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Laplace transforms of functions 1/√t

Short answer

A sufficient condition: f piecewise continuous and of exponential order |f(t)| ≤ M e^(kt). L{1/√t} = √(π/s); L{(1-cos2t)/t} = (1/2) ln((s^2+4)/s^2), using integration of the transform.

State the existence condition for the Laplace transform. Obtain the Laplace transform of a) 1/√t b) (1 − cos 2t)/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 is of exponential order, that is |f(t)| ≤ M e^(kt) for all t ≥ T. (These conditions are sufficient, not necessary.)

a) L{1/√t}

Use L{tⁿ} = Γ(n + 1)/s^(n+1) with n = −1/2:

L{t^(−1/2)} = Γ(1/2)/s^(1/2) = √π / √s = √(π/s)

b) L{(1 − cos 2t)/t}

Use L{f(t)/t} = ∫ₛ^∞ F(u) du, with F(s) = L{1 − cos 2t} = 1/s − s/(s² + 4). (The limit of f(t)/t as t → 0 exists and equals 0.)

∫ₛ^∞ [1/u − u/(u² + 4)] du = [ln u − ½ ln(u² + 4)]ₛ^∞
 = [ln (u/√(u² + 4))]ₛ^∞ = 0 − ln (s/√(s² + 4))
 = ½ ln((s² + 4)/s²)

Answers: a) √(π/s); b) ½ ln((s² + 4)/s²) = ln(√(s² + 4)/s).

✓ Answer checked by Claude Opus 5.5 (high effort)

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