What is the Laplace Transform of, with $t\in\mathbb{R}^+$:
$$\int_{0}^{t}\text{U}_{\text{in}}(t)\space\text{d}t$$
I know that the Laplace Transform of a indefinite integral is:
$$\mathcal{L}_{t}\left[\int\text{U}_{\text{in}}(t)\space\text{d}t\right]_{(s)}=\frac{\text{U}_{\text{in}}(s)-\text{U}_{\text{in}}(0)}{s}$$
Best Answer
Integrate by parts with $u=\int_0^t U_{in}(t')\,dt'$ and $v=-\frac1s e^{-st}$. Then, we have
$$\int_0^\infty \left(\int_0^t U_{in}(t')\,dt'\right)e^{-st}\,dt=\frac1s \int_0^\infty U_{in}(t)e^{-st}\,dt=\frac1s \mathscr{L}\{U_{in}\}(s)$$