Many differential equations textbooks (“First course” books) limit themselves to taking Laplace transforms of functions of exponential order. That is a reasonable thing to do. However I’ll present an example of a function NOT of exponential order that has a valid (if not very useful) Laplace transform.

Consider the following function:

Now note the following: is unbounded on , does not exist and

One can think of the graph of as a series of disjoint “rectangles”, each of width and height The rectangles get skinnier and taller as goes to infinity and there is a LOT of zero height in between the rectangles.

Needless to say, the “boxes” would be taller and skinnier.

Note: this ~~is an example~~ can be easily modified to provide an example of a function which is (square integrable) which is unbounded on . Hat tip to Ariel who caught the error.

It is easy to compute the Laplace transform of :

. The transform exists if, say, by routine comparison test as for that range of and the calculation is easy:

Note: if one wants to, one can see that the given series representation converges for by using the ratio test and L’Hoptial’s rule.

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“Note: this is an example of a function which is l^2” ___ In my opinion is a L^1 example, g is not elevated by two.

and there is a tinny error, in the line where the serie is shown, 1 + 1/10 +(1/10)^2 + …… etc. is the correct form.

but despite all, great example….

Comment by Ariel — May 8, 2014 @ 10:57 am

Thanks Ariel. Note: I should have said: “g can easily be modified to be made L^2” and I will correct the error.

Regards

Comment by blueollie — May 14, 2014 @ 10:37 am

[…] that only functions of exponential order could have a Laplace transform; I replied that, while many texts restricted Laplace transforms to such functions, that was not mathematically necessary (though it is a reasonable restriction for an applied first course). (briefly: imagine a function […]

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