Why does $\int_0^ae^\frac1x x^pdx$ diverge?

Question

Let $a$ be positive and $p$ be real.

Why does the improper integral $$\int_{0}^{a}{\rm e}^{1/x} x^{p}\,{\rm d}x$$ diverge ?

Direct integration over $\left[b,a\right]$ for positive $b$ is hard. On the other hand, although I know that $\displaystyle\int_{0}^{a}{{\rm d}x \over x^{k}}$ diverges for all positive-integral $k$, I can't find a good comparison (i.e. $\frac 1{x^k}<e^\frac1x x^p$, i.e. $e^\frac1x x^{p+k}>1$) to prove what I want.

Answer 1

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \begin{align} \int_{0}^{a}\expo{1/x}x^{p}\,\dd x &= \int_{\infty}^{1/a}\expo{x}x^{-p}\, \pars{-\,{\dd x \over x^{2}}} = \int^{\infty}_{1/a}\expo{x}x^{-p - 2}\,\dd x\quad\mbox{diverges !!!} \end{align}

Answer 2

I think if we apply Quotient Test as follows then it diverges:

$$p\ge0\to k=-p\le0<1 \to \lim_{x\to 0^+}x^kf(x)=\infty$$ And $$p<0\to\lim_{x\to 0^+}x^{\color{red}{1}}f(x)=\infty$$