calculus - Let $a_n = intlimits_{0}^{n} e^{-x^4} dx$. Does ${ a_n }_{n rightarrow infty}$ converge?

Let $a_n = \int\limits_{0}^{n} e^{-x^4} dx$. Does $\{ a_n \}_{n \rightarrow \infty}$ converge?

$\{ a_n \} =\{ \int\limits_{0}^{1} e^{-x^4} dx, \int\limits_{0}^{2} e^{-x^4} dx, ..., \int\limits_{0}^{\infty} e^{-x^4} dx \}$

So we need need only to check that the definite integral

$\int\limits_{0}^{\infty} e^{-x^4} dx$ converges

By using Wolfram Alpha,

$\int\limits_{0}^{\infty} e^{-x^4} dx \} = \Gamma \left( \frac54 \right) \approx 0.906402$

where $\Gamma$ is the Gamma function

Therefore $\{ a_n \}$ converges, to $\Gamma \left( \frac54 \right)$.

But what if I can't use Wolfram Alpha? How can I solve for this integration by hand? Sorry if this makes me look stupid, I don't recall learning any techniques from Calculus that can help me solve this integration.

Thanks in advance!

Answer

We do not need an explicit expression to show that an improper integral converges.

The sequence $(a_n)$ is obviously increasing. It is bounded above by $\int_0^1 e^{-x^4}\,dx+\int_1^\infty e^{-x}\,dx$. To be explicit, the first integral is less than $1$, and the second is $e^{-1}$, so the sequence $(a_n)$ is bounded above by $1+e^{-1}$.

Any increasing sequence which is bounded above converges.