EXAMPLE OF LAPLACE TRANSFORM
Let f(t)=sin(at), then ℒ{f(t)}=∫[0-,∞]e-stsin(at)dt, by definition of Laplace function.
Let u=f(t)=sin(at), du=f'(t)dt=acos(at)dt,
dv=e-stdt, v=-(1/s)e-st.
Integrating by parts:
ℒ{f(t)}=-e-stf(t)/s+(1/s)∫e-stf'(t)dt=ℒ{sin(at)}=-e-stsin(at)/s+(a/s)∫e-stcos(at)dt for t∈[0-,∞],
Therefore, ℒ{sin(at)}=-e-stsin(at)/s+(a/s)ℒ{cos(at)}=(a/s)ℒ{cos(at)}, because the first term is zero when the limits are applied.
Now we have to evaluate ℒ{cos(at)}, so let u=cos(at), du=-asin(at).
ℒ{cos(at)}=-e-stcos(at)/s-(a/s)∫e-stsin(at)dt for t∈[0-,∞]=(1/s)-(a/s)ℒ{sin(at)}, when limits are applied.
ℒ{sin(at)}=(a/s)ℒ{cos(at)}=(a/s)[(1/s)-(a/s)ℒ{sin(at)}].
ℒ{sin(at)}+(a2/s2)ℒ{sin(at)}=a/s2,
ℒ{sin(at)}(1+(a2/s2))=a/s2,
ℒ{sin(at)}=a/(s2+a2), which is the Laplace Transform of sin(at).