Exchanging an integral with a distributional limit I have an analytic function $F(z_1,z_2)$ defined on $\{(z_1,z_2) \in \mathbb C^2 ~|~ 0 < \Im[z_1] < \Im[z_2]\}$ with distributional boundary values s.t.
$$F(f_1, f_2) = \lim_{\epsilon_2 \to 0} \int dx_2  \lim_{\epsilon_1 \to 0} \int dx_1 F(x_1 + i\epsilon_1, x_2 + i\epsilon_2) f_1(x_1) f_2(x_2) < \infty $$
for $f_1, f_2 \in C_0^\infty(\mathbb R)$. The limits are ordered such that $\epsilon_1 < \epsilon_2$ where F is analytic.
Let's say I know how to evaluate $\int dx_2 F(x_1 + i\epsilon_1, x_2 + i\epsilon_2) f_2(x_2)$ for some reasonable $0 <\epsilon_1 < \epsilon_2$ and $x_2 \in \mathbb R$.
Is it possible to exchange the integrals without exchanging the limits, evaluating $\int dx_2$ first:
$$ F(f_1, f_2) \overset{?}{=} \lim_{\epsilon_2 \to 0}  \lim_{\epsilon_1 \to 0} \int dx_1  f_1(x_1) \int dx_2 F(x_1 + i\epsilon_1, x_2 + i\epsilon_2) f_2(x_2)  $$
and are there any conditions?
It does not seem obvious to me that one can exchange an integral with a distributional limit, especially if that integral is tied to a limit as well.
 A: The answer is rather straightforward:
The $\lim_{\epsilon_1 \to 0}$ can be pulled out of the integral because $G_{\epsilon_1}(x_2) :=  f(x_2) \int dx_1~ F(x_1 + i\epsilon_1, x_2 + i\epsilon_2) f_1(x_1)$ converges uniformly in $x_2$ as $\epsilon_1 \to 0$ (it is a compactly supported analytic function and the limit exists).
So we have:
$$ F(f_1, f_2) = \lim_{\epsilon_2 \to 0} \lim_{\epsilon_1 \to 0} \int dx_2 \int dx_1~ F(x_1 + i\epsilon_1, x_2 + i\epsilon_2) f_1(x_1) f_2(x_2) < \infty$$
Now, for fixed $0 < \epsilon_1 < \epsilon_2$, also the partial integrals exists (they are given by integration of an analytic and a compactly supported smooth function):
$$ \int dx_1~ F(x_1 + i\epsilon_1, x_2 + i\epsilon_2) f_1(x_1) < \infty,~ \int dx_2~ F(x_1 + i\epsilon_1, x_2 + i\epsilon_2) f_1(x_2) < \infty $$
This means (by application of Fubini's Theorem) that the order of integration is arbitrary:
$$ F(f_1,f_2) = \lim_{\epsilon_2 \to 0} \lim_{\epsilon_1 \to 0} \int dx_1~ f_1(x_1) \int dx_2 F(x_1 + i\epsilon_1, x_2 + i\epsilon_2) f(x_2) $$
This only works if both limits are taken last. It does not hold that:
$$ \lim_{\epsilon_2\to 0}\lim_{\epsilon_1 \to 0} \int dx_1~ F(x_1 + i\epsilon_1, x_2 + i\epsilon_2) f_1(x_1) < \infty,~ \int dx_2~ F(x_1 + i\epsilon_1, x_2 + i\epsilon_2) f_1(x_2) < \infty $$
