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I am physicist so sorry for being not very formal.

One normally finds the formulas for following Laplace Transforms:

$\mathcal{L}(1)=\frac{1}{u}$

$\mathcal{L}(\delta(t))=1$,

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

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?

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