I am attempting to generate a power series in $v$ as $v\rightarrow0$ for 

$$g(x,v)\equiv ln\int_{-\infty}^{\infty}e^{-y\left(z\right)}N\left(x-z,v\right)dz
 $$

where $y(x)\equiv\sum_{i=1}^{n}c_{i}x^{i}$. Assume that n is even and $c_{n}>0$ so the intergral converges. N(x,v) is a Normal distribution in x with variance v.

The key mathematical step that I can't reasonably expect Mathematica to figure out is:

$$\frac{d}{dv}\int f(z)N(z-x,v)dz=\int f(z)\frac{d}{dv}N(z-x,v)dz=\int f(z)\frac{1}{2}\frac{d^{2}}{d^{2}x}N(z-x,v)dz=\frac{1}{2}\frac{d^{2}}{d^{2}x}\int f(z)N(z-x,v)dz\rightarrow\frac{1}{2}\frac{d^{2}}{d^{2}x}\int f(z)\delta(z-x)=\frac{1}{2}\frac{d^{2}}{d^{2}x}f(x)$$

The final result will necessarily be in terms of the $c$ coefficients and of
$$t_{j}\equiv\int_{-\infty}^{\infty}x^{j}e^{-y\left(x\right)}dx$$

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, not a general expression.