# A proof for the Final Value theorem using Dominated convergence theorem

I'm going over the proof for the Final Value theorem using the Dominated Convergence theorem on Wikipedia, I don't understand how from the equation

$$sF\left(s\right)=\int_{0}^{\infty}f\left(\frac{t}{s}\right)e^{-t}dt$$

and the fact that $$\left|f\left(\frac{t}{s}\right)e^{-t}\right|$$ is dominated by $$Me^{-t}$$ they get to the limit

$$\lim_{s\searrow0}sF\left(s\right)=\int_0^{\infty}{\alpha}e^{-t}dt=\alpha$$


Final value theorem. Suppose that the following conditions are satisfied:

1. $$f$$ is continuously differentiable and both $$f$$ and $$f'$$ have a Laplace transform
2. $$f'$$ is absolutely integrable, that is, $$\int_0^\infty|f'(\tau)|\dd\tau$$ is finite,
3. $$\lim_{t\to\infty}f(t)$$ exists and is finite,

Then,

$$$$\lim_{t\to \infty}f(t) = \lim_{s\to 0^+}sF(s).$$$$

Proof. We know that the Laplace transform of the derivative is

$$$$sF(s) - f(0^+) {}={} \lap\{f'(t)\}(s) {}={} \int_{0^+}^\infty f'(\tau) e^{-s\tau}\dd\tau.$$$$ Therefore, \begin{align} \lim_{s\to 0^+} sF(s) {}={}& f(0^+) + \lim_{s{}\to{}0^+} \int_{0^+}^{\infty}f'(\tau)e^{-s\tau}\dd\tau \notag \end{align}

We want to use the dominated convergence theorem ; Define $$\phi_s(\tau) = f'(\tau)e^{-s\tau}$$. We have $$|\phi_s(\tau)|\leq f'(\tau)$$ which is assumed to be absolutely integrable (Assumption 2). Therefore,

\begin{align} \lim_{s\to 0^+} sF(s) {}={}&f(0^+) + \int_{0^+}^{\infty} \lim_{s{}\to{}0^+} f'(\tau)e^{-s\tau}\dd\tau \notag \\ {}={}&f(0^+) + \int_{0^+}^{\infty} f'(\tau)\dd\tau \notag \\ {}={}& {f(0^+)} + \lim_{t\to\infty}f(t) - {f(0^+)}, \\ {}={}& \lim_{t\to\infty} f(t). \end{align} This completes the proof. $$\Box$$

Comment on the Wikipedia article (8 May, 2019): it is stated that with a change of variables, $$\xi = st$$, (in fact, the same symbol is used for $$t$$ and $$\xi$$), the integral becomes

$$\int_0^\infty \lim_{s\to 0^+}f(\xi/s) e^{-\xi}\dd\xi.$$

As the OP noted, there is some confusion in that proof; in fact, $$\xi$$ and $$s$$ are not independent variables (by definition), so the limit $$\lim_{s\to 0^+}f(\xi/s)$$ is not equal to $$\lim_{t\to\infty}f(t)$$.

• How did you make the step $\int_0^{\infty}\lim_{s\to 0^+} f\left(\frac{\xi}{s}\right)e^{-\xi}d\xi = \lim_{s\to 0^+} f\left(\frac{\xi}{s}\right)\int_0^{\infty}\lim_{s\to 0^+}$ if the limit depends on $\xi$? – SIMEL May 8 at 16:16
• @SIMEL Thank you for accepting my answer. Let me rewrite it in a more clear way. I'll also add the statement of FVT for completeness. – Pantelis Sopasakis May 8 at 16:45