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So, my question is whether the integral $\int_{0}^{\infty} e^{-x^5}dx$ is convergent or not?

My work-

So, I tried to use the comparison test. I see that $\frac{1}{e^{x^5}} \leq \frac{1}{x^5}$ but unfortunately the integral $\int_{0}^{\infty}\frac{1}{x^5}dx$ is not convergent. So, then I thought maybe adding $1$ will do the work, i.e., I checked that $\frac{1}{e^{x^5}} \leq \frac{1}{x^5+1}$. But it is quite difficult to calculate the integral $\int_{0}^{\infty}\frac{1}{x^5+1}dx$. But I used the integral calculator and saw that this indeed works. The integral $\int_{0}^{\infty}\frac{1}{x^5+1}dx$ is actually convergent.

So, is there any easier bound of $\frac{1}{e^{x^5}}$ whose integral is easy to calculate and of course convergent? And is there any other test which can be used to solve this problem?

Thank you.

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Since $\int_0^1e^{-x^5}$ converges, you can compare with ${1\over x^5}$ on $[1,+\infty)$

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  • $\begingroup$ Oh, I like this one. Break the domain into two parts and use the function that is suitable for each one. Thank you for your answer. $\endgroup$ – Saikat Aug 18 '19 at 10:13
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Compare the integrand with a simpler function:

for all $x\geq 1$, we have $e^{-x^5}\leq e^{-x}$, and $$\int_1^{+\infty}e^{-x}dx=\frac 1 e <+\infty$$ so the integral is convergent.

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It is $$\Gamma \left(\frac{6}{5}\right)$$ so the integral does converge.

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  • $\begingroup$ that's cheeky :D $\endgroup$ – Alvin Lepik Aug 18 '19 at 10:06

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