Laplace transforms of functions te^(−t)sint

Short answer

a) t e^{-t} sin t: 2(s+1)/(( (s+1)^2+1)^2. b) Integral from s: ½ ln(s^2+36) - ln(s-a), s>a.

Find the Laplace transforms of: (a) t e^(−t) sin t (b) (e^(at) − cos 6t)/t.

L[sin t] = 1/(s²+1); L[t sin t] = −d/ds[1/(s²+1)] = 2s/(s²+1)². Shift s → s + 1:

Answer: 2(s + 1)/((s + 1)² + 1)²

F(u) = 1/(u − a) − u/(u² + 36).

∫ₛ^∞ F(u) du = [ ln(u − a) − ½ ln(u² + 36) ]ₛ^∞
             = [ ln( (u − a)/√(u² + 36) ) ]ₛ^∞
             = 0 − ln( (s − a)/√(s² + 36) )

Answer: ln( √(s² + 36) / (s − a) ) = ½ ln(s² + 36) − ln(s − a), for s > a.

✓ Answer checked by Claude Opus 5.5 (high effort)

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