Why does $\int_0^ae^\frac1x x^pdx$ diverge? 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.
 A: $\newcommand{\+}{^{\dagger}}%
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\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}
A: 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 $$p<0\to\lim_{x\to 0^+}x^{\color{red}{1}}f(x)=\infty$$
