Convolution theorem and inverse Laplace

Convolution theorem

If L⁻¹{F(s)} = f(t) and L⁻¹{G(s)} = g(t), then

L⁻¹{F(s) G(s)} = (f * g)(t) = ∫₀^t f(u) g(t − u) du.

Application

Take F(s) = s/(s² + 1) and G(s) = 1/(s² + 4). Then f(t) = cos t and g(t) = (sin 2t)/2.

L⁻¹{s/((s² + 1)(s² + 4))} = ∫₀^t cos u · ½ sin 2(t − u) du

Use cos u sin(2t − 2u) = ½[sin(2t − u) + sin(2t − 3u)]:

∫₀^t sin(2t − u) du  = [cos(2t − u)]₀^t = cos t − cos 2t
∫₀^t sin(2t − 3u) du = [cos(2t − 3u)/3]₀^t = (cos t − cos 2t)/3

Result = ½ · ½ · [(cos t − cos 2t) + (cos t − cos 2t)/3]
       = ¼ · (4/3)(cos t − cos 2t)
       = (1/3)(cos t − cos 2t)

Answer: L⁻¹{s/((s² + 1)(s² + 4))} = (cos t − cos 2t)/3. (Check by partial fractions: s/((s² + 1)(s² + 4)) = (1/3)[s/(s² + 1) − s/(s² + 4)].)