# Prove that $\int\limits_1^\infty\frac{e^x}{x^{x^2}}dx$ converges

I need to prove that the following improper integral converges: $$\int\limits_1^\infty\frac{e^x}{x^{x^2}}dx$$

I can't find a function to compare with it in order to use the integral comparison test or the limit comparison test.

• Hint: $f(x)=e^{\ln(f(x))}$ Sep 12, 2020 at 13:20
• Hint: For $x>e$, $x^{x^2}>e^{x^2}$ Sep 12, 2020 at 13:39

Hint: For $$x \ge e$$, $$\dfrac {e^x} {x^{x^2}} < e^{-x}$$.
Let $$c$$ be such that $$c^c=e$$. Then, $$\int_1^\infty e^x/x^{x^2}dx=\int_1^\infty (e/x^x)^xdx=\int_1^{c+\varepsilon} (e/x^x)^xdx+\int_{c+\varepsilon}^\infty (e/x^x)^xdx$$ converges, since for $$x\ge c+\epsilon$$, we have $$e/x^x\le e/(c+\epsilon)^{c+\epsilon}<1$$.