Does Leibniz's rule hold for improper integrals?

No, the equality does not hold in general.

EXAMPLE $1$

For a first example, the integral $I(x)$ as given by

$$I(x)=\int_{-\infty}^\infty\frac{\sin(xt)}{t}\,dt$$

converges uniformly for all $|x|\ge \delta>0$. But the integral of the derivative with respect to $x$, $\int_{-\infty}^\infty \cos(xt)\,dt$ diverges for all $x$.

EXAMPLE $2$

As another example, let $J(x)$ be the integral given by

$$J(x)=\int_0^\infty x^3e^{-x^2t}\,dt$$

Obviously, $J(x)=x$ for all $x$ and hence $J'(x)=1$. However,

$$\int_0^\infty (3x^2-2x^4t)e^{-x^2t}\,dt=\begin{cases}1&,x\ne 0\\\\0&,x=0\end{cases}$$

Thus, formal differentiation under the integral sign leads to an incorrect result for $x=0$ even though all integrals involved are absolutely convergent.

Sufficient Conditions for Differentiating Under the Integral

If $f(x,t)$ and $\frac{\partial f(x,t)}{\partial x}$ are continuous for all $x\in [a,b]$ and $t\in \mathbb{R}$, and if $\int_{-\infty}^\infty f(x,t)\,dt$ converges for some $x_0\in[a,b]$ and $\int_{-\infty}^\infty \frac{\partial f(x,t)}{\partial x}\,dt$ converges uniformly for all $x\in [a,b]$, then

$$\frac{d}{dx}\int_{-\infty}^\infty f(x,t)\,dt=\int_{-\infty}^\infty \frac{\partial f(x,t)}{\partial x}\,dt$$