[crackle]

I would use a 20k resistor in series with the supply to the cap. This will allow the capacitor to automatically reform at its own rate. The voltage drop across the resistor caused by the initial high ESR will slowly drop to near zero as the capacitor becomes fully charged.
If the final current (V drop/R of the series resistance) at the full working voltage of the cap is less than 0.5mA then I class it as more or less being reformed.
(Obviously the cap has to be isolated from the rest of the circuit.)

Mike