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Laplace transforms of functions

Short answer

The Laplace transform is L[f(t)] = ∫_0^∞ e^(-st) f(t) dt. Existence requires f piecewise continuous on [0, A] and exponential order |f(t)| ≤ M e^(kt). For (1-cos 2t)/t, apply L[f/t] = ∫_s^∞ F(u) du to get ½ ln((s^2+4)/s^2).

Define the Laplace transform of a function and state the criteria of existence of the Laplace transform of a function. Find the Laplace transform of f(t) = (1 − cos 2t)/t.

Definition

Let f(t) be defined for t ≥ 0. The Laplace transform of f(t) is

L[f(t)] = F(s) = ∫₀^∞ e^(−st) f(t) dt

provided the integral converges for some values of s.

Conditions of existence

L[f(t)] exists for s > k if

  1. f(t) is piecewise continuous on every finite interval [0, A], and
  2. f(t) is of exponential order: |f(t)| ≤ M e^(kt) for all t ≥ T, for some constants M, k, T.

(These are sufficient conditions.)

L[(1 − cos 2t)/t]

Use L[f(t)/t] = ∫ₛ^∞ F(u) du. Here F(u) = 1/u − u/(u² + 4).

∫ₛ^∞ [1/u − u/(u² + 4)] du = [ ln u − ½ ln(u² + 4) ]ₛ^∞
                            = [ ln( u/√(u² + 4) ) ]ₛ^∞
                            = 0 − ln( s/√(s² + 4) )

Answer: ½ ln((s² + 4)/s²) = ln(√(s² + 4)/s)

✓ Answer checked by Claude Opus 5.5 (high effort)

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