boost inductor peak current
Limiting the inrush current can be accomplished by means of other components, not part of the boost circuit, and certinly not by increasing the ESR of the output capacitor.
As for the value of the output capacitor, it is not really that important, you select it based on the ESR and the ripple current, in order to meet the output voltage ripple spec:
ESR=Voutripple/Iripple
I will try to answer your question, though. Let's try first qualitatively, instead of mathematically.
When the switch is on, you know that the inductor is connected right across the input power and current flows through the inductor towards ground. What happens when the switch opens? You expect the voltage at the drain to rise. Well, it does, but only about 0.6V, when the diode turns on and it clamps the voltage, because the cathode of the diode is really at ground potential, because the output capacitor is discharged and it cannot charge up instantly.
So, when the transistor turns off, the inductor is still connected across the input voltage, the only difference being that there is now a diode in series to "ground". The situation is almost unchaged, so the curent continues to RISE in the inductor. As the voltage across the cap rises, things get better, but not much. Let's say the output voltage is now 1V. So when the transistor turns off, the drain voltage rises, but when it reaches 1.6V (that is, Vout+Vdiode), the diode turns on and the current still rises.
What changes is the SLOPE of the current, but it is still POSITIVE.
Mathematically:
di/dt=(Vin-1.6V)/L
When the transistor was on, the slope was dit/dt=Vin/L, also positive.
It is only when the output voltage is higher than the input, that the inductor current slope becomes negative, and so the current can actually decrease:
di/dt=(Vin-(Vd+Vout))/L this is negative when Vin<Vd+Vout