The Laplace transform converts a function of a real variable $t \ge 0$ — typically representing time — into a function of a complex variable $p$, replacing [differentiation](/page/Derivative) with multiplication and initial-value problems with algebraic equations. It is the principal tool of **operational calculus**: the systematic reduction of linear differential equations with constant coefficients to polynomial algebra in the transform variable. Unlike the [Fourier transform](/page/Fourier%20Transform), which requires global [integrability](/page/Integral) or square-integrability over the entire real line, the Laplace transform accommodates exponentially growing signals by restricting integration to the half-line $[0, \infty)$ and introducing an exponential damping factor $e^{-pt}$. This makes it the natural transform for causal systems — those whose behaviour depends only on the present and the past. The theory rests on three pillars. First, the **forward transform** maps a function $f$ to its image $\hat{f}(p)$, which is holomorphic in a half-plane determined by the growth rate of $f$. Second, the **operational rules** express derivatives, integrals, convolutions, and shifts of $f$ as algebraic operations on $\hat{f}$. Third, the **inversion formula** recovers $f$ from $\hat{f}$ via a contour integral in the complex plane, connecting the Laplace transform to the theory of residues and [complex analysis](/page/Cambridge%20IB%20Complex%20Analysis). [motivation] ## Motivation ### Direct Methods for Constant-Coefficient ODEs Consider a linear ordinary differential equation with constant coefficients, such as the second-order initial-value problem \begin{align*} y'' + 3y' + 2y &= e^{-3t}, \quad t > 0, \\ y(0) &= 1, \quad y'(0) = 0. \end{align*} The classical method proceeds in two stages: find the general solution to the homogeneous equation $y'' + 3y' + 2y = 0$ via the characteristic polynomial $\lambda^2 + 3\lambda + 2 = (\lambda + 1)(\lambda + 2) = 0$, then find a particular solution to the non-homogeneous equation by undetermined coefficients or [variation of parameters](/theorems/835), and finally impose initial conditions to determine the constants. This works, but it has two weaknesses. First, the initial conditions are imposed at the very end, after the general solution has been constructed — meaning the work of finding the full general solution is performed even though only one specific solution is needed. Second, the method scales poorly: for higher-order equations, systems of equations, or equations with discontinuous or impulsive forcing terms (such as the Heaviside step function or the Dirac delta), the machinery of undetermined coefficients becomes cumbersome or inapplicable entirely. What we want is a method that incorporates the initial conditions from the outset and reduces the entire problem — differential equation plus initial data — to algebra in one step. ### The Fourier Transform Attempt A natural candidate is the [Fourier transform](/page/Fourier%20Transform). If $f: \mathbb{R} \to \mathbb{C}$ is integrable, its Fourier transform $\hat{f}(\xi) = \int_{\mathbb{R}} f(t) e^{-i\xi t} \, d\mathcal{L}^1(t)$ converts differentiation into multiplication by $i\xi$, so a differential equation becomes a polynomial equation in frequency space. However, the Fourier transform requires $f \in L^1(\mathbb{R}, \mathcal{L}^1)$ (or $L^2(\mathbb{R}, \mathcal{L}^1)$ via the Plancherel extension), and the solutions to initial-value problems on $[0, \infty)$ typically grow — or at least do not decay — as $t \to \infty$. For example, the solution $y(t) = e^{-t}$ to the homogeneous problem $y' + y = 0$, $y(0) = 1$ does lie in $L^1([0,\infty))$, but the solution $y(t) = e^t$ to $y' - y = 0$, $y(0) = 1$ does not, and neither does the constant solution $y(t) = 1$ to $y' = 0$, $y(0) = 1$. The Fourier integral simply diverges for these functions. There is a second, subtler difficulty. Even when the integral converges, the Fourier transform is defined over $(-\infty, \infty)$ and treats the function as a two-sided signal. It has no mechanism for encoding initial conditions at $t = 0$ — the very data that distinguishes one solution of a linear ODE from another. Any transform that hopes to solve initial-value problems must be sensitive to the starting point $t = 0$. ### Exponential Damping and the Half-Line Both difficulties are resolved by a single modification: restrict the domain of integration to $[0, \infty)$ and introduce a decaying exponential weight $e^{-pt}$ with $\operatorname{Re} p$ large enough to kill the growth of $f$. If $f$ grows at most as fast as $e^{\sigma_0 t}$ for some $\sigma_0 \in \mathbb{R}$, then the product $f(t) e^{-pt}$ decays exponentially whenever $\operatorname{Re} p > \sigma_0$, and the integral \begin{align*} \hat{f}(p) = \int_0^\infty f(t) e^{-pt} \, d\mathcal{L}^1(t) \end{align*} converges absolutely. This is the Laplace transform. The half-line integration is not merely a convenience — it is the mechanism by which initial conditions enter the calculus. When we integrate $f'(t) e^{-pt}$ by parts over $[0, \infty)$, the boundary term at $t = 0$ produces exactly the initial value $f(0)$, so the derivative rule becomes $\mathcal{L}\{f'\}(p) = p\hat{f}(p) - f(0)$. The initial data is encoded automatically in the algebraic relation. [/motivation] ## The Laplace Transform ### Exponential Order Before defining the transform, we must identify the class of [functions](/page/Function) for which the defining integral converges. The key condition is that $f$ does not grow faster than some exponential — without this constraint, no exponential damping factor $e^{-pt}$ can restore integrability. [definition:Exponential Order] A function $f: [0, \infty) \to \mathbb{C}$ is of **exponential order** if there exist constants $M > 0$ and $\sigma_0 \in \mathbb{R}$ such that \begin{align*} |f(t)| \le M e^{\sigma_0 t} \quad \text{for all } t \ge 0. \end{align*} The infimum of all such $\sigma_0$ is called the **abscissa of exponential growth** of $f$. [/definition] Functions of exponential order include all polynomials (with $\sigma_0 > 0$ arbitrarily small), all exponentials $e^{at}$ (with $\sigma_0 = \operatorname{Re} a$), all bounded functions (with $\sigma_0 = 0$), and products of these. [example:Failure Of Exponential Order] The function $f: [0, \infty) \to \mathbb{R}$ defined by $f(t) = e^{t^2}$ is not of exponential order. To see this, suppose for contradiction that $|f(t)| \le M e^{\sigma_0 t}$ for all $t \ge 0$. Then $e^{t^2} \le M e^{\sigma_0 t}$, which gives $e^{t^2 - \sigma_0 t} \le M$ for all $t$. But \begin{align*} t^2 - \sigma_0 t = \left(t - \frac{\sigma_0}{2}\right)^2 - \frac{\sigma_0^2}{4} \to \infty \quad \text{as } t \to \infty, \end{align*} so $e^{t^2 - \sigma_0 t} \to \infty$, contradicting the bound. The super-exponential growth of $e^{t^2}$ cannot be tamed by any linear exponential $e^{\sigma_0 t}$. Consequently, the integral $\int_0^\infty e^{t^2} e^{-pt} \, d\mathcal{L}^1(t)$ diverges for every $p \in \mathbb{C}$: for any fixed $\operatorname{Re} p$, the integrand $e^{t^2 - (\operatorname{Re} p) t}$ grows without bound, and the Laplace transform of $e^{t^2}$ does not exist anywhere in the complex plane. [/example] ### Definition With exponential order in hand, we can define the transform itself. The construction follows directly from the motivation: integrate $f$ against the kernel $e^{-pt}$ over the half-line, with $\operatorname{Re} p$ large enough to ensure convergence. [definition:Laplace Transform] Let $f: [0, \infty) \to \mathbb{C}$ be a locally integrable function of exponential order with abscissa of exponential growth $\sigma_0$. The **Laplace transform** of $f$ is the function \begin{align*} \mathcal{L}\{f\}: \{p \in \mathbb{C} : \operatorname{Re} p > \sigma_0\} &\to \mathbb{C} \\ p &\mapsto \int_0^\infty f(t) e^{-pt} \, d\mathcal{L}^1(t). \end{align*} We write $\hat{f}(p) = \mathcal{L}\{f\}(p)$ for the transform of $f$ evaluated at $p$. [/definition] The transform is defined on a half-plane $\operatorname{Re} p > \sigma_0$, not on a line as in the Fourier case. The larger $\operatorname{Re} p$, the stronger the damping — so the integral converges more easily. The [boundary](/page/Boundary) $\operatorname{Re} p = \sigma_0$ is the critical line: on it the integral may or may not converge, depending on the finer behaviour of $f$. [example:Elementary Transforms] The simplest transforms can be computed directly from the definition. **The exponential.** Let $a \in \mathbb{C}$ and define $f: [0, \infty) \to \mathbb{C}$ by $f(t) = e^{at}$. Then $f$ is of exponential order with $\sigma_0 = \operatorname{Re} a$. For $\operatorname{Re} p > \operatorname{Re} a$: \begin{align*} \hat{f}(p) &= \int_0^\infty e^{at} e^{-pt} \, d\mathcal{L}^1(t) = \int_0^\infty e^{-(p-a)t} \, d\mathcal{L}^1(t) = \left[ \frac{e^{-(p-a)t}}{-(p-a)} \right]_0^\infty = \frac{1}{p - a}. \end{align*} The boundary term at $t = \infty$ vanishes because $\operatorname{Re}(p - a) > 0$ ensures $|e^{-(p-a)t}| = e^{-\operatorname{Re}(p-a) \cdot t} \to 0$. Setting $a = 0$ gives $\mathcal{L}\{1\}(p) = 1/p$ for $\operatorname{Re} p > 0$. **Powers of $t$.** Let $n \in \mathbb{N}$ and define $f(t) = t^n$. Repeated integration by parts yields \begin{align*} \mathcal{L}\{t^n\}(p) &= \int_0^\infty t^n e^{-pt} \, d\mathcal{L}^1(t) = \frac{n}{p} \int_0^\infty t^{n-1} e^{-pt} \, d\mathcal{L}^1(t) = \cdots = \frac{n!}{p^{n+1}}, \end{align*} valid for $\operatorname{Re} p > 0$. At each stage, the integration by parts uses [Integration By Parts](/theorems/210) with $u = t^k$ and $dv = e^{-pt} \, dt$; the boundary term at infinity vanishes because the exponential decay dominates any polynomial growth. **The sine function.** Let $b \in \mathbb{R} \setminus \{0\}$ and define $f(t) = \sin(bt)$. Since $|\sin(bt)| \le 1$, this function is of exponential order with $\sigma_0 = 0$. Using the identity $\sin(bt) = (e^{ibt} - e^{-ibt})/(2i)$ and the exponential transform: \begin{align*} \mathcal{L}\{\sin(bt)\}(p) &= \frac{1}{2i}\left(\frac{1}{p - ib} - \frac{1}{p + ib}\right) = \frac{1}{2i} \cdot \frac{2ib}{p^2 + b^2} = \frac{b}{p^2 + b^2}, \end{align*} valid for $\operatorname{Re} p > 0$. An analogous computation gives $\mathcal{L}\{\cos(bt)\}(p) = p/(p^2 + b^2)$. **The Heaviside step.** Define $H_a: [0, \infty) \to \mathbb{R}$ by $H_a(t) = \mathbb{1}_{[a, \infty)}(t)$ for $a > 0$. Then $H_a$ is bounded, so $\sigma_0 = 0$. For $\operatorname{Re} p > 0$: \begin{align*} \mathcal{L}\{H_a\}(p) &= \int_a^\infty e^{-pt} \, d\mathcal{L}^1(t) = \left[\frac{e^{-pt}}{-p}\right]_a^\infty = \frac{e^{-ap}}{p}. \end{align*} The factor $e^{-ap}$ encodes the delay: shifting a signal by $a$ units in the time domain multiplies its transform by $e^{-ap}$ in the frequency domain. [/example] ## Analytic Properties The Laplace transform is not merely a formal device — it produces a holomorphic function in the half-plane of convergence, and this holomorphicity is the source of its power. ### Holomorphicity A function defined by an integral depending on a complex parameter is often holomorphic in that parameter, provided the integrand is sufficiently well-behaved. The Laplace integral is a paradigmatic example: the integrand $f(t) e^{-pt}$ is entire in $p$ for each fixed $t$, and the exponential order condition provides the uniform domination needed to differentiate under the integral sign. [theorem:Holomorphicity Of The Laplace Transform] Let $f: [0, \infty) \to \mathbb{C}$ be locally integrable and of exponential order with abscissa $\sigma_0$. Then $\hat{f}$ is holomorphic on the half-plane $\{p \in \mathbb{C} : \operatorname{Re} p > \sigma_0\}$, and its derivative is given by \begin{align*} \hat{f}'(p) = -\int_0^\infty t f(t) e^{-pt} \, d\mathcal{L}^1(t) = -\mathcal{L}\{t f(t)\}(p). \end{align*} More generally, for each $n \in \mathbb{N}$: \begin{align*} \hat{f}^{(n)}(p) = (-1)^n \mathcal{L}\{t^n f(t)\}(p). \end{align*} [/theorem] The proof proceeds by verifying the hypotheses of the [Dominated Convergence Theorem](/theorems/4): for any compact subset $K$ of the half-plane $\operatorname{Re} p > \sigma_0$, there exists $c > \sigma_0$ such that $\operatorname{Re} p \ge c$ on $K$, and the integrand $|f(t) e^{-pt}| \le M e^{(\sigma_0 - c)t}$ provides an integrable dominating function. Differentiating under the integral sign is then justified, and each differentiation pulls down a factor of $-t$. This result has an important consequence: $\hat{f}(p)$ is automatically infinitely differentiable — indeed, analytic — in the half-plane. No smoothness of $f$ is required; even discontinuous or [distributional](/page/Distribution) inputs produce holomorphic outputs. This regularising effect is one of the most striking features of the transform. [example:Differentiation In The Transform Variable] We can use holomorphicity to compute transforms that would be awkward to evaluate directly. Since $\mathcal{L}\{e^{at}\}(p) = 1/(p-a)$, differentiating both sides with respect to $p$ gives: \begin{align*} \mathcal{L}\{-t e^{at}\}(p) = \frac{d}{dp}\left(\frac{1}{p-a}\right) = -\frac{1}{(p-a)^2}. \end{align*} Therefore $\mathcal{L}\{t e^{at}\}(p) = 1/(p-a)^2$. Differentiating again: \begin{align*} \mathcal{L}\{t^2 e^{at}\}(p) = \frac{2}{(p-a)^3}. \end{align*} More generally, by induction: $\mathcal{L}\{t^n e^{at}\}(p) = n!/(p-a)^{n+1}$. This gives the entire family of transforms $t^n e^{at}$ without performing any integration. [/example] ### Decay at Infinity Holomorphicity alone does not control the behaviour of $\hat{f}$ at infinity — a holomorphic function on a half-plane can grow without bound (e.g., $e^p$ is entire and grows exponentially). For the inversion formula to work, we must close contours in the left half-plane, and the integral over the closing arc must vanish. This requires $\hat{f}(p) \to 0$ as $|p| \to \infty$ in the half-plane — a decay estimate that holomorphicity alone does not provide. [theorem:Riemann Lebesgue For Laplace Transforms] Let $f: [0, \infty) \to \mathbb{C}$ be locally integrable and of exponential order with abscissa $\sigma_0$. Then for any fixed $c > \sigma_0$: \begin{align*} \hat{f}(p) \to 0 \quad \text{as } |p| \to \infty \text{ with } \operatorname{Re} p \ge c. \end{align*} [/theorem] This is the Laplace analogue of the Riemann–Lebesgue lemma for Fourier transforms. It guarantees that the transform vanishes at infinity in any closed half-plane strictly to the right of the abscissa, which is essential for closing contours in the inversion formula. The proof reduces to the classical Riemann–Lebesgue lemma applied to the function $g(t) = f(t) e^{-ct} \in L^1([0,\infty), \mathcal{L}^1)$ with Fourier variable $\operatorname{Im} p$. ### Why Uniqueness Matters A natural question is whether the transform is injective: if two functions have the same Laplace transform, must they be equal? Without injectivity, the entire operational calculus collapses — algebraic manipulations in the transform domain would not determine a unique solution, and reading off inverse transforms from a table would be ambiguous. [theorem:Lerch Uniqueness Theorem] Let $f, g: [0, \infty) \to \mathbb{C}$ be continuous and of exponential order. If $\hat{f}(p) = \hat{g}(p)$ for all $p$ in some half-plane $\operatorname{Re} p > \sigma$, then $f(t) = g(t)$ for all $t \ge 0$. [/theorem] The uniqueness theorem justifies the use of transform tables: if we recognise the algebraic expression $\hat{f}(p) = p/(p^2 + b^2)$ as the transform of $\cos(bt)$, then $\cos(bt)$ is the *only* continuous function with that transform. For locally integrable functions, the conclusion weakens to $f = g$ almost everywhere with respect to $\mathcal{L}^1$. [example:Uniqueness Determines Inverse Transforms] Suppose we have solved an ODE in the transform domain and obtained $\hat{y}(p) = (2p + 1)/(p^2 + 1)$. We decompose: \begin{align*} \frac{2p + 1}{p^2 + 1} = 2 \cdot \frac{p}{p^2 + 1} + 1 \cdot \frac{1}{p^2 + 1}. \end{align*} From the elementary transforms, $\mathcal{L}\{\cos(t)\}(p) = p/(p^2 + 1)$ and $\mathcal{L}\{\sin(t)\}(p) = 1/(p^2 + 1)$. By linearity of the transform and the Lerch uniqueness theorem, the *only* continuous function with this transform is: \begin{align*} y(t) = 2\cos(t) + \sin(t). \end{align*} Without uniqueness, we could not conclude that this is the solution — there might be other functions with the same image. [/example] ## Operational Rules The power of the Laplace transform lies not in computing integrals one at a time, but in its **operational rules**: algebraic identities that express operations on functions (differentiation, integration, convolution, shifting) as operations on their transforms. These rules convert differential equations into algebraic equations and are the engine of operational calculus. ### The Derivative Rule The most important operational rule is the one that originally motivated the transform: differentiation in the time domain becomes multiplication — plus an initial-condition correction — in the transform domain. Without this rule, the Laplace transform would be merely an integral curiosity; with it, the transform becomes a systematic tool for solving initial-value problems. The key mechanism is [Integration By Parts](/theorems/210) on the half-line, which produces boundary terms encoding the initial data. [quotetheorem:362] The structure of this formula deserves careful attention. The leading term $p\hat{f}(p)$ represents the algebraic multiplication that replaces differentiation — this is the feature shared with the Fourier transform. The correction term $-f(0)$ is what distinguishes the Laplace transform: it injects the initial condition directly into the algebraic equation. For the $n$-th derivative formula, the sum $\sum_{k=0}^{n-1} p^{n-1-k} f^{(k)}(0)$ collects all initial values $f(0), f'(0), \ldots, f^{(n-1)}(0)$, each multiplied by the appropriate power of $p$. This is precisely the data needed to specify a unique solution of an $n$-th order linear ODE — so the transform is perfectly adapted to initial-value problems. ### The Role of Exponential Order in the Derivative Rule The hypothesis that $f$ and its derivatives be of exponential order is essential: without it, the boundary terms at $t = \infty$ in the integration by parts would not vanish. Specifically, the integration by parts identity gives: \begin{align*} \int_0^R f'(t) e^{-pt} \, d\mathcal{L}^1(t) = f(R) e^{-pR} - f(0) + p \int_0^R f(t) e^{-pt} \, d\mathcal{L}^1(t). \end{align*} For the limit $R \to \infty$ to produce the derivative rule, we need $f(R) e^{-pR} \to 0$. If $|f(R)| \le M e^{\sigma_0 R}$, then $|f(R) e^{-pR}| \le M e^{(\sigma_0 - \operatorname{Re} p)R} \to 0$ whenever $\operatorname{Re} p > \sigma_0$. The requirement of continuous differentiability can be relaxed — $f$ need only be absolutely continuous — but the exponential order condition cannot. [example:Transform Of A Second Derivative] Let $f: [0, \infty) \to \mathbb{R}$ be twice continuously differentiable with $f, f', f''$ of exponential order, $f(0) = 2$, and $f'(0) = -1$. Applying the derivative rule twice: \begin{align*} \mathcal{L}\{f'\}(p) &= p\hat{f}(p) - f(0) = p\hat{f}(p) - 2, \\ \mathcal{L}\{f''\}(p) &= p \cdot \mathcal{L}\{f'\}(p) - f'(0) = p(p\hat{f}(p) - 2) - (-1) = p^2 \hat{f}(p) - 2p + 1. \end{align*} This matches the general formula with $n = 2$: $p^2 \hat{f}(p) - p f(0) - f'(0) = p^2 \hat{f}(p) - 2p + 1$. Each initial value appears exactly once, multiplied by the appropriate power of $p$. [/example] ### Frequency Shifting and Exponential Modulation A shift in the transform variable corresponds to multiplication by an exponential in the time domain — and vice versa. This rule extends the transform table dramatically: once the transform of a single function is known, the transforms of all its exponentially modulated variants follow immediately. [theorem:First Shifting Rule] Let $f: [0, \infty) \to \mathbb{C}$ be locally integrable and of exponential order with abscissa $\sigma_0$, and let $a \in \mathbb{C}$. Then $e^{at} f(t)$ is of exponential order with abscissa $\sigma_0 + \operatorname{Re} a$, and \begin{align*} \mathcal{L}\{e^{at} f(t)\}(p) = \hat{f}(p - a) \quad \text{for } \operatorname{Re} p > \sigma_0 + \operatorname{Re} a. \end{align*} [/theorem] The proof is a direct substitution: $\int_0^\infty e^{at} f(t) e^{-pt} \, d\mathcal{L}^1(t) = \int_0^\infty f(t) e^{-(p-a)t} \, d\mathcal{L}^1(t) = \hat{f}(p - a)$. Despite its simplicity, this rule is indispensable. [example:Damped Oscillations Via Shifting] Knowing that $\mathcal{L}\{\cos(bt)\}(p) = p/(p^2 + b^2)$, the first shifting rule with $a = -\alpha$ ($\alpha > 0$) immediately gives: \begin{align*} \mathcal{L}\{e^{-\alpha t}\cos(bt)\}(p) = \frac{p + \alpha}{(p + \alpha)^2 + b^2}, \end{align*} valid for $\operatorname{Re} p > -\alpha$. Similarly: \begin{align*} \mathcal{L}\{e^{-\alpha t}\sin(bt)\}(p) = \frac{b}{(p + \alpha)^2 + b^2}. \end{align*} These are the transforms of damped oscillations — functions of central importance in circuit theory and vibration analysis. The shift $p \mapsto p + \alpha$ in the denominator encodes the damping rate $\alpha$, while the parameters $b$ and $p + \alpha$ control the oscillation frequency and the decay envelope respectively. [/example] ### The Convolution Rule In the time domain, many systems are described by convolution integrals: the output of a linear time-invariant system with impulse response $g$ driven by input $f$ is $(f * g)(t) = \int_0^t f(t - t') g(t') \, d\mathcal{L}^1(t')$. Computing such convolutions directly is often difficult, but in the transform domain, convolution becomes pointwise multiplication. The Laplace convolution differs from the standard [convolution](/page/Convolution) on $\mathbb{R}$: the [limits](/page/Limit) of integration run from $0$ to $t$ rather than from $-\infty$ to $\infty$, reflecting the causal nature of the functions involved (both vanish for $t < 0$). [quotetheorem:363] The convolution theorem converts a nonlocal operation (integration over $[0, t]$) into a local one (pointwise multiplication). From the theoretical side, it shows that the space of causal functions of exponential order, equipped with the convolution product $*$, forms a commutative algebra, and the Laplace transform is an algebra homomorphism from this convolution algebra to the algebra of holomorphic functions under pointwise multiplication. From the practical side, it provides a mechanism for solving integral equations and for computing the response of linear systems: if $\hat{g}(p)$ is the transfer function of a system and $\hat{f}(p)$ is the transform of the input, then $\hat{g}(p) \cdot \hat{f}(p)$ is the transform of the output. The proof uses the [Fubini Theorem](/theorems/513) to exchange the order of integration. One writes the double integral $\int_0^\infty \left(\int_0^t f(t - t') g(t') \, d\mathcal{L}^1(t')\right) e^{-pt} \, d\mathcal{L}^1(t)$, changes the order of integration over the triangular region $\{(t, t') : 0 \le t' \le t < \infty\}$, and substitutes $s = t - t'$ to separate the integrals into a product $\hat{f}(p) \cdot \hat{g}(p)$. The exponential order hypotheses guarantee the absolute convergence needed to apply Fubini. [example:Convolution To Solve An Integral Equation] Consider the integral equation \begin{align*} y(t) = t + \int_0^t y(t') \, d\mathcal{L}^1(t'). \end{align*} The integral on the right is $(y * 1)(t)$, the convolution of $y$ with the constant function $1$. Taking Laplace transforms of both sides and using the [Convolution Theorem for Laplace Transforms](/theorems/363): \begin{align*} \hat{y}(p) &= \frac{1}{p^2} + \hat{y}(p) \cdot \frac{1}{p}. \end{align*} Solving for $\hat{y}(p)$: \begin{align*} \hat{y}(p) - \frac{\hat{y}(p)}{p} &= \frac{1}{p^2}, \\ \hat{y}(p) \left(1 - \frac{1}{p}\right) &= \frac{1}{p^2}, \\ \hat{y}(p) \cdot \frac{p - 1}{p} &= \frac{1}{p^2}, \\ \hat{y}(p) &= \frac{1}{p(p - 1)}. \end{align*} Decomposing into partial fractions: $\frac{1}{p(p-1)} = \frac{1}{p-1} - \frac{1}{p}$. Since $\mathcal{L}\{e^t\}(p) = 1/(p-1)$ and $\mathcal{L}\{1\}(p) = 1/p$, uniqueness gives \begin{align*} y(t) = e^t - 1. \end{align*} One can verify directly: $y'(t) = e^t = 1 + (e^t - 1) = 1 + y(t)$, so $y(t) = t + \int_0^t y(t') \, d\mathcal{L}^1(t')$ by the [fundamental theorem of calculus](/theorems/632). [/example] ## Inversion Computing the Laplace transform of a given function is, in principle, a matter of evaluating a one-dimensional integral. The deeper question is the **inverse problem**: given a holomorphic function $\hat{f}(p)$ defined in a half-plane, how does one recover the original function $f(t)$? For simple rational functions, partial fractions and a table of known transforms suffice. But for general transforms — especially those arising from integral equations, boundary-value problems, or systems with transcendental transfer functions — a systematic inversion formula is indispensable. ### The Bromwich Contour Integral The fundamental inversion formula expresses $f(t)$ as a contour integral of $\hat{f}(p) e^{pt}$ along a vertical line in the complex plane. The line must lie entirely within the half-plane of convergence — to the right of all singularities of $\hat{f}$ — so that the exponential growth of $e^{pt}$ is controlled by the decay of $\hat{f}(p)$. [quotetheorem:360] The Bromwich formula is remarkable for several reasons. First, it provides an *exact* inversion — not an approximation or a formal identity, but a pointwise recovery of $f(t)$ at every point of [continuity](/page/Continuity). Second, the formula does not depend on the particular choice of $c > \sigma_0$; any vertical line to the right of all singularities gives the same result, a consequence of Cauchy's theorem: deforming the contour within the region of holomorphicity does not change the integral. Third, the condition $f(t) = 0$ for $t < 0$ is not a restriction but a reflection of causality: the inverse Laplace transform of a function holomorphic in a right half-plane is automatically causal. ### Residue Computation of the Inverse Transform In practice, the Bromwich integral is rarely evaluated directly as a line integral. Instead, one closes the contour by appending a large semicircular arc in the left half-plane and applies the residue theorem. When $\hat{f}$ has finitely many isolated singularities and decays sufficiently fast on the closing arc, the inverse transform reduces to a finite sum of residues. [quotetheorem:361] This result transforms the analytic problem of evaluating a contour integral into the algebraic problem of computing residues — a dramatic simplification. The hypothesis that $\hat{f}(p) \to 0$ uniformly as $|p| \to \infty$ in the left half-plane ensures that the integral over the closing semicircular arc vanishes (by Jordan's lemma or a direct estimate). For rational transforms $\hat{f}(p) = P(p)/Q(p)$ where $\deg P < \deg Q$, the decay condition is automatically satisfied, and the singularities are the roots of $Q$. This covers the vast majority of transforms arising from linear ODEs with constant coefficients. [example:Inversion By Residues] Let us invert \begin{align*} \hat{f}(p) = \frac{2p + 3}{(p + 1)(p + 2)^2}. \end{align*} The singularities are $p_1 = -1$ (simple pole) and $p_2 = -2$ (double pole), both in the left half-plane. By the [Residue Inversion for Laplace Transforms](/theorems/361), for $t > 0$: \begin{align*} f(t) = \operatorname{Res}_{p = -1}\left(\frac{(2p+3) e^{pt}}{(p+1)(p+2)^2}\right) + \operatorname{Res}_{p = -2}\left(\frac{(2p+3) e^{pt}}{(p+1)(p+2)^2}\right). \end{align*} **Residue at $p = -1$ (simple pole).** The residue is \begin{align*} \operatorname{Res}_{p = -1} &= \lim_{p \to -1} (p + 1) \cdot \frac{(2p+3) e^{pt}}{(p+1)(p+2)^2} = \frac{(2(-1)+3) e^{-t}}{(-1+2)^2} = \frac{1 \cdot e^{-t}}{1} = e^{-t}. \end{align*} **Residue at $p = -2$ (double pole).** For a pole of order $2$, the residue is \begin{align*} \operatorname{Res}_{p = -2} &= \lim_{p \to -2} \frac{d}{dp}\left[(p+2)^2 \cdot \frac{(2p+3) e^{pt}}{(p+1)(p+2)^2}\right] = \lim_{p \to -2} \frac{d}{dp}\left[\frac{(2p+3) e^{pt}}{p+1}\right]. \end{align*} Let $h(p) = (2p+3) e^{pt} / (p+1)$. By the quotient rule: \begin{align*} h'(p) &= \frac{[2e^{pt} + (2p+3) t e^{pt}](p+1) - (2p+3) e^{pt}}{(p+1)^2}. \end{align*} Evaluating at $p = -2$: the numerator becomes \begin{align*} [2e^{-2t} + (-1) t e^{-2t}](-1) - (-1) e^{-2t} &= (-2 + t) e^{-2t} \cdot (-1) + e^{-2t} \\ &= (2 - t) e^{-2t} + e^{-2t} \\ &= (3 - t) e^{-2t}, \end{align*} and the denominator is $(-2 + 1)^2 = 1$. Thus the residue at $p = -2$ is $(3 - t) e^{-2t}$. Combining: \begin{align*} f(t) = e^{-t} + (3 - t) e^{-2t}, \quad t > 0. \end{align*} [/example] ### Failure Of Inversion For Non-Rational Transforms The residue method applies cleanly to rational transforms. When $\hat{f}(p)$ has infinitely many singularities or essential singularities, the situation is more delicate. [example:Branch Cut Inversion] Consider $\hat{f}(p) = 1/\sqrt{p}$, where $\sqrt{p}$ denotes the principal branch with [branch cut](/page/Branch%20Cuts) along $(-\infty, 0]$. This is the Laplace transform of $f(t) = 1/\sqrt{\pi t}$ (for $t > 0$). The function $\hat{f}$ has no isolated singularities — instead, it has a branch point at $p = 0$. The residue theorem does not apply directly. To invert, one must deform the Bromwich contour around the branch cut: starting from the vertical line $\operatorname{Re} p = c > 0$, one closes the contour with a large arc in the left half-plane, a keyhole around the negative real axis, and a small circle around $p = 0$. On the negative real axis, $p = -s$ with $s > 0$, and $\sqrt{p}$ takes the values $\pm i\sqrt{s}$ above and below the cut. The contributions from the two sides of the cut combine to give: \begin{align*} f(t) &= \frac{1}{2\pi i} \int_0^\infty \left(\frac{1}{i\sqrt{s}} - \frac{1}{-i\sqrt{s}}\right) e^{-st} \, d\mathcal{L}^1(s) = \frac{1}{\pi} \int_0^\infty \frac{e^{-st}}{\sqrt{s}} \, d\mathcal{L}^1(s). \end{align*} Substituting $u = \sqrt{s}$ (so $s = u^2$, $ds = 2u \, du$): \begin{align*} f(t) = \frac{2}{\pi} \int_0^\infty e^{-u^2 t} \, d\mathcal{L}^1(u) = \frac{2}{\pi} \cdot \frac{\sqrt{\pi}}{2\sqrt{t}} = \frac{1}{\sqrt{\pi t}}. \end{align*} This confirms $\mathcal{L}\{(\pi t)^{-1/2}\}(p) = p^{-1/2}$ and illustrates that branch-cut integrals, not residues, are the correct tool when $\hat{f}$ is not meromorphic. [/example] ## Application to Differential Equations The raison d'être of the Laplace transform is the solution of initial-value problems for linear ODEs (and systems thereof) with constant coefficients. The procedure is uniform and mechanical: apply the transform to both sides, use the derivative rule to convert derivatives to polynomial expressions, solve the resulting algebraic equation for the transform of the unknown, and invert. [example:Solving A Second Order Initial Value Problem] Consider the initial-value problem from the motivation: \begin{align*} y'' + 3y' + 2y &= e^{-3t}, \quad t > 0, \\ y(0) &= 1, \quad y'(0) = 0. \end{align*} **Step 1: Apply the Laplace transform.** Taking $\mathcal{L}$ of both sides and using the [Laplace Transform of Derivatives](/theorems/362): \begin{align*} \mathcal{L}\{y''\}(p) + 3\mathcal{L}\{y'\}(p) + 2\hat{y}(p) &= \mathcal{L}\{e^{-3t}\}(p). \end{align*} Substituting the derivative rules with $y(0) = 1$ and $y'(0) = 0$: \begin{align*} [p^2 \hat{y}(p) - p \cdot 1 - 0] + 3[p \hat{y}(p) - 1] + 2\hat{y}(p) &= \frac{1}{p + 3}. \end{align*} **Step 2: Solve for $\hat{y}(p)$.** Collecting terms: \begin{align*} (p^2 + 3p + 2)\hat{y}(p) - p - 3 &= \frac{1}{p + 3}, \\ (p + 1)(p + 2)\hat{y}(p) &= p + 3 + \frac{1}{p + 3} = \frac{(p+3)^2 + 1}{p + 3} = \frac{p^2 + 6p + 10}{p + 3}. \end{align*} Therefore: \begin{align*} \hat{y}(p) = \frac{p^2 + 6p + 10}{(p + 1)(p + 2)(p + 3)}. \end{align*} **Step 3: Partial fraction decomposition.** We write \begin{align*} \frac{p^2 + 6p + 10}{(p+1)(p+2)(p+3)} = \frac{A}{p+1} + \frac{B}{p+2} + \frac{C}{p+3}. \end{align*} Multiplying through by $(p+1)(p+2)(p+3)$: \begin{align*} p^2 + 6p + 10 = A(p+2)(p+3) + B(p+1)(p+3) + C(p+1)(p+2). \end{align*} Setting $p = -1$: $1 - 6 + 10 = A(1)(2)$, so $A = 5/2$. Setting $p = -2$: $4 - 12 + 10 = B(-1)(1)$, so $B = -2$. Setting $p = -3$: $9 - 18 + 10 = C(-2)(-1)$, so $C = 1/2$. Thus: \begin{align*} \hat{y}(p) = \frac{5/2}{p + 1} - \frac{2}{p + 2} + \frac{1/2}{p + 3}. \end{align*} **Step 4: Invert.** Since $\mathcal{L}\{e^{at}\}(p) = 1/(p - a)$, we read off term by term: \begin{align*} y(t) = \frac{5}{2} e^{-t} - 2 e^{-2t} + \frac{1}{2} e^{-3t}, \quad t \ge 0. \end{align*} **Step 5: Verification.** One checks $y(0) = 5/2 - 2 + 1/2 = 1$ ✓ and $y'(0) = -5/2 + 4 - 3/2 = 0$ ✓. Computing $y'' + 3y' + 2y$ and simplifying confirms that it equals $e^{-3t}$. [/example] The procedure above illustrates the key advantage of the Laplace method: the initial conditions are encoded in Step 1 and carried through automatically. There is no need to find a general solution and then fit constants. [example:Discontinuous Forcing With The Heaviside Function] Consider the initial-value problem with a forcing term that switches on at $t = 2$: \begin{align*} y' + y &= 3 H_2(t), \quad t > 0, \\ y(0) &= 0, \end{align*} where $H_2(t) = \mathbb{1}_{[2, \infty)}(t)$ is the Heaviside step function shifted to $t = 2$. **Step 1: Transform.** Using the [Laplace Transform of Derivatives](/theorems/362) and the earlier computation $\mathcal{L}\{H_2\}(p) = e^{-2p}/p$: \begin{align*} p \hat{y}(p) - 0 + \hat{y}(p) = \frac{3 e^{-2p}}{p}. \end{align*} **Step 2: Solve.** $(p + 1) \hat{y}(p) = 3 e^{-2p}/p$, so: \begin{align*} \hat{y}(p) = \frac{3 e^{-2p}}{p(p + 1)}. \end{align*} **Step 3: Partial fractions and the second shifting theorem.** Decompose $3/(p(p+1)) = 3/p - 3/(p+1)$. The factor $e^{-2p}$ corresponds to a time delay of $2$ units: if $\mathcal{L}\{g(t)\}(p) = \hat{g}(p)$, then $\mathcal{L}\{g(t - 2) H_2(t)\}(p) = e^{-2p} \hat{g}(p)$. The function $g(t)$ whose transform is $3/p - 3/(p+1)$ is $g(t) = 3 - 3e^{-t}$. Therefore: \begin{align*} y(t) = \big(3 - 3e^{-(t-2)}\big) H_2(t) = \begin{cases} 0 & \text{if } 0 \le t < 2, \\ 3 - 3e^{-(t-2)} & \text{if } t \ge 2. \end{cases} \end{align*} **Step 4: Verification.** For $0 \le t < 2$: $y = 0$, $y' = 0$, and $y' + y = 0 = 3 \cdot 0$ ✓. For $t > 2$: $y' = 3e^{-(t-2)}$ and $y' + y = 3e^{-(t-2)} + 3 - 3e^{-(t-2)} = 3$ ✓. At $t = 2$, $y$ is continuous: $y(2^-) = 0$ and $y(2^+) = 3 - 3 = 0$ ✓. This example illustrates a situation where classical methods require separate treatment of the two intervals $[0, 2)$ and $[2, \infty)$ with matching conditions at $t = 2$, while the Laplace method handles the discontinuity through the exponential factor $e^{-2p}$ in a single computation. [/example] ## Problems [problem] Compute the Laplace transform $\hat{f}(p) = \mathcal{L}\{t^2 e^{3t}\}(p)$ and state the half-plane of convergence. [/problem] [solution] **Step 1: Identify the base transform.** From the elementary transforms, $\mathcal{L}\{t^2\}(p) = 2/p^3$ for $\operatorname{Re} p > 0$. **Step 2: Apply the first shifting rule.** By the [First Shifting Rule](/theorems/0) with $a = 3$: \begin{align*} \mathcal{L}\{t^2 e^{3t}\}(p) = \mathcal{L}\{t^2\}(p - 3) = \frac{2}{(p - 3)^3}. \end{align*} **Step 3: Determine the half-plane.** Since $\mathcal{L}\{t^2\}(p)$ converges for $\operatorname{Re} p > 0$, the shifted transform converges for $\operatorname{Re}(p - 3) > 0$, i.e., $\operatorname{Re} p > 3$. This matches the abscissa of exponential growth: $|t^2 e^{3t}| \le M e^{(3 + \varepsilon)t}$ for any $\varepsilon > 0$, so $\sigma_0 = 3$. [/solution] [problem] Solve the initial-value problem \begin{align*} y'' - 4y &= 8, \quad t > 0, \\ y(0) &= 0, \quad y'(0) = 0 \end{align*} using the Laplace transform. [/problem] [solution] **Step 1: Transform the equation.** Taking $\mathcal{L}$ of both sides and using the [Laplace Transform of Derivatives](/theorems/362) with $y(0) = 0$, $y'(0) = 0$: \begin{align*} [p^2 \hat{y}(p) - 0 - 0] - 4\hat{y}(p) = \frac{8}{p}. \end{align*} **Step 2: Solve for $\hat{y}(p)$.** \begin{align*} (p^2 - 4)\hat{y}(p) &= \frac{8}{p}, \\ \hat{y}(p) &= \frac{8}{p(p^2 - 4)} = \frac{8}{p(p-2)(p+2)}. \end{align*} **Step 3: Partial fractions.** Write \begin{align*} \frac{8}{p(p-2)(p+2)} = \frac{A}{p} + \frac{B}{p-2} + \frac{C}{p+2}. \end{align*} Setting $p = 0$: $8 = A(-2)(2) = -4A$, so $A = -2$. Setting $p = 2$: $8 = B(2)(4) = 8B$, so $B = 1$. Setting $p = -2$: $8 = C(-2)(-4) = 8C$, so $C = 1$. Thus $\hat{y}(p) = -2/p + 1/(p-2) + 1/(p+2)$. **Step 4: Invert.** Using $\mathcal{L}\{1\}(p) = 1/p$ and $\mathcal{L}\{e^{at}\}(p) = 1/(p-a)$: \begin{align*} y(t) = -2 + e^{2t} + e^{-2t} = -2 + 2\cosh(2t). \end{align*} **Step 5: Verify.** $y(0) = -2 + 2\cosh(0) = -2 + 2 = 0$ ✓. $y'(t) = 4\sinh(2t)$, so $y'(0) = 0$ ✓. $y''(t) = 8\cosh(2t)$, and $y'' - 4y = 8\cosh(2t) - 4(-2 + 2\cosh(2t)) = 8\cosh(2t) + 8 - 8\cosh(2t) = 8$ ✓. [/solution] [problem] Let $f(t) = \sin(t)$ and $g(t) = e^{-t}$ for $t \ge 0$. Compute the convolution $(f * g)(t) = \int_0^t \sin(t - t') e^{-t'} \, d\mathcal{L}^1(t')$ using the [Convolution Theorem for Laplace Transforms](/theorems/363). [/problem] [solution] **Step 1: Compute the individual transforms.** From the elementary transforms: \begin{align*} \hat{f}(p) = \mathcal{L}\{\sin(t)\}(p) = \frac{1}{p^2 + 1}, \quad \hat{g}(p) = \mathcal{L}\{e^{-t}\}(p) = \frac{1}{p + 1}. \end{align*} **Step 2: Multiply transforms.** By the convolution theorem, $\mathcal{L}\{f * g\}(p) = \hat{f}(p) \cdot \hat{g}(p)$: \begin{align*} \mathcal{L}\{f * g\}(p) = \frac{1}{(p^2 + 1)(p + 1)}. \end{align*} **Step 3: Partial fraction decomposition.** Write \begin{align*} \frac{1}{(p^2 + 1)(p+1)} = \frac{A}{p + 1} + \frac{Bp + C}{p^2 + 1}. \end{align*} Multiplying through: $1 = A(p^2 + 1) + (Bp + C)(p + 1)$. Setting $p = -1$: $1 = A(2)$, so $A = 1/2$. Expanding the right side with $A = 1/2$: \begin{align*} 1 &= \frac{1}{2}(p^2 + 1) + Bp^2 + Bp + Cp + C \\ &= \left(\frac{1}{2} + B\right)p^2 + (B + C)p + \left(\frac{1}{2} + C\right). \end{align*} Matching coefficients: $p^2$: $1/2 + B = 0$, so $B = -1/2$. $p^0$: $1/2 + C = 1$, so $C = 1/2$. Check $p^1$: $B + C = -1/2 + 1/2 = 0$ ✓. Thus: \begin{align*} \mathcal{L}\{f * g\}(p) = \frac{1/2}{p + 1} + \frac{-p/2 + 1/2}{p^2 + 1} = \frac{1}{2} \cdot \frac{1}{p+1} - \frac{1}{2} \cdot \frac{p}{p^2+1} + \frac{1}{2} \cdot \frac{1}{p^2+1}. \end{align*} **Step 4: Invert.** Reading off from the table: \begin{align*} (f * g)(t) = \frac{1}{2} e^{-t} - \frac{1}{2}\cos(t) + \frac{1}{2}\sin(t). \end{align*} This is the convolution $\int_0^t \sin(t - t') e^{-t'} \, d\mathcal{L}^1(t')$ computed without performing any direct integration. [/solution] [problem] Invert the Laplace transform \begin{align*} \hat{f}(p) = \frac{p}{(p^2 + 1)^2} \end{align*} using (a) the differentiation rule $\hat{f}^{(n)}(p) = (-1)^n \mathcal{L}\{t^n f(t)\}(p)$, and (b) the convolution theorem. [/problem] [solution] **(a) Differentiation method.** **Step 1: Recognise the derivative.** Observe that $\frac{d}{dp}\left(\frac{1}{p^2 + 1}\right) = \frac{-2p}{(p^2 + 1)^2}$, so: \begin{align*} \frac{p}{(p^2 + 1)^2} = -\frac{1}{2} \cdot \frac{d}{dp}\left(\frac{1}{p^2 + 1}\right). \end{align*} **Step 2: Apply the differentiation rule.** Since $\mathcal{L}\{\sin(t)\}(p) = 1/(p^2 + 1)$ and $\hat{g}'(p) = -\mathcal{L}\{t g(t)\}(p)$: \begin{align*} \frac{d}{dp}\left(\frac{1}{p^2 + 1}\right) = -\mathcal{L}\{t \sin(t)\}(p). \end{align*} Therefore: \begin{align*} \frac{p}{(p^2 + 1)^2} = -\frac{1}{2} \cdot (-\mathcal{L}\{t\sin(t)\}(p)) = \frac{1}{2}\mathcal{L}\{t\sin(t)\}(p). \end{align*} By uniqueness: $f(t) = \frac{1}{2} t \sin(t)$. **(b) Convolution method.** **Step 1: Factor the transform.** Write $\hat{f}(p) = \frac{p}{(p^2+1)^2} = \frac{p}{p^2+1} \cdot \frac{1}{p^2+1}$. This is $\mathcal{L}\{\cos(t)\}(p) \cdot \mathcal{L}\{\sin(t)\}(p)$. **Step 2: Apply the convolution theorem.** By the [Convolution Theorem for Laplace Transforms](/theorems/363): \begin{align*} f(t) = (\cos * \sin)(t) = \int_0^t \cos(t - t') \sin(t') \, d\mathcal{L}^1(t'). \end{align*} **Step 3: Evaluate the convolution.** Using the product-to-sum identity $\cos(A)\sin(B) = \frac{1}{2}[\sin(A+B) - \sin(A-B)]$ with $A = t - t'$, $B = t'$: \begin{align*} \cos(t-t')\sin(t') = \frac{1}{2}[\sin(t) - \sin(t - 2t')]. \end{align*} Integrating: \begin{align*} f(t) &= \frac{1}{2}\int_0^t \sin(t) \, d\mathcal{L}^1(t') - \frac{1}{2}\int_0^t \sin(t - 2t') \, d\mathcal{L}^1(t') \\ &= \frac{1}{2} t \sin(t) - \frac{1}{2}\left[\frac{\cos(t - 2t')}{2}\right]_0^t \\ &= \frac{1}{2} t \sin(t) - \frac{1}{4}[\cos(-t) - \cos(t)] \\ &= \frac{1}{2} t \sin(t), \end{align*} confirming the result from method (a). [/solution] ## References - Widder, D.V., *The Laplace Transform* (1941). - Doetsch, G., *Introduction to the Theory and Application of the Laplace Transformation* (1974). - Schiff, J.L., *The Laplace Transform: Theory and Applications* (1999).