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 $g(0)=0$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.