Laplace Transform of Distributions

Abstract

The classic Laplace transform is closely related to the Fourier one and has similar properties. In a way it can be seen as a modification of the latter in such a way that it can handle exponentially growing functions.

1 Definition

The classic Laplace transform is closely related to the Fourier one and has similar properties. In a way it can be seen as a modification of the latter in such a way that it can handle exponentially growing functions. To achieve this the transformable functions are required to be bounded on the left.¹ Concretely, the classic one-dimensional Laplace transform is defined on so-called original functions  $f: {\mathbb {R}}\rightarrow {\mathbb {C}}$ with the following properties:

• 1. f(t) = 0    for    t < 0.

• 2. There is a real number $\sigma _0$ such that $f(t) \textrm{e}^{-\sigma _0 t}$ is absolutely integrable over ${\mathbb {R}}$.

The greatest lower bound $\sigma _f$ satisfying the last property is called the abscissa of convergence of f.

The Laplace transform of an original function f is a function of the complex variable $s :=\sigma + \jmath \omega$ ($\sigma ,\omega \in {\mathbb {R}}$) defined by

$$\begin{aligned} F(s) :={\mathcal {L}}\left\{ f\right\} (s) :=\int _0^\infty f(t) \textrm{e}^{-st} dt \qquad \Re \{s\} > \sigma _f\,. \end{aligned}$$

(5.1)

We adopt the common convention of denoting the Laplace transform of a function f with the same letter, but capitalized. That is, for this example $F = {\mathcal {L}}\left\{ f\right\}$.

We want to extend this definition to distributions with support contained in $[0, \infty )$, that is to right-sided distributions $T \in {\mathcal {D_+'}}$. To this end note that the integrand in the above definition is the product of an original function f with an indefinitely differentiable function with unbounded support. If we multiply the latter by any indefinitely differentiable function $\gamma (t)$ with support bounded on the left and equal to 1 on a neighborhood of $[0, \infty )$, then we obtain the product of two functions with support bounded on the left without changing the value of the integral. Then, since f is an original function, for any $\sigma _0 > \sigma _f$ the product $f(t) \textrm{e}^{-\sigma _0 t}$ is a (regular) tempered distribution and $\gamma (t) \textrm{e}^{-s t} \textrm{e}^{\sigma _0 t}$, for $\Re \{s\} > \sigma _0$, a test function of fast descent. We have thus obtained a way to define the Laplace transform for a restricted class of distributions.

Definition 5.1

(Laplace transformable) A distribution $T \in {\mathcal {D_+'}}$ is said to be Laplace transformable if there exists a constant $\sigma _0 \in {\mathbb {R}}$ such that

$$\begin{aligned} T(t) \, \textrm{e}^{-\sigma _0 t} \end{aligned}$$

is a distribution in ${\mathcal {S'}}({\mathbb {R}})$. The greatest lower bound $\sigma _T$ is called the abscissa of convergence of T.

Definition 5.2

(Laplace transform) The Laplace transform of a Laplace transformable distribution T is defined by

$$\begin{aligned} {\mathcal {L}}\left\{ T\right\} :=\langle T(t) \, \textrm{e}^{-\sigma _0 t},\gamma (t) \textrm{e}^{-(s - \sigma _0) t} \rangle \quad \text {for} \quad \Re \{s\} > \sigma _0 > \sigma _T \end{aligned}$$

with $\gamma$ any indefinitely differentiable function with support bounded on the left and equal to 1 on a neighborhood of the support of T. It is commonly abbreviated by

$$\begin{aligned} \langle T(t),\textrm{e}^{-s t} \rangle \quad \text {for} \quad \Re \{s\} > \sigma _T\,. \end{aligned}$$

The right-half plane $\Re \{s\} > \sigma _T$ is called region of convergence (ROC).

If T is Laplace transformable, its transform is a well-defined number for any value of the complex parameter s with $\Re \{s\} > \sigma _T$. In other words, it is a function of s. Since s only appears as a parameter of the test function of fast descent $\gamma (t) \textrm{e}^{-(s - \sigma _0) t}$, with the continuity and linearity of distributions it is easy to see that

$$\begin{aligned} D_s \langle T(t),\textrm{e}^{-s t} \rangle = \langle T(t),-t \textrm{e}^{-s t} \rangle = \langle -t T(t),\textrm{e}^{-s t} \rangle \,. \end{aligned}$$

In addition, since $\textrm{e}^{-s t}$ is an entire analytic function, in the right half-plane $\Re \{s\} > \sigma _T$, ${\mathcal {L}}\left\{ T\right\}$ is a holomorphic function. This is an important result as it allows to use many results from complex analysis.

Higher order derivatives are obtained by iterating the above result so that we have

$$\begin{aligned} D^k_s {\mathcal {L}}\left\{ T\right\} (s) = {\mathcal {L}}\left\{ (-t)^k T\right\} (s)\,. \end{aligned}$$

(5.2)

Note that the abscissa of convergence of the derivatives is the same as the one of T.

Example 5.1: Laplace Transform of $\boldsymbol{\delta }$

The Laplace transform of $\delta$, of $\delta (t - a)$ and of $D^{k}\delta$ are

$$\begin{aligned} \langle \delta ,\textrm{e}^{-s t} \rangle = & {} 1 \\ \langle \delta (t - a),\textrm{e}^{-s t} \rangle = & {} \textrm{e}^{-s a} \\ \langle D^k\delta ,\textrm{e}^{-s t} \rangle = & {} \langle \delta ,(-1)^kD^k\textrm{e}^{-s t} \rangle = s^{k}\,. \end{aligned}$$

In all cases the region of convergence is the entire complex plane ${\mathbb {C}}$.

Example 5.2: Laplace Transform of $\boldsymbol{1_{+}(t) \, t^k/k! \, {\textrm{e}}^{at}}$

Let a be a complex number. The Laplace transform of the regular distribution $\textsf{1}_{+}(t) \, \textrm{e}^{at}$ is

$$\begin{aligned} \langle \textrm{e}^{at},\textrm{e}^{-s t} \rangle = \int _0^\infty \textrm{e}^{-(s - a) t} \, dt = \frac{1}{s - a} \end{aligned}$$

with abscissa of convergence $\sigma _{\exp (at)} = a$.

From this and (5.2), the Laplace transform of

$$\begin{aligned} \textsf{1}_{+}(t) \, \frac{t^k}{k!} \, \textrm{e}^{at} \end{aligned}$$

is readily found to be

$$\begin{aligned} {\mathcal {L}}\left\{ \textsf{1}_{+}(t) \, \frac{t^k}{k!} \, \textrm{e}^{at}\right\} = \frac{(-1)^k}{k!} D^k\frac{1}{s - a} = \frac{1}{(s - a)^{k+1}}\,. \end{aligned}$$

2 Properties

The Laplace transform is a linear operation: given two Laplace transformable distribution S and T with abscissa of convergence $\sigma _S$ resp. $\sigma _T$, the transform of their weighted sum is

$$\begin{aligned} {\mathcal {L}}\left\{ c_1 S + c_1 T\right\} = c_1 {\mathcal {L}}\left\{ S\right\} + c_2 {\mathcal {L}}\left\{ T\right\} \quad \text {for} \quad \Re \{s\} > \max (\sigma _S,\sigma _T)\,. \end{aligned}$$

Let a be a complex number. Then, from the definition, the Laplace transform of $\textrm{e}^{at} T(t)$

$$\begin{aligned} {\mathcal {L}}\left\{ \textrm{e}^{-at} T\right\} (s) = \langle \textrm{e}^{-at} T(t),\textrm{e}^{-st} \rangle = \langle T(t),\textrm{e}^{-(s+a)t} \rangle = {\mathcal {L}}\left\{ T\right\} (s + a) \end{aligned}$$

with region of convergence $\Re \{s\} > \sigma _T - \Re \{a\}$.

We saw in Example 3.6 that the convolution of distributions in ${\mathcal {D_+'}}$ is always well-defined. Therefore, the convolution of two Laplace transformable distributions S and T is well-defined. Then, for $\Re \{s\} > \sigma _0 = \max (\sigma _S,\sigma _T)$, the transform of their convolution product is by definition

$$\begin{aligned} {} & {} {{\mathcal {L}}\left\{ S *T\right\} (s)} \\ {} & {} = \langle (S *T) \textrm{e}^{-\sigma _0 t},\gamma (t) \textrm{e}^{-(s - \sigma _0)t} \rangle \\ {} & {} = \langle S(t) \otimes T(\lambda ) \textrm{e}^{-\sigma _0 \lambda },\gamma (t + \lambda ) \textrm{e}^{-(s - \sigma _0)(t + \lambda )} \rangle \,. \end{aligned}$$

By noting that, over a neighborhood of the support of $S \otimes T$, $\gamma (t + \lambda ) = 1 = \gamma (t)\gamma (\lambda )$ we can proceed further and obtain

$$\begin{aligned} {} & {} {{\mathcal {L}}\left\{ S *T\right\} (s)} \\ {} & {} = \langle S(t) \textrm{e}^{-\sigma _0 t} \otimes T(\lambda ) \textrm{e}^{-\sigma _0 \lambda },\gamma (t) \textrm{e}^{-(s - \sigma _0)t} \gamma (\lambda ) \textrm{e}^{-(s - \sigma _0)\lambda } \rangle \\ {} & {} = \langle S(t) \textrm{e}^{-\sigma _0 t},\langle T(\lambda ) \textrm{e}^{-\sigma _0 \lambda },\gamma (\lambda ) \textrm{e}^{-(s - \sigma _0)\lambda } \rangle \gamma (t) \textrm{e}^{-(s - \sigma _0)t} \rangle \\ {} & {} = \langle S(t) \textrm{e}^{-\sigma _0 t},\gamma (t) \textrm{e}^{-(s - \sigma _0)t} \rangle \langle T(\lambda ) \textrm{e}^{-\sigma _0 \lambda },\gamma (\lambda ) \textrm{e}^{-(s - \sigma _0)\lambda } \rangle \\ {} & {} = {\mathcal {L}}\left\{ S\right\} (s) {\mathcal {L}}\left\{ T\right\} (s) \end{aligned}$$

which is well-defined since, in the specified ROC, the Laplace transforms of S and T are holomorphic functions. We thus see that, as the Fourier transform, the Laplace transform converts convolutions into products

$$\begin{aligned} {\mathcal {L}}\left\{ S *T\right\} = {\mathcal {L}}\left\{ S\right\} {\mathcal {L}}\left\{ T\right\} \quad \text {for} \quad \Re \{s\} > \max (\sigma _S,\sigma _T)\,. \end{aligned}$$

(5.3)

A key advantage over the Fourier transform is that here the multiplication is between functions that are holomorphic in the specified open right-half plane.

In a similar way as we did for the Fourier transform, we can use this property to derive several additional properties in a straightforward way. Specifically, using the properties of the convolution product and the Laplace transform of the Dirac $\delta$ and related distributions (Example 5.1), we immediately obtain the properties in Table 5.1 that we have not yet discussed.

Table 5.1 Properties of the Laplace transformation

Example 5.3: Convolution of exp’s

Let k and l be natural numbers and a a complex constant. We want to calculate the following convolution product

$$\begin{aligned} \textsf{1}_{+}(t) \frac{t^k}{k!} \, \textrm{e}^{at} *\textsf{1}_{+}(t) \frac{t^l}{l!} \, \textrm{e}^{at}\,. \end{aligned}$$

From the convolution property of the Laplace transform and the results of Example 5.2 the transform of the above convolution product is

$$\begin{aligned} \frac{1}{(s - a)^k} \,\frac{1}{(s - a)^l} = \frac{1}{(s - a)^{k+l}}\,. \end{aligned}$$

With it, we find the desired result as

$$\begin{aligned} \textsf{1}_{+}(t) \frac{t^{k+l}}{(k+l)!} \, \textrm{e}^{at}\,. \end{aligned}$$

3 Inverse Laplace Transform

The Laplace transform isn’t only similar to the Fourier one, it can also be formally related to it. Consider first an original function f. By writing its Laplace transform as

$$\begin{aligned} \int _0^\infty f(t) \textrm{e}^{-\sigma t} \textrm{e}^{-\jmath \omega t} \, dt \end{aligned}$$

we see that, for every value of $\sigma > \sigma _f$, it can be interpreted as the Fourier transform of the function $f(t) \textrm{e}^{-\sigma t}$.

This relation between the two transforms can be extended to distributions. Consider a Laplace transformable distribution T. We have established that, for $\Re \{s\} > \sigma _T$, its Laplace transform is a holomorphic function of s. In addition, by definition, for every value $\sigma _0 > \sigma _T$, $T(t) \, \textrm{e}^{-\sigma _0 t}$ is a distribution of slow growth and for $\sigma > \sigma _0$, $\gamma (t) \textrm{e}^{-(\sigma - \sigma _0) t} \textrm{e}^{-\jmath \omega t}$ is a test function of fast descent for every value of $\omega$. We conclude that the Laplace transform of T considered as a function of $\omega$ for fixed $\sigma$ must be a regular distribution of slow growth. Hence, the following integral is well-defined

$$\begin{aligned} \int _{\mathbb {R}}\langle T(t) \, \textrm{e}^{-\sigma _0 t},\gamma (t) \textrm{e}^{-(\sigma - \sigma _0) t} \textrm{e}^{-\jmath \omega t} \rangle \phi (\omega ) \, d\omega \end{aligned}$$

for any $\phi (\omega ) \in {\mathcal {S}}$. This integral can be recognized as the tensor product $1(\omega ) \otimes T(t) \, \textrm{e}^{-\sigma _0 t}$ and, using Fubini’s theorem, it can be rearranged to become

$$\begin{aligned} \langle T(t) \, \textrm{e}^{-\sigma _0 t},\gamma (t) \textrm{e}^{-(\sigma - \sigma _0) t} & \int _{\mathbb {R}}\textrm{e}^{-\jmath \omega t} \phi (\omega ) \, d\omega \rangle \\ & = \langle T(t) \, \textrm{e}^{-\sigma t},\hat{\phi }(t) \rangle = \langle {\mathcal {F}}\{T(t) \, \textrm{e}^{-\sigma t}\},\phi (\omega ) \rangle \,. \end{aligned}$$

We thus obtain the claimed relation between Laplace and Fourier transforms

$$\begin{aligned} {\mathcal {L}}\left\{ T\right\} = {\mathcal {F}}\{\textrm{e}^{-\sigma t} T\} \quad \text {for} \quad \sigma > \sigma _T \end{aligned}$$

(5.4)

which gives us a formal way to invert the Laplace transform.

A first consequence of this relation is that, given the Laplace transform of a distribution T with abscissa of convergence $\sigma _T < 0$, we can immediately find its Fourier transform by setting $s = \jmath \omega$

$$\begin{aligned} \hat{T}(\omega ) = {\mathcal {L}}\left\{ T\right\} (\jmath \omega )\,. \end{aligned}$$

Another important consequence of (5.4) is that, if ${\mathcal {L}}\left\{ T\right\} = 0$ on a vertical line with $\Re \{s\} > \sigma _T$, then $T = 0$. In fact, with the above result we have that

$$\begin{aligned} 0 = \langle {\mathcal {L}}\left\{ T\right\} ,\phi \rangle = \langle {\mathcal {F}}\{\textrm{e}^{-\sigma t} T\},\phi \rangle = \langle \textrm{e}^{-\sigma t} T,\hat{\phi } \rangle \end{aligned}$$

from which we conclude that $\textrm{e}^{-\sigma t} T$ and hence T must vanish. In addition, with $T = S - U$ this implies that, if ${\mathcal {L}}\left\{ S\right\} = {\mathcal {L}}\left\{ U\right\}$ on a vertical line of the region of convergence, then $S = U$. In other words, if a function, holomorphic in an open right-half plane, is the Laplace transform of a distribution in ${\mathcal {D_+'}}$, then it is the transform of a unique distribution.

The next logical question to ask is: which holomorphic functions are transforms of a distribution? To answer this question, consider first an holomorphic function F bounded by

$$\begin{aligned} |F(s)| \le \frac{C}{|s|^2} \quad \text {for} \quad \Re \{s\} \ge \sigma _0 > 0 \end{aligned}$$

with C a constant. Then

$$\begin{aligned} \int _{-\infty }^\infty \left| F(\sigma _0 + \jmath \omega ) \textrm{e}^{\jmath \omega t} \right| \, d\omega \le \int _{-\infty }^\infty \frac{C}{\sigma _0^2 + \omega ^2} \, d\omega < \infty \end{aligned}$$

and the inverse Fourier integral

$$\begin{aligned} \frac{1}{2\pi } \int _{-\infty }^\infty F(\sigma _0 + \jmath \omega ) \textrm{e}^{\jmath \omega t} \, d\omega \end{aligned}$$

exists and defines a continuous function that we may write as $\textrm{e}^{-\sigma _0 t}f(t)$. The thus defined function f is therefore

$$\begin{aligned} f(t) = & {} \frac{\textrm{e}^{\sigma _0 t}}{2\pi } \int _{-\infty }^\infty F(\sigma _0 + \jmath \omega ) \textrm{e}^{\jmath \omega t} \, d\omega \nonumber \\ = & {} \frac{1}{2\pi } \int _{-\infty }^\infty F(\sigma _0 + \jmath \omega ) \textrm{e}^{(\sigma _0 + \jmath \omega ) t} \, d\omega \nonumber \\ = & {} \frac{1}{2\pi \jmath } \int _{\sigma _0 -\jmath \infty }^{\sigma _0 + \jmath \infty } F(s) \, \textrm{e}^{s t} \, ds \end{aligned}$$

(5.5)

and corresponds to the integral of an holomorphic function along the vertical line defined by $\Re \{s\} = \sigma _0$. If we write the variable s in its polar representation $s = R\textrm{e}^{\jmath \varphi }$ it’s easy to verify that in the right-half plane and for $t < 0$

$$\begin{aligned} |F(s) \textrm{e}^{st}| \le \frac{C}{R^2}\,. \end{aligned}$$

Therefore, if we close the integration path of the above integral by first making the line finite, then closing it along the half-circle shown in Fig. 5.1 and then taking the limit $R \rightarrow \infty$, the value of the integral remains unchanged. In fact

$$\begin{aligned} \lim _{R \rightarrow \infty } \left| \frac{1}{2\pi \jmath } \int _{\Gamma _2} F(s) \, \textrm{e}^{s t} \, ds \right| \le \lim _{R \rightarrow \infty } \frac{C}{2\pi \, R^2} \, R \pi = 0\,. \end{aligned}$$

Having closed the integration path we can now use Cauchy’s theorem and conclude that for $t < 0$, $f(t) = 0$.

Fig. 5.1

Integration path for $t < 0$

Cauchy’s theorem can also be used to show that the value of the integral is the same along any vertical line with $\Re \{s\} > 0$. To show this, we integrate along two vertical segments and close the path with horizontal ones. Since the contribution of the horizontal paths vanishes as we extend the length of the vertical ones toward infinity, we conclude that the value of the integral along the two vertical lines must be the same.

We have thus established that the function f doesn’t depend on the value of $\sigma _0$, is continuous and vanishes for $t < 0$. These characteristics make f an original function and hence a Laplace transformable distribution in ${\mathcal {D_+'}}$. Furthermore, from the definition of f and (5.4) we see that F is its Laplace transform.

Now consider the more general function $G(s) = s^k \, F(s - \sigma _G)$ with $\sigma _G \in {\mathbb {R}}$, $k \in {\mathbb {N}}$ and F as before. From the properties of the Laplace transform we know that it is the transform of the distribution $g = D^k(\textrm{e}^{\sigma _G t} \, f)$ which is also clearly in ${\mathcal {D_+'}}$. We therefore conclude that every function G that, for some $\sigma _G \in {\mathbb {R}}$, is holomorphic in the open right-half plane $\Re \{s\} > \sigma _G$ and is bounded above by a polynomial P

$$\begin{aligned} |G(s) | \le P(|s|) \qquad \Re \{s\} > \sigma _G \end{aligned}$$

(5.6)

is the Laplace transform of a distribution in ${\mathcal {D_+'}}$.

Without going into the details we also mention that the converse is also true. That is, every Laplace transformable distribution $T \in {\mathcal {D_+'}}$ is a derivative of some regular distribution associated with a continuous original function [16].

With the transforms of Examples 5.1 and  5.2 we can find the inverse Laplace transform of any rational function of s by partial fraction expansion. Note also that (5.5) corresponds to the classic inverse Laplace transform for functions.

Example 5.4: Laplace versus Fourier Transform of $\boldsymbol{1_{+}}$

In this example we calculate the Laplace transform of the Heaviside step function $\textsf{1}_{+}$. While it’s easy to obtain it directly from the definition, we calculate it from our previous results $D\textsf{1}_{+}= \delta$ (Example 2.8) and ${\mathcal {L}}\left\{ D\delta \right\} = s$ (Example 5.1). This is to compare it with the methods used in Examples 4.8 and 4.9 to obtain its Fourier transform.

By using

$$\begin{aligned} D\textsf{1}_{+}= D\delta *\textsf{1}_{+}\end{aligned}$$

and the convolution property of the Laplace transform, we obtain the following equation for the Laplace transform of $\textsf{1}_{+}$

$$\begin{aligned} s \, {\mathcal {L}}\left\{ \textsf{1}_{+}\right\} = 1\,. \end{aligned}$$

Then, since all Laplace transforms are holomorphic functions in an open right-half plane and only the zero distribution has zero as its transform, we can conclude that

$$\begin{aligned} {\mathcal {L}}\left\{ \textsf{1}_{+}\right\} = \frac{1}{s} \qquad \Re \{s\} > 0\,. \end{aligned}$$

As an extra step we want to obtain the Fourier transform of $\textsf{1}_{+}$ from its Laplace transform. The abscissa of convergence of ${\mathcal {L}}\left\{ \textsf{1}_{+}\right\}$ is not smaller than zero. Therefore, we can’t obtain it by simply setting $s = \jmath \omega$. However, in cases like this, where the abscissa of convergence is zero, given the continuity of distributions, it’s still possible to obtain the Fourier transform as a limit, so that

$$\begin{aligned} \langle {\mathcal {F}}\{\textsf{1}_{+}\},\phi \rangle = \lim _{\Re \{s\} \downarrow 0} \int _\Gamma \frac{\phi (\Im \{s\})}{s} \, \frac{ds}{\jmath } \end{aligned}$$

with $\Gamma$ a vertical line in the ROC@. The limit is only problematic around the origin, where we can integrate along a small half-circle of radius $\epsilon$

$$\begin{aligned} {} & {} {\lim _{\Re \{s\} \downarrow 0} \frac{1}{\jmath }\int _\Gamma \frac{\phi (\Im \{s\})}{s} \, ds} \\ {} & {} = \lim _{\epsilon \downarrow 0} \frac{1}{\jmath } \left\{ \int _{|\omega |>\epsilon } \frac{\phi (\omega )}{\jmath \omega } \, \jmath \, d\omega + \int _{-\pi /2}^{\pi /2} \frac{\phi (\epsilon \sin (\varphi ))}{\epsilon \textrm{e}^{\jmath \varphi }} \, \jmath \epsilon \, \textrm{e}^{\jmath \varphi } \, d\varphi \right\} \\ {} & {} = \lim _{\epsilon \downarrow 0} \left\{ \int _{|\omega |>\epsilon } \frac{\phi (\omega )}{\jmath \omega } \, d\omega \right\} + \int _{-\pi /2}^{\pi /2} \phi (0) \, d\varphi \\ {} & {} = \langle \textrm{pv} \, \frac{1}{\jmath \omega } + \pi \delta ,\phi \rangle \,. \end{aligned}$$

Example 5.5: Exploiting Continuity

To simplify the notation we assume that all appearing functions (more correctly, regular distributions) disappear for $t < 0$. For example, we write $t^k$ for $\textsf{1}_{+}(t) \, t^k$.

From the results of Example 5.2, by setting $a=0$, we can note a dualism between positive and negative powers

$$\begin{aligned} {\mathcal {L}}\left\{ \frac{t^k}{k!}\right\} = \frac{1}{s^{k+1}} \qquad \Re \{s\} > 0\,. \end{aligned}$$

If we sum the first N powers of t we obtain

$$\begin{aligned} {\mathcal {L}}\left\{ \sum _{k=0}^{N-1} \frac{t^k}{k!}\right\} = \sum _{k=0}^{N-1} \frac{1}{s^{k+1}} = \frac{1}{s} \sum _{k=0}^{N-1} \frac{1}{s^{k}}\,. \end{aligned}$$

Using the continuity of distributions we can let N tend to infinity. The original distribution converges to the exponential function, while the transform becomes a geometric series (plus a factor)

$$\begin{aligned} {\mathcal {L}}\left\{ \textrm{e}^t\right\} = \frac{1}{s} \sum _{k=0}^\infty \frac{1}{s^{k}} \end{aligned}$$

that converges for $|s| > 1$. This means that there is a right-half plane $\Re \{s\} > 1$ where the series converges and can be summed to obtain the expected result

$$\begin{aligned} {\mathcal {L}}\left\{ \textrm{e}^t\right\} = \frac{1}{s} \, \frac{1}{1 - 1/s} = \frac{1}{s - 1} \qquad \Re \{s\} > 1\,. \end{aligned}$$

Note that the last expression can also be expressed as a geometric series of positive powers of s

$$\begin{aligned} \frac{1}{s - 1} = - \sum _{k=0}^\infty s^k\,. \end{aligned}$$

However, this series only converges if $|s| < 1$. Consequently, there is no right-half plane where the series is holomorphic and hence it isn’t the Laplace transform of a distribution.

4 Extension to Several Variables

The Laplace transform can be extended to functions of several variables in the same way as we did for the Fourier transform. Let $\tau$ be an n-tuple in ${\mathbb {R}}^n$. A multi-variable original function $f: {\mathbb {R}}^n \rightarrow {\mathbb {C}}$ is a function that is an original function with respect to each variable independently, that is

• 1. $f(\tau ) = 0$ if any $\tau _i < 0, i=1,\ldots ,n$.

• 2. There is a $\sigma _0 \in {\mathbb {R}}^n$ such that $f(\tau )\textrm{e}^{-\left( \sigma _0,\tau \right) }$ is absolutely integrable over ${\mathbb {R}}^n$ .

The classic  Laplace transform is then defined as

$$\begin{aligned} {\mathcal {L}}\left\{ f\right\} (s) = \int _{{\mathbb {R}}_+^n} f(\tau ) \, \textrm{e}^{-\left( s,\tau \right) } \, d^n\tau \end{aligned}$$

(5.7)

with $s \in {\mathbb {C}}^n$ and ${\mathbb {R}}_+^n$ the n-dimensional Cartesian product of the half-line $[0,\infty )$.

With this definition we see that the definition of the Laplace transform for distributions extends to higher dimensions essentially without modification. By interpreting k as a multi-index and all variables as n-dimensional ones, all properties of Table 5.1 remain valid. The only exception is the classic inverse Laplace transform integral (5.5). As we have seen, this integral is based on the inverse Fourier transform and the factor $2\pi \jmath$ has therefore to be replaced by $(2\pi \jmath )^n$.