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Laplace transforms of functions te^(−t)sint

Short answer

Existence: f is piecewise continuous and of exponential order. a) L{e^(−t) sin t}=1/(( s+1)²+1), multiply by t: −d/ds gives 2(s+1)/(s²+2s+2)². b) L{(e^(−at)−e^(−bt))/t} = ∫ₛ^∞ (1/(u+a)−1/(u+b)) du = ln((s+b)/(s+a)).

State the existence theorem for Laplace transform. Obtain the Laplace transform of a) t e^(−t) sin t b) (e^(−at) − e^(−bt))/t.

Existence theorem

If f(t) is piecewise continuous on every finite interval 0 ≤ t ≤ A and of exponential order (|f(t)| ≤ M e^(kt) for t > T), then L[f(t)] = ∫₀^∞ e^(−st) f(t) dt exists for all s > k.

(a) L[t e^(−t) sin t]

Step 1: L[sin t] = 1/(s² + 1). Step 2 (multiply by t): L[t sin t] = −d/ds [1/(s²+1)] = 2s/(s² + 1)². Step 3 (first shift, s → s + 1):

L[t e^(−t) sin t] = 2(s + 1) / ((s + 1)² + 1)²

Answer: 2(s + 1)/(s² + 2s + 2)²

(b) L[(e^(−at) − e^(−bt))/t]

Here F(u) = 1/(u + a) − 1/(u + b). Using L[f/t] = ∫ₛ^∞ F(u) du:

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

Answer: ln((s + b)/(s + a))

✓ Answer checked by Claude Opus 5.5 (high effort)

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