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

Question

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 1

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.