I am attempting to generate a power series in $v$ as $v\rightarrow0$ for
$$g(x,v)\equiv ln\int_{-\infty}^{\infty}e^{-y\left(z\right)}N\left(x-z,v\right)dz $$$$g(x,v)\equiv \ln\int_{-\infty}^{\infty}e^{-y\left(z\right)}N\left(x-z,v\right)\,\mathrm{d}z $$
where $y(x)\equiv\sum_{i=1}^{n}c_{i}x^{i}$, n$n$ is even, and $c_{n}>0$ so the integral converges. N(x,v)$N(x,v)$ is athe PDF of the Normal distribution centered in x$x$ with variance v$v$.
The key mathematical step that I can't reasonably expect Mathematica to figure out is:
$$\frac{d}{dv}\int f(z)N(z-x,v)dz=\int f(z)\frac{d}{dv}N(z-x,v)dz=\int f(z)\frac{1}{2}\frac{d^{2}}{d^{2}x}N(z-x,v)dz=\frac{1}{2}\frac{d^{2}}{d^{2}x}\int f(z)N(z-x,v)dz\rightarrow\frac{1}{2}\frac{d^{2}}{d^{2}x}\int f(z)\delta(z-x)=\frac{1}{2}\frac{d^{2}}{d^{2}x}f(x)$$
I've developed some truly ugly code that does the expansion of $g[x,v]$ in $v$, but it's an unreadable mess that relies on my doing most of the manipulations on paper beforehand, and I have to rewrite most of it every time I want to change the calculation a bit. How would a Mathematicagician more experienced than me do this calculation?
I only need the first several terms in the expansion in v$v$, not a general expression. But I'll be impressed all day if someone comes up with a general expression.
