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Laplace transforms of functions te^(−t)sint

Short answer

i) L{e^(−t) sin t} = 1/(( s+1)²+1); multiply by t: −d/ds yields 2(s+1)/(s²+2s+2)². ii) Use L{f/t} = ∫ₛ^∞ F(u) du with F(s)=s/(s²+4)−s/(s²+9) to get ½ ln((s²+9)/(s²+4)).

Find the Laplace transform of: i) t e^(−t) sin t ii) (cos 2t − cos 3t)/t.

i) L{t e^(−t) sin t}

L{sin t} = 1/(s² + 1)
Shift s → s + 1:  L{e^(−t) sin t} = 1/((s + 1)² + 1)
Multiply by t:  −d/ds [((s + 1)² + 1)⁻¹] = 2(s + 1)/((s + 1)² + 1)²

So L{t e^(−t) sin t} = 2(s + 1)/(s² + 2s + 2)².

ii) L{(cos 2t − cos 3t)/t}

Use L{f(t)/t} = ∫ₛ^∞ F(u) du with F(s) = s/(s² + 4) − s/(s² + 9).

∫ₛ^∞ [u/(u² + 4) − u/(u² + 9)] du = ½ [ln((u² + 4)/(u² + 9))]ₛ^∞
 = −½ ln((s² + 4)/(s² + 9))
 = ½ ln((s² + 9)/(s² + 4))

Answers: i) 2(s + 1)/(s² + 2s + 2)²; ii) ½ ln((s² + 9)/(s² + 4)).

✓ Answer checked by Claude Opus 5.5 (high effort)

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