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Laplace transforms of functions t²cosat

Short answer

For t² cos at, use L[t² f(t)] = F''(s) with F(s)=s/(s²+a²); result: 2s(s²-3a²)/(s²+a²)³. For sin t / t, use L[ f(t)/t] = ∫_s^∞ F(u) du; L[sin t]=1/(s²+1), integral gives π/2 - tan⁻¹ s = cot⁻¹ s.

Find the Laplace transform of: i) t² cos at ii) sin t / t.

i) L[t² cos at]

Use L[t² f(t)] = d²F/ds² with F(s) = L[cos at] = s/(s² + a²).

F′(s)  = (a² − s²)/(s² + a²)²
F″(s)  = [−2s(s² + a²)² − (a² − s²)·2(s² + a²)·2s] / (s² + a²)⁴
       = [−2s(s² + a²) − 4s(a² − s²)] / (s² + a²)³
       = (2s³ − 6a²s) / (s² + a²)³

Answer: 2s(s² − 3a²)/(s² + a²)³

ii) L[sin t / t]

L[sin t] = 1/(s² + 1). By L[f/t] = ∫ₛ^∞ F(u) du:

∫ₛ^∞ du/(u² + 1) = [tan⁻¹ u]ₛ^∞ = π/2 − tan⁻¹ s

Answer: cot⁻¹ s = tan⁻¹(1/s)

✓ Answer checked by Claude Opus 5.5 (high effort)

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