Laplace Transformation of sin(t+1)*DiracDelta(t)

Question

why is Laplace transformation of sin(t+1)*DiracDelta(t) = sin(1)? I thought:

L[f(t)-u(t-a)] = e^(-as)*L[f(t+a)] and according to this formula it gave me e^(2s) and why wolfram answers back sin(1)?

Answer 1

The Dirac delta satisfies $$\int_0^\infty g(t)\,\delta(t)\,dt=g(0).$$ Then $$ \mathcal L[\sin(t+1)\,\delta(t)](s)=\int_0^\infty \sin(t+1)\,e^{-st}\,\delta(t)\,dt=\sin(0+1)\,e^{-0}=\sin1 $$ (which is not zero). In your other formula you don't have a Dirac delta, but a unit impulse function.