Proof: $X\ge 0, r>0\Rightarrow E(X^r)=r\int_0^{\infty}x^{r-1}P(X>x)dx$

As the title states, the problem at hand is proving the following:

$X\ge 0, r>0\Rightarrow E(X^r)=r\int_0^{\infty}x^{r-1}P(X>x)dx$

Attempt/thoughts on a solution

I am guessing this is an application of Fubini's Theorem, but wouldn't that require writing $P(X>x)$ as an expectation? If so, how is this accomplished?

Thoughts and help are appreciated.

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It is $\int_0^\infty \Pr(X^r \gt x)\,dx$, which is $\int_0^\infty \Pr(X \gt x^{1/r})\,dx$. Make the change of variable $u^r=x$. –  André Nicolas Aug 4 '12 at 21:06
@Justin: I'm quite curious about the reference of this exercise. Could you tell me in which book you encountered it? –  Jack Oct 15 '12 at 22:59

Proof: Consider the expectation of the identity $$X^r=r\int_0^{X}x^{r-1}\,\mathrm dx=r\int_0^{+\infty}x^{r-1}\mathbf 1_{X>x}\,\mathrm dx.$$