Using method of steepest descent to find leading-order asymptotic behavior of the error function The error function is defined as $$erf(z) = \frac{2}{\sqrt{\pi}}\int_{0}^z e^{-s^2} ds$$
and I want to apply the method of steepest descent to find the leading-order asymptotic behavior of this as $|z|\to\infty$, given that $z$ only stays inside the first quadrant, i.e. I can substitute $z = re^{i\theta}$ where $r\to\infty$ and $0 < \theta <\pi/2$.
To use the method of steepest descent, first I find the stationary point of the function $-s^2$, which is just $0$. Then, I find the steepest descent direction, but it turns out to be $[0, 0]^T$. I'm not sure what this means and how I can draw the contour passing this stationary point in steepest descent direction.
 A: *

*The leading order of the error function is not very surprisingly 
$$ \begin{align}{\rm erf}(x)
~:=~&\frac{1}{\sqrt{\pi}}\int_{-x}^x\! ds~e^{-s^2}\cr
~\stackrel{s=xt}{=}&~\frac{x}{\sqrt{\pi}}\int_{-1}^1\! dt~e^{-S(t)},\qquad S(t):=x^2t^2,\cr
~=~&\frac{x}{\sqrt{\pi}}\sqrt{\frac{2\pi}{S^{\prime\prime}(0)}}e^{-S(0)}+o(1/x)\cr
~=~&1+o(1/x)\quad\text{for}\quad x~\to~\infty ,\end{align} $$
cf. Laplace's method.

*More interestingly, the complementary error function has asymptotic series 
$$ \begin{align}{\rm erfc}(x)
~:=~&1-{\rm erf}(x)\cr
~=~&\frac{2}{\sqrt{\pi}}\int_x^{\infty}\! ds~e^{-s^2}\cr
~\stackrel{t=s^2-x^2}{=}&~\frac{e^{-x^2}}{x\sqrt{\pi}}\int_{\mathbb{R}_+}\! dt\frac{e^{-t}}{\sqrt{1+t/x^2}}\cr
~=~&\frac{e^{-x^2}}{x\sqrt{\pi}}\int_{\mathbb{R}_+}\! dt~e^{-t}\sum_{n\in\mathbb{N}_0}\frac{(1/2)_n}{n!}\left(-\frac{t}{x^2}\right)^n\cr
~=~&\frac{e^{-x^2}}{x\sqrt{\pi}}\sum_{n\in\mathbb{N}_0}\frac{(1/2)_n}{(-x^2)^n}\cr
~=~&\frac{e^{-x^2}}{x\sqrt{\pi}}\sum_{n\in\mathbb{N}_0}\frac{(2n-1)!!}{(-2x^2)^n}  
\quad\text{for}\quad x~\to~\infty .\end{align} $$
Here $(s)_n=\frac{\Gamma(s+n)}{\Gamma(s)}$ is the Pochhammer symbol, cf. Abramowitz and Stegun, eq. (6.1.22), p. 256.
