I am trying to decide if this integral converges or diverges, so I'm trying to find a function that behaves like the integrand at infinity to use the limit comparison test since the integrand is positive.
I couldn't find a function to compare my integral with, I know that at infinity the highest power is what the function behaves like ($\frac {1}{x^2}$ and $\frac {1}{1+x^2}$ behave the same at infinity for example), and I have been solving a couple of improper integral questions with comparison test with a little list I made about what functions to compare with, but none was close to this integrand. So here's what I tried:
$\int^\infty_1 \frac {e^x}{x^{x^2}}=\int_1^\infty\frac {e^x}{e^{x^2ln(x)}}$, I know that $\int^\infty_2\frac {1}{xln^\alpha(x)}$ converges if and only if $\alpha>1$.
But here I got stuck and I don't know how to continue or if the function I'm trying to compare with isn't the right one.
I would appreciate any help in how to find a function that works for comparison test for this integrand, also if anyone knows where I can find a list of known improper integrals that converges or diverges that would help me alot.
Thanks in advance
Best Answer
It converges.
HINT: For $x>e^2$ note that $x^{x^2} \ge e^{2x^2} \ge e^{2x}$.