Second shifting theorem

Statement

If L[f(t)] = F(s) and

g(t) = f(t − a)   for t > a
g(t) = 0          for t < a        (a > 0)

i.e. g(t) = f(t − a) u(t − a) with u the unit step function, then

L[g(t)] = e^(−as) F(s)

Equivalently, L⁻¹[e^(−as) F(s)] = f(t − a) u(t − a).

Proof

L[g(t)] = ∫₀^∞ e^(−st) g(t) dt
        = ∫₀^a e^(−st)·0 dt + ∫ₐ^∞ e^(−st) f(t − a) dt

In the second integral put τ = t − a. Then t = τ + a, dt = dτ, and the limits become τ = 0 to ∞:

L[g(t)] = ∫₀^∞ e^(−s(τ + a)) f(τ) dτ
        = e^(−as) ∫₀^∞ e^(−sτ) f(τ) dτ
        = e^(−as) F(s)

∎

Example

L[u(t − 2)(t − 2)] = e^(−2s)/s². Also, L[u(t − a)] = e^(−as)/s (take f(t) = 1).