# Sophomore’s dream

The integral

 $\displaystyle I:=\int_{0}^{1}x^{x}\,dx$ (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

 $\displaystyle I=\int_{0}^{1}e^{x\ln x}\,dx=\int_{0}^{1}\sum_{n=0}^{\infty}% \frac{(x\ln x)^{n}}{n!}\,dx.$ (2)

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

 $\displaystyle I=\sum_{n=0}^{\infty}\int_{0}^{1}\frac{x^{n}(\ln x)^{n}}{n!}dx.$ (3)

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

 $\displaystyle I=\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(n\!+\!1)^{n+1}},$ (4)

i.e.,

 $\displaystyle\int_{0}^{1}x^{x}\,dx\;=\;1-\frac{1}{2^{2}}+\frac{1}{3^{3}}-\frac% {1}{4^{4}}+-\ldots$ (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 SophomoresDream 2014-07-20 10:46:23 2014-07-20 10:46:23 pahio (2872) pahio (2872) 11 pahio (2872) Derivation msc 26A24