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Laplace transforms of functions cos³2t

Short answer

Existence condition: f(t) piecewise continuous and of exponential order, |f(t)| ≤ M e^{kt}. a) L{cos^3 2t} = (1/4)[3s/(s^2+4) + s/(s^2+36)]. b) L{(cos at - cos bt)/t} = (1/2) ln((s^2+b^2)/(s^2+a^2).

State the condition for existence of the Laplace transform. Obtain the Laplace transform of: a) cos³ 2t b) (cos at − cos bt)/t.

Existence condition. The Laplace transform L{f(t)} = ∫₀^∞ e^(−st) f(t) dt exists for s > k if f(t) is piecewise continuous on every finite interval [0, T] and of exponential order, that is |f(t)| ≤ M e^(kt) for all t ≥ T. (These conditions are sufficient, not necessary.)

a) L{cos³ 2t}

Use cos³x = (3 cos x + cos 3x)/4 with x = 2t: cos³ 2t = (3 cos 2t + cos 6t)/4.

L{cos³ 2t} = ¼ [3s/(s² + 4) + s/(s² + 36)]

b) L{(cos at − cos bt)/t}

Use L{f(t)/t} = ∫ₛ^∞ F(u) du with F(s) = s/(s² + a²) − s/(s² + b²).

∫ₛ^∞ [u/(u² + a²) − u/(u² + b²)] du = ½ [ln((u² + a²)/(u² + b²))]ₛ^∞
 = −½ ln((s² + a²)/(s² + b²))
 = ½ ln((s² + b²)/(s² + a²))

Answers: a) ¼[3s/(s² + 4) + s/(s² + 36)]; b) ½ ln((s² + b²)/(s² + a²)).

✓ Answer checked by Claude Opus 5.5 (high effort)

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