Energy absorbed by the condenser
Lower case i usually represents i(t) at some specific instant in time. 

If i(t) = 0 at some specific time t, that tells you nothing about the voltage across the capacitor. All you know is that the voltage across the capacitor plus the voltage across the inductor must add up to the voltage across the two lower terminals (whatever that is). 

Suppose the capacitor is charged to 1V. Then the energy stored in the capacitor is C/2 with i(t) = 0 for all t. This is the arbitrary constant that you left out when you (incorrectly) did the integration. 

Energy in the capacitor is \$CV^2/2\$
Energy in the inductor is \$LI^2/2\$

If you know that Vc(t) = 0 at time = 0 and that i(t) = 0 for 0 \$\le\$ t &lt; \$\infty\$ then you can say that the energy stored in the capacitor is 0 for any t \$\ge\$ 0;