TITLE: Why the Laplace transform of the integral is 1/s?
QUESTION [5 upvotes]: I got this interpretation:
If we have $y'(t)=u(t)$ , it's like have $y(t)=\int{u(t)dt} $
If we solve this simple equation, we obtain:
$$sY(s)=U(s)$$
$$Y(s)=\frac{U(s)}{s}$$
So, we have the $U(s)$ multiplied for $\frac{1}{s}$
Is this conjunction correct? It doesn't seem quite complete. What do you think?

REPLY [2 votes]: Laplace transform of f(t) is defined as $\text{F(s)=}\int _0^{+\infty }f(t)e^{-\text{st}}dt$
Thus, the Laplace transform of $\frac{df}{dt}$ can be derived as below.
$$
\int _0^{+\infty }\left(\frac{d}{dt}f(t)\right)e^{-\text{st}}dt=\int_0^{+\infty } e^{-\text{st}} \, df=f(t)e^{-\text{st}}|_0^{+\infty }+\int _0^{+\infty }f(t)\text{se}^{-\text{st}}dt
$$
Note that $\text{$\exists $ F(s)=}
\int _0^{+\infty }f(t)e^{-\text{st}}dt\Rightarrow f(t)e^{-\text{st}}|_{t=+\infty }=0$,  we have
$$
\int _0^{+\infty }\left(\frac{d}{dt}f(t)\right)e^{-\text{st}}dt=\text{sF}(s)-f(0)
$$