Laplace transforms of functions t³e^(−3t)and

(a) L{t³ e^(−3t)}

By the first shifting theorem, L{e^(−3t) f(t)} = F(s + 3), with L{t³} = 3!/s⁴ = 6/s⁴:

L{t³ e^(−3t)} = 6/(s + 3)⁴

(b) L{(e^(at) − cos 6t)/t}

Use L{f(t)/t} = ∫ₛ^∞ F(u) du, with F(s) = 1/(s − a) − s/(s² + 36). (As t → 0, f(t)/t → a, finite.)

∫ₛ^∞ [1/(u − a) − u/(u² + 36)] du = [ln (u − a) − ½ ln(u² + 36)]ₛ^∞
 = [ln ((u − a)/√(u² + 36))]ₛ^∞
 = 0 − ln ((s − a)/√(s² + 36))
 = ln (√(s² + 36)/(s − a))        (s > a)

Answers: (a) 6/(s + 3)⁴; (b) ln(√(s² + 36)/(s − a)) = ½ ln(s² + 36) − ln(s − a).