{dx}}{{\mathrm{x}}^{10}+1}$$$$\mathrm{J}=\mathrm{I}-\mathrm{I}=0$$$$$$

Answered by mathmax by abdo last updated on 06/Feb/21

$$\mathrm{J}={\int }_0^{\mathrm{\infty }}\frac{{\mathrm{x}}^8}{1+{\mathrm{x}}^{10}}\mathrm{dx}-{\int }_0^{\mathrm{\infty }}\frac{\mathrm{dx}}{1+{\mathrm{x}}^{10}}\mathrm{but}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{x}}^8}{1+{\mathrm{x}}^{10}}\mathrm{dx}{=}_{{\mathrm{x}}^{10}=\mathrm{t}}\frac{1}{10}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{t}}^{\frac{8}{10}}}{1+\mathrm{t}}{\mathrm{t}}^{\frac{1}{10}-1}$$$$=\frac{1}{10}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{t}}^{\frac{9}{10}-1}}{1+\mathrm{t}}\mathrm{dt}=\frac{1}{10}.\frac{\pi }{\sin \left(\frac{9\pi }{10}\right)}\mathrm{also}{\int }_0^{\mathrm{\infty }}\frac{\mathrm{dx}}{1+{\mathrm{x}}^{10}}{=}_{{\mathrm{x}}^{10}=\mathrm{t}}\frac{1}{10}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{t}}^{\frac{1}{10}-1}}{1+\mathrm{t}}\mathrm{dt}$$$$=\frac{1}{10}.\frac{\pi }{\sin \left(\frac{\pi }{10}\right)}\Rightarrow \mathrm{J}=\frac{\pi }{10}\{\frac{1}{\sin \left(\frac{9\pi }{10}\right)}-\frac{1}{\sin \left(\frac{\pi }{10}\right)}\}\mathrm{but}\sin \left(\frac{9\pi }{10}\right)=\sin \left(\frac{\pi }{10}\right)\Rightarrow$$$$\mathrm{J}=0$$

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