I am physicist so sorry for being not very formal.

One normally finds the formulas for following Laplace Transforms:



$\mathcal{L}(\int\limits_0^t f(s)ds)=\frac{1}{u}F(u)$,

with $\mathcal{L}( f(t))=F(u)$ and $\delta(t)$ Dirac Delta function (distribution). From this I would conclude $\mathcal{L}(\int\limits_0^t \delta(s)ds)=\frac{1}{u}$.

However, I always use a convention, that $\int\limits_0^t f(s)\delta(s)ds=\frac{1}{2}f(0)$ and consequently

$\mathcal{L}(\int\limits_0^t \delta(s)ds)=\mathcal{L}(\frac{1}{2})=\frac{1}{2u}$.

Does $\mathcal{L}(\delta(t))=1$ assume other convention and it is "fine" always to set $\mathcal{L}(\delta(t))=\frac{1}{2}$ in my calculations or could it "spoil" something?


Your Answer

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

Browse other questions tagged or ask your own question.