In Charge Tunneling Rates in Ultrasmall Junctions section 2.1, the authors consider the problem of charge relaxation in a simple circuit shown in Figure A. The implicitly use an assumption made about the respresentation of voltage in the Laplace domain that I do not understand. To pose the question, let us first lay out the point of the calculation.
Figure A: A capacitor coupled to an arbitrary impedance $\tilde{Z}(\omega)$ and a dc voltage source $V$.
Denote the equilibrium charge across the capacitor is $Q_e$. At time $t=0$ we drop some extra charge on the capacitor, making the total capacitor charge $Q_0$. We'd like to solve for the time dependent charge $Q(t)$ on the capacitor.
From the definition of impedance we have $$\hat{V}(p) = \hat{Z}(p) \hat{I}(p)$$ where $V$ is voltage, $Z$ is impedance, and $I$ is current. Using the derivative rule for Laplace transforms (straightforward algebra/calculus), we find $$\hat{I}(p) = p \hat{Q}(p) - Q_0$$ which gives us $$\hat{V}(p) = \hat{Z}(p)(p\hat{Q}(p) - Q_0) \, .$$
To finish solving the problem, we need to express $\hat{V}$ in terms of $\hat{Q}$. In the paper Equation (9), the authors simply write $$\frac{Q_e}{pC} = \frac{\hat{Q}(p)}{C} + \hat{Z}(p)(p \hat{Q}(p) - Q_0)\ \, .$$ which implies that $$\hat{V}(p) = \frac{Q_e}{pC} - \frac{\hat{Q}(p)}{C} \, , \tag{$\star$}$$ but I don't understand why that's the case.
Why is equation $(\star)$ correct? Does it come from some considerations of the boundary conditions or some other physical reasoning, or from a simple mathematical consideration?
In this question, $\tilde{x}$ indicates a Fourier transform defined as $$\tilde{x}(\omega) \equiv \int_{-\infty}^\infty x(t) \exp(-i \omega t) \, dt$$ and $\hat{x}(p)$ indicates a Laplace transform defined as $$\hat{x}(p) \equiv \int_0^\infty x(t) \exp(-pt) \, dt \, .$$
With this Fourier convention:
A causal linear response function $Z$ has $Z(t<0)=0$.
The Fourier and Laplace transforms are related by $\hat{x}(p) = \tilde{x}(-ip)$ (again assuming causal functions).