Integral of complementary error function times exponential with polynomial argument

Question

I try to understand the behavior of the following integral as $a\rightarrow\infty$ $$I(\delta)=\int_{-\delta}^\delta{\rm erfc}\Big(-\frac t{\sqrt a}\Big){\rm e}^{bt-t^2/a}{\rm d} t,$$ where ${\rm erfc(x)}$ is the complementary error function and $b$ can be written as a power of $a$. According to Mathematica, I already know that $$J(\delta)=\int_{-\delta}^{\delta}{\rm e}^{bt-t^2/a}{\rm d} t=\frac12\sqrt{\pi a}{\rm e}^{ab^2/4}\Big\{{\rm erf}\Big(\frac{ab+2\delta}{2\sqrt a}\Big)-{\rm erf}\Big(\frac{ab-2\delta}{2\sqrt a}\Big)\Big\},$$ but Mathematica doesn't compute $I(\delta)$. It appears from the above estimation that, with $\delta=ab$, we have $$J(\delta)=_{a\to\infty}\sqrt{\pi a}{\rm e}^{ab^2/4}\{1+O(a^{-m})\},\ m\in\mathbb{N}.$$ Do we have a similar approximation for $I(\delta)$? Is it possible to find a closed formula for $J(\delta)$ similar to the one for $I(\delta)$?

Answer 1

$\newcommand\de\delta\newcommand\erfc{\operatorname{erfc}}$If $0<\de=o(\sqrt a)$, then $\erfc(-\frac t{\sqrt a})\to1$ and $e^{-t^2/a}\to1$ uniformly in $t\in[-\de,\de]$, whence $$\begin{aligned} I(\de)&\sim \int_{-\de}^{\de}e^{bt}\,dt =\frac2b\,\sinh(b\de) \end{aligned}$$ if $b\ne0$, and $I(\de)\sim2\de$ if $b=0$.

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Now consider the following limit conditions, which seem to be close to what the OP had in mind: $$a\to\infty,\quad b\ge1,\quad(\de-ab/2)/\sqrt a\to\infty. $$ Then we can find a varying $C$ such that $$C\to\infty,\quad C-b\sqrt a/4\to-\infty. $$ Write $$I(\de)=I_1+I_2,$$ where $$I_1:=\int_{-\de}^{C\sqrt a}\erfc(-t/\sqrt a)e^{bt-t^2/a}dt,\quad I_2:=\int_{C\sqrt a}^\de\erfc(-t/\sqrt a)e^{bt-t^2/a}dt,$$ Next, \begin{equation} I_1\le2\int_{-\de}^{C\sqrt a}e^{bt}dt=O(e^{Cb\sqrt a})=o(e^{ab^2/4}), \end{equation} \begin{equation} I_2\sim2\int_{C\sqrt a}^\de e^{bt-t^2/a}dt =2e^{ab^2/4}\sqrt a\,\int_{C-b\sqrt a/2}^{(\de-ab/2)/\sqrt a} e^{-z^2}dz \sim2e^{ab^2/4}\sqrt a\,\int_{-\infty}^\infty e^{-z^2}dz =2\sqrt{\pi a}\, e^{ab^2/4}. \end{equation} So, \begin{equation} I(\de)\sim2\sqrt{\pi a}\, e^{ab^2/4}. \end{equation}