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Integral of x^x Series Representation

  1. Nov 28, 2008 #1
    1. The problem statement, all variables and given/known data

    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

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

    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...

    [tex]\int_0^1{x^xdx}=\int_0^1{e^{xln(x)}dx}[/tex]

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

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

    [tex](-1)^{n+1}{\int_0^{\infty}e^{-u(n+1)}u^ndu[/tex]

    Here's the problem, I'm supposed to get [tex](-1)^n[/tex] 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 [tex]v=u(n+1),du=dv/(n+1)[/tex] and then I get

    [tex]\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[/tex].

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

    [tex]\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}}[/tex].

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

    [tex]\int_0^1x^xdx=\sum_{k=1}^{\infty}\frac{(-1)^k}{k^k}[/tex]. The problem is that I'm supposed to get [tex](-1)^{k+1}[/tex] in that final sum rather than [tex](-1)^k[/tex].
     
    Last edited: Nov 28, 2008
  2. jcsd
  3. Nov 28, 2008 #2
    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.
     
  4. Nov 28, 2008 #3
    Hrm, that's what I did (edited with full derivation above) and I'm off by one power of -1 at my conclusion.
     
  5. Nov 28, 2008 #4
    Actually, with your substitution we have

    [tex]\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[/tex]
     
  6. Nov 28, 2008 #5
    Argh! I didn't look closely at my bounds...thank you for your assistance Unco.
     
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