Analytic Continuation

It turns out that the domain of definition of the Laplace transform can be extended by means of analytic continuation [14, p. 259]. Analytic continuation is carried out by expanding a function of $s\in{\bf C}$ about all points in its domain of definition, and extending the domain of definition to all points for which the series expansion converges.

In the case of our exponential example

$$X(s) = \frac{A}{\alpha - s}, \quad ( \left\{s\right\}>\alpha) \protect$$ (B.1)

the Taylor series expansion of $X(s)$ about the point $s=s_0$ in the $s$ plane is given by

$$\begin{eqnarray*} X(s) &=& X(s_0) + (s-s_0) X^\prime (s_0) + (s-s_0)^2\frac{X^... ...\ &\isdef & \sum_{n=0}^\infty (s-s_0)^n\frac{X^{(n)}(s_0)}{n!} \end{eqnarray*}$$

where, writing $X(s)$ as $(\alpha-s)^{-1}$ and using the chain rule for differentiation,

$$\begin{eqnarray*} X^\prime (s_0) &\isdef & X^{(1)}(s_0) \isdef \left.\frac{d X(s... ...pha-s)^{-4}(-1)\right\vert _{s=s_0} = \frac{3!}{(\alpha-s)^4}\\ \end{eqnarray*}$$

and so on. We also used the factorial notation $n!\isdeftext n(n-1)(n-2)\cdots 3\cdot 2\cdot 1$, and we defined the special cases $0!\isdeftext 1$ and $X^{(0)}(s_0)\isdeftext X(s_0)$, as is normally done. The series expansion of $X(s)$ can thus be written

$$X(s) = \frac{1}{\alpha-s_0} + \frac{s-s_0}{(\alpha-s_0)^2} + \frac{(s-s_0)^2}{(\alpha-s_0)^3} + \cdots$$

$$= \sum_{n=0}^\infty \frac{(s-s_0)^n}{(\alpha-s_0)^{n+1}}. \protect$$ (B.2)

We now ask for what values of $s$ does the series Eq. (B.2) converge? The value $s=\alpha$ is particularly easy to check, since

$$X(\alpha) = \sum_{n=0}^\infty \frac{(\alpha-s_0)^n}{(\alpha-s_0)^{n+1}} = \sum_{n=0}^\infty \frac{1}{\alpha-s_0} = \infty \frac{1}{\alpha-s_0}.$$

Thus, the series clearly does not converge for $s=\alpha$, no matter what our choice of $s_0$ might be. We must therefore accept the point at infinity for $H(\alpha)$. This is eminently reasonable since the closed form Laplace transform we derived, $H(s) = 1/(\alpha - s)$ does ``blow up'' at $s=\alpha$. The point $s=\alpha$ is called a pole of $H(s) = 1/(\alpha - s)$.

More generally, let's apply the ratio test for the convergence of a geometric series. Since the $n$th term of the series is

$$\frac{(s-s_0)^n}{(\alpha-s_0)^{n+1}}$$

the ratio test demands that the ratio of term $n+1$ over term $n$ have absolute value less than $1$. That is, we require

$$1 > \left\vert \left. \frac{(s-s_0)^{n+1}}{(\alpha-s_0)^{n+2}} \r... ...lpha-s_0)^{n+1}}} \right\vert = \left\vert\frac{s-s_0}{\alpha-s_0}\right\vert,$$

or,

$$\zbox {\left\vert s-s_0\right\vert < \left\vert\alpha-s_0\right\vert.}$$

We see that the region of convergence is a circle about the point $s=s_0$ having radius approaching but not equal to $\vert\alpha-s_0\vert$. Thus, the circular disk of convergence is centered at $s=s_0$ and extends to, but does not touch, the pole at $s=\alpha$.

The analytic continuation of the domain of Eq. (B.1) is now defined as the union of the disks of convergence for all points $s_0\neq \alpha$. It is easy to see that a sequence of such disks can be chosen so as to define all points in the $s$ plane except at the pole $s=\alpha$.

In summary, the Laplace transform of an exponential $x(t)=A e^{\alpha t}$ is

$$X(s) = \frac{A}{s-\alpha}$$

and the value is well defined and finite for all $s\neq \alpha$.

Analytic continuation works for any finite number of poles of finite order,^B.2 and for an infinite number of distinct poles of finite order. It breaks down only in pathological situations such as when the Laplace transform is singular everywhere on some closed contour in the complex plane. Such pathologies do not arise in practice, so we need not be concerned about them.