An integrator design
First I'm surprised that circuit works at all. The input of the first OpAmp is at ground, and usually in ICs ground is 0V and there's no voltage lower than that. It's difficult for an inverting amplifier to operate around 0V, because the output can't go below that voltage. Maybe you are using a macro model, that would explain the behavior.

Your integrator has two 15M resistors and an 8pF capacitor. If we omit the capacitor for a second, this is a simple inverting amplifier with a gain of -1. For a step of 1V at the input the output will be at -1V. The opamp sinks a current of 1V/15M at its output.

Ok, now we put back the capacitor and assume that it is completely discharged. The opamp still sinks 1V/15M but now the current charges the capacitor. As the voltage across the capacitor rises, more and more current flows through the resistor. The voltage exponentially approaches the final value of -1V.

The conclusion is, that a switch is required instead of a resistor to initially discharge the capacitor.

A typical circuit would look like this: