# Pulse-excited integrator reaching infinite potential

analysiscircuit analysiskirchhoffs-lawsoperational-amplifiervoltage

I was given the following Op-Amp circuit to analyse and to find the potential V_out as a function of time, with V_in defined as the Heaviside step function:

$$V_{\text{in}}=\begin{cases}0\mathrm{V} & t<0 \\ 1\mathrm{V} & t \geq0\end{cases}$$

simulate this circuit – Schematic created using CircuitLab

I first begin by finding Vo as a function of V_in, by equating currents and using that the current going into the inverting input of OA1 must be 0A:

$$\frac{V_{\text{in}}}{R}=C\frac{\mathrm{d}V_{0}}{\mathrm{d}t}\implies V_{0}=\frac{1}{RC}\int_{-\infty}^{t'}V_{\text{in}}\:\mathrm{d}t$$

Then again at OA2:

$$\frac{V_{\text{out}}}{R}=\frac{V_0}{R}+\frac{V_{\text{ref}}}{R} \implies V_{\text{out}}=V_0+V_{\text{ref}}$$

Combining the two equations gives:

$$V_\text{out}=\frac{1}{RC}\int_{-\infty}^{t'}V_{\text{in}}\:\mathrm{d}t+V_{\text{ref}}$$

When we use V_in as defined above and with R = 100kΩ, C = 10µF and V_ref = 10V, we get:

$$V_{\text{out}}=\begin{cases}(t+10)\mathrm{V} & t\geq0 \\ 10\mathrm{V} & t<0\end{cases}$$

However, this means that we end up with:

$$\lim_{t\to\infty}V_{\text{out}}=\infty\mathrm{V}$$

Which suggests that I've done something very wrong here? Or is this just a consequence of using the ideal Op-Amp assumption?