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Inverse Laplace transforms e^(−πs)(s+1)/(s²+2s+2)

Short answer

i) Complete square to (s+1)²+1, shift, then apply second shifting theorem: e^(−(t−π)) cos(t−π) u(t−π) = −e^(π−t) cos t u(t−π). ii) Differentiate tan⁻¹(2/s) to −2/(s²+4); using L{t f(t)} = −F'(s) gives f(t) = sin 2t / t.

Find the inverse Laplace transform of: i) e^(−πs)(s + 1)/(s² + 2s + 2) ii) tan⁻¹(2/s).

i) L⁻¹{e^(−πs)(s + 1)/(s² + 2s + 2)}

Complete the square: s² + 2s + 2 = (s + 1)² + 1. So (s + 1)/((s + 1)² + 1) is the shift of s/(s² + 1):

L⁻¹{(s + 1)/((s + 1)² + 1)} = e^(−t) cos t

By the second shifting theorem with a = π:

L⁻¹ = e^(−(t−π)) cos(t − π) u(t − π) = −e^(π−t) cos t · u(t − π)

where u is the unit step function (cos(t − π) = −cos t).

ii) L⁻¹{tan⁻¹(2/s)}

Let F(s) = tan⁻¹(2/s). Then F'(s) = −2/(s² + 4), and using L{t f(t)} = −F'(s):

L{t f(t)} = 2/(s² + 4)  → t f(t) = sin 2t
f(t) = sin 2t / t

Answers: i) e^(−(t−π)) cos(t − π) u(t − π) = −e^(π−t) cos t u(t − π); ii) sin 2t / t.

✓ Answer checked by Claude Opus 5.5 (high effort)

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