Nylah Church

2022-01-23

How to calculate ${\int }_{0}^{1}{x}^{x}dx$ using series? I read from a book that
${\int }_{0}^{1}{x}^{x}dx=1-\frac{1}{{2}^{2}}+\frac{1}{{3}^{3}}+\dots +{\left(-1\right)}^{n}\frac{1}{{\left(n+1\right)}^{n+1}}+\dots$

Palandriy0

Just write
${x}^{x}={e}^{x\mathrm{ln}x}=\sum _{n=0}^{\mathrm{\infty }}\frac{{\left(x\mathrm{ln}x\right)}^{n}}{n!}$
and use that
${\int }_{0}^{1}{\left(x\mathrm{ln}x\right)}^{n}dx={\left(-1\right)}^{n}{\int }_{0}^{\mathrm{\infty }}{y}^{n}{e}^{-\left(n+1\right)y}dy$,
To show the last formula, make the change of variables $x={e}^{-y}$ so that
${\int }_{0}^{1}{\left(x\mathrm{ln}x\right)}^{n}dx={\left(-1\right)}^{n}{\int }_{0}^{\mathrm{\infty }}{y}^{n}{e}^{-\left(n+1\right)y}du$,
which is clearly expressible in terms of the Gamma function.

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