$$\lim_{\epsilon \to 0} \frac{1}{2\pi} \int_{-\infty}^\infty \int_{-\infty}^\infty f(x,y) \frac{y}{\epsilon^2} \exp\left(-\frac{x^2+y^2}{2\epsilon}\right) \, \mathrm{d}x \, \mathrm{d}y$$
here $f$ and $\nabla f$ are assumed to be continuous and bounded.
I've been trying for a bit with no luck. I think I would want to use the dominated convergence theorem, but I can't seem to bound the integrand by an integrable function.
I also tried a few different change of variables, which weren't so insightful.
Any ideas are welcome.
Thanks,
Idea. The intuition behind this integral is that
$$ \delta_{\epsilon}(x) := \frac{1}{\sqrt{2\pi\epsilon}} e^{-x^2/2\epsilon} $$
approximates the unit Dirac mass $\delta_0(x)$ as $\epsilon \downarrow 0$ by some suitable approximation to the identity argument. Thus writing the integral in terms of $\delta_x$, $\delta_y$ and performing integration by parts,
$$ -\iint f(x, y) \, \partial_y \delta_{\epsilon}(x) \delta_{\epsilon}(y) \, dxdy = \iint (\partial_y f(x, y)) \, \delta_{\epsilon}(x) \delta_{\epsilon}(y) \, dxdy. $$
Letting $\epsilon \downarrow 0$, we expect that the limit is $\partial_y f(0, 0)$.
Solution. Let $I_{\epsilon}$ denote the integral inside the limit. Since the integrand is integrable, we can interchange the order of integration. Then by the integration by parts,
$$ \int_{-\infty}^{\infty} f(x, y) \frac{y}{\epsilon}e^{-y^2/2\epsilon} \, dy = \int_{-\infty}^{\infty} \frac{\partial f}{\partial y}(x, y) e^{-y^2/2\epsilon} \, dy. $$
So if we let $g = \partial f/\partial y$, then we can write
\begin{align*} I_{\epsilon} &= \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} g(x, y) \frac{1}{2\pi \epsilon}e^{-(x^2+y^2)/2\epsilon} \, dydx \\ &= \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} g(\sqrt{\epsilon}x, \sqrt{\epsilon}y) \frac{1}{2\pi}e^{-(x^2+y^2)/2} \, dydx \end{align*}
where we applied the substitution $(x, y) \mapsto \sqrt{\epsilon}(x, y)$ in the second line. Since $g$ is bounded and continuous, by the dominated convergence theorem,
$$ \lim_{\epsilon \downarrow 0} I_{\epsilon} = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} g(0, 0) \frac{1}{2\pi}e^{-(x^2+y^2)/2} \, dydx = g(0, 0) = \frac{\partial f}{\partial y}(0, 0). $$