Transform of a piecewise continuous function

[Hiche]

We know that the \mathcal L\{f(t)\} = \int^{\infty}_0 e^{-st}f(t) dt.

Say we want to, for example, solve the following IVP: y&#039;&#039; + y = f(t) where f(t) = \begin{cases}<br /> 0 &amp; 0 \leq t &lt; \pi \\<br /> 1 &amp; \pi \leq t &lt; 2\pi\\<br /> 0 &amp; 2\pi \leq t<br /> \end{cases}

and y(0) = 0 , y&#039;(0) = 0

We apply Laplace on both side of the DE, and we get (s^2 + 1)Y(s) = \mathcal L\{f(t)\}. Using the cases above, do we divide the integral from 0 to \infty into three integrals?

I did that and \mathcal L\{f(t)\} = \int^{\pi}_0 e^{-st}(0) dt + \int^{2\pi}_{\pi} e^{-st}(1) dt + \int^{\infty}_{2\pi} e^{-st}(0) dt. The first and third integrals are zeros so we need to integrate the second one. We get -(1/s)[e^{-2\pi s} - e^{-\pi s}]. Right?

Back to the DE, Y(s) = -[e^{-2\pi s} - e^{-\pi s}] / s(s^2 + 1). How exactly do we find \mathcal L^{-1}\{Y(s)\}?

 

[Hiche]

Okay, so I tried solving it. 1/s(s^2 + 1) = 1/s - s/(s^2 + 1), and after a litle work, Y(s) = e^{-\pi s}/s - e^{-\pi s}/(s^2 + 1) - e^{-2\pi s}/s + e^{-2\pi s}/(s^2 + 1). Is this correct? Now, the Laplace inverse of Y(s) is \delta(t - \pi) - \delta(t - \pi)sin(t - \pi) - \delta(t - 2\pi) + \delta(t - 2\pi)cos(t - 2\pi). I'm not sure about this, but I think it's fine?