# The integral $\int_{0}^{\infty} e^{-x^5}dx$ is convergent or not?

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.

Since $$\int_0^1e^{-x^5}$$ converges, you can compare with $${1\over x^5}$$ on $$[1,+\infty)$$
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.
It is $$\Gamma \left(\frac{6}{5}\right)$$ so the integral does converge.