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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.

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    $\begingroup$ Hint: $f(x)=e^{\ln(f(x))}$ $\endgroup$
    – QC_QAOA
    Sep 12, 2020 at 13:20
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    $\begingroup$ Hint: For $x>e$, $x^{x^2}>e^{x^2}$ $\endgroup$
    – K.defaoite
    Sep 12, 2020 at 13:39

2 Answers 2

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Hint: For $x \ge e$, $\dfrac {e^x} {x^{x^2}} < e^{-x}$.

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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 $e/x^x\le e/(c+\epsilon)^{c+\epsilon}<1$.

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