Analyticity of the Laplace Transform on a Finite Interval

To err is human, to blame it on someone else is even more human, Jacob’s Law

Proposition. Let f(t) be a complex-valued continuous function defined on a real interval [a, b]. The Laplace transform $\mathbb{F}(z) = \int_a^b e^{-zt}f(t)dt$ is analytic for any $z \in \mathbb{C}$.

Proof.

We aim to show that F(z) is complex differentiable (holomorphic) for any $z \in \mathbb{C}$.

Fix an arbitrary $z \in \mathbb{C}$. Consider a small non-zero perturbation $h \in \mathbb{C} \setminus \{0\}$. We examine the difference quotient: $\frac{F(z+h) - F(z)}{h}$.

$\frac{F(z+h) - F(z)}{h} = \frac{\int_a^b e^{-(z+h)t}f(t)dt - \int_a^b e^{-zt}f(t)dt}{h} = \frac{\int_a^b e^{-zt}e^{-ht}f(t)dt - \int_a^b e^{-zt}f(t)dt}{h} =[\text{Combine the integrals using linearity: }] \frac{1}{h}[\int_a^b e^{-zt}f(t)(e^{-ht} -1)dt]$

Since h is independent from the variable of integration:

$\frac{1}{h}[\int_a^b e^{-zt}f(t)(e^{-ht} -1)dt] = \int_a^b e^{-zt}f(t)(\frac{e^{-ht} -1}{h})dt$

We need to compute the limit as $h \to 0, \lim_{h \to 0}\frac{F(z+h) - F(z)}{h} = \lim_{h \to 0} \int_a^b e^{-zt}f(t)(\frac{e^{-ht} -1}{h}) = \int_a^b e^{-zt}f(t) \lim_{h \to 0} \left( \frac{e^{-ht} - 1}{h} \right) dt$

The crucial step is to rigorously justify why we can interchange the limit and the integral.

First, let’s compute $\lim_{h \to 0} \frac{e^{-ht} - 1}{h}$. For any differentiable function f(h), $\lim _{h\rightarrow 0}\frac{f(h)-f(0)}{h}=f'(0)$. Consider $f(h)=e^{-ht}$. Since $f(0)=e^0=1$, the limit becomes $\lim _{h\rightarrow 0}\frac{e^{-ht}-1}{h},$ which is exactly the difference quotient for f at h = 0. $\lim_{h \to 0} \frac{e^{-ht} - 1}{h} = \frac{d}{dh} (e^{-ht}) \Big|_{h=0} = -te^{-ht}\Big|_{h=0} = -t e^{-0} = -t$

The previous interchange is valid if the expression $D_h(t) = \frac{e^{-ht} - 1}{h}$ converges uniformly to its limit -t on the interval $[a, b]$.

Consider the real function $g(u) = e^{-ut}$ (treating $t$ as a constant parameter). By the Mean Value Theorem, for any real $h \neq 0$, there exists a number $\theta \in (0, 1)$ such that :$\frac{g(h) - g(0)}{h} = g'(\theta h)$.

Substituting $g(u) = e^{-ut}$ and $g'(u) = -te^{-ut}$, we get: $\frac{e^{-ht} - 1}{h} = -t e^{-(\theta h)t}$

Now, compare this to the limit -t: $\left| \frac{e^{-ht} - 1}{h} - (-t) \right| = \left| -t e^{-\theta h t} - (-t) \right| = |t| \left| 1 - e^{-\theta h t} \right|$

Let $t \in [a, b]$, set $M = \max(|a|, |b|)$, so $|t| \le M$. From the Mean Value Theorem applied to $e^x$, for any real x, $e^x-1=e^{\xi }x$ for some $\xi$ between 0 and x. Hence $|e^x-1|=|e^{\xi }||x|\leq e^{|x|}|x|$. Now restrict x to a bounded interval, say $|x|\leq K$. Then, $e^{|x|}\leq e^K$, so $|e^x-1|\leq e^K|x|, \forall |x|\leq K.$

Since $x=-\theta ht$, with $\theta \in (0,1)$. Thus, $|x|=|\theta ht|\leq |h|\, |t|\leq |h|M.$

For $x$ in a bounded range, $|1 - e^x| \le C|x|$ where C = $e^k M$. Choose h small enough so that $|h|M \le K \implies |x| \le K$, and we get $|1-e^{-\theta ht}|=|e^x-1|\leq e^K|x|\leq e^KM|h| = C|h|$.

So we obtain a uniform bound (in $t \in [a,b]$): $\sup _{t\in [a,b]}|1-e^{-\theta ht}|\leq C|h|$, and plugging this back into: $\left| \frac{e^{-ht}-1}{h}+t\right| =|t|\left| e^{-\theta ht}-1\right|$ gives $\left| \frac{e^{-ht}-1}{h}+t\right| \leq |t|\cdot C|h|\leq MC|h| = e^kM^2|h|$.

The right-hand side $e^kM^2|h|$ depends only on $h$ and the fixed bounds $a, b$, but not on the specific choice of $t$. As $h \to 0$, this bound goes to 0 uniformly for all $t \in [a, b]$. Since $D_h(t) \to -t$ uniformly, and the factor $e^{-zt}f(t)$ is continuously bounded on $[a, b]$, the product converges uniformly. Uniform convergence on a finite interval allows you to pass the limit inside the integral: $\lim _{h\rightarrow 0}\int _a^be^{-zt}f(t)\frac{e^{-ht}-1}{h}\, dt=\int _a^bf(t)e^{-zt}(-t)\, dt.$

$e^{-zt}f(t)$ is continuous on the compact interval [a, b], hence bounded. Let’s denote $g(t)=e^{-zt}f(t),g_h(t)=g(t)D_h(t).$ Since g is continuous on a compact interval, it is bounded: $|g(t)|\leq M, \forall t\in [a,b].$ Uniform convergence of $D_h$ means: $\sup_{t \in [a, b]}|D_h(t)+t|\rightarrow 0.$ Now estimate the difference between the product and its limit: $|g_h(t)-g(t)(-t)| = |g(t)| |D_h(t)+t| \leq M|D_h(t)+t| \implies$ $\sup_{t \in [a,b]}|g_h(t)-g(t)(-t)| \leq M\sup_{t\in [a,b]}|D_h(t)+t|.$ But the right-hand side goes to 0 because $D_h \rightarrow -t$ uniformly. Therefore, the product the product $g(t)D_h(t)$ converges uniformly to g(t)(-t).

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If $g_h\rightarrow g$ uniformly on a finite interval [a, b], and each $g_h$ is integrable, then $\lim _{h\rightarrow 0}\int _a^bg_h(t)dt =\int _a^b\lim _{h\rightarrow 0}g_h(t)dt = \int _a^bg(t)dt$.

Uniform convergence means: $\sup _{t\in [a,b]}|g_h(t)-g(t)|\rightarrow 0$.

$$ \begin{aligned} \left| \int_a^b g_h(t)dt -\int_a^b g(t)dt\right| &=\left| \int _a^b(g_h(t)-g(t))\, dt\right| \\[2pt] &=\text{Triangle inequality} \\[2pt] &\leq \int_a^b |g_h(t)-g(t)|dt \\[2pt] &\text{Since the convergence is uniform, we can pull out the supremum} \\[2pt] &\leq (b-a) \sup_{t\in [a,b]}|g_h(t)-g(t)|. \end{aligned} $$

But the supremum goes to 0, so the whole expression goes to 0. Therefore: $\int _a^bg_h(t)dt \rightarrow \int_a^b g(t)dt$.

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Since we can pass the limit inside the integral: $F'(z) = \int_a^b f(t) e^{-zt} \left( \lim_{h \to 0} \frac{e^{-ht} - 1}{h} \right) dt = \int_a^b f(t) e^{-zt} (-t) dt = - \int_a^b f(t) e^{-zt}tdt$

We must verify this integral exists and is finite. Let’s define $G(t, z) = -t f(t) e^{-zt}$

1. f(t) is continuous on $[a, b]$, -t and $e^{-zt}$ are continuous with respect to t. Therefore, the product $G(t, z)$ is a continuous function of t on [a, b].
2. A continuous function on a compact interval is bounded.
3. Therefore, there exists some constant $K_z$ (depending on $z$) such that $|G(t, z)| = |-t f(t) e^{-zt}| \le K_z, \forall t \in [a, b]$.
4. Since the integrand is continuous and bounded on a finite interval $[a, b]$, it is Riemann integrable. The integral exists and is finite.