7
$\begingroup$

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 ?

$\endgroup$
4
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.