I'm trying to show that
$$\lim_{t \to \infty } \int_{0}^t w(\tau) \mathrm{ d } \tau = \lim_{t \to \infty } \int_{0}^t u(\tau) \mathrm{ d } \tau $$
When $w(t)$ is the solution of the ODE,
$$\dot{w}(t)=-w(t)+u(t), \quad w(0)=0$$
for any continuous, finite signal $u(t)$.
I would like to know if anyone disagrees with my (inelegant) proof:
We start with the solution of $\dot{w}(t)$:
$$w(t)=\int_0^t e^{-(t-\tau)}u(\tau) \mathrm{d}\tau~,$$ remember that the initial condition is zero
We proceed with taking the Laplace transform of the convolution integral where,
$$w(t)=(h \star u)(t)~, \qquad\text{ and}\qquad h(t)=e^{-t}$$
\begin{align} \mathcal{L}[w(t)]&=\mathcal{L}[(h \star u)(t)]\\ W(s) &= H(s)U(s), \quad \mathcal{L}[e^{-t}]=H(s)=\frac{1}{s+1}\\ &=\frac{1}{s+1}U(s) \end{align} now we integrate by dividing by the Laplace operator $s$
\begin{align} \mathcal{L}[\int_0^t w(\tau) \mathrm{d}\tau] &=\frac{1}{s}W(s)\\ &=\frac{1}{s}\frac{1}{s+1}U(s)\\ &=\frac{1}{s^2+s}U(s)\\ &=G(s)U(s), \quad G(s)=\frac{1}{s^2+s}\\ \end{align}
We now go back to the time domain
\begin{align} \mathcal{L}^{-1}[\frac{1}{s}W(s)] &=\mathcal{L}^{-1}[G(s)U(s)] \\ \int_0^t w(\tau) \mathrm{d}\tau & = (g\star u)(t) \\ &=\int_0^t (1-e^{-(t-\tau)}) u(\tau) \mathrm{d} \tau, \quad \mathcal{L}^{-1}[\frac{1}{s^2+s}]=1-e^{-t}\\ &=\int_0^t u(\tau) \mathrm{d} \tau - \int_0^t e^{-(t-\tau)}u(\tau)\mathrm{d}\tau \end{align}
We already know that
\begin{equation} \int_0^t e^{-(t-\tau)}u(\tau)\mathrm{d}\tau=w(t) \end{equation}
So we can substitute it into the previous equation such that
\begin{align} \int_0^t w(\tau) \mathrm{d}\tau & = \int_0^t u(\tau) \mathrm{d} \tau - w(t) \end{align}
Before we take the limit of the above equation, we look on the steady state of $\dot{w}(t)$
\begin{align} \dot{w}(t)&=-w(t)+u(t) \\ 0&=-w(t)+u(t) \\ w(t)&=u(t) \implies \lim_{t \to \infty} w(t)=u(t) \end{align}
We can now proceed with taking the limit as $t$ approaches infinity
\begin{align} \lim_{t \to \infty} \int_0^t w(\tau) \mathrm{d}\tau &= \lim_{t \to \infty} (\int_0^t u(\tau) \mathrm{d} \tau - w(t))\\ &=\lim_{t \to \infty} \int_0^t u(\tau) \mathrm{d} \tau - \lim_{t \to \infty} w(t), \quad \lim_{t \to \infty} w(t)=0 \\ &=\lim_{t \to \infty} \int_0^t u(\tau) \mathrm{d} \tau \end{align}
Since $u(t)$ is a finite signal that eventually will be zero as time approaches infinity, the limit of $w(t)$ is also zero.
Now that you've read my "proof", can you see any cheating/incorrect operations? Even if it could be shown in an easier way, is my way of reasoning correct?
Thanks in advance!
UPDATE, problem is solved
After discussion with a friend, the proof indeed has a lot simpler and elegant solution. It is as follows integrate both sides of the ODE and let $t$ go to infinity: \begin{equation} \lim_{t \to \infty} \int_0^t \dot{w}(\tau)\mathrm{d}\tau=- \lim_{t \to \infty} \int_0^t w(\tau)\mathrm{d}\tau+\lim_{t \to \infty} \int_0^t u(\tau)\mathrm{d}\tau \end{equation}
We know already that the ODE becomes stationary as $t$ approaches infinity, hence we can set the left hand side equal to zero and rearrange the equation and we obtain
\begin{equation} \lim_{t \to \infty} \int_0^t w(\tau)\mathrm{d}\tau=\lim_{t \to \infty} \int_0^t u(\tau)\mathrm{d}\tau \end{equation}