Chapter 3: Convergence of the Laplace Transform

Theorem (Absolute Convergence of the Laplace Transform)

Let \(f\in \mathcal{C}_{rd}([s, \infty)) \) be of exponential order \(\alpha \). Then the Laplace transform \(\mathcal{L}(f)(\cdot, s) \) exists on \(\mathbb{C}_{\mu_*(s)}(\alpha) \) and converges absolutely.

Proof

Since \(f\in \mathcal{C}_{rd}([s, \infty)) \) is of exponential order \(\alpha \) on \([s, \infty) \), there exists a constant \(K>0 \) such that
\[
|f(t)| \leq K e_{\alpha}(t, s),\quad t\in [s, \infty).
\notag\]
Note that
\[
\text{Re}_{\mu(t)} (z)\geq \text{Re}_{\mu_*(s)}(z),\quad t\in [s, \infty).
\notag\]
Thus,
\[\begin{aligned}
|1+\mu(t) z| =& 1+\mu(t) \text{Re}_{\mu(t)}(z) \\
\geq& 1+\mu(t) \text{Re}_{\mu_*(s)}(z),\quad t\in [s, \infty),.
\end{aligned}\notag\]
Hence, we obtain
\[\begin{aligned}
\left|\int_s^t f(\tau) e_{\ominus z}(\sigma(\tau), s)\Delta \tau\right|\leq& \int_s^t |f(\tau) e_{\ominus z}(\sigma(\tau), s)|\Delta\tau \\
\leq& K \int_s^t |e_{\alpha}(\tau, s)e_{\ominus z}(\sigma(\tau), s)|\Delta \tau \\
=& K\int_s^t \frac{1}{|1+\mu(\tau) z|}|e_{\alpha\ominus z}(\tau, s)| \Delta \tau \\
\leq& K\int_s^t \frac{e_{\alpha\ominus \text{Re}_{\mu_*(s)}(z)}(\tau, s)}{1+\mu(\tau)\text{Re}_{\mu_*(s)}(z)}\Delta \tau \\
=& \frac{K}{\alpha- \text{Re}_{\mu_*(s)}(z)}\int_s^t \frac{\alpha-\text{Re}_{\mu_*(s)}(z)}{1+\mu(\tau)\text{Re}_{\mu_*(s)}(z)}e_{\alpha\ominus \text{Re}_{\mu_*(s)}(z)}(\tau, s)\Delta \tau \\
=& \frac{K}{\alpha-\text{Re}_{\mu_*(s)}(z)}\int_s^t (\alpha\ominus \text{Re}_{\mu_*(s)}(z))e_{\alpha\ominus \text{Re}_{\mu_*(s)}(z)}(\tau, s)\Delta \tau \\
=& \frac{K}{\text{Re}_{\mu_*(s)}(z)-\alpha}\left(1-e_{\alpha\ominus \text{Re}_{\mu_*(s)}(z)}(t, s)\right) \\
\to& \frac{K}{\text{Re}_{\mu_*(s)}(z)-\alpha},\quad \text{as}\quad t\to\infty,\quad z\in \mathbb{C}_{\mu_*(s)}(\alpha).
\end{aligned}\notag\]
Thus,
\begin{equation}
\label{50.1}
\begin{array}{lll}
\left|\int_s^{\infty}f(\tau) e_{\ominus z}(\sigma(\tau), s) \Delta\tau\right|\leq& \int_s^{\infty} |f(\tau)e_{\ominus z}(\sigma(\tau), s)|\Delta \tau \\
\leq& \frac{K}{\text{Re}_{\mu_*(s)}(z)-\alpha},\quad z\in \mathbb{C}_{\mu_*(s)}(\alpha).
\end{array}
\end{equation}
Consequently the Laplace transform of the function \(f \) converges absolutely on \(\mathbb{C}_{\mu_*(s)}(z) \). This completes the proof.

Corollary

Let \(f\in \mathcal{C}_{rd}([s, \infty)) \) be of some exponential order. Then, by \eqref{50.1}, we have that
\[
\lim_{\text{Re}(z)\to \infty}\mathcal{L}(f)(z, s)=0.
\notag\]
Therefore
\[
\lim_{|z|\to\infty}\mathcal{L}(f)(z, s)=0
\notag\]
in the case when \(\mu_*(s)>0 \).

Theorem (Uniform Convergence of the Laplace Transform)

Let \(f \in \mathcal{C}_{rd}([s, \infty)) \) be of exponential order \(\alpha \). Then, the Laplace transform \(\mathcal{L}(f) \) converges uniformly in the half plane \(\mathbb{C}_{\mu_*(s)}(\beta) \), where \(\beta>\alpha \).

Proof

Since \(f \) is of exponential order \(\alpha \) on \([s, \infty) \), there exists a constant \(K>0 \) such that
\[
|f(t)| \leq K e_{\alpha}(t, s),\quad t\in [s, \infty).
\notag\]
For \(t\geq s \), we have
\[
\int_t^{\infty} |f(\tau) e_{\ominus z}(\sigma(\tau), s)|\Delta \tau\leq \frac{K}{\text{Re}_{\mu_*(s)}(z)-\alpha}e_{\alpha\ominus \text{Re}_{\mu_*(s)}(z)}(t, s).
\notag\]
Thus,
\begin{equation}
\label{50.2}
\int_t^{\infty} |f(\tau) e_{\ominus z}(\sigma(\tau), s)|\Delta \tau\leq \frac{K}{\beta-\alpha}e_{\alpha\ominus \text{Re}_{\mu_*(s)}(z)}(t, s),\quad t\in [s, \infty).
\end{equation}
Because
\[
1+\mu(\tau)\text{Re}_{\mu_*(s)}(z)\geq 1+\mu(\tau)\alpha,\quad \tau\in [s, t),
\notag\]
we have
\[
m_{\text{Re}_{\mu_*(s)}(z)}(t, s)\leq m_{\alpha}(t, s),\quad z\in \mathbb{C}_{\mu_*(s)}(\alpha),\quad t\in [s, \infty).
\notag\]
Hence, we arrive at
\[\begin{aligned}
e_{\alpha\ominus \text{Re}_{\mu_*(s)}(z)}(t, s)\leq& e^{\int_s^t\left(\alpha\ominus \text{Re}_{\mu_*(s)}(z)\right)(\tau)\Delta\tau} \\
=& e^{-\left(\text{Re}_{\mu_*(s)}(z)-\alpha\right)\int_s^t\frac{\Delta\tau}{1+\mu(\tau)\text{Re}_{\mu_*(s)}(z)}} \\
=& e^{-\left(\text{Re}_{\mu_*(s)}(z)-\alpha\right)m_{\text{Re}_{\mu_*(s)}(z)}(t, s)} \\
\to& 0,\quad \text{as}\quad t\to\infty.
\end{aligned}\notag\]
Therefore, for any \(\epsilon>0 \), there exists an \(r\in [s, \infty) \) such that
\[
\left|\int_t^{\infty} f(\tau) e_{\ominus z}(\sigma(\tau), s)\Delta\tau\right|\leq \epsilon,\quad t\in [r, \infty),\quad z\in \mathbb{C}_{\mu_*(s)}(\beta).
\notag\]
This completes the proof.