How to calculate ∫01xxdx using series? I read from a book that ∫01xxdx=1−122+133+…+(−1)n1(n+1)n+1+…

Nylah Church

Answered question

2022-01-23

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

Answer & Explanation

Palandriy0

Beginner2022-01-24Added 14 answers

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