$\displaystyle l^{-1}(1)$ = ??!
The question is a source of some little controversial... according to the definition of $\displaystyle \delta (*)$ given in Wolfram MathWorld , probably the most 'reliable' mathematical site in the Web, is...
http://mathworld.wolfram.com/DeltaFunction.html
$\displaystyle \delta (t)= \frac {d}{dt} \mathcal {H} (t)$ (1)
... where $\displaystyle \mathcal {H} (*)$ is the 'Heaviside Step Function' defined as...
http://mathworld.wolfram.com/HeavisideStepFunction.html
$\displaystyle \mathcal{H} (t)= \left\{\begin{array}{cc} 0,&\mbox { if } t<0\\ \frac{1}{2},&\mbox{ if } t=0\\1,&\mbox{ if } t>0\end{array}\right.$ (2)
Now, according to the basic properties of the Laplace Transform...
http://mathworld.wolfram.com/LaplaceTransform.html
..., we have...
$\displaystyle \mathcal{L} \{\mathcal{H} (t)\}= \frac {1}{s}$ (3)
... and for every $\displaystyle f(*)$ ...
$\displaystyle \mathcal{L} \{\frac{d}{dt} f(t)\}= s\cdot \mathcal{L} \{f(t)\} - f(0)$ (4)
Now if we combine (1), (2), (3) and (4) we obtain...
$\displaystyle \mathcal {L} \{\delta(t)\}= s\cdot \frac{1}{s} -\frac{1}{2} = \frac{1}{2}$ (5)
... so that it would be...
$\displaystyle \mathcal {L}^{-1} \{1\}= 2\cdot \delta(t)$ (6)
... just a little surprising! ...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$
All you're proving here, is that delta can not be considered as a function. The inverse laplace transform of 1 does not exist as a function. However, the laplace inverse of 1 is $\displaystyle \delta$, which is the Dirac distribution, in the sense of distributions.
When you say:
this is not true for distributions, where the formula doesn't have the $\displaystyle -f(0)$ term: $\displaystyle \mathcal{L}(T')(s)=s \mathcal{L}(T)$. Indeed, this border term is included in the differentiation: steps in the function $\displaystyle f$ give Dirac terms in the differential of $\displaystyle f$ in the sense of distributions. If you use distributions in the result, you must use a derivative in the sense of distributions for everything to make sense.... and for every $\displaystyle f(*)$ ...
$\displaystyle \mathcal{L} \{\frac{d}{dt} f(t)\}= s\cdot \mathcal{L} \{f(t)\} - f(0)$ (4)
By the way, as a distribution, the Heaviside function doesn't have a specific value at 0. It is customary to choose it to be 0 for symmetry reasons, but it could be anything when we consider $\displaystyle \mathcal{H}$ as a distribution (for instance, when we differentiate it).
And Pomp's justification was also correct (except for the use of the word "function" and a slight abuse of notation in writing delta in an integral like physicists do...)
Actually I don't think it is, he ought to more careful since $\displaystyle 0$ is one of the limits of integration and what one thinks:
$\displaystyle \int_0^{\infty} e^{-st}\delta(t) \; dt$
is short hand for is now not at all clear. If we take it as the limit of integrals where $\displaystyle \delta$ is replaced by functions whose support becomes more and more concentrated near the origin and with integrals all $\displaystyle 1$ what we get depends on the form of these functions and in this case could give any value between $\displaystyle 0$ and $\displaystyle 1$.
The sort of intuitive use that physicists and engineers make of $\displaystyle \delta$ fails here and one has to be more careful.
CB
Sure, that's the kind of thing I meant talking about "abuses of notation"... The integral should run from $\displaystyle -\infty$ (or any $\displaystyle -\varepsilon$) to $\displaystyle +\infty$ to be "a little" less controversial. Laplace transform is defined for distributions that are supported by $\displaystyle [0,\infty)$ (but not for any of these) hence the values on the negative axis don't matter. In fact, it is really not easy to define the Laplace transform of distributions properly; formally, it should be $\displaystyle \mathcal{L}(T)(z)=\langle T, t\mapsto e^{-tz}\rangle$ when one is able to make this make sense.