Answer:
the Laplace transform of f(t) , ℒ{f(t)](s) = 0
Explanation:
the Laplace transform for the function f(t)=cos t, 0 ≤ t < π and 0, t ≥ π is
ℒ{f(t)](s) =∫ f(t) *e^(-s*t) dt from 0 to ∞ = ∫ cos t * e^(-s*t) dt from 0 to π
then
ℒ{f(t)](s) = ∫cos (t) * e^(-s*t) dt = sin(t)*e^(-s*t) /₀π - ∫ cos(t)* (-s)e^(-s*t) dt = 0 +s*∫ cos(t)* e^(-s*t) dt
thus
∫cos (t) * e^(-s*t) dt - s*∫ cos(t)* e^(-s*t) dt = 0
(1-s)* ∫ cos(t)* e^(-s*t) dt = 0
therefore since s can be different from 0
∫ cos(t)* e^(-s*t) dt = 0 and thus ℒ{f(t)](s) = 0