Integral of x^x Series Representation

[Cistra]

Homework Statement

This has been driving me insane, and I'm sure it's something mind-boggling obvious but I can't seem to find it. I'll go through the work through here, I'm trying to prove that

\int_0^1{x^xdx}=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n^n}.

2. The attempt at a solution

Basically I'm having a negative somewhere that I'm not accounting for in my derivation of this. I'll just go through what I do step by step here until where I'm positive my error occurs...

\int_0^1{x^xdx}=\int_0^1{e^{xln(x)}dx}

e^{xln(x)}dx=\sum_{n=0}^{\infty}\frac{1}{n!}x^nln^n(x)

\sum_{n=0}^{\infty}\frac{1}{n!}\int_0^1x^nln^n(x)dx and then I made the substitution u=-ln(x) or e^{-u}=x,dx=-e^{-u}du and I get after changing the bounds of integration

(-1)^{n+1}{\int_0^{\infty}e^{-u(n+1)}u^ndu

Here's the problem, I'm supposed to get (-1)^n as the constant in the integrand at this point but I can't get that extra -1 to go away. Ignoring that constant the rest of the proof goes out without a hitch, just using the Gamma function etc. but I can't make this silly negative go away! Can anyone see my mistake? =(

Thank you!

*Edit: The rest of the derivation from the last step. I made another substitution v=u(n+1),du=dv/(n+1) and then I get

\frac{(-1)^{n+1}}{n+1}\int_0^{\infty}e^{-v}\left(\frac{v}{n+1}\right)^ndv=\frac{(-1)^{n+1}}{(n+1)^{n+1}}\int_0^{\infty}e^{-v}v^{n}dv.

That last formula is the one for \Gamma(n+1)=n! so then I get after substituting back into the sum,

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

Now I do make that substitution into the sum, setting n+1=k I get

\int_0^1x^xdx=\sum_{k=1}^{\infty}\frac{(-1)^k}{k^k}. The problem is that I'm supposed to get (-1)^{k+1} in that final sum rather than (-1)^k.

 

[Unco]

You'll kick yourself, Cistra: your sum is from n=0, the required sum is from k=1 (replacing their n with k), so at the end just put k=n+1.

 

[Cistra]

You'll kick yourself, Cistra: your sum is from n=0, the required sum is from k=1 (replacing their n with k), so at the end just put k=n+1.

Hrm, that's what I did (edited with full derivation above) and I'm off by one power of -1 at my conclusion.

 

[Unco]

\sum_{n=0}^{\infty}\frac{1}{n!}\int_0^1x^nln^n(x)dx and then I made the substitution u=-ln(x) or e^{-u}=x,dx=-e^{-u}du and I get after changing the bounds of integration

(-1)^{n+1}{\int_0^{\infty}e^{-u(n+1)}u^ndu

Actually, with your substitution we have

\int_0^1x^nln^n(x)dx = \int_{\infty}^{0}e^{-nu}(-u)^n(-e^{-u}\, du) \,\, = \, \, -(-1)^n\int_{\infty}^{0}e^{-u(n+1)}u^n\, du \, \, = \, \, (-1)^n\int_{0}^{\infty}e^{-u(n+1)}u^n\, du

 

[Cistra]

Argh! I didn't look closely at my bounds...thank you for your assistance Unco.