If $y(t)=\int_{0}^t f(t)dt$ & the Laplace transform of $f(t)$ is $\mathcal{L}\{f(t)\}=\int_{0}^\infty e^{-st}f(t)dt$,then prove that $\mathcal{L}\{y(t)\}=(1/s)\mathcal{L}\{f(t)\}$.
My attempt:Since, $\mathcal{L}\{f(t)\}=\int_{0}^\infty e^{-st}f(t)dt$,then $\mathcal{L}\{y(t)\}=\int_{0}^\infty e^{-st}y(t)dt=\int_{0}^\infty e^{-st}(\int_{0}^t f(t)dt)dt$.
From,here i got confused in dealing with variables.Please provide some suggestion for completing this problem.
Let $y(t)=\int_0^t f(x)\,dx$. Then, we have
$$\begin{align} \mathscr{L}\left(y \right)(s)&=\int_0^\infty \int_0^t f(x)\,dx\,e^{-st}\,dt\\\\ &=\int_0^\infty f(x) \int_x^\infty e^{-st}\,dt\,dx\\\\ &=\frac1s \int_0^\infty f(x)e^{-sx}\,dx\\\\ &=\frac1s\mathscr{L}\left(f\right)(s) \end{align}$$
as was to be shown!