Follow-up to Max Power Transfer

Circuit equations:

$100\sin(100t) = IR_1 + L\dot{I} + \frac{Q_1}{C}$ $\frac{Q_1}{C} = I_2R_2$ $I = I_1 + I_2$ $I_1 = \dot{Q}_1$

Laplace on both sides:

$\frac{10000}{s^2+10000} = I(s)R_1 + sLI(s) + \frac{Q_1(s)}{C}$ $\frac{Q_1(s)}{C} = I_2(s)R_2$ $I(s) = I_1(s) + I_2(s)$ $I_1(s) = sQ_1(s)$

Solving for $I(s)$ gives:

$I(S) = \frac{2000000\,\left(s+100\right)}{\left(s^2+10000\right)\,\left(s^2+200\,s+30000\right)}$

Inverse Laplace transform:

$I(t) = 100\,\sin\left(100\,t\right)-50\,\sqrt{2}\,{\mathrm{e}}^{-100\,t}\,\sin\left(100\,\sqrt{2}\,t\right)$

It is easy to spot, then, that:

$V(t) - I(t) = 50\,\sqrt{2}\,{\mathrm{e}}^{-100\,t}\,\sin\left(100\,\sqrt{2}\,t\right)$

Therefore, the required integral is:

$\boxed{\int_{0}^{\infty} 50\,\sqrt{2}\,{\mathrm{e}}^{-100\,t}\,\sin\left(100\,\sqrt{2}\,t\right) \ dt = \frac{1}{3}}$

@Karan Chatrath has shown how to derive the result. I got the inspiration for this problem by noting that the complex impedance of the circuit is $1 \Omega$ (purely resistive) at the frequency of interest. This means that in AC steady-state, the source voltage and source current are exactly the same. Only in the beginning transient period are they different. This is why the integral of the difference between the two is finite, even when calculated over an infinite span of time.

By same proces as @Karan Chatrath I acquired V(s) and I(s). I than integrated in the s-domain and used the Final value theorem (Wikipedia) to evaluate. I'm gonna leave out the final algebra: Please yell at me if that's unacceptable or the following is flawed :

$\int_{0}^{t} (v(t) - i(t)) dt =\mathcal{L}\Rightarrow \frac{1}{s} \times (V(s) - I(s))$

By final value theorem:

$\lim_{t\to\infty}\int_{0}^{t} (v(t) - i(t)) dt =\lim_{s\to0} s \times \frac{1}{s} \times (V(s) - I(s)) = 1 - \frac{1}{R_1 + R_2}$

Inserting the values

$\int_{0}^{\infty} (v(t) - i(t)) dt = 1 - \frac{1}{1 + 0.5} = \frac{1}{3}$