Question and Answers Forum

All Questions      Topic List

Differentiation Questions

Previous in All Question      Next in All Question      

Previous in Differentiation      Next in Differentiation      

Question Number 205200 by mathlove last updated on 12/Mar/24

$${\int }_1^{\mathrm{\infty }}\frac{x}{x^3+lnx}dx=?$$

Answered by mathzup last updated on 12/Mar/24

$$changementlnx=tgive$$$$I={\int }_0^{\mathrm{\infty }}\frac{e^t}{e^{3t}+t}{e}^{t}dt={\int }_0^{\mathrm{\infty }}\frac{e^{2t}}{e^{3t}+t}dt$$$$={\int }_0^{\mathrm{\infty }}\frac{e^{-t}}{1+{te}^{-3t}}dt={\int }_0^{\mathrm{\infty }}{e}^{-t}\sum _{n=0}^{\mathrm{\infty }}{(-1)}^n{t}^n{e}^{-3nt}dt$$$$=\sum _{n=0}^{\mathrm{\infty }}{(-1)}^n{\int }_0^{\mathrm{\infty }}{t}^n{e}^{-(3n+1)t}dt$$$$=\sum _{n=0}^{\mathrm{\infty }}{(-1)}^n{A}_n$$$$A_n{=}_{(3n+1)t=z}{\int }_0^{\mathrm{\infty }}{\left(\frac{z}{3n+1}\right)}^n{e}^{-z}\frac{dz}{(3n+1)}$$$$=\frac{1}{{(3n+1)}^n}{\int }_0^{\mathrm{\infty }}{z}^n{e}^{-z}dz$$$$=\frac{\mathrm{\Gamma }(n+1)}{{(3n+1)}^n}=\frac{n!}{{(3n+1)}^n}\Rightarrow$$$$I=\sum _{n=0}^{\mathrm{\infty }}{(-1)}^n\frac{n!}{{(3n+1)}^n}$$$$ifyouwantaaproximatevaluego$$$$ton=10inthisserie...$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com