# Laplace transform

1. Sep 3, 2011

### Jimmy Snyder

I am reading the book QFT II: Quantum Electrodynamics by Eberhard Zeidler. On page 32 he defines the Laplace transform as:

$$F(s) = \int_0^{\infty}f(t)e^{ist}dt$$

Where f is a smooth function and F is it's Laplace transform. I have changed his notation, but not the content of the formula. I have looked on the net for some background on this formula but could only find a similar one:

$$F(s) = \int_0^{\infty}f(t)e^{-st}dt$$

Can anyone point me to a book or web page that treats the Laplace transform and uses Zeidler's definition?

2. Sep 3, 2011

### I like Serena

There must be a typo.
Your first formula is not the Laplace transform, but the Fourier transform.
Your second formula is indeed the Laplace transform.

3. Sep 3, 2011

### Jimmy Snyder

Thanks for taking a look at this. Professor Zeidler's book is full of typos. It's published by Springer from which I had come to expect higher standards. The formula is similar to, but different from the Fourier transform in that the lower limit of integration is 0. Using the definition in the book, it is not difficult to verify that for
$$f(t) = e^t,$$
$$F(s) = \frac{i}{s - i}, \Im(s) > 1$$
I can't find this formula in tables of Fourier transforms, nor in tables of Laplace transforms either. Does someone know of a transform that takes this particular f to this particular F?

Last edited: Sep 3, 2011
4. Sep 3, 2011

### I like Serena

Well, apparently Zeidler is defining a Laplace/Fourier-like transform here.
Note that anyone can define his own transform if he wants to.
He should spend some text on that however.

Note that:
$$\int_0^{\infty} e^t e^{i s t} dt = {i \over {s - i}}, \textrm{ if } \Im(s) > 1$$

5. Sep 3, 2011

### Jimmy Snyder

Thanks again. In addition to uncountably many typos, professor Zeidler also uses a lot of non-standard notation and definitions. I doubt he would just haul off and make his own transform and then attribute it to Laplace. So I am hoping that this is just a non-standard way of defining the Laplace transform. Can anyone confirm that? Or is there something else afoot. By the way, I have tried communicating with professor Zeidler directly, but that has come to nothing, he does not respond.

6. Sep 3, 2011

### hunt_mat

The Fourier transform is from $-\infty$ to $\infty$, not from 0 to $\infty$.

7. Sep 4, 2011

### Jimmy Snyder

Apparently, there is something called a Fourier-Laplace transform. I can find references to it on the web, but I haven't been able to find a definition of it. Has anyone else heard of it and perhaps even knows how it is defined?

8. Sep 5, 2011

### hunt_mat

I have not heard of it. Perhaps you could play with the definition you gave to work out some properties of it.

9. Sep 5, 2011

### Jimmy Snyder

I did a little of that. See post #3 in this thread.

10. Sep 5, 2011

### I like Serena

In physics it is not unusual to define a function to be treated as zero for t < 0.
This is especially true for responses to a step or impulse function.
An integration from $-\infty$ to $\infty$ collapses then to 0 to $\infty$.

A Laplace transform for complex values is then basically the same as the Fourier transform.

11. Sep 5, 2011

### Jimmy Snyder

But the lower limit of zero is crucial to the computation. The Fourier transform of the function of post #3
$$f=e^t$$
is divergent for all s except $\Im s = 1$, where is it $\delta(\Re s - t)$.

12. Sep 5, 2011

### I like Serena

Where do you want to go with this?
Is there a specific problem involved?

What I can say is that multiplication by eat results in a frequency shift in the s-domain with "a" as you can already see.
You can find this in the properties of the Laplace transform and also in the properties of the Fourier transform.

As a side effect restrictions must be observed to keep the function integrable.
In physics such things are often neglected or postponed, until they become a bother.

Edit: btw, "t" should not appear in your transform. It should be a constant.

13. Sep 5, 2011

### Jimmy Snyder

The quick answer is that I think you are correct that the phrase "Laplace transform' is a typo, but I want to know what the correct text should be. I don't think Fourier transform is correct either.

The book is interesting to me because it introduces me to a lot of high powered mathematics that I am not familiar with. The author's design is to create "A Bridge Between Mathematicians and Physicists". I think the book is a huge success in achieving that purpose. However, a drawback of the book is that there are a huge number of typos, non-standard definitions, poor English (the author is German) and inconsistencies. In fact, it seems that whenever there is a commonality of ideas in two places in the book, there lacks a corresponding commonality of presentation. To make matters worse, the author refuses to correspond with me even though I have tried to contact him directly and through the Max Planck Institute in Germany where he works. Someone there told me that they gave my e-mail address to him but nothing has come of it.

As for the Fourier transform of $f=e^t$, you are correct, it has the value $\sqrt{2\pi}$ when $\Im s = 1$

I am writing an errata page for the book which you can find at the url erratapage. I would list this as a typo if I knew what he really should have writen. He calls it a Laplace transform, but it isn't. It's not a Fourier transform either. It might be a Fourier-Laplace transform, I can't tell because even though such a thing exists, I can't find a definition for it. I won't list it as a typo until I can be sure of what the correct text should be.

14. Sep 5, 2011

### I like Serena

If you define:
$$G(\sigma)=F(i \sigma)$$
then
$$G(\sigma) = \int_0^{\infty}f(t)e^{i (i \sigma) t}dt = \int_0^{\infty}f(t)e^{-\sigma t}dt$$

Then it's a Laplace transform, just with respect to a different variable.

Edit: if you look at the pages about Fourier transforms, you'll see that there are a lot of different conventions for the variables and normalization constants.
I'm not aware of this specific convention, but it works.

Last edited: Sep 5, 2011
15. Sep 15, 2011

### Jimmy Snyder

I have found the answer and it is not exactly a typo, the author is just using some terminology that he made up on his own. In the first volume of the series QFT I: Basics in Mathematics and Physics pages 93 and 94 (I have the second printing from 2009, the page numbers might be different in the first printing from 2006), in the section entitled "The basic trick of the Laplace transformation". Here the same transform is used and called the Laplace transform. However, although it is hidden, it is none the less clear that a factor of $\theta(t)$, the Heaviside function, is multiplied in and the notation suppressed. Then comes the payoff, a few lines later when he calls the inverse transform "the inverse Fourier transform′. What a confusing way to write. By the way, in the text that I quoted in the OP, he specifically called the inverse transform ′the inverse Laplace transform′, so I was lacking this crucial clue. In addition, he did not point out that the function was multiplied by the Heaviside function. Now I feel I can say with some confidence that what he means is the following:

Let $f(t)=\theta(t)g(t)$, Then the Fourier-Laplace transform (another name for the Fourier transform) of f(t) is

$$F(s)=\int_{-\infty}^{\infty}f(t)e^{ist}dt=\int_{0}^{\infty}f(t)e^{ist}dt$$

That he means this is made clear by his definition of the inverse Fourier transform in volume I

$$f(t) = \frac{1}{2\pi} PV \int_{-\infty}^{\infty}F(s)e^{-ist}ds$$

Where PV stands for Cauchy's principal value.

16. Sep 15, 2011

### I like Serena

Nice!

Seems my observations were not so far off (they were "dead-on").

Welcome to the confusing way some physics volumes are written.

17. Sep 18, 2011

### Jimmy Snyder

I spent more than an hour in the library at Temple U yesterday looking for this transform. There were several shelves of books on integral transforms, and none of them made reference to this particular one. However, several spoke of a two-sided Laplace transform where the lower limit of integration is negative infinity. I naturally thought of one-sided Fourier transform which is by analogy what I should be looking for. I could not find any reference to it in any of the books I pulled down. However, this morning on the net I did find a definition of 'unilateral Fourier transform' and gave a definition which is exactly what this thing is. Although I found no reference to it on the web, I think 'one-sided Fourier transform' would do as well. In any case, I could not find anyone willing to call it a Laplace transform.

Zeidler gives the definition once in Vol I, and once again in Vol II and has the word Laplace transform in several places in the book where he uses it. In other words, it is not a typo, he thinks correctly or otherwise that this thing is called the Laplace transform. He references another of his own books and that was in the library too. It did not shed any light on why he is calling this a Laplace transform. It only has the standard definitions of Laplace and Fourier transforms. He also references a book by Widder which was in the library in a different section. It shed no light.

18. Sep 18, 2011

### I like Serena

Does Zeidler have a purpose in defining his transform in this slightly different way?
Does it make some calculation or concept easier to do or understand?
Or is he just blissfully unaware of current scientific conventions and making his own up as he goes along without any purpose?

19. Sep 18, 2011

### Jimmy Snyder

My characterization of his transform in a previous post is not correct. There I implied that the transform was only applied to functions that were zero for values of t less than 0. In fact, his transform involves multiplying an arbitrary function by the Heaviside function first, and then applying the Fourier transform. There is no doubt that he wants the transform he is using. It is forced on him by the fact that he will apply this transform to functions related to forces that are turned on at time t = 0, i.e. are zero for times less than t = 0, but which formally have definitions that are not zero for times less than t = 0. As to why he calls it a Laplace transform, only the professor can say, and I can't reach him. The only issue in my mind when I started this investigation was that in order to write up the error for my errata page, I needed to know what the correct terminology was. Now I have found it. I appreciate all the help I got from you and others in this search.

20. Oct 5, 2011

### jeffreylaible

does anyon know the laplace transform of f(t)=(sin(t)+1)/cosh(t)