Values of $s$ for which $\mathcal{L}[f(t)](s)$ is valid?

Question

$$\mathcal{L}[1] = \int_0^\infty \exp(-st) \, dt = 1/s$$

I understand that in order for the integral to be defined $s$ must satisfy $\operatorname{Re}(s) > 0$.

My lecturer then said "As you can see from the formula, this can be extended to other values of $s,$ not just on the right half of the plane. In fact, for all $s, s \neq 0$." I'm a bit confused on this, as surely then the integral would not be defined for values of s.t $\operatorname{Re}(s) < 0$.

How does one determine the values of $s$ for which a Laplace transform is valid? Is it the case that once you have the formula for the transform, you can determine the values of s for which it is valid, regardless of how it affects the integral?

A second example was given $\mathcal{L}[\exp(\alpha t)](s) = \int_0^\infty \exp(-(s-\alpha)t) \, dt = 1/(s-\alpha)$ and for this example it was valid for $\operatorname{Re}(s) > \alpha$. Why could it not be extended here?

Another example: $\mathcal{L}[\exp(i\omega t)](s) = \int_0^\infty \exp(i\omega t - st) \, dt = 1/(s-i\omega)$ provided $\operatorname{Re}(s-i\omega)>0$ This was then extended to all $s\neq i\omega$

Answer 1

The result of a Laplace transform is always a holomorphic function. As you noted, the integral expression only defines this function on a subset of the complex plane ($\Re(s)>0$ in your example). But one of the very nice properties of holomorphic functions is that they can be extended to a bigger domain (this is called analytic continuation). The basic idea is, that there is a (nearly) unique extension of the function to (nearly) the complete conplex plane.

• Some isolated holes in the domain often remain, these are called poles of the function (in your example the point $s=0$.) and the function cannot possibly be defined there.
• There can also be some non-uniqueness in case of branch-cuts like in the complex-logarithm. Nothing like that is happening in your example, so Im not explaining this further.

The theory of analytic continuation can be quite complicated. But in practice there are many functions (like yours), where it is very easy. If you already have an expression for the function (like $f(s)=1/s$), that

• makes sense for a larger domain, and
• that function is holomorphic (all functions you can write with $+,-,\cdot,/,\exp,\sin,\cos,...$ are holomorphic. Notable exception is $|\cdot|$, which is non-holomorphic) Then that expression must already define the (essentially unique) analytic continuation of the function.

Because you know that the Laplace-Transform is always holomorphic, it makes sense to define the Laplace-Transform as the result of analytic continuation. This does however not mean that the integral expression is valid for all $s$. It is not.

Not sure if this precisely answers your question, but maybe it helps? :)