Show that $\int_0^{a} e^{1/x}x^p dx $ diverges for all $p$

Question

I'm trying to solve this problem using the convergence test: noting that for $a \ge 1 $ we have $0 \le e^{1/x} \le e^{1/x}x^p$ on the interval $[0,a]$ and so $\int_0^{a} e^{1/x} dx \le \int_0^{a} e^{1/x}x^p dx$. The integral on the left diverges since $e^{1/x}$ is unbounded and positive on this interval, so $\int_0^{a} e^{1/x}x^p dx$ must also diverge ($\forall p$). However, the comparison test doesn't seem to achieve this generality for $a < 1$ since $0 \le x^p \le e^{1/x}x^p$ and $\int_0^a x^pdx$ diverges for all $p \le -1$. Is there another approach to showing this result that does not involve breaking $[0,a]$ into 2 cases? Is there a modification of my argument that will work?

Answer 1

Choose $n> p+1.$ Then $$e^{1/x} = 1 + 1/x +\cdots 1/(n!x^n) + \cdots > 1/(n!x^n) \implies e^{1/x}x^p > 1/(n!x^{n-p}) > 1/(n!x)$$ for $0<x<1.$ Since $\int_0^1(1/x)\,dx = \infty,$ we have $\int_0^1e^{1/x}x^p\,dx=\infty$

Answer 2

You just need to notice that: $$ \int_{\frac{1}{n+1}}^{\frac{1}{n}}e^{1/x}x^p\,dx \geq \frac{1}{(n+1)^p}\int_{\frac{1}{n+1}}^{\frac{1}{n}}e^{1/x}\,dx \geq \frac{e^n}{n(n+1)^{p+1}}$$ and the series $\sum_{n\geq 1}\frac{e^n}{n(n+1)^p}$ cannot be convergent since $\frac{e^n}{n(n+1)^{p+1}}$ grows unbounded as $n\to +\infty$.