This is a general fact: assume that $X$ is a positive random variable and that, for a given nonnegative function $g$, $\displaystyle E(\mathrm{e}^{-y/X})=\int_y^{\infty}g(x)dx$ E(\mathrm{e}^{-y/X})=\int_y^{\infty}g(x)\mathrm{d}x$for every positive$y$. Then$E(X^k)=\displaystyle\int_0^{\infty}\frac{x^k}{k!}g(x)dx$\Gamma(s+1)E(X^s)=\displaystyle\int_0^{\infty}x^sg(x)\mathrm{d}x$ for every positive $k\ge1$.s$. To prove this, integrate the relation equality$\displaystyle\int_0^{\infty}\mathrm{e}^{-y/x}y^{k-1}dy=(k-1)!x^{k}$\displaystyle\int_0^{\infty}\mathrm{e}^{-y/x}y^{s-1}\mathrm{d}y=\Gamma(s)x^s$ over $x>0$ with respect to the distribution of $X$ and change the order of integration in the LHS.
This is a general fact: assume that $X$ is a positive random variable and that, for a given nonnegative function $g$, $\displaystyle E(\mathrm{e}^{-y/X})=\int_y^{\infty}g(x)dx$ for every positive $y$. Then $E(X^k)=\displaystyle\int_0^{\infty}\frac{x^k}{k!}g(x)dx$ for every $k\ge1$.
To prove this, integrate the relation $\displaystyle\int_0^{\infty}\mathrm{e}^{-y/x}y^{k-1}dy=(k-1)!x^{k}$ over $x>0$ with respect to the distribution of $X$ and change the order of integration in the LHS.