I have the following parametric integral
$$F(a)=\int_0^a e^{-x^2\operatorname{erf}(x)}dx$$
where $\operatorname{erf}(x)$ is the error function.
I'm interested in asymptotic behaviour of this integral for the cases $a\to 0$ and $a\to \infty$.
Could someone clarify to me how to obtain an asymptotic expansion of $F(a)$ ?
Thanks a lot.
For the case where $a$ is small, you can approximate $$e^{-x^2 \text{erf}(x)}$$ by a simple $[n,n]$ Padé approximant $P_n$ built around $x=0$. For example $$P_3=\frac{1+\frac 13 x^2 -\frac 1 {\sqrt \pi}x^3}{1+\frac 13 x^2 +\frac 1 {\sqrt \pi}x^3 }$$ whose error is $\frac {x^7} {45\sqrt \pi}$ which is easy to integrate using partial fraction decomposition; the denominator has only one real root and we shall face one logarithm and an arctangent.
For example, for $a=0.5$, numerical integration gives $0.48390392$ while the approximation of the integrand by $P_3$ gives $0.48389086$.