Quantum channels describe stochastic events, so it should be completely legit to consider continuous sets of events. Integrating over a continuous set of quantum channels with some distribution give you a valid quantum channel, describing the "average" effect in some sense.
Here's an example. Consider an $X$-rotation of $90^\circ$ (i.e., $\pi/2$-pulse) on a single qubit system,
\begin{equation}
X_{\pi/2} = \exp\left(-\frac{i}2\cdot\frac{\pi}2X \right).
\end{equation}
When implementing this pulse in experiment, due to imperfect control, the rotating angle might be something like $\pi/2+\delta$ instead of $\pi/2$, where $\delta$ is an independent random variable for every instance of realization of the pulse, satisfying certain distribution $p(\delta)$ such as a zero-mean Gaussian. In this case, the average effect of the experimental implementation of $X_{\pi/2}$ can be modeled as
$$
\begin{aligned}
\mathcal E(\rho) &=\int d\delta~p(\delta)~X_{\pi/2+\delta}\rho X_{\pi/2+\delta}^\dagger\\
&=\int d\delta~p(\delta)~\exp\left(-\frac{i}2\left(\frac{\pi}2+\delta\right)X \right)\rho\exp\left(\frac{i}2\left(\frac{\pi}2+\delta\right)X \right),
\end{aligned}
$$
which can be understood as integration over the continuous parameterized set of unitary quantum channels $\mathcal E_\delta(\rho) \equiv X_{\pi/2+\delta}(\rho)X^\dagger_{\pi/2+\delta}$ according to $p(\delta)$. Note that, $\mathcal E$ is no longer a unitary channel.
PS: By expanding $\rho$ in the eigenbasis of Pauli $X$ you can simplify the expression of $\mathcal E$.
