Can you get the asymptotics of the following integral?

Question

I am interested in the big $N$ asymptotics of the following integral $$ \int_0^{\infty}dx\,e^{-2xN}\frac{1-(2x)^N}{1-2x} $$ I have considered applying Laplace's method in some way, but I cannot make it work. Can anybody do better than me?

Answer 1

$\dfrac{1-(2x)^N}{1-2x} = \sum_{j=0}^{N-1} (2x)^j$ so your integral is $$ \sum_{j=0}^{N-1} \int_0^\infty dx\; e^{-2xN} (2x)^j = \sum_{j=0}^{N-1} \dfrac{j!}{2N^{1+j}}$$ Write this as $\dfrac{1}{2N}\sum_{j=0}^\infty a_j(N)$, so $a_0(N) = 1$. Note that for each $j$, $a_j(N) \le a_j(j+1) = j!/(j+1)^j$, and $\sum_{j=0}^\infty j!/(j+1)^j < \infty$, so by Dominated convergence $$\lim_{N \to \infty} \sum_{j=0}^\infty a_j(N) = \sum_{j=0}^\infty \lim_{N \to \infty} a_j(N) = 1$$ and your integral is asymptotic to $1/(2N)$.