Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 141133 by mnjuly1970 last updated on 16/May/21

$$$$$$\mathbf{I}:={\int }_0^{\mathrm{\infty }}\frac{(x^n-1)(x-1)}{x^{n+3}-1}dx=??$$$$$$

Answered by mathmax by abdo last updated on 17/May/21

$$\mathrm{at}\mathrm{form}\mathrm{of}\mathrm{series}\mathrm{\Psi }={\int }_0^1\frac{{\mathrm{x}}^{\mathrm{n}+1}-{\mathrm{x}}^{\mathrm{n}}-\mathrm{x}+1}{{\mathrm{x}}^{\mathrm{n}+3}-1}\mathrm{dx}+{\int }_1^{\mathrm{\infty }}\frac{{\mathrm{x}}^{\mathrm{n}+1}-{\mathrm{x}}^{\mathrm{n}}-\mathrm{x}+1}{{\mathrm{x}}^{\mathrm{n}+3}-1}\mathrm{dx}$$$$={\mathrm{\Psi }}_1+{\mathrm{\Psi }}_2$$$${\mathrm{\Psi }}_1=-{\int }_0^1({\mathrm{x}}^{\mathrm{n}+1}-{\mathrm{x}}^{\mathrm{n}}-\mathrm{x}+1)\sum _{\mathrm{k}=0}^{\mathrm{\infty }}{\mathrm{x}}^{\mathrm{k}(\mathrm{n}+3)}$$$$=-\sum _{\mathrm{k}=0}^{\mathrm{\infty }}{\int }_0^1({\mathrm{x}}^{\mathrm{n}+1+\mathrm{k}(\mathrm{n}+3)}-{\mathrm{x}}^{\mathrm{k}(\mathrm{n}+3)+\mathrm{n}}+{\mathrm{x}}^{\mathrm{k}(\mathrm{n}+3)}\mathrm{dx}$$$$=-\sum _{\mathrm{k}=0}^{\mathrm{\infty }}{[\frac{1}{\mathrm{n}+2+(\mathrm{n}+3)\mathrm{k}}{\mathrm{x}}^{\mathrm{n}+2+(\mathrm{n}+3)\mathrm{k}}-\frac{1}{\mathrm{n}+1+(\mathrm{n}+3)\mathrm{k}}{\mathrm{x}}^{\mathrm{n}+1+(\mathrm{n}+3)\mathrm{k}}+\frac{1}{1+(\mathrm{n}+3)\mathrm{k}}{\mathrm{x}}^{1+(\mathrm{n}+3)\mathrm{k}}]}_0^1$$$$=-\sum _{\mathrm{k}=0}^{\mathrm{\infty }}(\frac{1}{\mathrm{n}+2+(\mathrm{n}+3)\mathrm{k}}-\frac{1}{\mathrm{n}+1+(\mathrm{n}+3)\mathrm{k}}+\frac{1}{1+(\mathrm{n}+3)\mathrm{k}})$$$${\mathrm{\Psi }}_2{=}_{\mathrm{x}=\frac{1}{\mathrm{t}}}-{\int }_0^1\frac{\frac{1}{{\mathrm{t}}^{\mathrm{n}+1}}-\frac{1}{{\mathrm{t}}^{\mathrm{n}}}-\frac{1}{\mathrm{t}}+1}{\frac{1}{{\mathrm{t}}^{\mathrm{n}+3}}-1}\times (-\frac{\mathrm{dt}}{{\mathrm{t}}^2})$$$$={\int }_0^1{\mathrm{t}}^{\mathrm{n}+1}\left(\frac{\frac{1}{{\mathrm{t}}^{\mathrm{n}+1}}-\frac{1}{{\mathrm{t}}^{\mathrm{n}}}-\frac{1}{\mathrm{t}}+1}{1-{\mathrm{t}}^{\mathrm{n}+3}}\right)\mathrm{dt}$$$$={\int }_0^1\frac{1-\mathrm{t}-{\mathrm{t}}^{\mathrm{n}}+{\mathrm{t}}^{\mathrm{n}+1}}{1-{\mathrm{t}}^{\mathrm{n}+3}}\mathrm{dt}={\int }_0^1\frac{-{\mathrm{t}}^{\mathrm{n}+1}+{\mathrm{t}}^{\mathrm{n}}+\mathrm{t}-1}{{\mathrm{t}}^{\mathrm{n}+3}-1}\mathrm{dt}\Rightarrow$$$${\mathrm{\Psi }}_1+{\mathrm{\Psi }}_2={\int }_0^1\frac{{\mathrm{x}}^{\mathrm{n}+1}-{\mathrm{x}}^{\mathrm{n}}-\mathrm{x}+1-{\mathrm{x}}^{\mathrm{n}+1}+{\mathrm{x}}^{\mathrm{n}}+\mathrm{x}-1}{{\mathrm{x}}^{\mathrm{n}+3}-1}\mathrm{dx}$$$$=0\Rightarrow \mathrm{\Psi }=0!$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com