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I want to prove that $$\int_0^1x^{-x}dx=\sum_{n=1}^\infty n^{-n}$$ And I know that $$x^{-x}=e^{-x\log x}$$ and $$e^u=\sum_{n=0}^\infty \frac{u^n}{n!}$$ I get that $$x^{-x}=\sum_{n=1}^\infty \frac{(-x\log x)^n}{n!}$$

Then I have no idea how to prove the rest of it. Any suggestion? Thanks!

Nikunj
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J.doe
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1 Answers1

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Exchange the order of integration and summation. Then, make the change of variables $ u = -ln(x)$.

William Krebs
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    To do that one would have to have an integration and a summation in the same term. Where do you see that in the question? – Alex M. Feb 28 '16 at 23:29
  • Do you mean to substitute the right-hand side of the last equation and substitute it into the integral on the left-hand side of the first equation? – Rory Daulton Feb 28 '16 at 23:43