Asymptotic expansions for $\int_x^\infty e^{-y^3} dy$ I would like to find asymptotic expansions of $F(x) = \int_x^\infty e^{-y^3}  dy$ as (a) $x \to 0$ and as (b) $x \to \infty$.
I solved (a) using the expansion for $e^{-y^3}$ around $0$:
$$F(x) = \int_0^\infty e^{-y^3} dy - \int_0^x\sum_{j= 0}^\infty (-1)^j \frac{y^{3j}}{j!} dy = \int_0^\infty e^{-y^3} dy - \sum_{j= 0}^\infty (-1)^j \frac{x^{3j+1}}{(3j+1)j!} \\ = C - x  + \frac{x^4}{4} - \frac{x^7}{14} + \ldots$$
I know that the series here is convergent for all $x$ but convergence is so slow for large $x$ that it does not provide a good asymptotic representation.
How can I get a better asymptotic expansion as $x \to \infty$?
I thought about changing variables $y \to 1/t$ to get:
$$F(x) = \int_0^{1/x}\frac{e^{-t^{-3}}}{t^2}dt = \int_0^{1/x}\sum_{j=0}^\infty(-1)^j \frac{t^{-3j-2}}{j!} dt$$
But this cannot be integrated term by term.

The Integral equals $\frac13 \Gamma(1/3,x^3)$ where $\Gamma(a,z)$ is the incomplete gamma function.
You can look up that the asymptotics around $0$ are $\frac{1}{3}\Gamma \left(\frac{1}{3}\right)-x+\frac{x^4}{4}+O\left(x^5\right)$,
and around $\infty$ is $e^{-x^3+O\left(\left(\frac{1}{x}\right)^6\right)} \left(\frac{1}{3 x^2}-\frac{2}{9 x^5}+O\left(\left(\frac{1}{x}\right)^6\right)\right)$.
