Let $a>0$ and $A>0$, I am looking for the decay rate of the integral $$ \int^\infty_{-\infty} \frac{x^2 \exp(-(x-a)^2)}{1+Aa^{-1}\exp(-x^2/(1+a^2))} dx \quad \text{ as $a\to 0$}.$$ There is no closed form answer for the integral. I have tried on Matlab that it should converge to zero much faster than power growth. I think the growth should be exponential types. Do we have some literature discussing this kind of issue? Thanks!
I have successfully obtained the growth rate of $$ \int^\infty_{M} +\int_{-\infty}^{-M} \frac{x^2 \exp(-(x-a)^2)}{1+Aa^{-1}\exp(-x^2/(1+a^2))} dx \quad \text{ as $a\to 0$}$$ be expanding the denominator in power series.
But I do not know to deal with the integral in $[-M,M]$.
Let us assume $a\ll1$. Approximating the function (or simply plotting it), we can see that the maximum of the integrand is at $|x|\geq 1$ (in fact it is $|x|=1$ for $A\ll a$ and becomes larger when increasing $A$).
Because if this we can approximate $1+a^2 \approx 1$ and $(x-a)^2 \approx x^2$ in the exponents of the integrand to obtain the leading order behavior.
Denoting the integral by $I(a,A)$, we obtain $$ I(a,A) \simeq I_0(a,A) =\int^\infty_{-\infty} \dfrac{x^2 \exp(-x^2)}{1+Aa^{-1}\exp(-x^2)} dx = -\frac{\sqrt\pi a}{2A} \operatorname{Li}_{3/2}(-A/a)\,.$$ Here, Li$_s$ is the polylogarithm function. In fact, $I_0$ is an excellent approximation to $I$ for $a\lesssim 0.1$.
The polylogarithm function has a known asymptotic expansions in terms of $\log x$. In particular, we have ($x\gg 1$) $$ \operatorname{Li}_s(-x) \sim - \frac{\log^{s}(x)}{\Gamma(s)} \Bigl(1 + O(\log^{-2}(x))\Bigr)\,.$$
As a result, we obtain $$ I(a,A) \simeq I_0(a,A) \sim \frac{2 a}{3A} \log^{3/2}(A/a)\,. $$
Note that this is not a rigorous mathematical answer. What remains is to show that $I- I_0$ is small for $a\to0$. Numerically, it seems to hold that $I- I_0 = o(a^2)$.