= ??!

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- Jan 13th 2010, 01:45 PMGeneralFind the inverse laplace of 1
= ??!

- Jan 13th 2010, 02:01 PMpomp
- Jan 14th 2010, 02:18 AMchisigma
The question is a source of some little controversial... according to the definition of given in Wolfram MathWorld , probably the most 'reliable' mathematical site in the Web, is...

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

(1)

... where is the 'Heaviside Step Function' defined as...

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

(2)

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

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

..., we have...

(3)

... and for every ...

(4)

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

(5)

... so that it would be...

(6)

... just a little surprising! (Itwasntme) ...

Kind regards

- Jan 14th 2010, 02:58 AMLaurent
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__, which is the Dirac**distribution**, in the sense of distributions.

When you say:

Quote:

... and for every ...

(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 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...) - Jan 14th 2010, 03:57 AMCaptainBlack
Actually I don't think it is, he ought to more careful since is one of the limits of integration and what one thinks:

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

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

CB - Jan 14th 2010, 04:27 AMLaurent
Sure, that's the kind of thing I meant talking about "abuses of notation"... The integral should run from (or any ) to to be "a little" less controversial. Laplace transform is defined for distributions that are supported by (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 when one is able to make this make sense.

- Jan 14th 2010, 06:06 AMpomp
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.