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Laplace transforms of functions te^(−t)cosht

Short answer

L{f(t)} = ∫0∞ e^{-st}f(t)dt. a) t e^{-t} cosh t = (t + t e^{-2t})/2, transform is (1/2)[1/s^2 + 1/(s+2)^2]. b) L{sin t sin 5t / t} = (1/4) ln((s^2+36)/(s^2+16)).

Define the Laplace transform of f(t). Find the Laplace transform of: a) t e^(−t) cosh t b) sin t sin 5t / t.

Definition

L{f(t)} = F(s) = ∫₀^∞ e^(−st) f(t) dt, for those s for which the integral converges.

a) L{t e^(−t) cosh t}

Since e^(−t) cosh t = (1 + e^(−2t))/2, we have t e^(−t) cosh t = (t + t e^(−2t))/2.

L{t} = 1/s²,   L{t e^(−2t)} = 1/(s + 2)²
L{t e^(−t) cosh t} = ½ [1/s² + 1/(s + 2)²]        (s > 0)

b) L{sin t sin 5t / t}

Use sin t sin 5t = ½(cos 4t − cos 6t). Then L{f(t)/t} = ∫ₛ^∞ F(u) du with F(s) = ½[s/(s² + 16) − s/(s² + 36)].

∫ₛ^∞ ½ [u/(u² + 16) − u/(u² + 36)] du = ¼ [ln((u² + 16)/(u² + 36))]ₛ^∞
 = 0 − ¼ ln((s² + 16)/(s² + 36))
 = ¼ ln((s² + 36)/(s² + 16))

Answers: a) ½[1/s² + 1/(s + 2)²]; b) ¼ ln((s² + 36)/(s² + 16)).

✓ Answer checked by Claude Opus 5.5 (high effort)

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