How calculate the transfert function of boost converter with input capacitor?
If the input voltage V is assumed constant, then there is no current through C1, and it has no effect on the circuit.

In the real world, whatever source supplies V won't be ideal, it will have an equivalent output resistance, so its output will drop depending how much current is drawn from it.

There will also be resistance and inductance in the connection from the actual power source to the input of the boost converter, which will cause V to vary as the drawn current varies.

In those cases, the input capacitor smoothes out the variation in current that needs to be drawn from the actual source, and reduces the voltage ripple at the converter's input.