# Thread: Find the inverse laplace of 1

1. ## Find the inverse laplace of 1

$l^{-1}(1)$ = ??!

2. Originally Posted by General
$l^{-1}(1)$ = ??!
It's the Dirac delta function.

$\mathcal{L} [\delta] = \int_0^{\infty} e^{-sx} \delta (x) dx = e^{-s . 0} = 1$ ,

since

$\int_{- \infty}^{\infty} \phi (x) \delta (x) dx = \phi(0)$

3. The question is a source of some little controversial... according to the definition of $\delta (*)$ given in Wolfram MathWorld , probably the most 'reliable' mathematical site in the Web, is...

http://mathworld.wolfram.com/DeltaFunction.html

$\delta (t)= \frac {d}{dt} \mathcal {H} (t)$ (1)

... where $\mathcal {H} (*)$ is the 'Heaviside Step Function' defined as...

http://mathworld.wolfram.com/HeavisideStepFunction.html

$\mathcal{H} (t)= \left\{\begin{array}{cc} 0,&\mbox { if } t<0\\ \frac{1}{2},&\mbox{ if } t=0\\1,&\mbox{ if } t>0\end{array}\right.$ (2)

Now, according to the basic properties of the Laplace Transform...

http://mathworld.wolfram.com/LaplaceTransform.html

..., we have...

$\mathcal{L} \{\mathcal{H} (t)\}= \frac {1}{s}$ (3)

... and for every $f(*)$ ...

$\mathcal{L} \{\frac{d}{dt} f(t)\}= s\cdot \mathcal{L} \{f(t)\} - f(0)$ (4)

Now if we combine (1), (2), (3) and (4) we obtain...

$\mathcal {L} \{\delta(t)\}= s\cdot \frac{1}{s} -\frac{1}{2} = \frac{1}{2}$ (5)

... so that it would be...

$\mathcal {L}^{-1} \{1\}= 2\cdot \delta(t)$ (6)

... just a little surprising! ...

Kind regards

$\chi$ $\sigma$

4. Originally Posted by chisigma
... so that it would be...

$\mathcal {L}^{-1} \{1\}= 2\cdot \delta(t)$ (6)

... just a little surprising! ...
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 $\delta$, which is the Dirac distribution, in the sense of distributions.

When you say:
... and for every $f(*)$ ...

$\mathcal{L} \{\frac{d}{dt} f(t)\}= s\cdot \mathcal{L} \{f(t)\} - f(0)$ (4)
this is not true for distributions, where the formula doesn't have the $-f(0)$ term: $\mathcal{L}(T')(s)=s \mathcal{L}(T)$. Indeed, this border term is included in the differentiation: steps in the function $f$ give Dirac terms in the differential of $f$ 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.

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 $\mathcal{H}$ 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...)

5. Originally Posted by Laurent

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 $0$ is one of the limits of integration and what one thinks:

$\int_0^{\infty} e^{-st}\delta(t) \; dt$

is short hand for is now not at all clear. If we take it as the limit of integrals where $\delta$ is replaced by functions whose support becomes more and more concentrated near the origin and with integrals all $1$ what we get depends on the form of these functions and in this case could give any value between $0$ and $1$.

The sort of intuitive use that physicists and engineers make of $\delta$ fails here and one has to be more careful.

CB

6. Originally Posted by CaptainBlack
what one thinks:

$\int_0^{\infty} e^{-st}\delta(t) \; dt$

is short hand for is now not at all clear.
Sure, that's the kind of thing I meant talking about "abuses of notation"... The integral should run from $-\infty$ (or any $-\varepsilon$) to $+\infty$ to be "a little" less controversial. Laplace transform is defined for distributions that are supported by $[0,\infty)$ (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 $\mathcal{L}(T)(z)=\langle T, t\mapsto e^{-tz}\rangle$ when one is able to make this make sense.

7. Seems I inadvertently opened a can of worms!

The notation I used was for convenience. Thanks for the pointers, I'll be more careful in future.