How can I evaluate the following integral:
$$\int_0^\infty e^{-\text{s}t}\left|\hat{\text{u}}\sin\left(\omega t+\varphi\right)\right|\space\text{d}t=\left|\hat{\text{u}}\right|\int_0^\infty e^{-\text{s}t}\left|\sin\left(\omega t+\varphi\right)\right|\space\text{d}t$$
Where all variables are real except $\text{s}$ because $\text{s}\in\mathbb{C}$.
Thanks in advance.
Assume that $\omega=1$, $\varphi=0$. Then we write $$\int_0^{\infty} e^{-st}|\sin(t)|dt=\sum_{n\geq 0}\int_{n\pi}^{n\pi+\pi}e^{-st}|\sin(t)|dt$$ Now if $n\geq 0$ we have $$\int_{n\pi}^{n\pi+\pi}e^{-st}|\sin(t)|dt=\int_0^{\pi}e^{-s(u+n\pi)}|\sin(u)|du=e^{-sn\pi}\int_0^{\pi}e^{-su}\sin(u)du$$ The integral is standard and can be computed using integration by parts or complex numbers and gives $$\int_0^{\pi}e^{-su}\sin(u)du=\dfrac{1+e^{-\pi s}}{1+s^2}$$ hence $$\int_0^{\infty} e^{-st}|\sin(t)|dt=\dfrac{1+e^{-\pi s}}{1+s^2}\sum_{n\geq 0}e^{-sn\pi}=\dfrac{1+e^{-\pi s}}{1+s^2}\dfrac{1}{1-e^{-\pi s}}$$ according to the formula for a geometric series.