13.3: Exponential Type
- Page ID
- 6553
The Laplace transform is defined when the integral for it converges. Functions of exponential type are a class of functions for which the integral converges for all \(s\) with \(\text{Re} (s)\) large enough.
We say that \(f(t)\) has exponential type \(a\) if there exists an \(M\) such that \(|f(t)| < Me^{at}\) for all \(t \ge 0\).
As we’ve defined it, the exponential type of a function is not unique. For example, a function of exponential type 2 is clearly also of exponential type 3. It’s nice, but not always necessary, to find the smallest exponential type for a function.
If \(f\) has exponential type \(a\) then \(\mathcal{L} (f)\) converges absolutely for \(\text{Re} (s) > a\).
- Proof
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We prove absolute convergence by bounding
\[|f(t) e^{-st}|. \nonumber \]
The key here is that \(\text{Re} (s) > a\) implies \(\text{Re} (a - s) < 0\). So, we can write
\[\int_{0}^{\infty} |f(t) e^{-st}|\ dt \le \int_{0}^{\infty} |Me^{(a - s)t}|\ dt = \int_{0}^{\infty} Me^{\text{Re} (a - s)t} \ dt \nonumber \]
The last integral clearly converges when \(\text{Re} (a - s) < 0\). \(\text{RED}\)
Here is a list of some functions of exponential type.
\[\begin{array} {rclcl} {f(t) = e^{at}} & : & {|f(t)| < 2e^{\text{Re} (a) t}} & \ & {\text{(exponential type Re} (a))} \\ {f(t) = 1} & : & {|f(t)| < 2 = 2e^{0 - t}} & \ & {\text{(exponential type 0)}} \\ {f(t) = \cos (\omega t)} & : & {|f(t)| \le 1} & \ & {\text{(exponential type 0)}} \end{array} \nonumber \]
In the above, all of the inequalities are for \(t \ge 0\).
For \(f(t) = t\), it is clear that for any \(a > 0\) there is an \(M\) depending on \(a\) such that \(|f(t)| \le Me^{at}\) for \(t \ge 0\). In fact, it is a simple calculus exercise to show \(M = 1/(ae)\) works. So, \(f(t) = t\) has exponential type \(a\) for any \(a > 0\).
The same is true of \(t^n\). It’s worth pointing out that this follows because, if \(f\) has exponential type \(a\) and \(g\) has exponential type \(b\) then \(fg\) has exponential type \(a + b\). So, if \(t\) has exponential type \(a\) then \(t^n\) has exponential type \(na\).