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Laplace transforms of functions t²sin2t

Short answer

a) Using L[t^2 f]=F''(s): 4(3s^2-4)/(s^2+4)^3. b) Integration of transform: L[(1-e^t)/t] = ln((s-1)/s), s>1.

Find the Laplace transform of (a) t² sin 2t and (b) (1 − eᵗ)/t.

(a) L[t² sin 2t]

L[sin 2t] = 2/(s² + 4). Use L[t² f] = F″(s).

F′(s) = −4s/(s² + 4)²
F″(s) = [−4(s²+4)² + 4s·2(s²+4)·2s]/(s²+4)⁴
      = [−4(s²+4) + 16s²]/(s²+4)³
      = (12s² − 16)/(s²+4)³

Answer: 4(3s² − 4)/(s² + 4)³

(b) L[(1 − eᵗ)/t]

F(u) = 1/u − 1/(u − 1) for u > 1.

∫ₛ^∞ [1/u − 1/(u−1)] du = [ln(u/(u−1))]ₛ^∞ = 0 − ln(s/(s−1))

Answer: ln((s − 1)/s), valid for s > 1.

✓ Answer checked by Claude Opus 5.5 (high effort)

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