Again from the Shepp and Lloyd paper "ordered cycle lengths in a random permutation", I found this puzzling equality. This one might require access to the paper itself since it's quite a mouthful:


In equation (15), they claimed it is straightforward that if there is an $F\_r$ such that 


$$\int\_0^1 \exp(-y/\xi) dF\_r(\xi) = \int\_y^{\infty} \frac{E(x)^{r-1}}{(r-1)!} \frac{\exp(-E(x) -x)}{x} dx $$


then $F\_r$ will have moments $G\_{r,m}$. 


Here


$$G\_{r,m} = \int\_0^{\infty} \frac{x^{m-1}}{m!} \frac{E(x)^{r-1}}{(r-1)!} \exp(-E(x)-x) dx$$


and 


$$E(x) = \int\_x^{\infty} \frac{e^{-y}}{y} dy$$


which is related to the thread [Reference request for a "well-known identity" in a paper of Shepp and Lloyd](https://mathoverflow.net/questions/19526/reference-request-for-a-well-known-identity-in-a-paper-of-shepp-and-lloyd)


It looks to me like some sort of Laplace transform, but I can't manage to get the algebra to work, because of the inverse exponent $y/\xi$ with respect to $\xi$. 


I will be happy enough if one can tell me why we are looking at the transform $\int\_0^1 \exp(-y/\xi) dF\_r(\xi)$ instead of the usual moment generating function $\int\_0^1 \exp(-y \xi) dF\_r(\xi)$, or maybe it's a typo?



method of moments and Laplace transform from Shepp and Lloyd
This is a general fact: assume that $X$ is a positive random variable and that, for a given nonnegative function $g$, $\displaystyle E(\mathrm{e}^{-y/X})=\int\_y^{\infty}g(x)\mathrm{d}x$ for every positive $y$. Then $\Gamma(s+1)E(X^s)=\displaystyle\int\_0^{\infty}x^sg(x)\mathrm{d}x$ for every positive $s$.


To prove this, integrate the equality $\displaystyle\int\_0^{\infty}\mathrm{e}^{-y/x}y^{s-1}\mathrm{d}y=\Gamma(s)x^s$ over $x>0$ with respect to the distribution of $X$ and change the order of integration in the LHS.


