# Prove that $\mathcal{L}\left( \int_{0}^t f(u)du \right)=\frac{1}{s}\mathcal{L}(f)$

Prove that $$\mathcal{L}\left( \int_{0}^t f(u)du \right)=\frac{1}{s}\mathcal{L}(f)$$

I started out with the following identity:

$$\frac{1}{s}\mathcal{L}(f)=\frac{1}{s}\int_{0}^\infty e^{-st}f(t)dt$$

then tried to integrate the RHS by parts, giving

$$\frac{1}{s}\int_{0}^\infty e^{-st}f(t)dt= \frac{1}{s}\left[\frac{-f(t)e^{-st}}{s}\bigg|_{0}^\infty+\frac{1}{s}\int_{0}^\infty e^{-st}f'(t)dt\right]$$

But I'm not sure if that means anything. I'm not really sure where the $f(u)$ comes from in the statement of the problem, so that's an issue. I feel like this might be a Fundamental Theorem of Calculus problem, or something, but I don't see it. If anyone has any idea, any help here would be appreciated. Thanks!

Given $g(t)$, we know that $\mathcal{L}(g'(t))=sG(s)-g(0)$, where $G$ is the laplace transform of $g$. Now let $g(t)=\int\limits_{0}^{t} f(\tau)d\tau$. The fundamental theorem of calculus tells us that $g'(t)=f(t)$, and $g(0)=0$.