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 theexponential 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 beintegrated 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 ofLeibniz’ 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 firstterms, the first omitted term is $-\frac{1}{{10}^{10}}$; thus theerror is negative and its absolute value less than ${10}^{-10}$.