I'm looking for the following limit: $$\lim_{\varepsilon\to 0^+}\int_{-\sqrt{\varepsilon}}^{\sqrt{\varepsilon}}\frac{1}{\sqrt{2\pi}\varepsilon^{3/2}}\left(-1+\frac{x^2}{\varepsilon}\right)e^{-\frac{x^2}{2\varepsilon}}l(a+x)dx=???$$ Where $l$ is a bounded and nice function ($l\in C^{\infty}$) with $l(a)\neq 0$. We remark that $$\frac{\partial^2}{\partial x^2}\frac{1}{\sqrt{2\pi\varepsilon}}e^{-\frac{x^2}{2\varepsilon}}=\frac{1}{\sqrt{2\pi}\varepsilon^{3/2}}\left(-1+\frac{x^2}{\varepsilon}\right)e^{-\frac{x^2}{2\varepsilon}}.$$
1 Answer
Letting $h:=\sqrt\varepsilon$ and denoting the standard normal pdf by $f$, we see that the integral under the limit sign is $$\frac1{h^3}\int_{-h}^h f''(x/h)l(a+x)\,dx =\frac1{h^2}\,I(h),$$ where $$I(h):=\int_{-1}^1 f''(u)l(a+hu)\,du\underset{h\downarrow0}\longrightarrow \int_{-1}^1 f''(u)l(a)\,du \\ =2\int_0^1 f''(u)l(a)\,du =2l(a)f'(1)=-l(a)\sqrt{\frac2{\pi e}};$$ here we used the dominated convergence theorem. So, the integral under the limit sign is $$-\frac{l(a)+o(1)}{h^2}\,\sqrt{\frac2{\pi e}} =-\frac{l(a)+o(1)}\varepsilon\,\sqrt{\frac2{\pi e}}.$$ So, the limit of this integral is $-\infty$ if $l(a)>0$ and $\infty$ if $l(a)<0$.