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Question Number 124116 by danielasebhofoh last updated on 30/Nov/20

Answered by Dwaipayan Shikari last updated on 01/Dec/20

$$\int x^{-x}dx$$$$=\underset{n⩾0}{\sum ^{\mathrm{\infty }}}{(-1)}^n\int \frac{{\left(xlogx\right)}^n}{n!}dx$$$$=\underset{n⩾0}{\sum ^{\mathrm{\infty }}}{(-1)}^n\frac{1}{n!}\int x^n{log}^{n}xd3$$$$=\underset{n⩾0}{\sum ^{\mathrm{\infty }}}{(-1)}^{\mathit{n}}\frac{1}{\mathit{n}!}\int e^{(n+1)t}{t}^{n}dtlogx=t$$$$=\underset{n⩾0}{\sum ^{\mathrm{\infty }}}{(-1)}^n\frac{1}{n!{(n+1)}^{n+1}}\int e^u{u}^{n}du(n+1)t=u$$$$=\underset{n⩾0}{\sum ^{\mathrm{\infty }}}{(-1)}^n\frac{1}{n!{(n+1)}^{n+1}}\underset{k⩾0}{\sum ^{\mathrm{\infty }}}\int \frac{u^{n+k}}{k!}du$$$$=\underset{n⩾0}{\sum ^{\mathrm{\infty }}}{(-1)}^n\frac{1}{n!{(n+1)}^{n+1}}\underset{k⩾0}{\sum ^{\mathrm{\infty }}}\frac{u^{n+k+1}}{k!(n+k+1)}$$$$=\underset{n⩾0}{\sum ^{\mathrm{\infty }}}\underset{k⩾0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n{u}^{n+k+1}}{n!k!(n+1)(n+k+1)}$$

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