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

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?