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Question Number 103343 by Dwaipayan Shikari last updated on 14/Jul/20

$${\int }_0^1{x}^{-x}dx$$

Answered by mathmax by abdo last updated on 15/Jul/20

$$\mathrm{at}\mathrm{form}\mathrm{of}\mathrm{serie}$$$$\mathrm{I}={\int }_0^1{\mathrm{x}}^{-\mathrm{x}}\mathrm{dx}={\int }_0^1{\mathrm{e}}^{-\mathrm{xlnx}}\mathrm{dx}={\int }_0^1\left(\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{(-\mathrm{xlnx})}^{\mathrm{n}}}{\mathrm{n}!}\right)\mathrm{dx}$$$$=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{\mathrm{n}!}{\int }_0^1{\mathrm{x}}^{\mathrm{n}}{\left(\mathrm{lnx}\right)}^{\mathrm{n}}\mathrm{dx}\mathrm{let}{\mathrm{A}}_{\mathrm{n}}={\int }_0^1{\mathrm{x}}^{\mathrm{n}}{\left(\mathrm{lnx}\right)}^{\mathrm{n}}\mathrm{dx}$$$$\mathrm{lnx}=-\mathrm{t}\Rightarrow {\mathrm{A}}_{\mathrm{n}}=-{\int }_0^{\mathrm{\infty }}{\left({\mathrm{e}}^{-\mathrm{t}}\right)}^{\mathrm{n}}{(-\mathrm{t})}^{\mathrm{n}}(-{\mathrm{e}}^{-\mathrm{t}})\mathrm{dt}$$$$={(-1)}^{\mathrm{n}}{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-(\mathrm{n}+1)\mathrm{t}}{\mathrm{t}}^{\mathrm{n}}\mathrm{dt}{=}_{(\mathrm{n}+1)\mathrm{t}=\mathrm{u}}{(-1)}^{\mathrm{n}}{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-\mathrm{u}}\frac{{\mathrm{u}}^{\mathrm{n}}}{{(\mathrm{n}+1)}^{\mathrm{n}}}\times \frac{\mathrm{du}}{\mathrm{n}+1}$$$$=\frac{{(-1)}^{\mathrm{n}}}{{(\mathrm{n}+1)}^{\mathrm{n}+1}}{\int }_0^{\mathrm{\infty }}{\mathrm{u}}^{\mathrm{n}}{\mathrm{e}}^{-\mathrm{u}}\mathrm{du}=\frac{{(-1)}^{\mathrm{n}}\mathrm{\Gamma }(\mathrm{n}+1)}{{(\mathrm{n}+1)}^{\mathrm{n}+1}}=\frac{{(-1)}^{\mathrm{n}}\mathrm{n}!}{{(\mathrm{n}+1)}^{\mathrm{n}+1}}\Rightarrow$$$$\mathrm{I}=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{\mathrm{n}!}\times \frac{{(-1)}^{\mathrm{n}}\mathrm{n}!}{{(\mathrm{n}+1)}^{\mathrm{n}+1}}=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{1}{{(\mathrm{n}+1)}^{\mathrm{n}+1}}=\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{1}{{\mathrm{n}}^{\mathrm{n}}}$$$$=\frac{1}{1^1}+\frac{1}{2^2}+\frac{1}{3^3}+....$$

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