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Inverse Laplace transforms tan⁻¹(1/s)

Short answer

a) sin t / t; b) 1/16 (1 - e^(4t)) + (t/4) e^(4t).

Find the inverse Laplace transform of a) tan⁻¹(1/s) b) 1/(s(s − 4)²).

a) L⁻¹{tan⁻¹(1/s)}

Let F(s) = tan⁻¹(1/s) and f = L⁻¹{F}. Using L{t f(t)} = −F'(s):

F'(s) = −1/(1 + s²)
L{t f(t)} = 1/(s² + 1)  → t f(t) = sin t
f(t) = sin t / t

b) L⁻¹{1/(s(s − 4)²)}

Partial fractions: 1/(s(s − 4)²) = A/s + B/(s − 4) + C/(s − 4)².

A = 1/16        (s = 0)
C = 1/4         (s = 4)
A + B = 0 (large s behaviour) → B = −1/16
L⁻¹ = 1/16 − (1/16) e^(4t) + (1/4) t e^(4t)

Answers: a) sin t / t; b) (1/16)(1 − e^(4t)) + (t/4) e^(4t).

✓ Answer checked by Claude Opus 5.5 (high effort)

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