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I have been given a function f(x) defined on the interval $$[0,\infty] $$ by the formula $$ \begin{array}{cc} \ f(x)= \{ & \begin{array}{cc}\ 0 & x= 0 \\ \ x\space log(x) & x≠0 \end{array} \end{array} $$ I have argumented for is continues on the interval on the interval $$[0,\infty] $$ and that $$ exp(-(f(x)) = x^{-x} for \space x>0 $$ also I have shown f is not differentiable in x = 0

how do I show $$ \int_{0}^{1} exp(-f(x))dx=\sum_{n=0}^{\infty} \frac{(-1)^n}{n!} \int_{0}^{\infty} f(x)^n dx $$

in advance, thank you

Aditya Dwivedi
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Signe Winkel
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1 Answers1

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Since

$\exp(x)=\sum_{n=0}^\infty \frac{x^n}{n!} $

given a function f with real codomain, we have

$\exp(-f(x))=\sum_{n=0}^\infty \frac{(-f(x))^n}{n!} = \sum_{n=0}^\infty \frac{(-1)^n(f(x))^n}{n!} $

Therefore we have for a continous function f that

$\int_0^1\exp(-f(x))dx = \int_0^1\sum_{n=0}^\infty \frac{(-1)^n(f(x))^n}{n!}dx$

Edit: The reason we can swap integral and sum is not because of uniform convergence, but because of Fubini's theorem, as discussed in the comment.

Richard Jensen
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  • How do we know that it converges uniformly on the interval [0,1]? – Benjamin Bech Jun 17 '20 at 19:22
  • Huh, you raise a fair point. I was sure it was obvious, but now I'm not even 100% sure that it is true. The limit is continuous, so I guess one can make appropriate bound to prove it.

    But it turns out that it doesn't matter, since the swap of sum and integral is Fubini's theorem with Lebesgue and counting measure. My bad!

     – Richard Jensen Jun 18 '20 at 09:10