By evaluating ∫ i^2 R \, dt , determine how the energy dissipated...

Let the capacitance be $C$, resistance be $R$, and the voltage of the battery be $V$.

The current in the circuit at any time $t$ during charging is given by:
$i = \dfrac{V}{R} e^{-t/RC}$

The energy dissipated as heat across the resistor is:
$H = \int_{0}^{\infty} i^2 R \, dt$

Substituting the expression for $i$:
$H = \int_{0}^{\infty} \left( \dfrac{V}{R} e^{-t/RC} \right)^2 R \, dt$

$H = \dfrac{V^2}{R} \int_{0}^{\infty} e^{-2t/RC} \, dt$

$H = \dfrac{V^2}{R} \left[ \dfrac{e^{-2t/RC}}{-2/RC} \right]_{0}^{\infty}$

H = \dfrac{V^2}{R} \left( 0 - \left(-\dfrac{RC}{2} \right) \right)

$H =$\dfrac{1}{2}$C V^2$

The energy stored in the capacitor when fully charged is:
$U =$\dfrac{1}{2}$C V^2$

Thus, the energy dissipated as heat is equal to the energy stored in the capacitor.