# Alternative version of the Final Value theorem for Laplace Transform

Let $$f:[0,\infty) \to \mathbb{C}$$ be a continuous and bounded function such that the limit $$\lim_{T\to\infty} \frac{1}{T} \int_{0}^{T}f(t)dt = d \quad\text{exists}.$$

Let $$\hat{f}$$ be the Laplace transform of f, i.e., $$\hat{f} = \int_{0}^{\infty}e^{-st}f(t)dt.$$

Prove that $$\lim_{s\to0,\:s>0} s\hat{f} = d.$$

I have tried different ways, but for now I still did not get the whole proof.

1. I started with the following idea (non rigorous). Define $$T=\frac{1}{s}$$, then $$\lim_{s\to0,\:s>0} s\hat{f} = \lim_{T\to\infty} \frac{1}{T}\hat{f}=\lim_{T\to\infty} \frac{1}{T} \int_{0}^{T}e^{-t/T}f(t)dt$$ where the integrand converges to $$f(t)$$ and therefore the dominant convergence theorem I guess could be used. However, there are two problems: (a) $$T \in \mathbb{C}$$ and therefore I am not sure I can do that, (b) I don't want to use the dominant convergence theorem since it was not introduced in class.

2. A different approach is to use the fundamental theorem of calculus since f continuous and define: $$F(T) = F(0) + \int_{0}^{T}f(t)dt \quad \forall \; T \in [0,\infty).$$ Then from here use the formula for the Laplace transform of the derivative, but I did not manage to move on.

• It does not make sense to say $d<\infty$, since in general $\int_0^T f(t)dt\notin\mathbb R$. Did you mean to write that the limit exists, i.e. $d$ is well defined? Or did you mean to integrate over $|f|$? – Pink Panther May 7 at 7:35
• Well, actually you're right, in the text there is written "d exists", which I assumed was the same of saying d finite. Could you please explain briefly why the two things are not related? The limit can be $\infty$, but it has to be well-defined for $T \to \infty$, is that what you mean? – ofir_13 May 7 at 7:38
• On the real numbers, you have a strict linear order $<$. On the complex numbers you do not have that ordering, meaning there is no such thing as saying "$z<w$" for complex numbers $z,w\in\mathbb C$. If you have a sequence $(x_i)_i\subset\mathbb R^+$ and say $x_i\rightarrow\infty$, you are really saying $\forall T\in\mathbb R$ there exists $i$ such that $x_i>T$, but for complex numbers you cannot do that. – Pink Panther May 7 at 7:49
• Thank you very much for the insight! Ps. I think you mistyped y'' above. – ofir_13 May 7 at 7:52

Let $$g(t)=\int_0^{t}f(s)ds$$. Integrate by parts to see that $$s \hat {f} (s)=s^{2}\int_0^{\infty} (te^{-st}) (\frac {g(t)} t)dt$$. Now $$s^{2}\int_0^{\infty} (te^{-st}) (\frac {g(t)} t)dt=s^{2}\int_0^{\infty} (te^{-st}) (\frac {g(t)} t-d)dt+d$$ because $$s^{2}\int_0^{\infty} (te^{-st})dt=1$$. Suppose $$|\frac {g(t)} t-d| <\epsilon$$ for $$t \geq t_0$$. Now split the integral into integrals over $$(0,t_0)$$ and $$(t_0,\infty)$$. Can you finish the proof now?
The integral form $$t_0$$ to $$\infty$$ is bounded by $$\epsilon \int_{t_0} ^\infty s^{2}te^{-st}dt \leq \epsilon \int_{0} ^{\infty} s^{2}te^{-st}dt=\epsilon$$. The intergal from $$0$$ to $$t_0$$ tends to $$0$$ as $$s \to 0$$ because $$g(t)-td$$ is bounded on this interval and $$s^{2} \int_{0} ^{t_0} e^{-st} dt \to 0$$ by direct calculation of the integral.
• I did the following to complete the proof: $$s\hat{f}(s)=s^2 \bigg[ \int_{0}^{t_0} te^{-st}\bigg(\frac{g(t)}{t} -d\bigg)dt + \int_{t_0}^{\infty} te^{-st}\bigg(\frac{g(t)}{t} -d\bigg)dt\bigg] + d,$$ but the term $$\int_{t_0}^{\infty} te^{-st}\bigg(\frac{g(t)}{t} -d\bigg)dt < \int_{t_0}^{\infty} \big |te^{-st}\big |\bigg|\frac{g(t)}{t} -d\bigg|dt < \epsilon \int_{t_0}^{\infty} \big |te^{-st}\big |dt = \epsilon \int_{t_0}^{\infty} te^{-st}dt = \epsilon \frac{(t_0-1)e^{-st_0}}{s},$$ which vanishes as $t_0 \to \infty$. – ofir_13 May 7 at 9:59
• We are then left with $$s\hat{f}(s)=s^2 \int_{0}^{t_0} te^{-st}\bigg(\frac{g(t)}{t} -d\bigg)dt + d,$$ and taking the $\lim_{s\to0}$ at both sides complete the proof. – ofir_13 May 7 at 9:59