The discontinuous function $g(t)$ is defined by $$g(t)=\begin{cases} 0, & \text{for } t<0\\1, & \text{for } t=0 \\ -1, & \text{for }t=\pi \\ 0, & \text{for } t>\pi \end{cases}$$
and satisfies the ODE $$g'(t)=g(\pi-t),$$ where $g(0)=1$ and $g(\pi)=-1.$
I want to consider this ODE using Laplace transforms. Here the Laplace transform is given by $$\bar{g}(s)=\int_0^\infty e^{-st} g(t)\, dt.$$
The Laplace transform of the LHS is given by $$s\bar{g}(s)-1.$$
For the RHS we have \begin{align*} \bar{g}(\pi-t)=\int_0^\infty e^{-st}g(\pi-t) \, dt.\end{align*}
Setting $y=\pi-t$ it follows that \begin{align*} \int_0^\infty e^{-st}g(\pi-t) \, dt & = -\int_\pi^{-\infty}e^{-s(\pi-y)}g(y) \, dy \\ & = -e^{-\pi s}\int_{\pi}^{-\infty} e^{sy}g(y) \, dy \end{align*}
I'm told that the Laplace transform of the ODE is given by $$s\bar{g}(s)-1=e^{-\pi s}+e^{-\pi s}\bar{g}(-s).$$
How do I get to this answer from where I am?