Derive the expectation of the r-th inverse moment

Question

$X$ is nonnegative, $\phi(t) = E[e^{-tX}]$ is finite for $t \geq 0 $.

Show that for any $r > 0$, $E(\frac{1}{X^r}) = \frac{1}{\Gamma(r)} \int_{0}^{\infty} t^{r-1} \phi(t) dt.$

Thanks in advance.

Answer 1

Hint:

$$ \int_0^t E[e^{-sX}]\,ds = E \int_0^t e^{-sX}\,ds = E\left[ \frac{-e^{-tX}}{X} + \frac{1}{X}\right] $$ as $t\to \infty$ you get...? Next consider integrating $r$ times and using the same approach. Also note that the case where $X=0$ on a set of positive measure should be treated separately (in this case $\phi$ doesn't decay fast enough for the integral to converge).