Sophomore’s dream

The integral

$I:={\displaystyle {\int }_0^1}{x}^x𝑑x$ (1)

may be expanded to a rapidly converging series as follows.

Changing the integrand to a power of $e$ and using the power series expansion of the exponential function gives us

$I={\displaystyle {\int }_0^1}{e}^{x\ln x}𝑑x={\displaystyle {\int }_0^1}{\displaystyle \sum _{n=0}^{\mathrm{\infty }}}{\displaystyle \frac{{(x\ln x)}^n}{n!}}dx.$ (2)

Here the series is uniformly convergent on $[0,1]$ and may be integrated termwise:

$I={\displaystyle \sum _{n=0}^{\mathrm{\infty }}}{\displaystyle {\int }_0^1}{\displaystyle \frac{x^n{(\ln x)}^n}{n!}}𝑑x.$ (3)

The last equation of the parent entry (http://planetmath.org/ExampleOfDifferentiationUnderIntegralSign) then gives in the case $m=n$ from (3) the result

$I={\displaystyle \sum _{n=0}^{\mathrm{\infty }}}{\displaystyle \frac{{(-1)}^n}{{(n+1)}^{n+1}}},$ (4)

i.e.,

${\displaystyle {\int }_0^1}{x}^{x}dx=\mathrm{\hspace{0.33em}1}-{\displaystyle \frac{1}{2^2}}+{\displaystyle \frac{1}{3^3}}-{\displaystyle \frac{1}{4^4}}+-\mathrm{\dots }$ (5)

Cf. the function $x^x$ (http://planetmath.org/FunctionXX).

Since the series (5) satisfies the conditions of Leibniz’ theorem for alternating series (http://planetmath.org/LeibnizEstimateForAlternatingSeries), one may easily estimate the error made when a partial sum of (5) is used for the exact value of the integral $I$.  If one for example takes for $I$ the sum of nine first terms, the first omitted term is $-\frac{1}{{10}^{10}}$; thus the error is negative and its absolute value less than ${10}^{-10}$.

Title	Sophomore’s dream
Canonical name	SophomoresDream
Date of creation	2014-07-20 10:46:23
Last modified on	2014-07-20 10:46:23
Owner	pahio (2872)
Last modified by	pahio (2872)
Numerical id	11
Author	pahio (2872)
Entry type	Derivation
Classification	msc 26A24