= Laplace transform
{c}
{title2=$\mathcal L$}
{wiki}
The unilateral <Laplace transform> of a <function> $f:[0,\infty)\to\mathbb C$ is the <integral transform>
$$
F(s)=\mathcal L\{f\}(s)=\int_0^\infty e^{-st}f(t)\,dt,\qquad s=\sigma+i\omega\in\mathbb C.
$$
Here $s$ is a <complex number>, $\sigma$ is its <real part>, and $i$ is the <imaginary unit>. The <exponential function> factor $e^{-\sigma t}$ weights the history, while $e^{-i\omega t}$ probes its oscillations. The <integral> starts at zero, so it uses only the history from that starting time onward. This section concerns ordinary <functions>; their value at the single endpoint does not change the <integral>. See https://ocw.mit.edu/courses/18-03sc-differential-equations-fall-2011/d40ad07a62996f6dad93b832f8de0e04_MIT18_03SCF11_s27_1text.pdf[MIT 18.03SC, definition and elementary examples].
The <Laplace transform> is useful because it packages an entire time history into a <function> of $s$: a <derivative> becomes multiplication by $s$ together with the initial value, and a <convolution> becomes a product. Thus a <linear differential equation> can become an algebraic equation that already remembers its <initial value problem>. The worked example below shows both descriptions of the same response.
If $f$ is a <piecewise continuous function> and has <exponential order> $a$, then its defining <integral> is <absolutely convergent> for $\operatorname{Re}s>a$. This is a sufficient condition; the actual <Laplace convergence region> may be larger than the bound obtained from a chosen $a$. https://ocw.mit.edu/courses/18-03sc-differential-equations-fall-2011/fff4420903d82c5c5c483e54566bc5ec_MIT18_03SCF11_s27_2text.pdf[MIT 18.03SC, domain and growth conditions]{ref}.
\Image[/laplace-transform-exponential-weight.png]
{title=<Exponential function> weighting controls the <Laplace convergence region>}
{description=Left: for $f(t)=e^{-2t}$, the weighted <function> $e^{-\sigma t}f(t)$ decays when $\sigma>-2$, stays at one when $\sigma=-2$, and grows when $\sigma<-2$. Right: the real-parameter <Laplace transforms> of $1$ and $e^{-2t}$ grow without bound as $s$ approaches their respective convergence boundaries from the right. Curves are drawn only where the defining <integrals> converge.}
The bound has a short explanation. For sufficiently large $t$, $|f(t)|\leq Me^{at}$, so $|e^{-st}f(t)|\leq Me^{-(\sigma-a)t}$, whose tail <integral> is finite for $\sigma>a$. The finite initial <interval> is covered by <piecewise continuity>.
Three calculations make the convergence condition concrete. For the input $f(t)=1$, evaluate the finite-history <integral> first:
$$
\int_0^T e^{-st}\,dt=\frac{1-e^{-sT}}s\quad(s\ne0),\qquad \mathcal L\{1\}(s)=\frac1s\quad(\operatorname{Re}s>0).
$$
At $s=0$ the finite-history <integral> is $T$. For the <exponential function> input $f(t)=e^{-2t}$, the same calculation replaces $s$ by $s+2$, giving
$$
\mathcal L\{e^{-2t}\}(s)=\frac1{s+2},\qquad \operatorname{Re}s>-2.
$$
In particular, $s=-1$ is allowed for this decaying input: its weighted <integral> is $\int_0^\infty e^{-t}\,dt=1$. For the <sine> input, <Euler's formula> gives $\sin t=(e^{it}-e^{-it})/(2i)$; <linearity> then yields
$$
\mathcal L\{\sin t\}(s)=\frac1{2i}\left(\frac1{s-i}-\frac1{s+i}\right)=\frac1{s^2+1},\qquad \operatorname{Re}s>0.
$$
The <Laplace convergence region> is part of the answer. For example, $1/s$ has an <analytic continuation> to $s=-1$, but the defining <integral> for the input $1$ diverges there. The <sine> calculation is also developed in https://ocw.mit.edu/courses/18-03sc-differential-equations-fall-2011/2343f70cc426b469402f731015ac9719_MIT18_03SCF11_s27_3text.pdf[MIT 18.03SC, further elementary transforms].
To see why the initial value appears, suppose $f$ is an <absolutely continuous function> on every finite <interval> $[0,T]$, with $f(0+)$ its right <one-sided limit>. At the chosen $s$, assume $e^{-st}f(t)$ and $e^{-st}f'(t)$ are <Lebesgue integrable functions> and $e^{-sT}f(T)\to0$. The <Laplace transform of a derivative> is then
$$
\mathcal L\{f'\}(s)=sF(s)-f(0+).
$$
For example, <exponential order> bounds on both $f$ and $f'$ guarantee these conditions whenever $\operatorname{Re}s$ exceeds both growth bounds. The initial value is the lower-end boundary contribution of <integration by parts>.
Consider the <initial value problem>
$$
y'(t)+2y(t)=1,\qquad y(0)=0,\qquad t\geq0.
$$
Write $Y=\mathcal L\{y\}$. The <Laplace transform of a derivative> and <linearity> give $sY-0+2Y=1/s$ for $\operatorname{Re}s>0$. A <partial fraction decomposition> therefore gives
$$
Y(s)=\frac1{s(s+2)}=\frac12\left(\frac1s-\frac1{s+2}\right),\qquad y(t)=\frac{1-e^{-2t}}2.
$$
We recognize the two elementary <Laplace transforms> already computed. Directly verify the result: its <derivative> is $y'(t)=e^{-2t}$, so
$$
y'(t)+2y(t)-1=e^{-2t}+1-e^{-2t}-1=0,\qquad y(0)=\frac{1-1}2=0.
$$
This verifies both the <ordinary differential equation> and the initial value, and shows that the resulting bounded <function> has the <Laplace transform> used above. The response rises from zero toward $1/2$: at that value the input $1$ balances the term $2y$. The general method is described in https://ocw.mit.edu/courses/18-03sc-differential-equations-fall-2011/pages/unit-iii-fourier-series-and-laplace-transform/laplace-transform-solving-initial-value-problems/[MIT 18.03SC, solving initial value problems].
\Image[/laplace-transform-step-response.png]
{title=A <Heaviside step function> input produces a response equal to the accumulated <convolution kernel>}
{description=Left: the solution $y(t)=(1-e^{-2t})/2$ and the <convolution kernel> $h(t)=e^{-2t}$ for $t\geq0$. Right: at $t=1$, the shaded <integral> $\int_0^1e^{-2(1-\tau)}\,d\tau=(1-e^{-2})/2\approx0.4323$ includes only the available history $0\leq\tau\leq1$. The rising profile on the right is $h(1-\tau)$; recent inputs have had less time to decay.}
The same solution is a <convolution>. For <functions> defined on $[0,\infty)$, the unilateral <convolution> is
$$
(f*g)(t)=\int_0^t f(\tau)g(t-\tau)\,d\tau,\qquad t\geq0.
$$
The main <convolution> article integrates over the whole <real line>; applying a <causal extension by zero> to both <functions> makes its formula identical to this one. Nonzero contributions require both $\tau\geq0$ and $t-\tau\geq0$, hence $0\leq\tau\leq t$. If $e^{-\sigma t}f(t)$ and $e^{-\sigma t}g(t)$ are <Lebesgue integrable functions> on $[0,\infty)$, the <convolution theorem for Laplace transforms> states
$$
\mathcal L\{f*g\}(s)=F(s)G(s),\qquad \operatorname{Re}s=\sigma.
$$
For <piecewise continuous functions> of <exponential order> $a$ and $b$, it holds throughout $\operatorname{Re}s>\max(a,b)$. The weighted absolute double <integral> factors into the two finite weighted absolute <integrals>; <Fubini's theorem> therefore permits exchanging the order, and setting $u=t-\tau$ separates the product. https://ocw.mit.edu/courses/18-03sc-differential-equations-fall-2011/0d4d90053f2bb24f7551c78e2d1b21a2_MIT18_03SCF11_s30_5text.pdf[MIT 18.03SC, convolution and system response]{ref}.
For the <linear time-invariant system> $y'+2y=f$ starting at zero, the <convolution kernel> is its <impulse response> $h(t)=e^{-2t}$ for $t\geq0$, extended by zero for $t<0$. With the <Heaviside step function> input, $f(t)=1$ for $t\geq0$, the accumulated response is
$$
y(t)=(h*f)(t)=\int_0^t e^{-2(t-\tau)}\,d\tau=\frac{1-e^{-2t}}2.
$$
Thus $Y(s)=(1/(s+2))(1/s)$ on $\operatorname{Re}s>0$ is both the algebraic solution and the <Laplace transform> of a finite-history <convolution>. Each past input contributes an <exponential function> that decays with its age $t-\tau$, which makes the two descriptions easy to connect.
Finally, compare the <Fourier transform> convention $\widehat g(\omega)=\int_{\mathbb R}g(t)e^{-i\omega t}\,dt$. If $f_0$ is the <causal extension by zero> of $f$, define $g_\sigma(t)=e^{-\sigma t}f_0(t)$. Whenever $g_\sigma$ is a <Lebesgue integrable function>,
$$
F(\sigma+i\omega)=\widehat{g_\sigma}(\omega).
$$
A vertical slice of the <Laplace transform> is therefore the <Fourier transform> of the zero-extended, weighted input. Positive $\sigma$ damps its positive-time history. A finite <window function> instead gives $g_{\sigma,T}(t)=e^{-\sigma t}f_0(t)\mathbf1_{[0,T]}(t)$ and $\widehat{g_{\sigma,T}}(\omega)=\int_0^Te^{-(\sigma+i\omega)t}f(t)\,dt$; this approaches $F(\sigma+i\omega)$ in the <Laplace convergence region>. Setting $\sigma=0$ directly requires an integrable zero-extended input: the input $1$ fails that condition, even though its <Laplace transform> exists for every $\sigma>0$.
The history of the <Laplace transform> spans several stages. Schiff's historical summary notes that Laplace's 1812 treatise acknowledged earlier work on <integral transforms> and describes later development of the modern <Laplace transform> method. https://oldversion.stu.edu.vn/uploads/documents/170509-091205.pdf#page=9[Schiff, The Laplace Transform: Theory and Applications, historical summary, pp. x–xi]{ref}.
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