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Inverse Laplace transforms log((s+1)/(s−1))

Short answer

(i) 2 sinh t / t = (e^t - e^(-t))/t; (ii) (1/13)(3 e^(3t) - 3 cos 2t + 2 sin 2t).

Find the inverse Laplace transform of (i) log((s+1)/(s−1)) (ii) s/((s − 3)(s² + 4)).

(i) L⁻¹{log((s + 1)/(s − 1))}

Let F(s) = ln(s + 1) − ln(s − 1) and f = L⁻¹{F}. Use L{t f(t)} = −F'(s):

F'(s) = 1/(s + 1) − 1/(s − 1)
L{t f(t)} = 1/(s − 1) − 1/(s + 1)
t f(t) = eᵗ − e^(−t) = 2 sinh t
f(t) = 2 sinh t / t

(ii) L⁻¹{s/((s − 3)(s² + 4))}

Write s/((s − 3)(s² + 4)) = A/(s − 3) + (Bs + C)/(s² + 4). Then s = A(s² + 4) + (Bs + C)(s − 3).

s = 3:  3 = 13A → A = 3/13
s²:     A + B = 0 → B = −3/13
const:  4A − 3C = 0 → C = 4/13

F(s) = (3/13)/(s − 3) + (1/13)(−3s + 4)/(s² + 4)
f(t) = (3/13) e^(3t) − (3/13) cos 2t + (2/13) sin 2t

Answers: (i) 2 sinh t / t = (eᵗ − e^(−t))/t; (ii) (1/13)(3e^(3t) − 3cos 2t + 2 sin 2t).

✓ Answer checked by Claude Opus 5.5 (high effort)

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