It can be mathematically proven that when we have a resistor in the negative feedback of an op-amp, the input voltages in the inverting and non-inverting terminals get equal.
I'm wondering if we can say the same when we have a capacitor rather than a resistor in the feedback (like the integrator op-amp circuit).
If yes, can you tell me how we could prove that mathematically?
First of all, in your diagram, the current through and voltage across the feedback capacitor are in correct. See my diagram.
simulate this circuit – Schematic created using CircuitLab
The voltage \$v_{P}-v_{N}\$ determined by the quality of the differential amplifier input stage, and the open loop gain of the amplifier. So even if the input stage is perfect, then mathematically$$v_{P}-v_{N}=\frac{v_{out}}{Avol}$$ whether there is a resistor, capacitor, inductor, or other network in the feedback path. The only condition is that the amplifier must be operating in its linear region.
This is not a proof, It is the way the amplifier is designed.
EDIT:
I'm wondering how the feedback can decrease the Vp-Vn to zero.
Sometimes I think the word "feedback" gets us off track. The input current divides into \$I_{N}\$ and \$I_{f}\$. The currents must flow somewhere. If the output cannot absorb the current \$I_{f}\$, then \$I_{N}\$ will increase causing \$V_{N}\$ to increase. This in turn drives the output voltage more negative. This allows the output to absorb more current thus reducing \$I_{N}\$. Notice that this mechanism is the same for any feedback impedance, not just capacitors.
With an ordinary 2 resistor feedback amplifier configuration, as frequency is increased, the difference voltage between the op amp's inputs increases. This happens because, as frequency increases, the op amp's open loop gain falls off at -20 dB/decade and so the difference voltage amplitude between the inputs must increase in order to try and keep the output at a constant amplitude.
With the integrator shown. As frequency increases, not only does the open loop gain fall off at -20 dB/decade but also the output falls off at -20 dB/decade. This means that the the difference voltage between the op amp's inputs stays at a fairly constant amplitude.
In the case of a conventionally configured 2 resistor feedback amplifier the peak to peak difference voltage between the inputs can become quite large.
For example, assuming the open loop gain is on its -20 dB/decade roll-off stage of its frequency response (-90 degrees).
Then, for a non-inverting configuration, at the closed loop -3 dB frequency, The peak to peak difference voltage between the inputs will be 0.707 times the peak to peak input signal.
For an inverting amplifier with the same feedback beta, at it closed loop -3 dB frequency, the open loop gain will have the same value as for the non-inverting configuration. For the same magnitude of input signal, its output will be 1-beta times smaller and because its open loop gain is the same at this frequency, the peak to peak difference voltage between its inputs will also be 1-beta times smaller or 0.707*(1-beta)*the magnitude of the input signal.
