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Laplace transforms of functions (1/t)(e^(at)−cos6t)

Short answer

Existence: f piecewise continuous and of exponential order. (a) ln( sqrt(s^2+36) / (s-a) ); (b) (1+e^(-π s))/(s^2+1) + e^(-π s)(π/s + 1/s^2).

State the condition for a function to exist its Laplace transform. Find the Laplace transform of: a) (1/t)(e^(at) − cos 6t) b) f(t) = sin t for 0 < t < π; t for t > π.

Existence condition. The Laplace transform L{f(t)} = ∫₀^∞ e^(−st) f(t) dt exists for s > k if f(t) is piecewise continuous on every finite interval [0, T] and of exponential order, that is |f(t)| ≤ M e^(kt) for all t ≥ T. (These conditions are sufficient, not necessary.)

a) L{(1/t)(e^(at) − cos 6t)}

Use L{f(t)/t} = ∫ₛ^∞ F(u) du with F(s) = 1/(s − a) − s/(s² + 36).

∫ₛ^∞ [1/(u − a) − u/(u² + 36)] du = [ln((u − a)/√(u² + 36))]ₛ^∞
 = 0 − ln((s − a)/√(s² + 36))
 = ln(√(s² + 36)/(s − a))        (s > a)

b) f(t) = sin t (0 < t < π); t (t > π)

Split the integral:

L{f} = ∫₀^π e^(−st) sin t dt + ∫_π^∞ t e^(−st) dt

First:  [e^(−st)(−s sin t − cos t)/(s² + 1)]₀^π = (1 + e^(−πs))/(s² + 1)
Second: [−t e^(−st)/s − e^(−st)/s²]_π^∞ = e^(−πs) (π/s + 1/s²)

Answers: a) ln(√(s² + 36)/(s − a)); b) (1 + e^(−πs))/(s² + 1) + e^(−πs)(π/s + 1/s²).

✓ Answer checked by Claude Opus 5.5 (high effort)

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