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

Analyticity of the Finite Laplace Transform

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

Proof.

Step 1: The Setup (The Difference Quotient)

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: $\frac{F(z+h) - F(z)}{h}$.

$\frac{F(z+h) - F(z)}{h} = \frac{\int_a^b e^{-(z+h)t}f(t)dt - \int_a^b e^{-zt}f(t)dt}{h} = \frac{\int_a^b e^{-zt}e^{-ht}f(t)dt - \int_a^b e^{-zt}f(t)dt}{h} =[\text{Using the linearity of the integral and factoring out common terms: }] \frac{1}{h}[\int_a^b e^{-zt}f(t)(e^{-ht} -1)dt]$

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

Step 2: The Core Mathematical Problem.

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

To legally move the limit inside the integral, we must prove Uniform Convergence. We have to prove that as h shrinks to 0, the expression inside the integral reaches its target limit smoothly and evenly across the entire interval from a to b, without any “wild spots” where the error blows up.

Step 3: Finding the Target Limit

Before proving the limit behaves well, we have to know what the limit actually is.

If we isolate the h part, we have: $\lim_{h \to 0} \frac{e^{-ht} - 1}{h}$.

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. $\lim_{h \to 0} \frac{e^{-ht} - 1}{h} = \frac{d}{dh} (e^{-ht}) \Big|_{h=0} = -te^{-ht}\Big|_{h=0} = -t e^{-0} = -t$

In other words, the fraction is literally the definition of the derivative of the function $e^{-ht}$ evaluated at h = 0.

Step 4: Proving Uniform Convergence

The previous interchange is valid if the expression (integrand) $D_h(t) = \frac{e^{-ht} - 1}{h}$ converges uniformly to its limit -t on the interval $[a, b]$. The goal is to measure the “error” (the difference between our messy h fraction and our clean target −t) and prove that this error is locked under a ceiling that squishes down to 0, regardless of what t is.

A. 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: $\frac{e^{-ht} - 1}{h} = -t e^{-(\theta h)t}$

B. Measuring the Error. We subtract the target (−t) from the new expression to find the distance between them: $\left| \frac{e^{-ht} - 1}{h} - (-t) \right| = \left| -t e^{-\theta h t} - (-t) \right| = |t| \left| 1 - e^{-\theta h t} \right|$.
We want to prove this distance goes to 0 everywhere.

C. Bounding t. Because our integral only runs from a to b, t cannot be infinity ($t \in [a, b]$). It has a maximum possible size, define $M = \max(|a|, |b|)$, so $|t| \le M$.

D. Bounding the Exponential.
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 $|e^x-1|=|e^{\xi }||x|\leq e^{|\xi|}|x| \leq e^{|x|}|x|$.
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, $|x|=|\theta ht|\leq |h|\, |t|\leq |h|M.$
Set K = |h|M. For $|x|\leq K$, the previous bound applies: $|e^x -1| \leq e^K |x| \leq e^{|h|M}\cdot |h|M$
Define C = $e^K M = e^{|h|M}M$, then $|e^x-1| \leq C|h|$

E. The Final Squeeze.

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 $\left| \frac{e^{-ht}-1}{h}+t\right| \leq |t|\cdot C|h|\leq MC|h| = e^kM^2|h|$. Hence, we have established a “worst-case scenario” for the error: $e^kM^2|h|$.

Why is this a massive victory? Look at the worst-case scenario: $e^kM^2|h|$.

There is no t in that expression. It depends only on $h$ and the fixed bounds $a, b$, but not on the specific choice of $t$. This means the maximum possible error for the entire interval is controlled solely by h. As h shrinks to 0, the worst-case error shrinks to 0. We have achieved uniform convergence! $D_h(t) \to -t$ uniformly.

Step 5. The Ripple Effect (The Bordered Section)

We proved the h part behaves uniformly. But the whole integral also has $e^{-zt}f(t)$ sitting in it. Does that ruin things?

The factor $e^{-zt}f(t)$ is continuously bounded on a compact interval $[a, b]$ (closed and bounded). A fundamental theorem of calculus states that a continuous function on a closed interval must have a maximum value —it cannot shoot off to infinity.

If you multiply a bounded (safe) function by a uniformly converging function, the result is still a uniformly converging function. We are completely cleared to swap the limit and the integral.

Step 6: The Reward (The Swap)

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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Uniform convergence on a finite interval [a, b] justifies interchanging the limit and integral.

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

Estimate of the integral difference:

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

But the supremum goes to 0, so the whole expression goes to 0 (the factor (b - a) is a finite constant). Therefore, by the Squeeze Theorem, $\int _a^bg_h(t)dt \rightarrow \int_a^b g(t)dt$.

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Step 7. The Grand Finale.

$F'(z) = \int_a^b f(t) e^{-zt} \left( \lim_{h \to 0} \frac{e^{-ht} - 1}{h} \right) dt = \int_a^b f(t) e^{-zt} (-t) dt = - \int_a^b f(t) e^{-zt}tdt$

Since we can pass the limit inside the integral, we have this final, beautiful formula for the derivative of our function, $\boxed{F'(z) = \int_a^b -tf(t) e^{-zt}dt}$

The very last step is a sanity check. We found a formula for the derivative, but does this integral actually output a real, finite number?

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. Since the integrand is continuous and bounded on a finite interval $[a, b]$, it can be integrated safely (it is Riemann integrable)e.

Because the derivative exists and is finite for absolutely any z we choose in the complex plane, the function F(z) is, by definition, an entire function