Convergence of $\int_{0}^{\infty} \frac{e^{-x}-1}{x^p} dx$

Question

Can anybody suggest for what values of $p$,

$\int_{0}^{\infty}\frac{e^{-x}-1}{x^{p}}dx$

converge ?

I have tried so far $\int_{0}^{\infty}\frac{e^{-x}-1}{x^p}dx=\int_{0}^{1}\frac{e^{-x}-1}{x^p}dx+\int_{1}^{\infty}\frac{e^{-x}-1}{x^p}dx$

For $\int_{0}^{1}\frac{e^{-x}-1}{x^p}dx$ I apply limit comparison test $\lim_{x\to 0}\frac{\frac{e^{-x}-1}{x^p}}{\frac{1}{x^{p-1}}}=\lim_{x\to 0}\frac{e^{-x}-1}{x} =-1$ hence it $\int_{0}^{1}\frac{e^{-x}-1}{x^p}dx$ converge iff $p<2$

For $\int_{1}^{\infty}\frac{e^{-x}-1}{x^p}dx$ i have applied again limit comparison test I found it converge iff $p>1$ by taking denominator function $\frac{1}{x^{p}}$

so the $\int_{0}^{\infty}\frac{e^{-x}-1}{x^p}dx$ converge iff $p\in [1,2]$

is it alright?

Answer 1

You can start by writing $$-\int_{0}^{\infty} \frac{1-e^{-x}}{x}x^{1-p} dx=-\int_{0}^{\infty} F(x)x^{1-p} dx$$

We already know how to deal with $x^{1-p}$. Now, as $x\to 0$, $F(x)\to 1$ and as $x\to \infty$, $F(x)\sim x^{-1}$, and $F$ is continuous, positive (in fact its "positive" range is $(0,1]$) and decreasing over $x>0$. You know that $$\int_0^1 x^{-\alpha} dx$$ converges for $\alpha<1$ and that $$\int_1^\infty x^{-\beta} dx$$ converges for $\beta >1$. Now, you integral is similar $ x^{1-p}$ near the origin and similar to $x^{-p}$ for large values of $x$. Can you make a guess as to where $p$ should fall and then prove it?

Answer 2

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\begin{align} {\expo{-x} - 1 \over x^{p}} &\sim\ -\,{1 \over x^{p - 1}}\quad\mbox{when}\quad x \sim 0\quad\imp\quad 2 - p > 0 \ \imp \ p < 2 \\[3mm] {\expo{-x} - 1 \over x^{p}} &\sim\ -\,{1 \over x^{p}}\quad\mbox{when}\quad x \gg 1\quad\imp\quad p - 1 > 0\ \imp\ p > 1 \end{align} $$\color{#0000ff}{\large% \int_{0}^{\infty}{\expo{-x} -1 \over x^{p}}\,\dd x\quad \color{#000000}{\tt converges}\ \mbox{whenever}\quad 1 < p < 2} $$