Laplace transforms of functions te^(−4t)sin3t

Definition

For a function f(t) defined for t ≥ 0, the Laplace transform is

L{f(t)} = F(s) = ∫₀^∞ e^(−st) f(t) dt,

provided the integral converges (for s large enough).

a) L{t e^(−4t) sin 3t}

L{sin 3t} = 3/(s² + 9)
First shifting (s → s + 4): L{e^(−4t) sin 3t} = 3/((s + 4)² + 9)
Multiply by t: L{t f(t)} = −dF/ds
 = −d/ds [3((s + 4)² + 9)⁻¹]
 = 6(s + 4)/((s + 4)² + 9)²

b) L{(1 − eᵗ)/t}

Use L{f(t)/t} = ∫ₛ^∞ F(u) du with F(s) = 1/s − 1/(s − 1).

∫ₛ^∞ [1/u − 1/(u − 1)] du = [ln(u/(u − 1))]ₛ^∞ = 0 − ln(s/(s − 1)) = ln((s − 1)/s)     (s > 1)

Answers: a) 6(s + 4)/((s + 4)² + 9)²; b) ln((s − 1)/s).