Laplace transform ($\mathcal{L}$)

The unilateral Laplace transform of a function $f:[0,\infty )\to \mathbb{C}$ is the integral transformHere $s$ 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 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 $\mathrm{Re}s>a$. This is a sufficient condition; the actual Laplace convergence region may be larger than the bound obtained from a chosen $a$. ^[ref].

The bound has a short explanation. For sufficiently large $t$, $|f(t)|\le M{e}^{at}$, so $|e^{-st}f(t)|\le M{e}^{-(\sigma -a)t}$, 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:At $s=0$ the finite-history integral is $T$. For the exponential function input $f(t)=e^{-2t}$, the same calculation replaces $s$ by $s+2$, givingIn 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 yieldsThe 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 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 $e^{-st}{f}^{\prime }(t)$ are Lebesgue integrable functions and $e^{-sT}f(T)\to 0$. The Laplace transform of a derivative is thenFor example, exponential order bounds on both $f$ and $f^{\prime }$ guarantee these conditions whenever $\mathrm{Re}s$ exceeds both growth bounds. The initial value is the lower-end boundary contribution of integration by parts.

Consider the initial value problemWrite $Y=\mathcal{L}\{y\}$. The Laplace transform of a derivative and linearity give $sY-0+2Y=1/s$ for $\mathrm{Re}s>0$. A partial fraction decomposition therefore givesWe recognize the two elementary Laplace transforms already computed. Directly verify the result: its derivative is $y^{\prime }(t)=e^{-2t}$, soThis 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 MIT 18.03SC, solving initial value problems.

The same solution is a convolution. For functions defined on $[0,\infty )$, the unilateral convolution isThe 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 $\tau \ge 0$ and $t-\tau \ge 0$, hence $0\le \tau \le t$. 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 statesFor piecewise continuous functions of exponential order $a$ and $b$, it holds throughout $\mathrm{Re}s>\max (a,b)$. 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. ^[ref].

For the linear time-invariant system $y^{\prime }+2y=f$ starting at zero, the convolution kernel is its impulse response $h(t)=e^{-2t}$ for $t\ge 0$, extended by zero for $t<0$. With the Heaviside step function input, $f(t)=1$ for $t\ge 0$, the accumulated response isThus $Y(s)=(1/(s+2))(1/s)$ on $\mathrm{Re}s>0$ 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 $\hat{g}(\omega )={\int }_{\mathbb{R}}g(t)e^{-i\omega t}dt$. 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,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 $g_{\sigma ,T}(t)=e^{-\sigma t}{f}_0(t)1_{[0,T]}(t)$ and $\widehat{g_{\sigma ,T}}(\omega )={\int }_0^T{e}^{-(\sigma +i\omega )t}f(t)dt$; 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. ^[ref].

A bounded function on $[0,\infty )$ has a Laplace transform satisfying the displayed bound, by integration of $|y(t)e^{-pt}|$. Consequently its Laplace transform cannot have a pole of order two or higher at $p=0$. This can eliminate the arbitrary constant introduced when multiplication by $t$ turns an ordinary differential equation into a first-order differential equation for its Laplace transform.

For example, transforming $t{y}^{\prime \prime }+(t-1)y^{\prime }-y=0$ with $y(0)=1$ gives $p(p+1)Y^{\prime }+(3p+2)Y=2$. An integrating factor $p^2(p+1)$ yieldsThe bound forces $C=0$, selecting $y=e^{-t}$. This selection can be justified without assuming unproved endpoint regularity for every solution: $e^{-t}$ and $t-1$ have nonzero Wronskian $t{e}^{-t}$ on $(0,\infty )$, so they form a fundamental system of solutions. Boundedness removes $t-1$, and the limit at zero fixes the remaining coefficient. The endpoint limit by itself does not select a unique solution.

If two piecewise continuous functions of exponential order on $[0,\infty )$ have the same unilateral Laplace transform for every sufficiently large positive real parameter, they agree at every common point of continuity and agree almost everywhere. For continuous functions, the equality holds everywhere. Values at an isolated discontinuity cannot be recovered from an ordinary integral transform.

For real $p>0$, the Laplace transform $G$ of the order-zero Bessel function of the first kind satisfies $(1+p^2)G^{\prime }+pG=0$. This follows by transforming $x{J}_0^{\prime \prime }+J_0^{\prime }+x{J}_0=0$, using $\mathcal{L}[xg]=-{\partial }_p\mathcal{L}[g]$ and $J_0(0)=1$, $J_0^{\prime }(0)=0$. Thus $G=C/\sqrt{1+p^2}$; the initial value theorem for Laplace transforms gives $pG(p)\to 1$ and fixes $C=1$.

For a function continuous at zero and of exponential order, the displayed limit recovers its initial value from its Laplace transform. Substitute $u=pt$ to obtain $p\mathcal{L}f(p)={\int }_0^{\infty }{e}^{-u}f(u/p)du$ and apply the dominated convergence theorem; an exponential-order bound controls the tail. The limit along positive real $p$ suffices for this version.

A function has exponential order $a\in \mathbb{R}$ if some $M>0$ and $T\ge 0$ give the displayed bound. Bounded functions, including sine, have exponential order zero; $e^{-2t}$ has exponential order $-2$. The bound need not use the smallest possible $a$. Together with piecewise continuity, it guarantees a Laplace transform for $\mathrm{Re}s>a$.

The set of complex numbers $s$ for which the displayed integral is finite is the region of absolute convergence of a Laplace transform. It depends only on the real part $\sigma =\mathrm{Re}s$. Once the condition holds for one $\sigma$, it holds for every larger real part. For the exponential function $e^{-2t}$ it is exactly $\mathrm{Re}s>-2$; in general, convergence on the boundary must be checked separately. An analytic continuation outside this set need not equal a convergent defining integral.

Extend a function given for nonnegative time to the real line by setting it to zero before the starting time. This turns unilateral convolution into whole-line convolution and identifies a Laplace transform slice with a Fourier transform after exponential function weighting.

For $t>0$, this Laplace transform converges at zero because $dx/[x{\log }^{2}x]$ is integrable, and at infinity because of exponential damping. Its large-$t$ behavior is governed by a logarithmic zero-endpoint singularity and begins with $1/\log t$, rather than a power of $t$.

The Gamma reflection formula gives $(\sin \pi z/\pi )\Gamma (z)=1/\Gamma (1-z)$, analytic at zero with value one. A finite-parameter Watson's lemma representation of Ramanujan's function therefore gives the displayed coefficients. In particular $c_0=1$, $c_1=-\gamma$ and $c_2={\gamma }^2-{\pi }^2/6$, with $\gamma$ Euler's constant. Obtaining this representation requires taking an imaginary part before sending its Mellin exponent to zero; the original complex integral is divergent at that boundary.

For bounded locally integrable $f$ on the nonnegative real line, suppose its Laplace transform $F$ extends holomorphically to an open neighborhood of the closed right half-plane. Then the ordinary improper integral converges to $F(0)$. Contour damping for bounded Laplace transforms bounds the difference from the finite-interval transform by $2\Vert f{\Vert }_{\infty }/R$ in the limit for every radius $R$. Extension merely to the open right half-plane is insufficient, as the constant function and transform $1/z$ show.

The displayed kernel has residue one at zero and vanishes at the endpoints of a radius-$R$ imaginary-axis segment. On either radius-$R$ semicircle, $|1+z^2/R^2|=2|\mathrm{Re}z|/R$, cancelling the inverse real-part bound on a bounded function's Laplace transform tail. Use the full transform only on a right semicircle and a thin analytic leftward path; deform the finite-interval entire transform to the large left semicircle. This proves the Newman Tauberian theorem without assuming analytic continuation across an entire left half-disk.

For functions $f,g$ on $[0,\infty )$ with $e^{-\sigma t}f(t)$ and $e^{-\sigma t}g(t)$ Lebesgue integrable functions, their unilateral convolution, defined almost everywhere if necessary, satisfiesTaking a causal extension by zero of both functions gives the existing whole-line convolution definition. Absolute convergence and Fubini's theorem justify the product rule: put $t=\tau +u$ in the double integral, so $e^{-st}=e^{-s\tau }{e}^{-su}$. For piecewise continuous functions of exponential order $a$ and $b$, the hypotheses hold for $\mathrm{Re}s>\max (a,b)$. The Laplace transform article works out the concrete example $h(t)=e^{-2t}$ and the Heaviside step function input, whose output is $(1-e^{-2t})/2$. ^[ref].

Under suitable regularity and growth conditions, a function with Laplace transform $F$ is recovered by the inverse contour integralwhere the vertical line lies to the right of every singularity of $F$.

A Laplace transform $G_T(p)={\int }_0^T{e}^{-ps}g(s)ds$ can grow exponentially as $\mathrm{Re}p\to -\infty$. At an evaluation time $t<T$, multiplying by $e^{pt}$ does not automatically permit closing the inverse contour to the left. Either retain the growth-bearing contour pieces, use a causal time convolution, or justify a deformation with the time-truncated transform. Resolvent-pole information alone is insufficient.

Terms singular at $s=0$ in a Laplace transform encode persistent late-time behaviour. In particular,contributes $At+B$ to the inverse transform; singularities farther into the left half-plane contribute decaying terms under the usual analyticity and growth assumptions.

Suppose $f$ is an absolutely continuous function on every finite interval $[0,T]$, both $e^{-st}f(t)$ and $e^{-st}{f}^{\prime }(t)$ are Lebesgue integrable functions, and $e^{-sT}f(T)\to 0$. Then integration by parts on $[0,T]$ followed by a limit givesHere $f(0+)$ is the right one-sided limit. Exponential order bounds on $f$ and $f^{\prime }$ suffice when $\mathrm{Re}s$ exceeds both bounds. For $f(t)=e^{-2t}$, the identity reads $-2/(s+2)=s/(s+2)-1$ on $\mathrm{Re}s>-2$. This is the ordinary derivative on nonnegative time; differentiating a causal extension by zero at its jump instead introduces a Dirac delta function. ^[ref].

For $a\ge 0$, $\mathcal{L}\{f(t-a)H(t-a)\}(s)=e^{-as}\mathcal{L}\{f\}(s)$.

The delayed unit step has

The Heaviside function is zero before its switching time and one after it, allowing delayed signals to be written algebraically.

If $g$ has period $T$, then

The Laplace transform of a matrix-valued function is obtained by transforming each entry. A constant matrix may be taken outside on the corresponding side of the integral; in particular, $\mathcal{L}\{AB\}=A\mathcal{L}\{B\}$ when $A$ is constant.

Forthe derivative rule givesand hence $Y(s)=(sI-A{)}^{-1}(y_0+G(s))$ wherever the transforms converge and $sI-A$ is invertible.