4 added 220 characters in body

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 http://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?

3 deleted 1 characters in body

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

$$4E(x) E(x) = \int_x^{\infty} \frac{e^{-y}}{y} dy$$

which is related to the thread http://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 $\xi$.

2 Fixed LaTeX.

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

\begin{align*} \int_0^1

\int_0^1 \exp(-y/\xi) dF_r(\xi) = \int_y^{\infty} (E(x))^{r-1}/(r-1)! \exp(-E(x) frac{E(x)^{r-1}}{(r-1)!} \frac{\exp(-E(x) -x)/x x)}{x} dx \end{align*}

then $F_r$ will have moments $G_{r,m}$.

Here

\begin{align*} G_{r,m}

G_{r,m} = \int_0^{\infty} x^{m-1}/m! (E(x))^{r-1}/(r-1)! \exp(-E(x)-x)dx frac{x^{m-1}}{m!} \end{align*} and frac{E(x)^{r-1}}{(r-1)!} \begin{align*} E(x) exp(-E(x)-x) dx

and

4E(x) = \int_x^{\infty} e^{-y}/y dy \end{align*}frac{e^{-y}}{y} dy

which is related to the thread http://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 $\xi$.

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