Question about application example of OP

I review the basic OP application example recently.
In these example, OP can be implemented as integration, differential function, by changing combination of resistor and capacitance connected to it.

The example is shown in attached image. There is a capacitance (C2) connected in the negative feedback path of OP AMP and a another capacitance(C1) is connected between VI and negative input of OP AMP.

My question is
"The relation of VI/VO is "VO = -VI*(C1/C2)" ??"
I am confused because my friend who good at analog circuit design said it is not right.

Anybody can help me ? Thanks~

The equation is correct and is valid only in AC, since the current will not flow through the capacitors. In theory (considering ideal op-amp) the circuits can works with any frequency independently from the value of each capacitor. In practice, the capacitors must be choosen considering the frequency, so that at the lowest input frequency they will have a much lower reactance with respect to that of the op-amp. Furthermore, in DC since the capacitors will behave as an open circuits, the circuit will have an infinite gain, driving the op-amp in saturation due to its offset voltage. So, to have the circuit working in practice, you have to add in parallel to C2 a resistor higher than the reactance of the capacitor at the lowest input frequency. In general I suggest not to use resistor higher than some megaohm, that means reactance of capacitors not higher than some hundred of kohm.

Thank you very much, albbg.