I try to understand why according to my notes from some lecture the following equality is satisfied

$$ \mathcal{L}\left[ {\exp\left(\int_0^t ds \frac{\dot{x}(s)}{x(s)} \right)} \right] =\frac{1}{u-\frac{\mathcal{L}\left[\dot{x}\right](u)}{\mathcal{L}[x](u)}}, $$

where $\mathcal{L}[\cdot]$ denotes Laplace transform. Is this equation fulfilled for all $x(t)$, and how one can derive it?


1 Answer 1


The key point is to note, that

$\frac{d}{dt} \ln x(t)= \frac{\dot{x}(t)}{x(t)}$,


$ \exp(\int\limits_0^t ds \frac{\dot{x}(s)}{x(s)})=\exp(\int\limits_0^t ds \frac{d}{ds} \ln x(s))=\frac{x(t)}{x(0)}$.

The rest can be obtained by using the formula for the Laplace Transform of the derivative.


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.