Integral theorem of Laplace transform

Statement

If L[f(t)] = F(s), then the Laplace transform of the integral of f is

L[ ∫₀ᵗ f(u) du ] = F(s)/s

(Integration in the t-domain corresponds to division by s.)

Proof

Let g(t) = ∫₀ᵗ f(u) du. Then g(0) = 0 and g′(t) = f(t). Take the Laplace transform of g′(t) = f(t) using L[g′] = s G(s) − g(0):

s G(s) − g(0) = F(s)
s G(s) = F(s)          (since g(0) = 0)
G(s) = F(s)/s

Hence L[∫₀ᵗ f(u) du] = F(s)/s. ∎

(Direct proof: G(s) = ∫₀^∞ e^(−st) [∫₀ᵗ f(u) du] dt. Integrate by parts with dv = e^(−st) dt: the boundary term vanishes, leaving (1/s) ∫₀^∞ e^(−st) f(t) dt = F(s)/s.)

Related form (division by t)

If L[f(t)] = F(s) and lim f(t)/t exists as t → 0, then L[f(t)/t] = ∫ₛ^∞ F(u) du.

Example

L[∫₀ᵗ sin u du] = L[1 − cos t] = 1/s − s/(s²+1) = 1/(s(s²+1)) = (1/s)·1/(s²+1) ✓.