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Laplace transforms of functions f(t)=(1−cost)/t

Short answer

For (1-cos t)/t, L = ∫_s^∞ (1/u - u/(u^2+1)) du = ½ ln((s^2+1)/s^2). For t e^(-t) sin t, L[t sin t]=2s/(s^2+1)^2, then replace s by s+1 giving 2(s+1)/(( s+1)^2+1)^2.

Find the Laplace transform of (i) f(t) = (1 − cos t)/t (ii) f(t) = t e^(−t) sin t.

(i) L[(1 − cos t)/t]

∫ₛ^∞ [1/u − u/(u² + 1)] du = [ ln( u/√(u² + 1) ) ]ₛ^∞ = −ln( s/√(s² + 1) )

Answer: ½ ln((s² + 1)/s²)

(ii) L[t e^(−t) sin t]

L[sin t] = 1/(s² + 1); L[t sin t] = −d/ds [1/(s²+1)] = 2s/(s²+1)². Shift s → s + 1:

Answer: 2(s + 1)/((s + 1)² + 1)²

✓ Answer checked by Claude Opus 5.5 (high effort)

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