Confusion over definition of distributional derivatives

Question

Suppose I want to differentiate the following integral pairing with respect to some scalar $s\in \mathbb{R}$: $\int \psi(x,s)\rho(x,s)d\nu(x)$. Here $\psi \in C^\infty (\mathbb{R}^n\times \mathbb{R})$, $\rho(\cdot,s)\in{ \mathcal D}′(\mathbb{R}^n)$ for all $s$ (some set of distributions on $\mathbb{R}^n$) and $\rho$ is differentiable in $s$.

$\nu$ is the Lebesgue measure over some set over which is the integral is defined.

There are two terms that would appear on differentiation wrt $s$. One of them is $∫∂_sψ(x,s)ρ(x,s)dν(x)$ and the other is $∫ψ(x,s)∂_sρ(x,s)dν(x).$

If $ρ∈{\mathcal D}′(\mathbb{R}^n\times \mathbb{R})$ however, to get the same derivative, should I define a new integral pairing in order to define the derivative?

Perhaps write, $\int \psi(x,s)\rho(x,s)d\nu(x) = d_s \int_{-\infty}^s \int \psi(x,s)\rho(x,s)d\nu(x)d\mu(s)$

where the additional integral is a Lebesgue integral in $\mathbb{R}$? And then define,

$\int \psi(x,s) \partial_s\rho(x,s)d\nu(x)d\mu(s) = \int \partial_s\psi(x,s) \rho_s(x,s)d\nu(x) d\mu(s)$ for all smooth $\psi$ as is usually done with distributional derivatives?

Finally, this would mean,

$d_s \int \psi \rho d\nu = ∫∂_sψ ρ dν + d_s \int_{-\infty}^s ∫\partial_s ψ ρ d\nu d\mu$

But, this does not feel right - could you please explain how you would differentiate? Thank you very much for spending your time to clarify!

Answer 1

Unfortunately, I don't have the reputation to comment. I think that Theorem 2.1.3 In Hormander's "The Analysis of Linear Partial Differential Operators I" is useful for you which I'll restate here.

Theorem

If $\phi(x,y) \in C^\infty (X \times Y)$ where $Y$ is an open set in $\mathbb{R}^m$ and if there is a compact set $K \subset X$ such that $\phi(x,y) = 0$ when $x \notin K$, then $$y \to u(\phi(\cdot, y))$$ is a $C^\infty$ function of $y$ if $u \in D'(X)$. Furthermore $$\partial^\alpha_yu(\phi(\cdot,y))= u(\partial^\alpha_y\phi(\cdot,y))$$