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Question Number 138159 by BHOOPENDRA last updated on 10/Apr/21

Answered by Dwaipayan Shikari last updated on 10/Apr/21

$${\int }_0^1{x}^{-x}dx=\mathrm{\aleph }$$$$={\int }_0^1{e}^{-xlog\left(x\right)}dx=\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n}{n!}{\int }_0^1{x}^n{log}^n\left(x\right)dx=\mathrm{\aleph }$$$${\int }_0^1{x}^{n}dx=\frac{1}{n+1}\Rightarrow {\int }_0^1{x}^n{log}^m\left(x\right)dx=\frac{{\partial }^m}{\partial n^m}\left(\frac{1}{n+1}\right)=\frac{m!{(-1)}^m}{{(n+1)}^{m+1}}$$$$Butherem=n$$$$so\mathrm{\aleph }=\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n}{n!}.\frac{{(-1)}^{n}n!}{{(n+1)}^{n+1}}=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n^n}$$$$$$

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