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.

Here is a method that works for both asymptotics. Consider the inverse function
$$e^{-y^3} \longrightarrow \sqrt[3]{-\log z}$$
Then by drawing a picture we get that
$$\int_x^\infty e^{-y^3}~dy = \int_0^{e^{-x^3}}\sqrt[3]{-\log z}~dz - xe^{-x^3}$$
where the term on the right is the area of the rectangle we need to subtract to get the original integral. Now we'll use integration by parts:
$$ = -xe^{-x^3} - z(\log z)^{\frac{1}{3}}\Bigr|_0^{e^{-x^3}} + \int_0^{e^{-x^3}} \frac{1}{3}(\log z)^{-\frac{2}{3}}~dz = \frac{1}{3}\int_0^{e^{-x^3}} (\log z)^{-\frac{2}{3}}~dz$$
by choosing $u =$ integrand and $dv = dx$. We can keep this pattern going to get
$$ = \frac{1}{3}z(\log z)^{-\frac{2}{3}}\Bigr|_0^{e^{-x^3}} + \frac{2}{9}\int_0^{e^{-x^3}} (\log z)^{-\frac{5}{3}}~dz = \frac{e^{-x^3}}{3x^2} + \frac{2}{9}\int_0^{e^{-x^3}} (\log z)^{-\frac{5}{3}}~dz$$
etc, the pattern will continue where the $z$ cancels out with the chain rule from the $\log$, getting a nice series for $x\to\infty$ since the integral term which is effectively acting as our error will go to $0$.
For $x\to 0$, notice that the inverse function is nearly vertical for $z\to 1$. This means that we can approximate the integral in the first line as being a constant (the full value) minus just the rectangle.
$$\int_0^1 \sqrt[3]{-\log z}~dz - x(1-x^3+\cdots)$$
This wasn't as fast as using integration by parts on the original integral, but the cancellation is nice.
