convergence of $\int_{x=0}^{\infty}x^{-\frac{M-1}{2}-N}(1-e^{-x})^{M-1}e^{-x}dx$

Question

I am trying to study the convergence of $$\int_{x=0}^{\infty}x^{-\frac{M-1}{2}-N}(1-e^{-x})^{M-1}e^{-x}dx,$$ where $M$ and $N$ are positive integers.

I've tried some $M$ and $N$, and it seems that when $M>N$ the integral converges, o.w. it does not.

But there is a way to prove or dis prove it?

Thanks.

Answer 1

Here's a start.

Around zero, $1-e^{-x} =1-(1-x+x^2/2-x^3/6+...) =x-x^2/2+x^3/6-... =x(1-x/2+x^2/6-...) $ so the integral up to $c$, where $c$ is small, is about

$\begin{array}\\ \int_{0}^{c}x^{-\frac{m-1}{2}-n}(1-e^{-x})^{m-1}e^{-x}dx &=\int_{0}^{c}x^{-\frac{m-1}{2}-n}(x(1-x/2+x^2/6-...))^{m-1}(1-x)dx\\ &\approx \int_{0}^{c}x^{-\frac{m-1}{2}-n}x^{m-1}(1-x/2+x^2/6-...))^{m-1}(1-x)dx\\ &\approx \int_{0}^{c}x^{\frac{m-1}{2}-n}(1-(m-1)x/2)(1-x)dx\\ &\approx \int_{0}^{c}x^{\frac{m-1}{2}-n}dx\\ &= \frac{x^{\frac{m-1}{2}-n+1}}{\frac{m-1}{2}-n+1}\big|_{0}^{c}\\ &= \frac{x^{\frac{m+1}{2}-n}}{\frac{m+1}{2}-n}\big|_{0}^{c}\\ \end{array} $

For this to converge, we must have $\frac{m+1}{2}-n > 0 $.