Is $z \mapsto \int f(x,z) K_z(x) dx$ analytic if $K_z(x)$ is a distribution analytic in the paramter $z$? Let $K_z(x) \in \mathcal D'(\mathbb R)$ be a distribution kernel depending analytically on $z \in O \subset \mathbb C$, i.e. for $f \in \mathcal D(\mathbb R)$ fixed,
$$
  \int K_z(x) f(x) dx 
$$
is analytic for $z$ on an open set $O$.
Q: Is this still true if we multiply $K_z(x)$ with an analytic function $a(x,z)$ that is bounded on $O$? I.e., is
$$
  z \mapsto \int f(x) a(x,z) K_z(x) dx
$$
still analytic in $O$?
Attempt:
Showing that the integral satisfies the Cauchy-Riemann equations would be a way to prove that it is analytic, but is it possible to

*

*Exchange integral and derivative $\frac{d}{dz} \int f(x) a(x,z) K_z(x) dx = \int \frac{d}{dz} f(x) a(x,z) K_z(x) dx$?

*Argue that the Kernel $K_z(x)$ satisfies the CR-equations and not just $\int f(x) K_z(x) dx$?

Is there another approach?

Update:
While the answer by @reuns provides some inside, it is still missing the crucial point, i.e. exchanging two limit operations (in this case an integral and a derivative). To make this point more obvious, one can also consider
$$
  (z,z') \mapsto \int f(x) a(x, z') K_z(x) dx
$$
which is certainly analytic in $z$. It remains to show that it is analytic in $z'$. This boils down to the question whether distribution are linear in derivatives w.r.t. a paramter of the test function, i.e.
$$
  \partial_z \int a(x,z) K(x) dx = \int \partial_z a(x,z) K(x) dx
$$
This is something I suspect to be true, but am unsure about.
 A: Take $\phi \in D(\Bbb{R}),\int \phi = 1,\phi_n(x)=n \phi(nx)$, let
$$H(y,z)= \int_\Bbb{R} f(x-y) a(x-y,z) K_z(x)dx$$
$$h_n(z) =
\int_\Bbb{R} \phi_n(y)H(y,z) = \int_\Bbb{R}f(y) a(y,z) (\int_\Bbb{R} \phi_n(x-y) K_z(x)dx)dy$$
where $y \to x-y$ was substituted.
$\int_\Bbb{R} \phi_n(x-y) K_z(x)dx$ is continuous in $y$ and analytic in $z$ so by Morera's theorem, considering the sequence of Riemman sums $$h_{n,M}(z)=\frac1M\sum_{m=-\infty}^\infty f(m/M) a(m/M,z) \int_\Bbb{R} \phi_n(m/M-x) K_z(x)dx$$ $h_n=\lim_{M\to \infty} h_{n,M}$ is analytic.
$H$ is continuous in $(y,z)$ so $h_n$ converges locally uniformly as $n\to \infty$.

Edit: not sure of the proof, it is somewhat the key step

Whence by Morera's theorem the limit $$h_\infty(z)=H(0,z)=\int_\Bbb{R} f(x) a(x,z) K_z(x)dx$$ is analytic.
A: Here is a proof assuming $K(z,x),a(x,z)$ are analytic in $z$ and $f(x) K(x,z)$ is absolutely integrable in $z,x$. I am aware that the question wants for dsitribution kernel. But i thought this would be useful.
Let $$g(z) = \int f(x) a(x,z) K(z,x) dx.$$
Now if $\int_{\gamma} g(z) dz = 0$ then by morera's theorem $g(z)$ is holomorphic.
Now $$\int_{\gamma} g(z) dz = \int_{\gamma} \int f(x) a(x,z) K(z,x) dx dz.$$
Assuming the integral can be swapped, since $a(x,z) K(z,x)$ is analytic: $\int_{\gamma} a(x,z) K(z,x) dz = 0$.
$$ =  \int f(x) \int_{\gamma} a(x,z) K(z,x) dz dx = 0.$$
So the proof is complete if we can swap the two iterated integrals. Since we assume $f(x)K(z,x)$ is absolutely integrable in $x,z$ with $a(x,z)$ bounded uniformly in $x,z$, we can swap the integrals by Fubini's theorem.
