Laplace transform of derivative/shift property

Assumptions: f(t) is continuous for t ≥ 0, of exponential order (|f(t)| ≤ M e^(kt)), and f′(t) is piecewise continuous, so that L[f′] exists.

Proof. By definition,

L[f′(t)] = ∫₀^∞ e^(−st) f′(t) dt

Integrate by parts with u = e^(−st), dv = f′(t) dt:

L[f′(t)] = [ e^(−st) f(t) ]₀^∞ + s ∫₀^∞ e^(−st) f(t) dt

Because |f(t)| ≤ M e^(kt), for s > k we have e^(−st) f(t) → 0 as t → ∞. So the upper limit gives 0, and the lower limit gives −f(0):

L[f′(t)] = 0 − f(0) + s F(s)

Hence L[f′(t)] = s F(s) − f(0). ∎

Note. Applying the result again to f″ gives L[f″] = s²F(s) − s f(0) − f′(0). If f has a jump at t = 0 the value f(0) is taken as f(0⁺).