# Convergence and Closed Form of an Integral Involving Bell Numbers

1. Does the following integral converge ?

$$\int_0^\infty \frac{b(x)}{B(x)} dx$$

where

$$b(x) = \sum_{n=1}^\infty \frac{n^x}{n^n} \qquad and \qquad B(x) = \sum_{n=1}^\infty \frac{n^x}{n!}$$

2. Does it possess a closed form, or some other alternative expression ?

## 1 Answer

The question on convergence is certainly not research level. I'm too lazy to give a detailed answer, but here is a direction in which I believe one can obtain a proof with a little effort.

Obviously, all we need is to estimate the asymptotics of $B(x)$ and $b(x)$ for $x \to \infty$. Using a variant of Laplace's method one can show that for large $x$ the value of the sums $b(x)$ and $B(x)$ is asymptotic to the maximal term in these sums, up to a polynomial factor. The maximal term is somewhere around $n(x) \sim \frac{x}{\log x}$, so $b(x) / B(x) \sim {n(x)!} / {n(x)^{n(x)}} \sim e^{-n(x)}$, up to a polynomial factor again. This $e^{-n(x)}$ is approximately $e^{-x / \log x}$, so it decays exponentially at infinity, therefore the integral converges.