Here's an example that came up in practice.

Theorem: (Cauchy-Pompeiu) Let $U \subset \mathbb{C}$ be a bounded open set with piecewise- $C^1$ boundary $\partial U$ oriented positively (see appendix B ), and let $f: \bar{U} \rightarrow \mathbb{C}$ be continuous with bounded continuous partial derivatives in $U$. Then for $z \in U$ : $$ f(z)=\frac{1}{2 \pi i} \int_{\partial U} \frac{f(\zeta)}{\zeta-z} d \zeta+\frac{1}{2 \pi i} \int_U \frac{\frac{\partial f}{\partial \bar{\zeta}}(\zeta)}{\zeta-z} \, \text{Leb}_{\mathbb C}(d \zeta) $$

In the linked source (a libretexts book written by Jiří Lebl), there is the following question:

Why can we not differentiate in $\bar z$ under the integral in the second term of the Cauchy–Pompeiu formula? Notice that it would lead to an impossible result.

[Namely, applying $\partial_{\bar z}$ to both sides, if it was possible to move $\partial_{\bar z}$ from the outside to the inside of the integral in the 2nd term, then we would get that 2nd term $=0$ and so we would conclude $\partial_{\bar z} f=0$, i.e. every $C^1$ function $f$ on $\bar U$ is holomorphic. Which is of course totally false.]

The standard differentiation under the integral result is

Let $X$ be an open subset of $\mathbb R$, and $\Omega$ be a measure space. Suppose ${\displaystyle g\colon X\times \Omega \to \mathbb R}$ satisfies the following conditions:

1. ${\displaystyle g(x,\omega )}$ is a Lebesgue-integrable function of ${\displaystyle \omega }$ for each ${\displaystyle x\in X}$.

2. For almost all ${\displaystyle \omega \in \Omega }$, the partial derivative ${\displaystyle g_{x}}$ exists for all ${\displaystyle x\in X}$.

3. There is an integrable function ${\displaystyle \theta \colon \Omega \to \mathbb {R} }$ such that ${\displaystyle |g_{x}(x,\omega)|\leq \theta (\omega )}$ for all ${\displaystyle x\in X}$ and almost every ${\displaystyle \omega \in \Omega }$.

Then, for all ${\displaystyle x\in X}$, $${\frac d{dx} \int _{\Omega }g(x,\omega )\,d\omega =\int _{\Omega}g_{x}(x,\omega )\,d\omega .}$$

The same proof (Dominated Convergence Theorem) shows the same thing if we replace $\mathbb R$ by $\mathbb C$ and $\partial_x$ by $\partial_{\bar z}$.

So in our Cauchy-Pompeiu setting, we have both $X,\Omega:= U$; and $g(z,\zeta):= \frac{\partial_{\bar \zeta} f(\zeta)}{\zeta-z} : U \times U \to \mathbb C$.

(1.) is satisfied, (2.) is almost satisfied: for all fixed $\zeta \in U$, the derivative $\partial_{\bar z} g$ exists (and equals $0$, and is in fact (3.) dominated by the integrable function $\theta \equiv 0$), for all $z\in U$ except at just one point $z=\zeta$!

So the Leibniz rule fails to apply for our Cauchy-Pompeiu integral, because a "for all" condition fails at just one point! I find this to be quite subtle: there are "for all" and "for almost all" conditions in the theorem, and it is of critical importance to remember which variable requires the "for all" condition!

(Just remember: the variable being differentiated needs "for all", and the variable being integrated just needs "for almost all". Which makes total sense.)

Here's an example that came up in practice.

Theorem: (Cauchy-Pompeiu) Let $U \subset \mathbb{C}$ be a bounded open set with piecewise- $C^1$ boundary $\partial U$ oriented positively (see appendix B ), and let $f: \bar{U} \rightarrow \mathbb{C}$ be continuous with bounded continuous partial derivatives in $U$. Then for $z \in U$ : $$ f(z)=\frac{1}{2 \pi i} \int_{\partial U} \frac{f(\zeta)}{\zeta-z} d \zeta+\frac{1}{2 \pi i} \int_U \frac{\frac{\partial f}{\partial \bar{\zeta}}(\zeta)}{\zeta-z} \, \text{Leb}_{\mathbb C}(d \zeta) $$

In the linked source (a libretexts book written by Jiří Lebl), there is the following question:

Why can we not differentiate in $\bar z$ under the integral in the second term of the Cauchy–Pompeiu formula? Notice that it would lead to an impossible result.

[Namely, applying $\partial_{\bar z}$ to both sides, if it was possible to move $\partial_{\bar z}$ from the outside to the inside of the integral in the 2nd term, then we would get that 2nd term $=0$ and so we would conclude $\partial_{\bar z} f=0$, i.e. every $C^1$ function $f$ on $\bar U$ is holomorphic. Which is of course totally false.]

The standard differentiation under the integral result is

Let $X$ be an open subset of $\mathbb R$, and $\Omega$ be a measure space. Suppose ${\displaystyle g\colon X\times \Omega \to \mathbb R}$ satisfies the following conditions:

1. ${\displaystyle g(x,\omega )}$ is a Lebesgue-integrable function of ${\displaystyle \omega }$ for each ${\displaystyle x\in X}$.

2. For almost all ${\displaystyle \omega \in \Omega }$, the partial derivative ${\displaystyle g_{x}}$ exists for all ${\displaystyle x\in X}$.

3. There is an integrable function ${\displaystyle \theta \colon \Omega \to \mathbb {R} }$ such that ${\displaystyle |g_{x}(x,\omega)|\leq \theta (\omega )}$ for all ${\displaystyle x\in X}$ and almost every ${\displaystyle \omega \in \Omega }$.

Then, for all ${\displaystyle x\in X}$, $${\frac d{dx} \int _{\Omega }g(x,\omega )\,d\omega =\int _{\Omega}g_{x}(x,\omega )\,d\omega .}$$

The same proof (Dominated Convergence Theorem) shows the same thing if we replace $\mathbb R$ by $\mathbb C$ and $\partial_x$ by $\partial_{\bar z}$.

So in our Cauchy-Pompeiu setting, we have both $X,\Omega:= U$; and $g(z,\zeta):= \frac{\partial_{\bar \zeta} f(\zeta)}{\zeta-z} : U \times U \to \mathbb C$.

(1.) is satisfied, (2.) is almost satisfied: for all fixed $\zeta \in U$, the derivative $\partial_{\bar z} g$ exists (and equals $0$, and is in fact (3.) dominated by the integrable function $\theta \equiv 0$), for all $z\in U$ except at just one point $z=\zeta$!

So the Leibniz rule fails to apply for our Cauchy-Pompeiu integral, because a "for all" condition fails at just one point! I find this to be quite subtle: there are "for all" and "for almost all" conditions in the theorem, and it is of critical importance to remember which variable requires the "for all" condition!

Perhaps due to the examples students are typically shown when learning this theorem, people are more likely to think about the "domination" condition, than the subtle "existence" condition. This is why this example stood out to me.

(Just remember: the variable being differentiated needs "for all", and the variable being integrated just needs "for almost all". Which makes total sense.)