Find p where Improper Integral converges

Question

I've had hard time to compute p where

$\int_0^\infty x^{-p} (e^{-x} -1) dx$

converges.

I already proved it diverges when $p \le 0 $ and converges when $ 0<p<1$ (or maybe it's wrong) by comparison test and change of variables.

but I can't compute it when $p>1$

Any Idea for that or complete different idea to solve this prob will be very thankful.

Answer 1

Let $f(x)=x^{-p}(e^{-x}-1)$. $f$ is locally integrable at $(0,+\infty)$ since it continuous. It keeps a constant sign, so we can use the limit comparison test.

near $x=0^+$.

$$f(x)\sim -x^{1-p} \;(x\to 0^+)$$

thus,

$\int_0 f(x)dx$ converges $\iff p-1<1$ .

near $x=+\infty$

$$f(x)\sim -x^{-p} \; (x\to +\infty)$$

thus,

$\int^{+\infty} f(x)dx$ converges $\iff p>1$.

We conclude that $\int_0^{+\infty}f(x)dx$ converges $$\iff 1<p<2.$$