If I make a circuit as shown using an ideal op-amp (i.e. open-loop gain ≈ ∞, unity-gain-frequency ≈ ∞ and output resistance ≈ 0), then it will behave as perfect integrator.
But, as the op-amps can not be ideal, what and all non-idealities are expected in the integrator? As per my calculations, I found that there will be zeros in the transfer function along with the pole (which is supposed to come at '0' frequency) coming at a finite non-zero frequency. And there is one more pole coming at high frequency.
If anybody can intuitively explain the behavior of the op-amp based integrator it will be very helpful for me.
As all the non-idealities are caused by the opamp the best way to see and to evaluate the deviations from the ideal behaviour is circuit simulation.
As the most important "error" the phase shift will be 90 deg at one frequency only!
This frequency always will be below the unity gain frequency (crossover through 0 dB).