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.
 A: Rewrite the integrand as
$$e^{-y^3}=-3y^2e^{-y^3}\cdot\frac1{-3y^2}$$
and integrate by parts to get
$$f(x)=\frac{e^{-x^3}}{3x^2}-\frac23\int_x^\infty\frac{e^{-y^3}}{y^3}~\mathrm dy$$
Repeating the process we get:
$$f(x)e^{x^3}=\frac1{3x^2}-\frac2{9x^5}+\frac{2\cdot5}{27x^8}-\frac{2\cdot5\cdot8}{81x^{11}}+\mathcal O(x^{-14})$$
Note that this is an asymptotic series (the coefficients tend to $\pm\infty$ faster than $x^{-3n}$ tends to zero as $n\to\infty$).
This process may also be clearer if you first make the substitution $y\to\sqrt[3]t$ to get
$$f(x)=\frac13\int_{x^3}^\infty t^{-2/3}e^{-t}~\mathrm dt$$
which makes the process a little easier as well as relates this to the incomplete Gamma function.
A: 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 $e^{-x^3+O\left(\left(\frac{1}{x}\right)^6\right)} \left(\frac{1}{3 x^2}-\frac{2}{9 x^5}+O\left(\left(\frac{1}{x}\right)^6\right)\right)$.
A: 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.
