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 55223 by peter frank last updated on 19/Feb/19

Commented by maxmathsup by imad last updated on 19/Feb/19

$${\int }_0^1\frac{dx}{x^x}={\int }_0^1{x}^{-x}dx={\int }_0^1{e}^{-xln\left(x\right)}dx={\int }_0^1\left(\sum _{n=0}^{\mathrm{\infty }}\frac{{(-xln\left(x\right))}^n}{n!}\right)dx$$$$=\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^{\mathit{n}}}{\mathit{n}!}{\int }_0^1{x}^n{\left(ln\left(x\right)\right)}^{n}dxlet{A}_n={\int }_0^1{x}^n{\left(lnx\right)}^{n}dx$$$$A_n{=}_{ln\left(x\right)=-t}-{\int }_0^{+\mathrm{\infty }}{(-t)}^n{e}^{-nt}(-e^{-t})dt={(-1)}^n{\int }_0^{+\mathrm{\infty }}{t}^n{e}^{-(n+1)t}dt$$$${=}_{(n+1)t=u}{(-1)}^n{\int }_0^{\mathrm{\infty }}\frac{u^n}{{(n+1)}^n}{e}^{-u}\frac{du}{(n+1)}=\frac{{(-1)}^n}{{(n+1)}^{n+1}}{\int }_0^{\mathrm{\infty }}{u}^n{e}^{-u}du$$$$wehave\mathrm{\Gamma }\left(x\right)={\int }_0^{\mathrm{\infty }}{t}^{x-1}{e}^{-t}dt\Rightarrow \mathrm{\Gamma }(n+1)={\int }_0^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}{t}^n{e}^{-t}dt=n!\Rightarrow$$$$A_n=\frac{{(-1)}^{n}n!}{{(n+1)}^{n+1}}\Rightarrow {\int }_0^1\frac{dx}{x^x}=\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n}{n!}\frac{{(-1)}^{n}n!}{{(n+1)}^{n+1}}$$$$=\sum _{n=0}^{\mathrm{\infty }}\frac{1}{{(n+1)}^{n+1}}=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^n}=\frac{1}{1^1}+\frac{1}{2^2}+\frac{1}{3^3}+....$$

Commented by maxmathsup by imad last updated on 19/Feb/19

$$erroroftyping\mathrm{\Gamma }(n+1)={\int }_0^{\mathrm{\infty }}{t}^n{e}^{-t}dt=n!$$

Commented by peter frank last updated on 19/Feb/19

$$\mathrm{thank}\mathrm{you}$$

Commented by peter frank last updated on 20/Feb/19

$$\mathrm{sorry}\mathrm{sir}\mathrm{how}\mathrm{did}\mathrm{u}\mathrm{get}$$$$\mathrm{that}\mathrm{lnx}=-\mathrm{t}\mathrm{when}\mathrm{x}\to 0$$$$\mathrm{t}\to 0\mathrm{and}\mathrm{x}\to 1\mathrm{t}\to +\propto$$

Commented by maxmathsup by imad last updated on 24/Feb/19

$$nosirwithchangementln\left(x\right)=-tsoifx\to 0^{+}ln\left(x\right)\to -\mathrm{\infty }\Rightarrow -t\to -\mathrm{\infty }\Rightarrow$$$$t\to +\mathrm{\infty }becarrefulsir...$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com