2803 Domain of F(s)

Complex $s$ and region of convergence

We will allow $s$ to be complex, using as needed the properties of the complex exponential we learned in unit 1.
Example 1. In the previous note we saw that $\mathcal{L}(1) = 1/s$, valid for all $s > 0$. Let's recompute $\mathcal{L}(1)(s)$ for complex $s$. Let $s = \alpha+i\beta$ $$\begin{aligned} \mathcal{L}(1)(s)&=\int_0^\infty e^{-st}dt\\ &=\lim_{T \to \infty}\frac{e^{-st}}{-s}\bigg|_0^T\\ &=\lim_{T \to \infty}\frac{e^{-\alpha t}(\cos(\beta t)+i\sin(\beta t))}{-s}\bigg|_0^T \end{aligned}$$ This converges if $\alpha > 0$ and diverges if $\alpha < 0$. Since $\alpha = \text{Re}(s)$ we have $$\mathcal{L}(1)=1/s, \text{ for Re($s$)} > 0$$ The region Re$(s) > 0$ is called the region of convergence of the transform. It is a right half-plane.

Frequency: The Laplace transform variable $s$ can be thought of as complex frequency. It will take us a while to understand this, but we can begin here. Euler's formula says $e^{i\omega t} = \cos(\omega t) + i\sin(\omega t)$ and we call $\omega$ the angular frequency. By analogy for any complex number exponent we call $s$ the complex frequency in $e^{st}. If$s = a + i\omega$then$s$is the complex frequency and its imaginary part$\omega$ is an actual frequency of a sinusoidal oscillation.

Piecewise continuous functions and functions of exponential order

If the integral fails to converge for any s then the function does not have a Laplace transform.
Example. It is easy to see that $f(t) = e^{t^2}$ has no Laplace transform.
The problem is the $e^{t^2}$ grows too fast as $t$ gets large. Fortunately, all of the functions we are interested in do have Laplace transforms valid for Re$(s) > a$ for some value $a$.
Functions of Exponential Order
The class of functions that do have Laplace transforms are those of exponential order. Fortunately for us, all the functions we use in 18.03 are of this type.
A function is said to be of exponential order if there are numbers $a$ and $M$ such that $|f(t)| < Me^{at}$. In this case, we say that $f$ has exponential order $a$.
Examples. $1, \cos(\omega t), \sin(\omega t), t^n$ all have exponential order 0. $e^{at}$ has exponential order $a$.

A function $f(t)$ is piecewise continuous if it is continuous everywhere except at a finite number of points in any finite interval and if at thesepoints it has a jump discontinuity (i.e. a jump of finite height).
Example. The square wave is piecewise continuous.
Theorem: If $f(t)$ is piecewise continuous and of exponential order $a$ then the Laplace transform $\mathcal{L}f(s)$ converges for all $s$ with Re$(s) > a$.
Proof: Suppose Re$(s) > a$ and $|f(t)| < Me^{at}$. Then we can write $s = (a + \alpha) + i\beta$, where $\alpha > 0$. Then, since $|e^{-i\beta t}| = 1$, $$|f(t)e^{-st}|=|f(t)e^{-(a+\alpha)t}e^{-i\beta t}|=|f(t)e^{-(a+\alpha)t}|<Me^{-\alpha t}$$ Since $\int_0^\infty Me^{-\alpha t}dt$ converges for $\alpha > 0$, the Laplace transform integral also converges.

Domain of $F(s)$: For $f(t)$ we have $F(s) = 1/s$ with region of convergence Re$(s) > 0$. But, the function $1/s$ is well defined for all $s \neq 0$. The process of extending the domain of $F(s)$ from the region of convergence is called analytic continuation. In this class analytic continuation will always consist of extending $F(s)$ to the complex plane minus the zeros of the denominator.