Here,

`f_1(t)=sin t*{u(t)-u(t-pi)}`

and the period, `p=pi`.

`Lap{f_1(t)}`

`= Lap{sin\ t*(u(t)-u(t-pi))}`

`= Lap{sin t*u(t)}+` ` Lap{sin(t-pi)*u(t-pi)}`

`=1/(s^2+1)+(e^(-pis))/(s^2+1)`

`=(1+e^(-pis))/(s^2+1)`

So the Laplace Transform of the periodic function is given by:

`Lap{f(t)}=(1+e^(-pis))/((s^2+1)(1-e^(-pis))`

Please support IntMath!