Closed form for a Gaussian exponential mixture

Given: Assume AC, $\rho >0$, $v\ge 0$, and $s\in \mathbb{R}$.

Standard normal and normal laws: $N(m,{\sigma }^2)$ is the law of the affine image $z\mapsto m+\sigma z$ of the standard normal law, whose density is $e^{-z^2\mathrm{/}2}\mathrm{/}\sqrt{2\pi }$.

Change of variables for expectation: Integrating a nonnegative measurable function against a random element's law equals its expectation on the underlying space.

The standard normal density has total mass one: The standard normal density $e^{-z^2\mathrm{/}2}\mathrm{/}\sqrt{2\pi }$ integrates to one.

A C^1 diffeomorphism satisfies the change-of-variables formula for nonnegative Lebesgue measurable functions: If $T:U\to V$ is a $C^1$ diffeomorphism between open subsets of ${\mathbb{R}}^n$ and $f:V\to [0,\mathrm{\infty }]$ is Lebesgue measurable, then, assuming countable choice, ${\int }_{V}f(y)d{\lambda }_n(y)={\int }_{U}f(T(x))\mathrm{\mid }\det DT(x)\mathrm{\mid }d{\lambda }_n(x)\mathrm{.}$

Continuous functions on Euclidean spaces are Borel measurable and Assuming countable choice, every Borel subset of ${\mathbb{R}}^n$ is Lebesgue measurable: Under countable choice, every continuous function on $\mathbb{R}$ is Borel measurable, and every Borel set is Lebesgue measurable.

Square roots exist: a unique $\sqrt{a}\ge 0$ with $(\sqrt{a}{)}^2=a$; the positives are $\{x^2:x\mathrm{\ne }0\}$: Every nonnegative real has a unique nonnegative square root.

The Axiom of Choice: AC asserts that every family of nonempty sets has a choice function; it is assumed here.

The Axiom of Countable Choice (${\mathrm{A}\mathrm{C}}_{\omega }$): AC implies countable choice, so the countable-choice premise of [F4] follows from the assumed AC.