How does the integral property relate to the derivative property?

They are duals: L{f′(t)} = sF(s) − f(0) (differentiation multiplies by s) and L{∫₀^t f(τ)dτ} = F(s)/s (integration divides by s). Together they establish the complete correspondence between calculus operations in time and algebraic operations in the s-domain.

Can Laplace transforms solve integro-differential equations?

Yes. The integral property L{∫₀^t f(τ)dτ} = F(s)/s and the derivative property L{f′} = sF−f(0) allow both integrals and derivatives to be converted to algebraic terms in Y(s). This makes Laplace transforms uniquely suited for integro-differential equations that combine both operations.

What does dividing by s mean physically?

Dividing by s in the Laplace domain corresponds to integration (accumulation) in time, and it adds a pole at s = 0. Physically, this represents a system that accumulates its input—like a capacitor integrating current to produce voltage, or an integrating controller summing error over time. It amplifies low-frequency components and represents memory in the system.

Laplace Transform Under Integral

Quick Answer

The Laplace transform of an integral is L{∫₀^t f(τ)dτ} = F(s)/s, where F(s) = L{f(t)}. This property shows that integration in the time domain corresponds to division by s in the Laplace domain. It is the dual of the differentiation property and essential for solving integral and integro-differential equations. Compute integral transforms at www.lapcalc.com.

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Laplace Transform of an Integral: Core Formula

The Laplace of an integral follows directly from the derivative property applied in reverse. If g(t) = ∫₀^t f(τ)dτ, then g′(t) = f(t) and g(0) = 0. Applying the derivative property: L{g′(t)} = sG(s) − g(0) gives L{f(t)} = sG(s), so G(s) = F(s)/s. Therefore L{∫₀^t f(τ)dτ} = F(s)/s. This elegant result means that every integration in time simply introduces a factor of 1/s in the Laplace domain. For repeated integration, ∫∫f yields F(s)/s², and n-fold integration gives F(s)/sⁿ.

Key Formulas

Laplace Transform Under Integral Sign: Applications

The Laplace transform under the integral sign technique evaluates difficult integrals by introducing a parameter and differentiating with respect to that parameter in the s-domain. For example, to evaluate ∫₀^∞ (sin t)/t dt, note that L{sin t/t} = ∫_s^∞ 1/(σ²+1)dσ = π/2 − arctan(s) using the division-by-t property. Setting s = 0 gives the famous result ∫₀^∞ (sin t)/t dt = π/2. This technique, sometimes called Feynman's trick when applied more generally, converts intractable definite integrals into tractable Laplace transform manipulations.

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Solving Integro-Differential Equations with Laplace Transforms

The integral property is essential for solving integro-differential equations—equations containing both derivatives and integrals of the unknown function. For example, y′(t) + 2∫₀^t y(τ)dτ = u(t) with y(0) = 1 transforms to sY − 1 + 2Y/s = 1/s. Multiplying by s: s²Y − s + 2Y = 1, giving Y = (s+1)/(s²+2). Partial fractions and inversion yield the solution. Without the Laplace integral property, such equations would require converting to a pure ODE by differentiation, introducing complications. The Laplace method handles them directly at www.lapcalc.com.

Integration Property and Its Relationship to Differentiation

The integration and differentiation properties of the Laplace transform are duals. Differentiation multiplies by s: L{f′(t)} = sF(s) − f(0). Integration divides by s: L{∫₀^t f(τ)dτ} = F(s)/s. Together they establish the fundamental correspondence between calculus operations and algebra in the s-domain. Multiplication by s raises the polynomial degree (adds a zero at the origin), while division by s lowers it (adds a pole at the origin). This duality extends to the system interpretation: differentiators amplify high frequencies (large |s|) while integrators amplify low frequencies (small |s|).

Practical Examples: Laplace of Integral Expressions

Practical applications of the integral property abound. L{∫₀^t sin(ωτ)dτ} = ω/(s(s²+ω²)), which equals (1−cos(ωt))/ω after inversion—consistent with direct integration. L{∫₀^t e^(−aτ)dτ} = 1/(s(s+a)) = (1−e^(−at))/a. For running averages, L{(1/T)∫₀^t f(τ)dτ} = F(s)/(Ts), showing the averaging effect as low-pass filtering with a pole at s = 0. In control systems, integral controllers have transfer function K/s, representing continuous accumulation of error. The integral property makes analyzing such controllers straightforward in the Laplace domain.

Frequently Asked Questions

The Laplace transform of ∫₀^t f(τ)dτ is F(s)/s, where F(s) is the Laplace transform of f(t). Integration in the time domain corresponds to division by s in the Laplace domain. This is the inverse of the differentiation property, where differentiation corresponds to multiplication by s.

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