Laplace transform

In mathematics, the Laplace transform, named after Pierre-Simon Laplace (/ləˈplɑːs/), is an integral transform that converts a function of a real variable (usually ${\displaystyle t}$, in the time domain) to a function of a complex variable ${\displaystyle s}$ (in the complex-valued frequency domain, also known as s-domain, or s-plane).

The transform is useful for converting differentiation and integration in the time domain into much easier multiplication and division in the Laplace domain (analogous to how logarithms are useful for simplifying multiplication and division into addition and subtraction). This gives the transform many applications in science and engineering, mostly as a tool for solving linear differential equations^[1] and dynamical systems by simplifying ordinary differential equations and integral equations into algebraic polynomial equations, and by simplifying convolution into multiplication.^[2][3] Once solved, the inverse Laplace transform reverts to the original domain.

The Laplace transform is defined (for suitable functions ${\displaystyle f}$) by the integral $${\displaystyle {\mathcal {L}}\{f\}(s)=\int _{0}^{\infty }f(t)e^{-st}\,dt,}$$ where s is a complex number. It is related to many other transforms, most notably the Fourier transform and the Mellin transform. Formally, the Laplace transform is converted into a Fourier transform by the substitution ${\displaystyle s=i\omega }$ where ${\displaystyle \omega }$ is real. However, unlike the Fourier transform, which gives the decomposition of a function into its components in each frequency, the Laplace transform of a function with suitable decay is an

, can be used for calculations.

The Laplace transform is named after mathematician and astronomer Pierre-Simon, Marquis de Laplace, who used a similar transform in his work on probability theory.^[4] Laplace wrote extensively about the use of generating functions (1814), and the integral form of the Laplace transform evolved naturally as a result.^[5]

Laplace's use of generating functions was similar to what is now known as the

From 1744, Leonhard Euler investigated integrals of the form $${\displaystyle z=\int X(x)e^{ax}\,dx\quad {\text{ and }}\quad z=\int X(x)x^{A}\,dx}$$ as solutions of differential equations, introducing in particular the gamma function.^[7] Joseph-Louis Lagrange was an admirer of Euler and, in his work on integrating probability density functions, investigated expressions of the form $${\displaystyle \int X(x)e^{-ax}a^{x}\,dx,}$$ which resembles a Laplace transform.^[8][9]

These types of integrals seem first to have attracted Laplace's attention in 1782, where he was following in the spirit of Euler in using the integrals themselves as solutions of equations.^[10] However, in 1785, Laplace took the critical step forward when, rather than simply looking for a solution in the form of an integral, he started to apply the transforms in the sense that was later to become popular. He used an integral of the form $${\displaystyle \int x^{s}\varphi (x)\,dx,}$$ akin to a

Laplace also recognised that

Poisson summation

, to the functional equation.

In 1929, Vannevar Bush and Norbert Wiener published Operational Circuit Analysis as a text for engineering analysis of electrical circuits, applying both Fourier transforms and operational calculus, and in which they included one of the first precessessors of the modern table of Laplace transforms. In 1934,

wrote the influential Introduction to the theory of the Fourier integral (1937).

The current widespread use of the transform (mainly in engineering) came about during and soon after World War II,^[17] replacing the earlier Heaviside operational calculus. The advantages of the Laplace transform had been emphasized by Gustav Doetsch,^[18] to whom the name Laplace transform is apparently due.

The Laplace transform of a function f(t), defined for all real numbers t ≥ 0, is the function F(s), which is a unilateral transform defined by

where s is a complex frequency-domain parameter $${\displaystyle s=\sigma +i\omega }$$ with real numbers σ and ω.

An alternate notation for the Laplace transform is ${\displaystyle {\mathcal {L}}\{f\}}$ instead of F,^[3] often written as ${\displaystyle F(s)={\mathcal {L}}\{f(t)\}}$ in an abuse of notation.

The meaning of the integral depends on types of functions of interest. A necessary condition for existence of the integral is that f must be

(${\displaystyle |f(t)|\leq Ae^{B|t|}}$), the integral can be understood to be a (proper) , and this is dealt with below.

One can define the Laplace transform of a finite Borel measure μ by the Lebesgue integral^[19] $${\displaystyle {\mathcal {L}}\{\mu \}(s)=\int _{[0,\infty )}e^{-st}\,d\mu (t).}$$

An important special case is where μ is a probability measure, for example, the Dirac delta function. In operational calculus, the Laplace transform of a measure is often treated as though the measure came from a probability density function f. In that case, to avoid potential confusion, one often writes $${\displaystyle {\mathcal {L}}\{f\}(s)=\int _{0^{-}}^{\infty }f(t)e^{-st}\,dt,}$$ where the lower limit of 0⁻ is shorthand notation for $${\displaystyle \lim _{\varepsilon \to 0^{+}}\int _{-\varepsilon }^{\infty }.}$$

This limit emphasizes that any point mass located at 0 is entirely captured by the Laplace transform. Although with the Lebesgue integral, it is not necessary to take such a limit, it does appear more naturally in connection with the Laplace–Stieltjes transform.

When one says "the Laplace transform" without qualification, the unilateral or one-sided transform is usually intended. The Laplace transform can be alternatively defined as the bilateral Laplace transform, or two-sided Laplace transform, by extending the limits of integration to be the entire real axis. If that is done, the common unilateral transform simply becomes a special case of the bilateral transform, where the definition of the function being transformed is multiplied by the Heaviside step function.

The bilateral Laplace transform F(s) is defined as follows:

An alternate notation for the bilateral Laplace transform is ${\displaystyle {\mathcal {B}}\{f\}}$, instead of F.

Two integrable functions have the same Laplace transform only if they differ on a set of

from one function space into another in many other function spaces as well, although there is usually no easy characterization of the range.

Typical function spaces in which this is true include the spaces of bounded continuous functions, the space

on (0, ∞). The Laplace transform is also defined and injective for suitable spaces of tempered distributions.

In these cases, the image of the Laplace transform lives in a space of analytic functions in the region of convergence. The inverse Laplace transform is given by the following complex integral, which is known by various names (the Bromwich integral, the Fourier–Mellin integral, and Mellin's inverse formula):

where γ is a real number so that the contour path of integration is in the region of convergence of F(s). In most applications, the contour can be closed, allowing the use of the

.

In practice, it is typically more convenient to decompose a Laplace transform into known transforms of functions obtained from a table and construct the inverse by inspection.

Probability theory

In pure and applied probability, the Laplace transform is defined as an expected value. If X is a random variable with probability density function f, then the Laplace transform of f is given by the expectation $${\displaystyle {\mathcal {L}}\{f\}(s)=\operatorname {E} \left[e^{-sX}\right],}$$ where ${\displaystyle \operatorname {E} [r]}$ is the expectation of random variable ${\displaystyle r}$.

By

.

Of particular use is the ability to recover the cumulative distribution function of a continuous random variable X by means of the Laplace transform as follows:^[20] $${\displaystyle F_{X}(x)={\mathcal {L}}^{-1}\left\{{\frac {1}{s}}\operatorname {E} \left[e^{-sX}\right]\right\}(x)={\mathcal {L}}^{-1}\left\{{\frac {1}{s}}{\mathcal {L}}\{f\}(s)\right\}(x).}$$

The Laplace transform can be alternatively defined in a purely algebraic manner by applying a

Region of convergence

If f is a locally integrable function (or more generally a Borel measure locally of bounded variation), then the Laplace transform F(s) of f converges provided that the limit $${\displaystyle \lim _{R\to \infty }\int _{0}^{R}f(t)e^{-st}\,dt}$$ exists.

The Laplace transform converges absolutely if the integral $${\displaystyle \int _{0}^{\infty }\left|f(t)e^{-st}\right|\,dt}$$ exists as a proper Lebesgue integral. The Laplace transform is usually understood as conditionally convergent, meaning that it converges in the former but not in the latter sense.

The set of values for which F(s) converges absolutely is either of the form Re(s) > a or Re(s) ≥ a, where a is an

.

Similarly, the set of values for which F(s) converges (conditionally or absolutely) is known as the region of conditional convergence, or simply the region of convergence (ROC). If the Laplace transform converges (conditionally) at s = s₀, then it automatically converges for all s with Re(s) > Re(s₀). Therefore, the region of convergence is a half-plane of the form Re(s) > a, possibly including some points of the boundary line Re(s) = a.

In the region of convergence Re(s) > Re(s₀), the Laplace transform of f can be expressed by integrating by parts as the integral $${\displaystyle F(s)=(s-s_{0})\int _{0}^{\infty }e^{-(s-s_{0})t}\beta (t)\,dt,\quad \beta (u)=\int _{0}^{u}e^{-s_{0}t}f(t)\,dt.}$$

That is, F(s) can effectively be expressed, in the region of convergence, as the absolutely convergent Laplace transform of some other function. In particular, it is analytic.

There are several Paley–Wiener theorems concerning the relationship between the decay properties of f, and the properties of the Laplace transform within the region of convergence.

In engineering applications, a function corresponding to a linear time-invariant (LTI) system is stable if every bounded input produces a bounded output. This is equivalent to the absolute convergence of the Laplace transform of the impulse response function in the region Re(s) ≥ 0. As a result, LTI systems are stable, provided that the poles of the Laplace transform of the impulse response function have negative real part.

This ROC is used in knowing about the causality and stability of a system.

The Laplace transform's key property is that it converts differentiation and integration in the time domain into multiplication and division by s in the Laplace domain. Thus, the Laplace variable s is also known as an operator variable in the Laplace domain: either the derivative operator or (for s⁻¹) the integration operator.

Given the functions f(t) and g(t), and their respective Laplace transforms F(s) and G(s), $${\displaystyle {\begin{aligned}f(t)&={\mathcal {L}}^{-1}\{F(s)\},\\g(t)&={\mathcal {L}}^{-1}\{G(s)\},\end{aligned}}}$$

the following table is a list of properties of unilateral Laplace transform:^[24]

Properties of the unilateral Laplace transform

Property	Time domain	s domain	Comment
Linearity	${\displaystyle af(t)+bg(t)\ }$	${\displaystyle aF(s)+bG(s)\ }$	Can be proved using basic rules of integration.
Frequency-domain derivative	${\displaystyle tf(t)\ }$	${\displaystyle -F'(s)\ }$	F′ is the first derivative of F with respect to s.
Frequency-domain general derivative	${\displaystyle t^{n}f(t)\ }$	${\displaystyle (-1)^{n}F^{(n)}(s)\ }$	More general form, nth derivative of F(s).
Derivative	${\displaystyle f'(t)\ }$	${\displaystyle sF(s)-f(0^{-})\ }$	f is assumed to be a differentiable function, and its derivative is assumed to be of exponential type. This can then be obtained by integration by parts
Second derivative	${\displaystyle f''(t)\ }$	${\textstyle s^{2}F(s)-sf(0^{-})-f'(0^{-})\ }$	f is assumed twice differentiable and the second derivative to be of exponential type. Follows by applying the Differentiation property to f′(t).
General derivative	${\displaystyle f^{(n)}(t)\ }$	${\displaystyle s^{n}F(s)-\sum _{k=1}^{n}s^{n-k}f^{(k-1)}(0^{-})\ }$	f is assumed to be n-times differentiable, with nth derivative of exponential type. Follows by mathematical induction.
Frequency-domain integration	${\displaystyle {\frac {1}{t}}f(t)\ }$	${\displaystyle \int _{s}^{\infty }F(\sigma )\,d\sigma \ }$	This is deduced using the nature of frequency differentiation and conditional convergence.
Time-domain integration	${\displaystyle \int _{0}^{t}f(\tau )\,d\tau =(u*f)(t)}$	${\displaystyle {1 \over s}F(s)}$	u(t) is the Heaviside step function and (u ∗ f)(t) is the convolution of u(t) and f(t).
Frequency shifting	${\displaystyle e^{at}f(t)}$	${\displaystyle F(s-a)\ }$
Time shifting	${\displaystyle f(t-a)u(t-a)}$ ${\displaystyle f(t)u(t-a)\ }$	${\displaystyle e^{-as}F(s)\ }$ ${\displaystyle e^{-as}{\mathcal {L}}\{f(t+a)\}}$	a > 0, u(t) is the Heaviside step function
Time scaling	${\displaystyle f(at)}$	${\displaystyle {\frac {1}{a}}F\left({s \over a}\right)}$	a > 0
Multiplication	${\displaystyle f(t)g(t)}$	${\displaystyle {\frac {1}{2\pi i}}\lim _{T\to \infty }\int _{c-iT}^{c+iT}F(\sigma )G(s-\sigma )\,d\sigma \ }$	The integration is done along the vertical line Re(σ) = c that lies entirely within the region of convergence of F.[25]
Convolution	${\displaystyle (f*g)(t)=\int _{0}^{t}f(\tau )g(t-\tau )\,d\tau }$	${\displaystyle F(s)\cdot G(s)\ }$
Circular convolution	${\displaystyle (f*g)(t)=\int _{0}^{T}f(\tau )g(t-\tau )\,d\tau }$	${\displaystyle F(s)\cdot G(s)\ }$	For periodic functions with period T.
Complex conjugation	${\displaystyle f^{*}(t)}$	${\displaystyle F^{*}(s^{*})}$
Periodic function	${\displaystyle f(t)}$	${\displaystyle {1 \over 1-e^{-Ts}}\int _{0}^{T}e^{-st}f(t)\,dt}$	f(t) is a periodic function of period T so that f(t) = f(t + T), for all t ≥ 0. This is the result of the time shifting property and the geometric series.
Periodic summation	${\displaystyle f_{P}(t)=\sum _{n=0}^{\infty }f(t-Tn)}$ ${\displaystyle f_{P}(t)=\sum _{n=0}^{\infty }(-1)^{n}f(t-Tn)}$	${\displaystyle F_{P}(s)={\frac {1}{1-e^{-Ts}}}F(s)}$ ${\displaystyle F_{P}(s)={\frac {1}{1+e^{-Ts}}}F(s)}$

Initial value theorem

${\displaystyle f(0^{+})=\lim _{s\to \infty }{sF(s)}.}$

Final value theorem

${\displaystyle f(\infty )=\lim _{s\to 0}{sF(s)}}$, if all

poles

of ${\displaystyle sF(s)}$ are in the left half-plane.

The final value theorem is useful because it gives the long-term behaviour without having to perform

partial fraction

decompositions (or other difficult algebra). If F(s) has a pole in the right-hand plane or poles on the imaginary axis (e.g., if ${\displaystyle f(t)=e^{t}}$ or ${\displaystyle f(t)=\sin(t)}$), then the behaviour of this formula is undefined.

The Laplace transform can be viewed as a continuous analogue of a power series.^[26] If a(n) is a discrete function of a positive integer n, then the power series associated to a(n) is the series $${\displaystyle \sum _{n=0}^{\infty }a(n)x^{n}}$$ where x is a real variable (see Z-transform). Replacing summation over n with integration over t, a continuous version of the power series becomes $${\displaystyle \int _{0}^{\infty }f(t)x^{t}\,dt}$$ where the discrete function a(n) is replaced by the continuous one f(t).

Changing the base of the power from x to e gives $${\displaystyle \int _{0}^{\infty }f(t)\left(e^{\ln {x}}\right)^{t}\,dt}$$

For this to converge for, say, all bounded functions f, it is necessary to require that ln x < 0. Making the substitution −s = ln x gives just the Laplace transform: $${\displaystyle \int _{0}^{\infty }f(t)e^{-st}\,dt}$$

In other words, the Laplace transform is a continuous analog of a power series, in which the discrete parameter n is replaced by the continuous parameter t, and x is replaced by e^−s.

The quantities $${\displaystyle \mu _{n}=\int _{0}^{\infty }t^{n}f(t)\,dt}$$

are the moments of the function f. If the first n moments of f converge absolutely, then by repeated

, $${\displaystyle (-1)^{n}({\mathcal {L}}f)^{(n)}(0)=\mu _{n}.}$$ This is of special significance in probability theory, where the moments of a random variable X are given by the expectation values ${\displaystyle \mu _{n}=\operatorname {E} [X^{n}]}$. Then, the relation holds $${\displaystyle \mu _{n}=(-1)^{n}{\frac {d^{n}}{ds^{n}}}\operatorname {E} \left[e^{-sX}\right](0).}$$

It is often convenient to use the differentiation property of the Laplace transform to find the transform of a function's derivative. This can be derived from the basic expression for a Laplace transform as follows: $${\displaystyle {\begin{aligned}{\mathcal {L}}\left\{f(t)\right\}&=\int _{0^{-}}^{\infty }e^{-st}f(t)\,dt\\[6pt]&=\left[{\frac {f(t)e^{-st}}{-s}}\right]_{0^{-}}^{\infty }-\int _{0^{-}}^{\infty }{\frac {e^{-st}}{-s}}f'(t)\,dt\quad {\text{(by parts)}}\\[6pt]&=\left[-{\frac {f(0^{-})}{-s}}\right]+{\frac {1}{s}}{\mathcal {L}}\left\{f'(t)\right\},\end{aligned}}}$$ yielding $${\displaystyle {\mathcal {L}}\{f'(t)\}=s\cdot {\mathcal {L}}\{f(t)\}-f(0^{-}),}$$ and in the bilateral case, $${\displaystyle {\mathcal {L}}\{f'(t)\}=s\int _{-\infty }^{\infty }e^{-st}f(t)\,dt=s\cdot {\mathcal {L}}\{f(t)\}.}$$

The general result $${\displaystyle {\mathcal {L}}\left\{f^{(n)}(t)\right\}=s^{n}\cdot {\mathcal {L}}\{f(t)\}-s^{n-1}f(0^{-})-\cdots -f^{(n-1)}(0^{-}),}$$ where ${\displaystyle f^{(n)}}$ denotes the nth derivative of f, can then be established with an inductive argument.

A useful property of the Laplace transform is the following: $${\displaystyle \int _{0}^{\infty }f(x)g(x)\,dx=\int _{0}^{\infty }({\mathcal {L}}f)(s)\cdot ({\mathcal {L}}^{-1}g)(s)\,ds}$$ under suitable assumptions on the behaviour of ${\displaystyle f,g}$ in a right neighbourhood of ${\displaystyle 0}$ and on the decay rate of ${\displaystyle f,g}$ in a left neighbourhood of ${\displaystyle \infty }$. The above formula is a variation of integration by parts, with the operators ${\displaystyle {\frac {d}{dx}}}$ and ${\displaystyle \int \,dx}$ being replaced by ${\displaystyle {\mathcal {L}}}$ and ${\displaystyle {\mathcal {L}}^{-1}}$. Let us prove the equivalent formulation: $${\displaystyle \int _{0}^{\infty }({\mathcal {L}}f)(x)g(x)\,dx=\int _{0}^{\infty }f(s)({\mathcal {L}}g)(s)\,ds.}$$

By plugging in ${\displaystyle ({\mathcal {L}}f)(x)=\int _{0}^{\infty }f(s)e^{-sx}\,ds}$ the left-hand side turns into: $${\displaystyle \int _{0}^{\infty }\int _{0}^{\infty }f(s)g(x)e^{-sx}\,ds\,dx,}$$ but assuming Fubini's theorem holds, by reversing the order of integration we get the wanted right-hand side.

This method can be used to compute integrals that would otherwise be difficult to compute using elementary methods of real calculus. For example, $${\displaystyle \int _{0}^{\infty }{\frac {\sin x}{x}}dx=\int _{0}^{\infty }{\mathcal {L}}(1)(x)\sin xdx=\int _{0}^{\infty }1\cdot {\mathcal {L}}(\sin )(x)dx=\int _{0}^{\infty }{\frac {dx}{x^{2}+1}}={\frac {\pi }{2}}.}$$

The (unilateral) Laplace–Stieltjes transform of a function g : ℝ → ℝ is defined by the

$${\displaystyle \{{\mathcal {L}}^{*}g\}(s)=\int _{0}^{\infty }e^{-st}\,d\,g(t)~.}$$

The function g is assumed to be of bounded variation. If g is the antiderivative of f:

$${\displaystyle g(x)=\int _{0}^{x}f(t)\,d\,t}$$

then the Laplace–Stieltjes transform of g and the Laplace transform of f coincide. In general, the Laplace–Stieltjes transform is the Laplace transform of the

Fourier transform

Let ${\displaystyle f}$ be a complex-valued Lebesgue integrable function supported on ${\displaystyle [0,\infty )}$, and let ${\displaystyle F(s)={\mathcal {L}}f(s)}$ be its Laplace transform. Then, within the region of convergence, we have

${\displaystyle F(\sigma +i\tau )=\int _{0}^{\infty }f(t)e^{-\sigma t}e^{-i\tau t}\,dt,}$

which is the Fourier transform of the function ${\displaystyle f(t)e^{-\sigma t}}$.^[28]

Indeed, the

of the function. This perspective has applications in probability theory.

Formally, the Fourier transform is equivalent to evaluating the bilateral Laplace transform with imaginary argument s = iω[29]^[30] when the condition explained below is fulfilled,

$${\displaystyle {\begin{aligned}{\hat {f}}(\omega )&={\mathcal {F}}\{f(t)\}\\[4pt]&={\mathcal {L}}\{f(t)\}|_{s=i\omega }=F(s)|_{s=i\omega }\\[4pt]&=\int _{-\infty }^{\infty }e^{-i\omega t}f(t)\,dt~.\end{aligned}}}$$

This convention of the Fourier transform (${\displaystyle {\hat {f}}_{3}(\omega )}$ in

or dynamical system.

The above relation is valid as stated if and only if the region of convergence (ROC) of F(s) contains the imaginary axis, σ = 0.

For example, the function f(t) = cos(ω₀t) has a Laplace transform F(s) = s/(s² + ω₀²) whose ROC is Re(s) > 0. As s = iω₀ is a pole of F(s), substituting s = iω in F(s) does not yield the Fourier transform of f(t)u(t), which contains terms proportional to the

δ(ω ± ω₀).

However, a relation of the form $${\displaystyle \lim _{\sigma \to 0^{+}}F(\sigma +i\omega )={\hat {f}}(\omega )}$$ holds under much weaker conditions. For instance, this holds for the above example provided that the limit is understood as a

.

Mellin transform

The Mellin transform and its inverse are related to the two-sided Laplace transform by a simple change of variables.

If in the Mellin transform $${\displaystyle G(s)={\mathcal {M}}\{g(\theta )\}=\int _{0}^{\infty }\theta ^{s}g(\theta )\,{\frac {d\theta }{\theta }}}$$ we set θ = e^−t we get a two-sided Laplace transform.

The unilateral or one-sided Z-transform is simply the Laplace transform of an ideally sampled signal with the substitution of $${\displaystyle z{\stackrel {\mathrm {def} }{{}={}}}e^{sT},}$$ where T = 1/f_s is the

).

Let $${\displaystyle \Delta _{T}(t)\ {\stackrel {\mathrm {def} }{=}}\ \sum _{n=0}^{\infty }\delta (t-nT)}$$ be a sampling impulse train (also called a Dirac comb) and $${\displaystyle {\begin{aligned}x_{q}(t)&{\stackrel {\mathrm {def} }{{}={}}}x(t)\Delta _{T}(t)=x(t)\sum _{n=0}^{\infty }\delta (t-nT)\\&=\sum _{n=0}^{\infty }x(nT)\delta (t-nT)=\sum _{n=0}^{\infty }x[n]\delta (t-nT)\end{aligned}}}$$ be the sampled representation of the continuous-time x(t) $${\displaystyle x[n]{\stackrel {\mathrm {def} }{{}={}}}x(nT)~.}$$

The Laplace transform of the sampled signal x_q(t) is $${\displaystyle {\begin{aligned}X_{q}(s)&=\int _{0^{-}}^{\infty }x_{q}(t)e^{-st}\,dt\\&=\int _{0^{-}}^{\infty }\sum _{n=0}^{\infty }x[n]\delta (t-nT)e^{-st}\,dt\\&=\sum _{n=0}^{\infty }x[n]\int _{0^{-}}^{\infty }\delta (t-nT)e^{-st}\,dt\\&=\sum _{n=0}^{\infty }x[n]e^{-nsT}~.\end{aligned}}}$$

This is the precise definition of the unilateral Z-transform of the discrete function x[n]

$${\displaystyle X(z)=\sum _{n=0}^{\infty }x[n]z^{-n}}$$ with the substitution of z → e^sT.

Comparing the last two equations, we find the relationship between the unilateral Z-transform and the Laplace transform of the sampled signal, $${\displaystyle X_{q}(s)=X(z){\Big |}_{z=e^{sT}}.}$$

The similarity between the Z- and Laplace transforms is expanded upon in the theory of

.

The integral form of the Borel transform $${\displaystyle F(s)=\int _{0}^{\infty }f(z)e^{-sz}\,dz}$$ is a special case of the Laplace transform for f an entire function of exponential type, meaning that $${\displaystyle |f(z)|\leq Ae^{B|z|}}$$ for some constants A and B. The generalized Borel transform allows a different weighting function to be used, rather than the exponential function, to transform functions not of exponential type. Nachbin's theorem gives necessary and sufficient conditions for the Borel transform to be well defined.

Since an ordinary Laplace transform can be written as a special case of a two-sided transform, and since the two-sided transform can be written as the sum of two one-sided transforms, the theory of the Laplace-, Fourier-, Mellin-, and Z-transforms are at bottom the same subject. However, a different point of view and different characteristic problems are associated with each of these four major integral transforms.

The following table provides Laplace transforms for many common functions of a single variable.^[31][32] For definitions and explanations, see the Explanatory Notes at the end of the table.

Because the Laplace transform is a linear operator,

• The Laplace transform of a sum is the sum of Laplace transforms of each term.$${\displaystyle {\mathcal {L}}\{f(t)+g(t)\}={\mathcal {L}}\{f(t)\}+{\mathcal {L}}\{g(t)\}}$$
• The Laplace transform of a multiple of a function is that multiple times the Laplace transformation of that function.$${\displaystyle {\mathcal {L}}\{af(t)\}=a{\mathcal {L}}\{f(t)\}}$$

Using this linearity, and various

, and complex number (etc.) properties and/or identities, some Laplace transforms can be obtained from others more quickly than by using the definition directly.

The unilateral Laplace transform takes as input a function whose time domain is the

reals, which is why all of the time domain functions in the table below are multiples of the Heaviside step function, u(t).

The entries of the table that involve a time delay τ are required to be causal (meaning that τ > 0). A causal system is a system where the impulse response h(t) is zero for all time t prior to t = 0. In general, the region of convergence for causal systems is not the same as that of anticausal systems.

Selected Laplace transforms

Function	Time domain ${\displaystyle f(t)={\mathcal {L}}^{-1}\{F(s)\}}$	Laplace s-domain ${\displaystyle F(s)={\mathcal {L}}\{f(t)\}}$	Region of convergence	Reference
unit impulse	${\displaystyle \delta (t)\ }$	${\displaystyle 1}$	all s	inspection
delayed impulse	${\displaystyle \delta (t-\tau )\ }$	${\displaystyle e^{-\tau s}\ }$	all s	time shift of unit impulse
unit step	${\displaystyle u(t)\ }$	${\displaystyle {1 \over s}}$	${\displaystyle \operatorname {Re} (s)>0}$	integrate unit impulse
delayed unit step	${\displaystyle u(t-\tau )\ }$	${\displaystyle {\frac {1}{s}}e^{-\tau s}}$	${\displaystyle \operatorname {Re} (s)>0}$	time shift of unit step
product of delayed function and delayed step	${\displaystyle f(t-\tau )u(t-\tau )}$	${\displaystyle e^{-s\tau }{\mathcal {L}}\{f(t)\}}$		u-substitution, ${\displaystyle u=t-\tau }$
rectangular impulse	${\displaystyle u(t)-u(t-\tau )}$	${\displaystyle {\frac {1}{s}}(1-e^{-\tau s})}$	${\displaystyle \operatorname {Re} (s)>0}$
ramp	${\displaystyle t\cdot u(t)\ }$	${\displaystyle {\frac {1}{s^{2}}}}$	${\displaystyle \operatorname {Re} (s)>0}$	integrate unit impulse twice
nth power (for integer n)	${\displaystyle t^{n}\cdot u(t)}$	${\displaystyle {n! \over s^{n+1}}}$	${\displaystyle \operatorname {Re} (s)>0}$ (n > −1)	integrate unit step n times
qth power (for complex q)	${\displaystyle t^{q}\cdot u(t)}$	${\displaystyle {\operatorname {\Gamma } (q+1) \over s^{q+1}}}$	${\displaystyle \operatorname {Re} (s)>0}$ ${\displaystyle \operatorname {Re} (q)>-1}$	[33][34]
nth root	${\displaystyle {\sqrt[{n}]{t}}\cdot u(t)}$	${\displaystyle {1 \over s^{{\frac {1}{n}}+1}}\operatorname {\Gamma } \left({\frac {1}{n}}+1\right)}$	${\displaystyle \operatorname {Re} (s)>0}$	Set q = 1/n above.
nth power with frequency shift	${\displaystyle t^{n}e^{-\alpha t}\cdot u(t)}$	${\displaystyle {\frac {n!}{(s+\alpha )^{n+1}}}}$	${\displaystyle \operatorname {Re} (s)>-\alpha }$	Integrate unit step, apply frequency shift
delayed nth power with frequency shift	${\displaystyle (t-\tau )^{n}e^{-\alpha (t-\tau )}\cdot u(t-\tau )}$	${\displaystyle {\frac {n!\cdot e^{-\tau s}}{(s+\alpha )^{n+1}}}}$	${\displaystyle \operatorname {Re} (s)>-\alpha }$	integrate unit step, apply frequency shift, apply time shift
exponential decay	${\displaystyle e^{-\alpha t}\cdot u(t)}$	${\displaystyle {1 \over s+\alpha }}$	${\displaystyle \operatorname {Re} (s)>-\alpha }$	Frequency shift of unit step
two-sided exponential decay (only for bilateral transform)	${\displaystyle e^{-\alpha |t|}\ }$	${\displaystyle {2\alpha \over \alpha ^{2}-s^{2}}}$	${\displaystyle -\alpha <\operatorname {Re} (s)<\alpha }$	Frequency shift of unit step
exponential approach	${\displaystyle (1-e^{-\alpha t})\cdot u(t)\ }$	${\displaystyle {\frac {\alpha }{s(s+\alpha )}}}$	${\displaystyle \operatorname {Re} (s)>0}$	unit step minus exponential decay
sine	${\displaystyle \sin(\omega t)\cdot u(t)\ }$	${\displaystyle {\omega \over s^{2}+\omega ^{2}}}$	${\displaystyle \operatorname {Re} (s)>0}$	[35]
cosine	${\displaystyle \cos(\omega t)\cdot u(t)\ }$	${\displaystyle {s \over s^{2}+\omega ^{2}}}$	${\displaystyle \operatorname {Re} (s)>0}$	[35]
hyperbolic sine	${\displaystyle \sinh(\alpha t)\cdot u(t)\ }$	${\displaystyle {\alpha \over s^{2}-\alpha ^{2}}}$	${\displaystyle \operatorname {Re} (s)>\left|\alpha \right|}$	[36]
hyperbolic cosine	${\displaystyle \cosh(\alpha t)\cdot u(t)\ }$	${\displaystyle {s \over s^{2}-\alpha ^{2}}}$	${\displaystyle \operatorname {Re} (s)>\left|\alpha \right|}$	[36]
exponentially decaying sine wave	${\displaystyle e^{-\alpha t}\sin(\omega t)\cdot u(t)\ }$	${\displaystyle {\omega \over (s+\alpha )^{2}+\omega ^{2}}}$	${\displaystyle \operatorname {Re} (s)>-\alpha }$	[35]
exponentially decaying cosine wave	${\displaystyle e^{-\alpha t}\cos(\omega t)\cdot u(t)\ }$	${\displaystyle {s+\alpha \over (s+\alpha )^{2}+\omega ^{2}}}$	${\displaystyle \operatorname {Re} (s)>-\alpha }$	[35]
natural logarithm	${\displaystyle \ln(t)\cdot u(t)}$	${\displaystyle -{1 \over s}\left[\ln(s)+\gamma \right]}$	${\displaystyle \operatorname {Re} (s)>0}$	[36]
Bessel function of the first kind, of order n	${\displaystyle J_{n}(\omega t)\cdot u(t)}$	${\displaystyle {\frac {\left({\sqrt {s^{2}+\omega ^{2}}}-s\right)^{\!n}}{\omega ^{n}{\sqrt {s^{2}+\omega ^{2}}}}}}$	${\displaystyle \operatorname {Re} (s)>0}$ (n > −1)	[37]
Error function	${\displaystyle \operatorname {erf} (t)\cdot u(t)}$	${\displaystyle {\frac {1}{s}}e^{s^{2}/4}\!\left(1-\operatorname {erf} {\frac {s}{2}}\right)}$	${\displaystyle \operatorname {Re} (s)>0}$	[37]
Explanatory notes: u(t) represents the Heaviside step function. δ represents the Dirac delta function. Γ(z) represents the gamma function. γ is the Euler–Mascheroni constant . t, a real number, typically represents time, although it can represent any independent dimension. s is the real part . α, β, τ, and ω are real numbers . n is an integer.

The Laplace transform is often used in

impedances.

Here is a summary of equivalents:

Note that the resistor is exactly the same in the time domain and the s-domain. The sources are put in if there are initial conditions on the circuit elements. For example, if a capacitor has an initial voltage across it, or if the inductor has an initial current through it, the sources inserted in the s-domain account for that.

The equivalents for current and voltage sources are simply derived from the transformations in the table above.

The Laplace transform is used frequently in engineering and physics; the output of a linear time-invariant system can be calculated by convolving its unit impulse response with the input signal. Performing this calculation in Laplace space turns the convolution into a multiplication; the latter being easier to solve because of its algebraic form. For more information, see control theory. The Laplace transform is invertible on a large class of functions. Given a simple mathematical or functional description of an input or output to a system, the Laplace transform provides an alternative functional description that often simplifies the process of analyzing the behavior of the system, or in synthesizing a new system based on a set of specifications.^[38]

The Laplace transform can also be used to solve differential equations and is used extensively in mechanical engineering and electrical engineering. The Laplace transform reduces a linear differential equation to an algebraic equation, which can then be solved by the formal rules of algebra. The original differential equation can then be solved by applying the inverse Laplace transform. English electrical engineer Oliver Heaviside first proposed a similar scheme, although without using the Laplace transform; and the resulting operational calculus is credited as the Heaviside calculus.

Let ${\displaystyle {\mathcal {L}}\left\{f(t)\right\}=F(s)}$. Then (see the table above)

$${\displaystyle \partial _{s}{\mathcal {L}}\left\{{\frac {f(t)}{t}}\right\}=\partial _{s}\int _{0}^{\infty }{\frac {f(t)}{t}}e^{-st}\,dt=-\int _{0}^{\infty }f(t)e^{-st}dt=-F(s)}$$

From which one gets:

$${\displaystyle {\mathcal {L}}\left\{{\frac {f(t)}{t}}\right\}=\int _{s}^{\infty }F(p)\,dp.}$$

In the limit ${\displaystyle s\rightarrow 0}$, one gets $${\displaystyle \int _{0}^{\infty }{\frac {f(t)}{t}}\,dt=\int _{0}^{\infty }F(p)\,dp,}$$ provided that the interchange of limits can be justified. This is often possible as a consequence of the final value theorem. Even when the interchange cannot be justified the calculation can be suggestive. For example, with a ≠ 0 ≠ b, proceeding formally one has $${\displaystyle {\begin{aligned}\int _{0}^{\infty }{\frac {\cos(at)-\cos(bt)}{t}}\,dt&=\int _{0}^{\infty }\left({\frac {p}{p^{2}+a^{2}}}-{\frac {p}{p^{2}+b^{2}}}\right)\,dp\\[6pt]&=\left[{\frac {1}{2}}\ln {\frac {p^{2}+a^{2}}{p^{2}+b^{2}}}\right]_{0}^{\infty }={\frac {1}{2}}\ln {\frac {b^{2}}{a^{2}}}=\ln \left|{\frac {b}{a}}\right|.\end{aligned}}}$$

The validity of this identity can be proved by other means. It is an example of a Frullani integral.

Another example is Dirichlet integral.

In the theory of

unit system). Symbolically, this is expressed by the differential equation $${\displaystyle i=C{dv \over dt},}$$ where C is the capacitance of the capacitor, i = i(t) is the

voltage

across the terminals of the capacitor, also as a function of time.

Taking the Laplace transform of this equation, we obtain $${\displaystyle I(s)=C(sV(s)-V_{0}),}$$ where $${\displaystyle {\begin{aligned}I(s)&={\mathcal {L}}\{i(t)\},\\V(s)&={\mathcal {L}}\{v(t)\},\end{aligned}}}$$ and $${\displaystyle V_{0}=v(0).}$$

Solving for V(s) we have $${\displaystyle V(s)={I(s) \over sC}+{V_{0} \over s}.}$$

The definition of the complex impedance Z (in ohms) is the ratio of the complex voltage V divided by the complex current I while holding the initial state V₀ at zero: $${\displaystyle Z(s)=\left.{V(s) \over I(s)}\right|_{V_{0}=0}.}$$

Using this definition and the previous equation, we find: $${\displaystyle Z(s)={\frac {1}{sC}},}$$ which is the correct expression for the complex impedance of a capacitor. In addition, the Laplace transform has large applications in control theory.

Consider a linear time-invariant system with transfer function $${\displaystyle H(s)={\frac {1}{(s+\alpha )(s+\beta )}}.}$$

The impulse response is simply the inverse Laplace transform of this transfer function: $${\displaystyle h(t)={\mathcal {L}}^{-1}\{H(s)\}.}$$

Partial fraction expansion

To evaluate this inverse transform, we begin by expanding H(s) using the method of partial fraction expansion, $${\displaystyle {\frac {1}{(s+\alpha )(s+\beta )}}={P \over s+\alpha }+{R \over s+\beta }.}$$

The unknown constants P and R are the

to the transfer function's overall shape.

By the residue theorem, the inverse Laplace transform depends only upon the poles and their residues. To find the residue P, we multiply both sides of the equation by s + α to get $${\displaystyle {\frac {1}{s+\beta }}=P+{R(s+\alpha ) \over s+\beta }.}$$

Then by letting s = −α, the contribution from R vanishes and all that is left is $${\displaystyle P=\left.{1 \over s+\beta }\right|_{s=-\alpha }={1 \over \beta -\alpha }.}$$

Similarly, the residue R is given by $${\displaystyle R=\left.{1 \over s+\alpha }\right|_{s=-\beta }={1 \over \alpha -\beta }.}$$

Note that $${\displaystyle R={-1 \over \beta -\alpha }=-P}$$ and so the substitution of R and P into the expanded expression for H(s) gives $${\displaystyle H(s)=\left({\frac {1}{\beta -\alpha }}\right)\cdot \left({1 \over s+\alpha }-{1 \over s+\beta }\right).}$$

Finally, using the linearity property and the known transform for exponential decay (see Item #3 in the Table of Laplace Transforms, above), we can take the inverse Laplace transform of H(s) to obtain $${\displaystyle h(t)={\mathcal {L}}^{-1}\{H(s)\}={\frac {1}{\beta -\alpha }}\left(e^{-\alpha t}-e^{-\beta t}\right),}$$ which is the impulse response of the system.

Convolution

The same result can be achieved using the convolution property as if the system is a series of filters with transfer functions 1/(s + α) and 1/(s + β). That is, the inverse of $${\displaystyle H(s)={\frac {1}{(s+\alpha )(s+\beta )}}={\frac {1}{s+\alpha }}\cdot {\frac {1}{s+\beta }}}$$ is $${\displaystyle {\mathcal {L}}^{-1}\!\left\{{\frac {1}{s+\alpha }}\right\}*{\mathcal {L}}^{-1}\!\left\{{\frac {1}{s+\beta }}\right\}=e^{-\alpha t}*e^{-\beta t}=\int _{0}^{t}e^{-\alpha x}e^{-\beta (t-x)}\,dx={\frac {e^{-\alpha t}-e^{-\beta t}}{\beta -\alpha }}.}$$

Time function	Laplace transform
${\displaystyle \sin {(\omega t+\varphi )}}$	${\displaystyle {\frac {s\sin(\varphi )+\omega \cos(\varphi )}{s^{2}+\omega ^{2}}}}$
${\displaystyle \cos {(\omega t+\varphi )}}$	${\displaystyle {\frac {s\cos(\varphi )-\omega \sin(\varphi )}{s^{2}+\omega ^{2}}}.}$

Starting with the Laplace transform, $${\displaystyle X(s)={\frac {s\sin(\varphi )+\omega \cos(\varphi )}{s^{2}+\omega ^{2}}}}$$ we find the inverse by first rearranging terms in the fraction: $${\displaystyle {\begin{aligned}X(s)&={\frac {s\sin(\varphi )}{s^{2}+\omega ^{2}}}+{\frac {\omega \cos(\varphi )}{s^{2}+\omega ^{2}}}\\&=\sin(\varphi )\left({\frac {s}{s^{2}+\omega ^{2}}}\right)+\cos(\varphi )\left({\frac {\omega }{s^{2}+\omega ^{2}}}\right).\end{aligned}}}$$

We are now able to take the inverse Laplace transform of our terms: $${\displaystyle {\begin{aligned}x(t)&=\sin(\varphi ){\mathcal {L}}^{-1}\left\{{\frac {s}{s^{2}+\omega ^{2}}}\right\}+\cos(\varphi ){\mathcal {L}}^{-1}\left\{{\frac {\omega }{s^{2}+\omega ^{2}}}\right\}\\&=\sin(\varphi )\cos(\omega t)+\cos(\varphi )\sin(\omega t).\end{aligned}}}$$

This is just the

of the arguments, yielding: $${\displaystyle x(t)=\sin(\omega t+\varphi ).}$$

We can apply similar logic to find that $${\displaystyle {\mathcal {L}}^{-1}\left\{{\frac {s\cos \varphi -\omega \sin \varphi }{s^{2}+\omega ^{2}}}\right\}=\cos {(\omega t+\varphi )}.}$$

In statistical mechanics, the Laplace transform of the density of states ${\displaystyle g(E)}$ defines the partition function.^[39] That is, the canonical partition function ${\displaystyle Z(\beta )}$ is given by $${\displaystyle Z(\beta )=\int _{0}^{\infty }e^{-\beta E}g(E)\,dE}$$ and the inverse is given by $${\displaystyle g(E)={\frac {1}{2\pi i}}\int _{\beta _{0}-i\infty }^{\beta _{0}+i\infty }e^{\beta E}Z(\beta )\,d\beta }$$

The wide and general applicability of the Laplace transform and its inverse is illustrated by an application in astronomy which provides some information on the spatial distribution of matter of an

, rather than relating the time domain with the spectrum (frequency domain).

Assuming certain properties of the object, e.g. spherical shape and constant temperature, calculations based on carrying out an inverse Laplace transformation on the spectrum of the object can produce the only possible model of the distribution of matter in it (density as a function of distance from the center) consistent with the spectrum.^[40] When independent information on the structure of an object is available, the inverse Laplace transform method has been found to be in good agreement.

Consider a random walk, with steps ${\displaystyle \{+1,-1\}}$ occurring with probabilities ${\displaystyle p,q=1-p}$.

, with parameter ${\displaystyle \lambda }$. Then the probability of the walk being at the lattice point ${\displaystyle n}$ at time ${\displaystyle t}$ is

${\displaystyle P_{n}(t)=\int _{0}^{t}\lambda e^{-\lambda (t-s)}(pP_{n-1}(s)+qP_{n+1}(s))\,ds\quad (+e^{-\lambda t}\quad {\text{when}}\ n=0).}$

This leads to a system of integral equations (or equivalently a system of differential equations). However, because it is a system of convolution equations, the Laplace transform converts it into a system of linear equations for

${\displaystyle \pi _{n}(s)={\mathcal {L}}(P_{n})(s),}$

namely:

${\displaystyle \pi _{n}(s)={\frac {\lambda }{\lambda +s}}(p\pi _{n-1}(s)+q\pi _{n+1}(s))\quad (+{\frac {1}{\lambda +s}}\quad {\text{when}}\ n=0)}$

which may now be solved by standard methods.

The Laplace transform of the measure ${\displaystyle \mu }$ on ${\displaystyle [0,\infty )}$ is given by

${\displaystyle {\mathcal {L}}\mu (s)=\int _{0}^{\infty }e^{-st}d\mu (t).}$

It is intuitively clear that, for small ${\displaystyle s>0}$, the exponentially decaying integrand will become more sensitive to the concentration of the measure ${\displaystyle \mu }$ on larger subsets of the domain. To make this more precise, introduce the distribution function:

${\displaystyle M(t)=\mu ([0,t)).}$

Formally, we expect a limit of the following kind:

${\displaystyle \lim _{s\to 0^{+}}{\mathcal {L}}\mu (s)=\lim _{t\to \infty }M(t).}$

are theorems relating the asymptotics of the Laplace transform, as ${\displaystyle s\to 0^{+}}$, to those of the distribution of ${\displaystyle \mu }$ as ${\displaystyle t\to \infty }$. They are thus of importance in asymptotic formulae of probability and statistics, where often the spectral side has asymptotics that are simpler to infer.^[42]

Two Tauberian theorems of note are the

, which is the following statement:

Let A(x) be a non-negative, monotonic nondecreasing function of x, defined for 0 ≤ x < ∞. Suppose that

${\displaystyle f(s)=\int _{0}^{\infty }A(x)e^{-xs}\,dx}$

converges for ℜ(s) > 1 to the function ƒ(s) and that, for some non-negative number c,

${\displaystyle f(s)-{\frac {c}{s-1}}}$

has an extension as a continuous function for ℜ(s) ≥ 1. Then the limit as x goes to infinity of e^−x A(x) is equal to c.

This statement can be applied in particular to the logarithmic derivative of Riemann zeta function, and thus provides an extremely short way to prove the prime number theorem.^[43]

References

Modern

• Bracewell, Ronald N. (1978), The Fourier Transform and its Applications (2nd ed.), McGraw-Hill Kogakusha,
  ISBN 978-0-07-007013-4
• Bracewell, R. N. (2000), The Fourier Transform and Its Applications (3rd ed.), Boston: McGraw-Hill,
  ISBN 978-0-07-116043-8
• MR 0270403
• Korn, G. A.;
  ISBN 978-0-07-035370-1
• Widder, David Vernon (1941), The Laplace Transform, Princeton Mathematical Series, v. 6,
  MR 0005923
• Williams, J. (1973), Laplace Transforms, Problem Solvers, George Allen & Unwin,
  ISBN 978-0-04-512021-5
• Takacs, J. (1953), "Fourier amplitudok meghatarozasa operatorszamitassal", Magyar Hiradastechnika (in Hungarian), IV (7–8): 93–96

Historical

• Euler, L. (1744), "De constructione aequationum" [The Construction of Equations], Opera Omnia, 1st series (in Latin), 22: 150–161
• Euler, L. (1753), "Methodus aequationes differentiales" [A Method for Solving Differential Equations], Opera Omnia, 1st series (in Latin), 22: 181–213
• ISBN 978-3764314743
  , Chapters 3–5
• Euler, Leonhard (1769), Institutiones calculi integralis [Institutions of Integral Calculus] (in Latin), vol. II, Paris: Petropoli, ch. 3–5, pp. 57–153
• ISBN 978-0-691-01185-1
• Lagrange, J. L.
  (1773), Mémoire sur l'utilité de la méthode, Œuvres de Lagrange, vol. 2, pp. 171–234

Further reading

• Arendt, Wolfgang; Batty, Charles J.K.; Hieber, Matthias; Neubrander, Frank (2002), Vector-Valued Laplace Transforms and Cauchy Problems, Birkhäuser Basel,
  ISBN 978-3-7643-6549-3
  .
• Davies, Brian (2002), Integral transforms and their applications (Third ed.), New York: Springer,
  ISBN 978-0-387-95314-4
• Deakin, M. A. B. (1981), "The development of the Laplace transform", Archive for History of Exact Sciences, 25 (4): 343–390,
  S2CID 117913073
• Deakin, M. A. B. (1982), "The development of the Laplace transform", Archive for History of Exact Sciences, 26 (4): 351–381,
  S2CID 123071842
• ISBN 978-0-387-06407-9
• Mathews, Jon; Walker, Robert L. (1970), Mathematical methods of physics (2nd ed.), New York: W. A. Benjamin,
  ISBN 0-8053-7002-1
• Polyanin, A. D.; Manzhirov, A. V. (1998), Handbook of Integral Equations, Boca Raton: CRC Press,
  ISBN 978-0-8493-2876-3
• MR 0052555
• ISBN 978-0-486-46662-0
  - See Chapter VI. The Laplace transform.
• Siebert, William McC. (1986), Circuits, Signals, and Systems, Cambridge, Massachusetts: MIT Press,
  ISBN 978-0-262-19229-3
• Widder, David Vernon (1945), "What is the Laplace transform?",
  MR 0013447
• J.A.C.Weidman and Bengt Fornberg: "Fully numerical Laplace transform methods", Numerical Algorithms, vol.92 (2023), pp. 985–1006. https://doi.org/10.1007/s11075-022-01368-x .

External links

Wikiquote has quotations related to Laplace transform.

Wikimedia Commons has media related to Laplace transformation.

• "Laplace transform", Encyclopedia of Mathematics, EMS Press, 2001 [1994]
• Online Computation of the transform or inverse transform, wims.unice.fr
• Tables of Integral Transforms at EqWorld: The World of Mathematical Equations.
• Weisstein, Eric W. "Laplace Transform". MathWorld.
• Good explanations of the initial and final value theorems Archived 2009-01-08 at the Wayback Machine
• Laplace Transforms at MathPages
• Computational Knowledge Engine allows to easily calculate Laplace Transforms and its inverse Transform.
• Laplace Calculator to calculate Laplace Transforms online easily.
• Code to visualize Laplace Transforms and many example videos.