Integrate x^x

laurenc89 · New Member Offline Joined: 01 Nov 2009 Posts: 1

Can somebody help me express the integral of x^x, using the taylor series? I know that you can not express x^x using elementary functions.

**Posted:** Sun Nov 01, 2009 11:55 am

jmerry

The Taylor series of $x^x$ ? Not practical. The series expansion you might use comes from writing $x^x=e^{x\ln x}$ , and using the series for the exponential: $x^x=1+x\ln x+\frac{x^2\ln^2 x}{2}+\frac{x^3\ln^3 x}{6}+\cdots$

Next comes integrating those $x^k\ln^k x$ terms, which is repeated integration by parts. $\frac1{k!}\int x^k\ln^k x\,dx=\frac1{k!}\frac{x^{k+1}}{k+1}\ln^k x-\frac1{k!}\int \frac{x^{k+1}}{k+1}\cdot\frac{k\ln^{k-1}x}{...$
$\cdots=x^{k+1}\left(\frac{\ln^k x}{k!\cdot (k+1)}-\frac{\ln^{k-1} x}{(k-1)!\cdot (k+1)^2}+\cdots +(-1)^k\frac1{(k+1)^{k+1}}\r...$

It's a horrible mess. In the one special case $\int_0^1 x^x\,dx$ , all of the terms with logarithms go away and you get the nice series expression $\int_0^1 x^x\,dx=1-2^{-2}+3^{-3}-4^{-4}+\cdots$ .

**Posted:** Sun Nov 01, 2009 12:29 pm

AndrewTom

I think I posted this already but can't find it:

$x^{x} = e^{x \ln x} = 1 + (x \ln x) + \frac{(x \ln x)^{2}}{2!} + \frac{(x \ln x)^{3}}{3!} + ...$

Integrating by parts, with $u=\ln ^{n} x, dv = x^{n}\ dx$ ,

$\int_{0}^{1} x^{n} \ln ^{n} x\ dx = [\frac{(x^{n} + 1) \ln^{n} x}{(n+1)}]_{0}^{1} - \frac {n}{(n+1)}\int_{0}^{1} x^{n} \ln^{n...$

Repeating this process, we eventually get to

$\frac{(-1)^{n} n!}{(n+1)^{n+1}} \int_{0}^{1} x^{n}\ dx$

$=\frac{(-1)^{n} n!}{(n+1)^{n+1}}$

Therefore,

$\int_{0}^{1} x^{x} \dx = \sum_{n=0}^{\infty} \frac{(-1)^{n}n!}{(n+1)^{n+1} n!}$

$= \sum_{0}^{\infty} \frac{(-1)^{n}}{(n+1)^{n+1}}$

$= 1 - \frac{1}{2^{2}} + \frac{1}{3^{3}} - \frac{1}{4^{4}} + ...$

**Posted:** Mon Nov 02, 2009 12:18 am