We know that $L\{\int_0^tf(t)dt\}=\frac{F(s)}{s}$. There's a proof based on it's definition, however, I was wondering if proving it with convolution theorem is possible. I've got a proof, but it's too easy, and I'm not sure abot it.
Let $g(t)=1$ and $G(s)=L\{g(t)\}=\frac{1}{s}$.
So, $L\{f(t)*g(t)\}=\int_0^t f(t-\tau)g(\tau)d\tau=\int_0^t f(\tau)g(t-\tau)dt=\int_0^tf(t)dt$ and, since $L\{f*g\}=F(s)G(s)$ we have $\int_0^tf(t)dt=\frac{F(s)}{s}$
The proof based on definition is $L\{\int_0^tf(t)dt\}=\int_0^{\infty}e^{-st}\bigg(\int_0^{\tau}f(\tau)d\tau\bigg)dt=\frac{e^{-st}}{s}\int_0^{\tau}f(\tau)d\tau\bigg|_0^{\infty}+\frac{1}{s}\int_0^{\infty}f(t)e^{-st}dt=\frac{1}{s}\int_0^{\infty}f(t)e^{-st}dt=\frac{F(s)}{s}$