According to a paper the equation with integral $\int_{-\infty}^{\infty}dx \rho_0(\lambda)e^{-b\lambda}=1/N$ (#1) where $\rho_0(\lambda)$ is a distribution function, $N$ is a natural number, can be written in other way :
$\sum_{n=1}^{\infty} \frac {(-b)^n} {n!} <\lambda^n>+lnN=0$ (#2)
Here $<\lambda^n>$ is the nth cumulant of the distribution function.
My question is how to transform the initial eq step by step?
The paper notes that 'by the fact that the Fourier transform of $rho_0(\lambda)$ is the exponential of the cumulant generating function [10], we finnd that Eq. implies' #2. However it's not so clear how to get #2.
This is the definition. With $X$ a random variable with pdf $\rho$ let $f(t) = \Bbb{E}[e^{-X t}]= \int_{-\infty}^\infty e^{-xt}\rho(x)dx$ if $\rho$ decays faster than $e^{r |x|}$ then $f$ is analytic around $t=0$ since $f(0)=1$ then $g(t) = \ln f(-t)$ is analytic around $t=0$ that is $g(t) = \sum_{n=1}^\infty \frac{t^n}{n!}g^{(n)}(0)$.
It is useful because if $X,Y$ are independent then $e^{-Xt},e^{-Yt}$ are independent so $\Bbb{E}[e^{-Xt}e^{-Yt}] =\Bbb{E}[e^{-Xt}]\Bbb{E}[e^{-Yt}]$ and $f_{X+Y}(t) = f_X(t)f_Y(t), g_{X+Y}(t) = g_X(t)+g_Y(t)$.
$g^{(n)}(0)$ is called the $n$-th cumulant. Here you are told one value $f(b) = 1/N$ so that $-\ln N =g(-b) = \sum_{n=1}^\infty \frac{(-b)^n}{n!}g^{(n)}(0)$