prove laplace trnansform f(t) cosbt= s/s^2+b^2

Question

how to prove if the Laplace transform f(t) cosbt = s/s^2+b^2

Answer 1

We start with the definition of the Laplace Transform L{f(t)} which is represented by the function F(s).

F(s)=∫e⁻ˢᵗf(t)dt where the integral is to be evaluated for t between 0 and ∞. Let f(t)=cos(bt).

Let u=e⁻ˢᵗ so du=-se⁻ˢᵗ and let dv=cos(bt)dt so v=sin(bt)/b. Integrating by parts:

F(s)=e⁻ˢᵗsin(bt)/b+(s/b)∫e⁻ˢᵗsin(bt)dt.

If we let dv=sin(bt)dt, then v=-cos(bt)/b, and we can integrate the above integral by parts:

∫e⁻ˢᵗsin(bt)dt=-e⁻ˢᵗcos(bt)/b-(s/b)∫e⁻ˢᵗcos(bt)dt, but the latter integral is identical to F(s) so:

F(s)=e⁻ˢᵗsin(bt)/b-(s/b²)e⁻ˢᵗcos(bt)-(s/b)²F(s).

Therefore, F(s)(1+(s/b)²)=e⁻ˢᵗsin(bt)/b-(s/b²)e⁻ˢᵗcos(bt).

We now apply the limits 0 and ∞: F(s)(1+(s/b)²)=0-(-s/b²)=s/b².

Multiply through by b²: F(s)(s²+b²)=s, so F(s)=L{cos(bt)}=s/(s²+b²) QED.