Editing Laplace transform (section)

== Properties and theorems ==
The Laplace transform's key property is that it converts [[derivative|differentiation]] and [[integral|integration]] in the time domain into multiplication and division by {{math|''s''}} in the Laplace domain. Thus, the Laplace variable {{math|''s''}} is also known as an ''operator variable'' in the Laplace domain: either the ''derivative operator'' or (for {{math|''s''<sup>−1</sup>)}} the ''integration operator''.

Given the functions {{math|''f''(''t'')}} and {{math|''g''(''t'')}}, and their respective Laplace transforms {{math|''F''(''s'')}} and {{math|''G''(''s'')}},
<math display=block>\begin{align}
f(t) &= \mathcal{L}^{-1}\{F(s)\},\\
g(t) &= \mathcal{L}^{-1}\{G(s)\},
\end{align}</math>

the following table is a list of properties of unilateral Laplace transform:<ref>{{harvnb|Korn|Korn|1967|pp=226&ndash;227}}</ref>

{| class="wikitable" id="291017_tableid"
|+ Properties of the unilateral Laplace transform
|-
 ! scope="col" | Property
 ! scope="col" | Time domain
 ! scope="col" | {{math|''s''}} domain
 ! scope="col" | Comment
|-
 ! scope="row" | [[Linearity]]
 | <math> a f(t) + b g(t) \ </math>
 | <math> a F(s) + b G(s) \ </math>
 | Can be proved using basic rules of integration.
|-
 ! scope="row" | Frequency-domain derivative
 | <math> t f(t) \ </math>
 | <math> -F'(s) \ </math>
 | {{math|''F''′}} is the first derivative of {{math|''F''}} with respect to {{math|''s''}}.
|-
 ! scope="row" | Frequency-domain general derivative
 | <math> t^{n} f(t) \ </math>
 | <math> (-1)^{n} F^{(n)}(s) \ </math>
 | More general form, {{math|''n''}}th derivative of {{math|''F''(''s'')}}.
|-
 ! scope="row" | [[Derivative]]
 | <math> f'(t) \ </math>
 | <math> s F(s) - f(0^{-}) \ </math>
 | {{math|''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
|-
 ! scope="row" | Second derivative
 | <math> f''(t) \ </math>
 | <math display="inline"> s^2 F(s) - s f(0^{-}) - f'(0^{-}) \ </math>
 | {{math|''f''}} is assumed twice differentiable and the second derivative to be of exponential type. Follows by applying the Differentiation property to {{math|''f''′(''t'')}}.
|-
 ! scope="row" | General derivative
 | <math> f^{(n)}(t)  \ </math>
 | <math> s^n F(s) - \sum_{k=1}^{n} s^{n-k} f^{(k-1)}(0^{-}) \ </math>
 | {{math|''f''}} is assumed to be {{math|''n''}}-times differentiable, with {{math|''n''}}th derivative of exponential type.  Follows by [[mathematical induction]].
|-
 ! scope="row" | Frequency-domain [[Integral|integration]]
 | <math> \frac{1}{t}f(t)  \ </math>
 | <math> \int_s^\infty F(\sigma)\, d\sigma \ </math>
 | This is deduced using the nature of frequency differentiation and conditional convergence.
|-
 ! scope="row" | Time-domain integration
 | <math> \int_0^t f(\tau)\, d\tau = (u * f)(t)</math>
 | <math> {1 \over s} F(s) </math>
 | {{math|''u''(''t'')}} is the Heaviside step function and {{math|(''u'' ∗ ''f'')(''t'')}} is the [[convolution]] of {{math|''u''(''t'')}} and {{math|''f''(''t'')}}.
|-
 ! scope="row" | Frequency shifting
 | <math> e^{at} f(t) </math>
 | <math> F(s - a) \ </math>
 |
|-
 ! scope="row" | Time shifting
 | <math> f(t - a) u(t - a) </math>
<math> f(t) u(t - a) \ </math>
 | <math> e^{-as} F(s) \ </math>
<math> e^{-as} \mathcal{L}\{f(t + a)\} </math>
 | {{math|''a'' > 0}}, {{math|''u''(''t'')}} is the Heaviside step function
|-
 ! scope="row" | Time scaling
 | <math>f(at)</math>
 | <math> \frac{1}{a} F \left ({s \over a} \right)</math>
 | {{math|''a'' > 0}}
|-
 ! scope="row" | [[Multiplication]]
 | <math>f(t)g(t)</math>
 | <math> \frac{1}{2\pi i}\lim_{T\to\infty}\int_{c - iT}^{c + iT}F(\sigma)G(s - \sigma)\,d\sigma \ </math>
 | The integration is done along the vertical line {{math|1=Re(''σ'') = ''c''}} that lies entirely within the region of convergence of {{math|''F''}}.<ref>{{harvnb|Bracewell|2000|loc=Table 14.1, p. 385}}</ref>
|-
 ! scope="row" | [[Convolution]]
 | <math> (f * g)(t) = \int_{0}^{t} f(\tau)g(t - \tau)\,d\tau</math>
 | <math> F(s) \cdot G(s) \ </math>
 | 
|-
 ! scope="row" | [[Circular convolution]]
 | <math> (f * g)(t) = \int_{0}^T f(\tau)g(t - \tau)\,d\tau</math>
 | <math> F(s) \cdot G(s) \ </math>
 | For periodic functions with period {{math|''T''}}.
|-
 ! scope="row" | [[Complex conjugation]]
 | <math> f^*(t) </math>
 | <math> F^*(s^*) </math>
 |
|-
 ! scope="row" | [[Periodic function]]
 | <math>f(t)</math>
 | <math>{1 \over 1 - e^{-Ts}} \int_0^T e^{-st} f(t)\,dt </math>
 | {{math|''f''(''t'')}} is a periodic function of period {{math|''T''}} so that {{math|1=''f''(''t'') = ''f''(''t'' + ''T'')}}, for all {{math|''t'' ≥ 0}}. This is the result of the time shifting property and the [[geometric series]].
|-
! scope="row" | [[Periodic summation]]
| <math>  f_P(t) = \sum_{n=0}^{\infty} f(t-Tn) </math>
<math> f_P(t) = \sum_{n=0}^{\infty} (-1)^n f(t-Tn) </math>
| <math>  F_P(s) = \frac{1}{1-e^{-Ts}} F(s) </math>
<math> F_P(s) = \frac{1}{1+e^{-Ts}} F(s) </math>
| 
|}

; [[Initial value theorem]]
:<math>f(0^+)=\lim_{s\to \infty}{sF(s)}.</math>
; [[Final value theorem]]
:<math>f(\infty)=\lim_{s\to 0}{sF(s)}</math>, if all [[Pole (complex analysis)|poles]] of <math>sF(s)</math> 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 {{math|''F''(''s'')}} has a pole in the right-hand plane or poles on the imaginary axis (e.g., if <math>f(t) = e^t</math> or <math>f(t) = \sin(t)</math>), then the behaviour of this formula is undefined.

=== Relation to power series ===
The Laplace transform can be viewed as a [[continuous function|continuous]] analogue of a [[power series]].<ref>Archived at [https://ghostarchive.org/varchive/youtube/20211211/zvbdoSeGAgI Ghostarchive]{{cbignore}} and the [https://web.archive.org/web/20141220033002/https://www.youtube.com/watch?v=zvbdoSeGAgI&gl=US&hl=en Wayback Machine]{{cbignore}}: {{cite web |last1=Mattuck |first1=Arthur |title=Where the Laplace Transform comes from |website=[[YouTube]] |date=7 November 2008 |url=https://www.youtube.com/watch?v=zvbdoSeGAgI}}{{cbignore}}</ref> If  {{math|''a''(''n'')}} is a discrete function of a positive integer {{math|''n''}}, then the power series associated to {{math|''a''(''n'')}} is the series
<math display=block>\sum_{n=0}^{\infty} a(n) x^n</math>
where {{math|''x''}} is a real variable (see ''[[Z-transform]]''). Replacing summation over {{math|''n''}} with integration over  {{math|''t''}}, a continuous version of the power series becomes
<math display=block>\int_{0}^{\infty} f(t) x^t\, dt</math>
where the discrete function {{math|''a''(''n'')}} is replaced by the continuous one {{math|''f''(''t'')}}.

Changing the base of the power from {{math|''x''}} to {{math|''e''}} gives
<math display=block>\int_{0}^{\infty} f(t) \left(e^{\ln{x}}\right)^t\, dt</math>

For this to converge for, say, all bounded functions {{math|''f''}}, it is necessary to require that {{math|ln ''x'' < 0}}. Making the substitution {{math|1=&minus;''s'' = ln ''x''}} gives just the Laplace transform:
<math display=block>\int_{0}^{\infty} f(t) e^{-st}\, dt</math>

In other words, the Laplace transform is a continuous analog of a power series, in which the discrete parameter {{math|''n''}} is replaced by the continuous parameter {{math|''t''}}, and {{math|''x''}} is replaced by {{math|''e''<sup>&minus;''s''</sup>}}.

=== Relation to moments ===
{{main article|Moment-generating function}}
The quantities
<math display=block>\mu_n = \int_0^\infty t^nf(t)\, dt</math>

are the ''moments'' of the function {{math|''f''}}.  If the first {{math|''n''}} moments of {{math|''f''}} converge absolutely, then by repeated [[differentiation under the integral]], 
<math display=block>(-1)^n(\mathcal L f)^{(n)}(0) = \mu_n .</math>
This is of special significance in probability theory, where the moments of a random variable {{math|''X''}} are given by the expectation values <math>\mu_n=\operatorname{E}[X^n]</math>.  Then, the relation holds
<math display=block>\mu_n = (-1)^n\frac{d^n}{ds^n}\operatorname{E}\left[e^{-sX}\right](0).</math>

=== Transform of a function's derivative ===
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:
<math display=block>\begin{align}
  \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{align}</math>
yielding
<math display=block>\mathcal{L} \{ f'(t) \} = s\cdot\mathcal{L} \{ f(t) \}-f(0^-), </math>
and in the bilateral case,
<math display=block> \mathcal{L} \{ f'(t) \} = s \int_{-\infty}^\infty e^{-st} f(t)\,dt  = s \cdot \mathcal{L} \{ f(t) \}. </math>

The general result
<math display=block>\mathcal{L} \left\{ f^{(n)}(t) \right\} = s^n \cdot \mathcal{L} \{ f(t) \} - s^{n - 1} f(0^-) - \cdots - f^{(n - 1)}(0^-),</math>
where <math>f^{(n)}</math> denotes the {{math|''n''}}th derivative of {{math|''f''}}, can then be established with an inductive argument.

=== Evaluating integrals over the positive real axis ===
A useful property of the Laplace transform is the following:
<math display=block>\int_0^\infty f(x)g(x)\,dx = \int_0^\infty(\mathcal{L} f)(s)\cdot(\mathcal{L}^{-1}g)(s)\,ds </math>
under suitable assumptions on the behaviour of <math>f,g</math> in a right neighbourhood of <math>0</math> and on the decay rate of <math>f,g</math> in a left neighbourhood of <math>\infty</math>. The above formula is a variation of integration by parts, with the operators 
<math>\frac{d}{dx}</math> and <math>\int \,dx</math> being replaced by <math>\mathcal{L}</math> and <math>\mathcal{L}^{-1}</math>. Let us prove the equivalent formulation:
<math display=block>\int_0^\infty(\mathcal{L} f)(x)g(x)\,dx = \int_0^\infty f(s)(\mathcal{L}g)(s)\,ds. </math>

By plugging in <math>(\mathcal{L}f)(x)=\int_0^\infty f(s)e^{-sx}\,ds</math> the left-hand side turns into:
<math display=block>\int_0^\infty\int_0^\infty f(s)g(x) e^{-sx}\,ds\,dx, </math>
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,
<math display=block>\int_0^\infty\frac{\sin x}{x}dx = \int_0^\infty \mathcal{L}(1)(x)\sin x dx = \int_0^\infty 1 \cdot \mathcal{L}(\sin)(x)dx = \int_0^\infty \frac{dx}{x^2 + 1} = \frac{\pi}{2}. </math>