# Talk:Laplace transform/Archive 1

## Confusing statement in the introduction

" In mathematics, the Laplace transform is a technique for analyzing linear time-invariant systems such as electrical circuits, harmonic oscillators, optical devices, and mechanical systems."

However, electrical circuits, harmonic oscillators, optical devices, and mechanical systems are subjects of physics and technical sciences, not of mathematics per se. It will be more appropriate to replace mathematics with physics. On the other hand, the main use of Laplace transform in mathematics proper is for solving differential equations. I do not see this in the introduction.Lantonov 14:31, 20 July 2007 (UTC)

I wrote a new intro/definition which can be understood by persons with high-school math background which immediately leads the reader to possible applications. Lantonov 05:30, 30 August 2007 (UTC)

The use of Laplace transform to solve differential equations really came out of the engineering applications. In English speaking countries, linear differential equations were until after World War II usually solved by the D-operator method due to Boole. True the Laplace transform was used but almost always in connexion with much more complicated problems involving tricky inversions of Laplace transforms.JFB80 (talk) 12:38, 3 February 2011 (UTC)

## Oliver Heaviside and Laplace Transform

"The method of using the Laplace Transform to solve differential equations was developed by the English electrical engineer Oliver Heaviside." The Oliver Heaviside wikipedia article also suggests this. Can anyone site a reference supporting this?

Oliver Heaviside developed an operational calculus for solving differential equations, while similar to using the Laplace transform it is not the same. This site goes into more detail. http://www.du.edu/~jcalvert/math/laplace.htm Yardleydobon 20:04, 30 April 2007 (UTC)

"Not the same" is a technically correct, but very obtuse argument, although a popular one among pure mathematicians who completely dismiss[ed] Heaviside. The introduction of the book of Debnath and Bhatta compares the development of Heaviside's operational calculus to that of calculus in the 17th century: neither were well-founded at the time of their conception, but both were later put on firm mathematical ground (19th century for calculus, 20th century for H's op. calculus.) They also compare the opposition to H's op. calculus among mathematicians to that against Dirac's delta (before it was properly formalized, of course). Tijfo098 (talk) 17:13, 24 September 2010 (UTC)
Contrary to popular belief Heaviside knew about the Laplace transform method. But he considered his own operational method superior because as he said it goes straight to the answer. I think he discusses this in his Electromagnetic Theory 1893 but am unable to check just now.JFB80 (talk) 12:02, 3 February 2011 (UTC)

## Alternative Name

I notice many texts refer to the Laplace "Transformation", particularly more classical texts. Would there be any problem with this minor addition? ... Laplace Transform (also known as Laplace Transformation)...Tparameter 06:00, 8 November 2006 (UTC)

## What is the point of LaPlace?

Say I'm not familiar with him, and am looking him up in the Wikipedia for that reason. I would then like to find out what is so important about this Transform. When can it be used, and how? What is, again, the point of the LaPlace transform? The article does a poor job of informing 'me'.-- Ec5618 11:32, Apr 21, 2005 (UTC)

First of all, it is Laplace, with small p. Its uses are explained in the second paragraph in the article. I am not sure its signifance can be explained in non-mathematical terms, as it is a sofisticated tool used for solving differential equations and such.
Let me know if you have further questions. Oleg Alexandrov 15:34, 21 Apr 2005 (UTC)
Also, this is an encyclopedia, not a textbook. For a non-specialist, this might be tricky to understand, but then, to understand it well, you might need to read a book or take a course, as again, this is a complex tool. Oleg Alexandrov 15:36, 21 Apr 2005 (UTC)
I understand that an encyclopedia cannot possibly contain all knowledge in a way that any person could understamd it, however, if you look at the CPR article, for example, you'll see that it's quite possible for wikipedia to be textbook-like. Perhaps this article should be simplified, and expanded somewhat. -- Gerriegijsen 15:58, May 13, 2005 (UTC)
I think the CPR article is a bad example. Firstly, you need to know a bit of mathematics to understand why the Laplace transform is useful; at least differential equations and integration. This is rather different from CPR, where not much beyond common sense is necessary to understand the gist of the procedure. Secondly, in my opinion the CPR article goes too much into detail for an encyclopedia (this is justified for that specific example, I think).
On the other hand, the Laplace transform article can certainly be improved. It shouldn't have a technical definition in the first sentence and it should include an example of solving a differential equation with Laplace transforms. Apparently, no editor has yet found the time and motivation to do this. -- Jitse Niesen 00:11, 14 May 2005 (UTC)

Yes, this article needs to be cleaned up. Maybe I'll do that sometime in the near future. I have no definite plans, but I might get around to it. If I do indeed get the gumption to do this, I'll do my best to include a way to solve a D.E. with it. In any case, I also wonder what this is used for. I know how to find the transform, and how to do the inverse. Pretty easy. I was taught in my D.E. class that it is useful in some physics problems, or at least in electrical engineering. So how can it be used in E.E.? I learn a lot better when I can comprehend how it will benefit me. Until then, it's trivial, much like the trivial solutions that no one cares about in D.E. Why should I care about it? D. F. Schmidt [[User_talk:D. F. Schmidt|(talk)]] 09:14, 16 November 2005 (UTC)

The problem you raise regarding the usefulness of the Laplace Transform is the result of poor teaching and has nothing to do with the actual usefulness of the technique. The Laplace Transform is an exceedingly powerful tool in understanding and analyzing all types of linear time-invariant (LTI) systems, not only in electrical engineering, but in virtually any branch of engineering and many of the physical sciences as well. Unfortunately, the teaching of Laplace, particularly in many mathematics courses, is rather abysmal. It is really a shame that you have learned the mechanics of the Laplace Transform but did not have any opportunity to see how it is applied in the real world -- in electrical engineering, for example. If I have some free time, I will try to add some better explanations and maybe a few examples to this article. -- 24.218.218.28 13:12, 16 November 2005 (UTC)
Regarding the practical use in Physics or EE: With a basic step function or Dirac Delta function, when combined with other disturbances, it can be useful to find the unknown function that describes these things as they act together - possibly eliminating the step function or reducing it when combined with the transforms of the other disturbances. DB 12/13/2005

Maybe a small text like this can solve a lot of headaches: "Laplace transformations are used as a tool to convert a complex equation in a very simple to use equation. This process involves 3 operations:

• converting into the laplace domain
• using basic math properties to 'clean up the equation'
• converting back into the normal world"

I don't speak english so someone should clean up this text and insert it in the article. Thx

I wrote a new intro/definition which can be understood by persons with high-school math background which immediately leads the reader to uses of the Laplace transform. Also ordered the sentences in the intro a bit. The intro is still too big, with an emphasis on physical and electronic applications. I do not find anything bad in this but think that they can be shortened a little without losing too much from their information content.Lantonov 05:31, 30 August 2007 (UTC)

I have always wanted to know how Laplace was led to the transform, but never could find much on the web. Anybody know of a direct account from Laplace? Here's something of interest perhaps. http://www.math.niu.edu/~rusin/known-math/97/laplace.fourier thanks Phil —Preceding unsigned comment added by 74.170.68.234 (talk) 04:08, 12 December 2007 (UTC)

Laplace used the method in his 1812 book on probability where he talked about the relation between generating functions and the corresponding probability distribution. Actually he did'nt use what is nowadays called the Laplace transform but what is called now called the z-transform, i.e. the discrete version. The continuous-time version is due to Abel 1820. The history given in the article quotes formulae previous to 1812 which don't look very convincing to me. JFB80 (talk) 13:08, 3 February 2011 (UTC)

## Engineering/physics notation

How is this equation different from the one at the top? Taral 16:13, 27 May 2005 (UTC)

## Natural Logarithm and definition of Gamma

I think it would be helpful for those unfamiliar with it to define ${\displaystyle \,\gamma }$.

Agree, disagree? Guardian of Light 16:43, 14 July 2005 (UTC)

Done. I discovered, somewhat by accident, that ${\displaystyle \gamma \,}$ is the Euler-Mascheroni constant. -- Metacomet 05:52, 7 December 2005 (UTC)

## Query

Somebody help find the laplace transform of exp(t-3) and exp(-(t-1))? Thanks

While this isn't really the place for hw help, and for all I know this request might have been posted years ago... Consider that exp(t-3)=exp(t)*exp(-3). Since exp(-3) is a constant the solution is now trivial.

## Another Query

Does anyone know more about the history of Laplace

Did you have a look at Pierre-Simon Laplace? -- Jitse Niesen (talk) 09:53, 8 November 2005 (UTC)

## World War II

Is there relationship between Laplace transformations and the engineering used in the construction of German weapons during World War II?

yeah, we're really lucky that the allies discovered that
${\displaystyle {\mathcal {L}}\{\,I_{n}(t)\}={\frac {\left(s+{\sqrt {s^{2}-1}}\right)^{-n}}{\sqrt {s^{2}-1}}}}$
before the Nazis, or they would have gotten the bomb first and we would all be speaking German now. r b-j 01:45, 10 November 2005 (UTC)

## Region of convergence

For the lists of common transforms, I believe there are certain restrictions on the values that the variable s can take (e.g. see http://mathworld.wolfram.com/LaplaceTransform.html). Can we incorporate these into the table(s)? --GregRM 23:16, 29 November 2005 (UTC)

I beleive the restrictions you are talking about are called the Region of Convergence or ROC. The article discusses the ROC in general terms in Section 2, but you are correct, the ROC should be specified in any tables of common Laplace transforms or their properties.
I have started working on a new table of common Laplace transforms (see below), but I have not had time to complete the table beyond a handful of examples. I would like to take all of the transform pairs from the main article and put them into this table so that they are all in a consistent format and they include the ROC. -- Metacomet 05:04, 30 November 2005 (UTC)
I agree, the region of convergence should definitely be included. Metacomet, thanks for all the work you have been doing and are still doing on the article. I hope you will replace or explain the phrase "for causal systems" in the header of the table, as I have no idea what that is supposed to mean. -- Jitse Niesen (talk) 13:22, 30 November 2005 (UTC)
I have added an explanation of causal systems to the Explanatory Notes at the bottom of the table, along with some other definitions and explanations. -- Metacomet 05:07, 5 December 2005 (UTC)
The table is now nearly complete, and I have moved it from this discussion page to the main article. -- Metacomet 05:07, 5 December 2005 (UTC)

I found an important error here.

"However, if the integral defining the Laplace transform does converge (possibly only as an improper integral) at s = s0, then it automatically converges absolutely for all s with Re{s} > Re{s0}."

This is false. Consider a function defined as follows (for positive reals): divide the interval ${\displaystyle [n-1,n]}$ in ${\displaystyle 8^{n}}$ equal intervals. Let ${\displaystyle f}$ take the values ${\displaystyle 4^{n}}$ and ${\displaystyle -4^{n}}$ alternately in those intervals. Fourier transform at 0 converges conditionally, while it doesn't converge absolutely at 1. However both of the following are valid:

(absolute) convergence at ${\displaystyle s=s_{0}}$ implies (absolute) convergence for all ${\displaystyle s}$ with ${\displaystyle Re(s)>Re(s_{0})}$ —Preceding unsigned comment added by 201.231.212.56 (talk) 06:25, 24 November 2009 (UTC)

I corrected the mistake. Sławomir Biały (talk) 14:02, 24 November 2009 (UTC)

## ROC for the first two entries

Does anyone know what the ROC is for the first two entries in the table below, the unit impulse and the unit step functions? I have filled in the table, but I am not sure if they are correct. Thanks. -- Metacomet 19:51, 30 November 2005 (UTC)

I believe the unit step requires s > 0. (I believe this can be seen by plugging the function into the Laplace transform definition; also, see http://mathworld.wolfram.com/LaplaceTransform.html). I'm not sure about the unit impulse.--GregRM 20:30, 30 November 2005 (UTC)

Please help: If anyone knows the correct Regions of Convergence (ROC) for any of the entries in the table where the ROC is missing, please feel free to add them to the appropriate cell in the table. Thanks. -- Metacomet 22:28, 8 December 2005 (UTC)

Special thanks to Jitse Niesen for providing the ROCs for the last 6 or 8 entries in the table. I greatly appreciate your assistance. -- Metacomet 18:35, 10 December 2005 (UTC)

## Bessel functions

Where did you get the transforms for the Bessel function from? I think the exponent should be n instead of −n, but I'm not at all sure. -- Jitse Niesen (talk) 18:49, 10 December 2005 (UTC)

I believe the form in the table is equivalent to the form given by MathWorld which has a positive n exponent (http://mathworld.wolfram.com/LaplaceTransform.html). I went through the algebra and it seemed that one was the same as the other. However, one might argue that the form with positive n exponent is simpler to use or in a more "simplified" form...I would not know however, as I do not have experience with this transformation or the Bessel function.--GregRM 21:37, 10 December 2005 (UTC) PS...there may be some incorrect operations during my algebraic "simplification" that I overlooked (e.g. the possibility of multiplying by 0/0), so someone might want to check this.--GregRM 21:58, 10 December 2005 (UTC)
You will find a very extensive set of tables at:
There is an entire page devoted to Bessel functions (pdf format).
I believe that the form in the Wikipedia table is correct. I have seen it where the sign on n is positive, but the entire term is in the denominator instead of the numerator, so that algebraically, it is the same thing. It makes no difference to me which way it appears. If you want to move it to the denominator and switch the sign on n, feel free.
-- Metacomet 22:17, 10 December 2005 (UTC)

The table here says that the transform of ${\displaystyle J_{n}(x)}$ is

${\displaystyle {\frac {\left(s+{\sqrt {s^{2}+1}}\right)^{-n}}{\sqrt {s^{2}+1}}}.}$

The table at MathWorld says that it is

${\displaystyle {\frac {\left({\sqrt {s^{2}+1}}-s\right)^{n}}{\sqrt {s^{2}+1}}}.}$

I can't see how they can be the same. -- Jitse Niesen (talk) 23:19, 10 December 2005 (UTC)

The table at Equation World says the transform of ${\displaystyle J_{n}(\omega x)\!}$ is:
${\displaystyle {\frac {\omega ^{n}}{{\sqrt {s^{2}+\omega ^{2}}}\left(s+{\sqrt {s^{2}+\omega ^{2}}}\right)^{n}}}}$
The form we are using in Wikipedia is clearly the same algebraically as the one in Equation World, but different from the one in MathWorld. I have no idea which one is correct. Clearly, we need to find out.
BTW, I did not put these expressions into the article origninally, although I did modify them as they were in the article based on Equation World.
-- Metacomet 00:49, 11 December 2005 (UTC)
It looks like eFunda agrees with MathWorld: eFunda Laplace transforms
But Wolfram research agrees with Equation World.
Is it possible that these two expressions are actually the same? Maybe by invoking difference of squares, or by combining exponents, or...?
GregRM, can you shed any light on this problem?
-- Metacomet 00:59, 11 December 2005 (UTC)
Starting with the Wikipedia version, I think what I did was multiply the c*(a+b)^(-n)/d fraction by (a-b)^(-n)/(a-b)^(-n) (I think this was part of my concern about multiplying by 0/0) which (I think) gives c*(a^2-b^2)^(-n)/(d*(a-b)^(-n)); this appeared to simplify to the desired form. Sorry for the rough formatting of the math, but I have not had the chance to learn the formatted form. Let me know if you need more details or if there are any problems with this.--GregRM 03:01, 11 December 2005 (UTC)
Indeed, we have
${\displaystyle {\frac {1}{s+{\sqrt {s^{2}+\omega ^{2}}}}}={\frac {{\sqrt {s^{2}+\omega ^{2}}}-s}{({\sqrt {s^{2}+\omega ^{2}}}+s)({\sqrt {s^{2}+\omega ^{2}}}-s)}}={\frac {{\sqrt {s^{2}+\omega ^{2}}}-s}{\omega ^{2}}}.}$
The factor ${\displaystyle {\sqrt {s^{2}+\omega ^{2}}}-s}$ is never zero (unless ω is zero). So the two forms are indeed the same. Thanks for clearing that up. -- Jitse Niesen (talk) 13:38, 11 December 2005 (UTC)

Well done. I had a feeling that the difference of squares rule would play a part! -- Metacomet 16:07, 11 December 2005 (UTC)

Would it be practical and useful to put both versions in the table? I would think that it might be helpful, particularly for doing inverse Laplace transforms, but it might also be viewed as being somewhat redundant. It is not really a pressing issue, but I figured I would bring it up.--GregRM 03:17, 19 December 2005 (UTC)

I have been mulling over the very same idea, and I had not come to any conclusions one way or the other. It might be a bit too much detail for this article. I am not sure how often Bessel functions come up in the context of Laplace transforms. It might be usefule to create a more detailed table of transforms to which this article could link. On the other hand, the external links section points readers to several excellent resources. I am not sure we should do anything just yet, maybe continue to think about it some more. As you said, there is no need to rush into anything in this case. -- Metacomet 03:27, 19 December 2005 (UTC)

## Background color inside the [itex] tag

Does anyone know how to change the background color from white to something else (in RGB) of a LaTeX equation inside the [itex] tag? Please let me know. Thanks. -- Metacomet 03:40, 1 December 2005 (UTC)

## Table of Selected Laplace Transforms

I have now added this table to the main article. -- Metacomet 16:42, 4 December 2005 (UTC)

Very impressive. I have noticed that the Laplace transform for the exponentially decaying cosine wave has numerator s. However, based on http://mathworld.wolfram.com/LaplaceTransform.html, I would think it would have numerator s + α .--GregRM 16:58, 6 December 2005 (UTC)
Yes, you're right. If you combine the transform of the cosine with the rule for frequency shifts, you also get s + α in the numerator. Well spotted. -- Jitse Niesen (talk) 17:42, 6 December 2005 (UTC)

Yes, I think you are both correct. It does follow directly from the rule for frequency shifting. Good catch. -- Metacomet 22:16, 6 December 2005 (UTC)

Should the table use the heading Frequency Domain? I wonder if this might cause confusion with the Fourier Transform which is more commonly described as a Time-Frequency transform than the Laplace Transform. Would "Laplace Domain" or "S Domain" be more appropriate? jackocleebrown 21:20 , 30 Aprl 2007 (UTC)

I think the transforms used in the example should be included in the table so there can be a chance of following them. 70.133.83.60 (talk) 20:50, 3 September 2010 (UTC)

The Laplace transform of 1 is 1/s but the Laplace Transform of the unit step function ${\displaystyle {u}_{c}(t)}$ is ${\displaystyle {\frac {{e}^{-sc}}{s}}}$., 16 October 2011

## Alternative Definition

Near the top of the article is a section on the "alternative defintions" of the Laplace transform,

${\displaystyle F(s)={\mathcal {L}}\left\{f(t)\right\}=s\int _{0^{-}}^{+\infty }e^{-st}f(t)\,dt,}$

Can someone elaborate on this? When and why are these alternative definitions used? -Monguin61 09:32, 9 December 2005 (UTC)

i am interested in the answer, also. i have never seen this definition with the leading s. and i think it is wrong and should be removed. r b-j 17:39, 9 December 2005 (UTC)
For what it is worth, I believe this is the edit in which the alternative definition was introduced: http://en.wikipedia.org/w/index.php?title=Laplace_transform&diff=10239127&oldid=10200910--GregRM 18:08, 9 December 2005 (UTC)
i see that the original contributor has been contacted. even though the alternate definition (with corresponding modification of the inverse Laplace transform - divide by s before the regular-old {LT}-1) is self-consistent and works i'll wait a day, but i have never seen this alternate definition anywhere else, and i'll wack it in about a day. even though this had been on my watchlist, i hadn't noticed this before. r b-j 22:55, 9 December 2005 (UTC)
I don't think you should remove it. If you read my comments below, and review the changes that I have made to the article, I think you will see that I have addressed the concerns that you and others have raised. I am not in love with it, but I have come across this form a few times. And, as I said, I believe I have eliminated any confusion with the edits that I made earlier today. -- Metacomet 00:54, 10 December 2005 (UTC)
you have "come across this form a few times"??! i have never seen it and cannot imagine, at present, why it would ever be necessary or beneficial. i would like to see a reference of a textbook or a bunch of uses of this form in the lit. it's not just a scaling issue or a bilateral/unilateral issue like we see in the LT or FT or ZT, it's a really different definition and having two really different definitions in the same context is bound to confuse. i'll give it 48 hours before i start shooting (hell, someone else might shoot it before i do). since we can raise it from the dead, i see no good reason to leave it very long while its entitlement to exist is suspect. r b-j 03:09, 10 December 2005 (UTC)

• Why are you attacking me personally?
why do you think i am?
• Why are you so angry?
i'm not angry, i just want to blow this Alternate definition that is inconsistent with the traditional definition away. and i have yet to see a justification for keeping it.
• Are you saying that I am lying?
where did i say that? why would you think i had?
• Are you saying that I have no idea what I am talking about?
no, i'm saying that i have never seen this definition and i am so highly dubious of its value in an encylopedia and i am concerned about the potential for this non-equivalent definition to confuse, that i think it should be gone.
• Do I owe you something that I am not aware of?
only a few good references, at least one, if you really think it should be kept.
• I agree that the way it was written was very confusing. But I have now made it so it is not confusing.
• I don't care if you wait 48 hours, 48 seconds, or 48 days. You are free to do whatever you want. Go ahead and shoot.
• On the other hand, keeping an open mind is usually a good practice.
-- Metacomet 03:18, 10 December 2005 (UTC)
i try to keep an open mind, but mathematics and physics and engineering are pretty exacting and unforgiving disciplines. e.g. those Mars landers that probably crashed instead of landing because someone used the wrong conversion factor because of multiple definitions for the same thing. we must minimize spurious definitions of common and salient concepts. sometimes we get stuck with different definition (like at least three for the F.T., but the differences are about scaling, that's all), but why introduce another incompatible definition to something so fundamental? please, Meta, don't take this personally. you're fine by me, but this alternate definition is not. r b-j 03:47, 10 December 2005 (UTC)
I agree with much of what you are saying. The F.T. example is particularly apt. The reason that these differences arise is that different communities have created their own standards and conventions independent of the others. For example, physicists tend to use c.g.s. units, while electrical engineers almost always use SI units. In my opinion, I think the physicists should abandon c.g.s because SI is the international standard and because it is objectively simpler and easier to use, particularly in electromagntism. And in fact, when I write or edit articles related to electrical engineering, I generally remove the c.g.s. units and replace them with SI. Of course, the physicists haven't exactly asked me for my opinion on the subject, and they are free to continue using c.g.s. units, and probably will. Too bad for them, but that is the reality.
In the case of the F.T., the issue has to do with normalization. The three definitions, which are really just different scaling conventions (as you point out), arose because mathematicians like to do things one way, physicists another, and electrical engineers yet another.
You are right to be concerned, and I agree that Wikipedia can be part of the solution. But, the answer is not for Wikipedia to ignore these different conventions and definitions or to pretend that they don't exist. That will not make them go away. People will continue to come across all three definitions of the F.T. out there in school, in books, at work, and what not. Some of them will get confused. Some of them will use the wrong definition in the wrong situation. And disasters like Mars (or worse) will be the result, unless people can find information that helps them understand things at a deeper, more fundamental level.
So the right answer is for Wikipedia to highlight these different conventions and definitions, to explain what they are and where they come from, to help people understand that they need to be careful in how they apply different concepts, to outline the advantages and disadvantages of each, and maybe in some cases to recommend one approach over another or to report that one is the accepted standard and that the others are out of favor or not necessarily useful.
In any case, I fully agree with you that we need to provide clear and unambiguous definitions of terms, particularly in math, science, and engineering. In the case of the Laplace transform, the way the Alternative Form was written as of a few days ago was very confusing. And you were right to bring it up as a concern. But I think the changes that I have made, and the additional changes that you have also made, have eliminated much of the confusion. And if someone were to come across this "alternative" definition of the L.T. out there in a textbook or physics course or something, and then turn to Wikipedia as a resource, we have now provided a pretty good explanation of where this so-called "alternative" form comes from and what it really means. And if that helps avoid confusion, then Wikipedia will have done a good thing. -- Metacomet 09:46, 10 December 2005 (UTC)

I think it should kept, but it needs to be clarified. I believe that it may have to do with the fact that multiplication by s in the frequency domain is equivalent to differentiating with respect to time in the time domain. So this alternative form gives the Laplace transformation not of the function f(t) itself, but rather of the derivative of f with respect to time t. Does that make sense? -- Metacomet 20:50, 9 December 2005 (UTC)
I have made this clarification in the article. I think with a bit more elaboration, we can keep it as a useful form. -- Metacomet 20:59, 9 December 2005 (UTC)
Please do give a reference. It does not seem useful. At the very least it should be flagged as idiosyncratic, in my opinion. As Rbj noted, I asked the editor who put it there orignally for clarification. We'll wait and see what he says. -- Jitse Niesen (talk) 01:27, 10 December 2005 (UTC)
• The example (below) illustrates that it is correct.
• I didn't raise this question. Several people asked for an explanation (including r-b-j). I happened to know one way to explain it. So I did. Quote from r-b-j: "i am interested in the answer" (see above).
• If you are not interested in the answer, then don't ask the question.
i want a really fast and persuasive anwer. it's like were talking at a party and someone in the conversation says something contrary to the known set of facts (like Saddam had WMD and Al Quaeda traning camps in Iraq and had a part in 9/11) my ears would perk up and i would like to hear that justified nearly immediately otherwise i don't want to waste any real estate in my cranium thinking about it. r b-j 03:09, 10 December 2005 (UTC)
Wikipedia is not a cocktail party, and we are not discussing current events. Wikipedia is an publicly written and edited encyclopedia. Articles do not get written in 30 seconds. It takes a lot of time, solid research, thoughtful discussion and reflection, and a lot of fine-tuning. I have learned that making snap judgements when editing Wikipedia usually leads to sub-optimal outcomes. But again, you are free to do what you want. I am sorry that the time and energy I put into trying to provide you a clear and well-explained answer was not helpful to you. Next time, I will try to give you an answer in a sound byte instead. -- Metacomet 03:34, 10 December 2005 (UTC)
• Also, please don't shoot the messenger (me).
i'm aiming at that "alternate definition" of the LT. he is evil and a cancer and has no right to live. Messenger, please step aside, out of the line of fire.
Do you own any actual guns? -- Metacomet 03:24, 10 December 2005 (UTC)
oh, no. actually i am a Christian pacifist (assholes like Jerry Falwell, Pat Robertson, etc. don't speak for Christianity as i see it), but i like to use the metaphor of battle sometimes in discussion (indeed, christian pacifists believe that Jesus was a pacifist and taught pacifism, but he also used warfare and militerism as conceptual vehicles in what he taught). multiple incompatible definitions of something so objective and concrete is bad. they're bad guys that can become malignant if the slip into usage. personally, i wish there was only one definition for the Fourier Transform (and not the one used in Wikipedia) because, with each definition, you have to have a different table of common transforms and a different table of properties/theorems because of the scaling inconsistency. but at least it's just a scaling difference, not one of form. when there is a competing alternate definition of different form of something fundamental, only bad stuff can result from using it. r b-j 03:59, 10 December 2005 (UTC)
• If you look at the Properties and Theorems section of the main article, you will see that the differentiation property does exactly what this "Alternative Form" says (again, please read the example below).
not the issue. i "conceded" right away that the alternate definition is self-consistent (but not consistent, of course, with the real definition) and "works", but it needs more than that to justify its existence.
I was not aware that it needs to do anything. It is an inanimate object that has no power on its own.
Also, I do not believe that the rules and principles of Wikipedia require that authors and editors need to justify the existence of the sections they write. -- Metacomet 03:24, 10 December 2005 (UTC)
• Any decent reference that includes a list of properties of the Laplace Transform will include this property. One such reference, by Siebert, is listed in the Reference section in the main article.
• I must disagree with your statement that it is not useful. Actually, it is an extremely useful property for finding unknown transform pairs from known transform pairs and their derivatives.
show me how it is productively useful at "finding unknown transform pairs from known transform pairs and their derivatives" in a qualitatively different manner than the traditional definition.
• The information is correct, but whether it is important enough to warrant its own section and equation, I really am not sure.
i'm pretty sure. the potential for confusion, IMO, vastly outweighs anything it can offer. r b-j 03:09, 10 December 2005 (UTC)
We can help eliminate the confusion that exists out there by explaining the differences between the correct or standard definition, and alternatives that others are using, perhaps incorrectly. -- Metacomet 10:02, 10 December 2005 (UTC)

• Do whatever you want. And I will too.
-- Metacomet 02:14, 10 December 2005 (UTC)
We have all read the example. You should distinguish between the differentiation property, which is a theorem (the Laplace transform of a derivative is the Laplace transform of the original function multiplied by s) and the statement under Alternate definition that the Laplace transform is sometimes defined with an extra factor s. The differentiation property is either correct or not (in this case, it is). This is not the case for a definition. For a definition, the only question of interest is whether it is in common use. What I contend, and r b-j apparently too, is that it is not. -- Jitse Niesen (talk) 02:40, 10 December 2005 (UTC)
Then take it out. I am sorry that I tried to be helpful by responding to a question that somebody asked.
I didn't put this section in the article. I noticed it about three months ago, and at first I thought it was incorrect. But, instead of jumping to conclusions and eliminating it immediately, I decided that maybe somebody put it there for a reason. So, after thinking about it with an open mind for a while, I realized what it was, so I left it there, even though it was not well presented.
Then, earlier today, in response to the question that others had asked, I decided that it was time to clean it up and clarify the issue. And, more or less, I have changed it from a definition of Laplace transform, to a statement of a property.
These discussions are endless and mindless. Like I said, I am really sorry. -- Metacomet 03:01, 10 December 2005 (UTC)

### Example

A good example is to look at the Laplace transforms for sine and cosine. Suppose that

${\displaystyle f(t)=\sin(\omega t)\!}$

Then we have

${\displaystyle f(0)=\sin(0)=0\!}$
${\displaystyle {df \over dt}=\omega \cdot \cos(\omega t)}$
${\displaystyle {\mathcal {L}}\{f(t)\}={\mathcal {L}}\{\sin(\omega t)\}={\omega \over s^{2}+\omega ^{2}}}$

and

${\displaystyle {\mathcal {L}}\left\{{df \over dt}\right\}={\mathcal {L}}\{\omega \cdot \cos(\omega t)\}=\omega \cdot {\mathcal {L}}\{\cos(\omega t)\}={s\omega \over s^{2}+\omega ^{2}}=s\cdot {\mathcal {L}}\{f(t)\}-f(0)}$

as expected. -- Metacomet 21:11, 9 December 2005 (UTC)

To return to the main point - the alternative definition. I believe it was defined like this in the early days (about 1930). See the book of Carleson on Operational methods. The change to the modern form came about with the publication of Doetsch's Theory and application of the Laplace transform (In German) which is also where the name Laplace transform comes from. Needs checking. (If you are free why not do some historical research?)JFB80 (talk) 12:30, 3 February 2011 (UTC)

## Gamma functions are the basis of Laplace transform

quote:

### Alternative Definition

Near the top of the article is a section on the "alternative defintions" of the Laplace transform,

${\displaystyle F(s)={\mathcal {L}}\left\{f(t)\right\}=s\int _{0^{-}}^{+\infty }e^{-st}f(t)\,dt,}$

Can someone elaborate on this? When and why are these alternative definitions used? -Monguin61 09:32, 9 December 2005 (UTC)

Quote:

### Natural Logarithm and definition of Gamma

I think it would be helpful for those unfamiliar with it to define ${\displaystyle \,\gamma }$.

Agree, disagree? Guardian of Light 16:43, 14 July 2005 (UTC)

:Done. I discovered, somewhat by accident, that ${\displaystyle \gamma \,}$ is the Euler-Mascheroni constant. -- Metacomet 05:52, 7 December 2005 (UTC)

RE:
Metacomet is right.
Gamma functions are the basis of Laplace transform :
${\displaystyle \Gamma (x)=\int _{0}^{\infty }{\frac {e^{-t}t^{x}dt}{t}}}$
--HydrogenSu 12:45, 22 December 2005 (UTC)

## Vote for new external link

Here's my website of example problems with Laplace transforms. Someone please put it in the external links section if you think it's helpful!

http://www.exampleproblems.com/wiki/index.php/PDE:Laplace_Transforms

## a L.T. question

Could anyone tell me why do

${\displaystyle {\mathcal {L}}\left\{{df \over dt}\right\}=s\int _{0^{-}}^{+\infty }e^{-st}f(t)\,dt-f(0)=s\cdot {\mathcal {L}}\{f(t)\}-f(0)}$

and

${\displaystyle {\mathcal {L}}\left\{{\ln t}\right\}=-{\frac {{\gamma }+\ln s}{s}}}$  ?

thanks...

the page of that simplied too much ,i need a lot of noting help to the result... -- HydrogenSu

For the first one, use integration by parts. For the second one, start by using the first integral in Euler-Mascheroni constant. If you don't manage, please tell me how far you got. -- Jitse Niesen (talk) 18:41, 8 January 2006 (UTC)
Thank you,I've got them.

## a L.T. question II

For the first one, use integration by parts. For the second one, start by using the first integral in Euler-Mascheroni constant. If you don't manage, please tell me how far you got. -- Jitse Niesen (talk) 18:41, 8 January 2006 (UTC)
Thank you,I've got them. Onece more for a question, that
${\displaystyle \int _{0}^{\infty }\delta (\tau -t)d{\tau }=1}$
Could you tell me why? Thank you first.--HydrogenSu 17:02, 9 January 2006 (UTC)
Perhaps see Dirac delta or the MathWorld article about the Delta Function. In particular, perhaps the property described by the equation with the integral from a - ε to a + ε (in the MathWorld link) explains your integral for t > 0 (perhaps with f(x) = 1, and a (in MathWorld) = t (in your equation)). However, I have never used the delta function, so my ability to help is limited.--GregRM 19:28, 22 January 2006 (UTC)
thank you, I'll try to do it..(I asked for my teacher,he said that "取樣"(sample taken to y axis ${\displaystyle {\mathcal {\tau *\delta }}=1)}$ )--HydrogenSu 12:09, 23 January 2006 (UTC) (I'm not quite sure for that.)

## formulas

I think a page of L.T. shoud give more proves below some formulas. If just showed "Formulas" might let readers confused and get some information of "just memorize them". If improving,it is nicer. :)--HydrogenSu 12:00, 14 February 2006 (UTC)

## Seems to be Not appropriate

@.Why the Applications part had to be put in an article of Laplace Trs.? It shall be put at the nearly ending.--GyBlop 08:41, 24 February 2006 (UTC)

Could you please give some specific objections that you have to a section on "Applications" in this article? Exactly why do you say that this section is "Not appropriate" and that it should be moved near the end of the article? Thanks -- Metacomet 12:29, 24 February 2006 (UTC)
BTW, the reason I wrote the "Applications" section several months ago was in response to criticisms of the article listed on this discussion page. Please see the very first section at the top of this page entitled "What is the point of Laplace?" and the section immediately preceding this one entitled "formulas", among others. -- Metacomet 12:31, 24 February 2006 (UTC)

## circuit elements transformed into s-domain

I think it would be very useful to have a table or list of how to transform circuit elements (like resistors capacitors inductors, sources, etc) into s-domain elements. For example, the transform of an inductor with an initial current of I is an inductor and a source in series with s-domain values... but I'm not quite sure what those are at the moment. Fresheneesz 08:53, 21 April 2006 (UTC)

If you're looking for the transforms, they are simple:
${\displaystyle {\frac {V}{I}}=sL}$ impedance of an inductor
${\displaystyle {\frac {V}{I}}={\frac {1}{sC}}}$ impedance of a capacitor
${\displaystyle {\frac {V}{I}}=R}$ impedance of a resistor
The rest is basically application of KVL and KCL. I don't know of a good circuit analysis page. The ones I found are a good overview, but they don't cover the process of how to analyze/transform a circuit and get meaningful results. I'm not sure if that even belongs here, maybe it should be part of a wikibook? Madhu 21:27, 21 April 2006 (UTC)
Thanks, I've actually found it in my text book. But I plan to add that information here. Perhaps a detailed explanation of it belongs in a wikibook - but an overview and simple reference would be good here. I'll make up the pics. Fresheneesz 21:03, 22 April 2006 (UTC)
"However, s-Domain impedances are valid for many more inputs than phasor impedances." Please give an example of such an input. Alfred Centauri 23:34, 28 April 2006 (UTC)

I had intended to re-direct a link to 'first order lag' to this article, but although Laplace transforms are intimately related to transfer functions, this has not proved as helpful as I had hoped. The article isn't bad as it stands, but a note in the introduction to the effect that the LT provides a means of transforming differential equations into algebraic equations might motivate the non-specialist to read further, rather than dismiss the whole thing as another example of Emperor's New Clothes. Gordon Vigurs 19:16, 14 May 2006 (UTC)

## Post's Inverse

Apparently this is a fairly recent discovery, and I haven't looked to see where it was published, but apparently the following holds:

If ${\displaystyle f(t)e^{-bt}\in L^{1}(0,\infty )}$, and ${\displaystyle F(s)={\mathcal {L}}\left\{f(t)\right\}}$, then at any point of continuity of f, ${\displaystyle f(t)=\lim _{k\to \infty }{\frac {(-1)^{k}}{k!}}\left({\frac {k}{t}}\right)^{k+1}F^{(k)}\left({\frac {k}{t}}\right)}$.

Of course, actually calculating that for arbitrary f is rather cumbersome, but it is better for complex-integralophobics. Confusing Manifestation 05:05, 16 May 2006 (UTC)

## Inverse Fourier transform of s/(a+s)

I keep comming across transfer functions that involve a term s/(a+s), for example a series RC circuit. Why is the transfer function not in the table here, and in fact not in any tables I've seen. I must be missing something key, but .. I don't know what that would be. Fresheneesz 23:29, 26 May 2006 (UTC)

The Laplace transform is a linear transform so generally, the transforms of sums of functions are not given in tables. The high pass filter transfer can be written as:
${\displaystyle 1-{\frac {\alpha }{s+\alpha }}}$
The inverse tranform is thus:
${\displaystyle \delta (t)-\alpha e^{-\alpha t}u(t)\,}$
We can also look at this way. We know the inverse transform for 1/(s + a) so the inverse transform of s/(s + a) is the time derivative of the inverse transform of 1/(s + a). Alfred Centauri 01:00, 27 May 2006 (UTC)
Hmm, how do you get ${\displaystyle 1-{\frac {\alpha }{s+\alpha }}}$ from s/(a+s)? Deriving the inverse transform of 1/(s+a) gives me -ae-at, so where does that delta funciton enter into the picture? Fresheneesz 11:14, 28 May 2006 (UTC)
Three ways to see this:
(1) Partial Fraction Expansion:
${\displaystyle {\frac {s}{s+\alpha }}=A+{\frac {B}{s+\alpha }}}$
set s = - α to see that B must be -α, then set s = 0 to see that A must be 1
(2) Recall that normalized LPF + HPF = 1:
${\displaystyle {\frac {1}{1+{\frac {s}{s_{0}}}}}+{\frac {\frac {s}{s_{0}}}{1+{\frac {s}{s_{0}}}}}=1}$
(3) Do the algebra...
${\displaystyle 1-{\frac {\alpha }{s+\alpha }}={\frac {s+\alpha }{s+\alpha }}-{\frac {\alpha }{s+\alpha }}={\frac {s+\alpha -\alpha }{s+\alpha }}={\frac {s}{s+\alpha }}}$
When you differentiate, don't forget the implicit u(t) in the one-sided Laplace transform...
${\displaystyle {\mathcal {L}}\left\{e^{-\alpha t}u(t)\right\}={\frac {1}{s+\alpha }}}$
${\displaystyle {\mathcal {L}}^{-1}\left\{{\frac {s}{s+\alpha }}\right\}={\frac {d}{dt}}(e^{-\alpha t}u(t))=e^{-\alpha t}\delta (t)-\alpha e^{-\alpha t}u(t)=e^{-\alpha 0}\delta (t)-\alpha e^{-\alpha t}u(t)=\delta (t)-\alpha e^{-\alpha t}u(t)\,}$
Alfred Centauri 11:51, 28 May 2006 (UTC)
wow, i never thought of doing PFE to that sort of expression. This is very useful thanks! One last question, did you set t = to 0 because the delta function is 0 everywhere else? Fresheneesz 20:09, 28 May 2006 (UTC)
That's correct. The delta function has what is called the sifting property in that it selects a particular value of any function that it is multiplied with. That's why the delta function is used to mathematically represent sampling. We have:
${\displaystyle f(t)\delta (t-t_{0})=f(t_{0})\delta (t-t_{0})\,}$
Alfred Centauri 22:11, 28 May 2006 (UTC)

Could someone post an example using a second order differential equation where the roots are complex? The sign changes associated with the real part of the root that are in the denominator of the partial fraction expansion are confusing and should be explained. Thanks. Statum (talk) 13:45, 25 January 2011 (UTC)

## Error in transform table ?

I think number 2 in the table is wrong ...

 delayed nth power with frequency shift  ==>  ${\displaystyle {\frac {t^{n}}{n!}}e^{-\alpha t}\cdot u(t-\tau )}$  ====>  ${\displaystyle {\frac {e^{-\tau s}}{(s+\alpha )^{n+1}}}}$ ||  ${\displaystyle s>0\,}$


It doesn't look right to me.

Amfortas

Isn't it supposed to be:
${\displaystyle {\frac {(t-\tau )^{n}}{n!}}e^{-\alpha (t-\tau )}\cdot u(t-\tau )}$ ====> ${\displaystyle {\frac {e^{-\tau s}}{(s+\alpha )^{n+1}}}}$ || ${\displaystyle s>0\,}$
Alfred Centauri 04:43, 5 June 2006 (UTC)
Yes , that should be it .... I'll change it right now Amfortas

## Notation u(t) for unit step in table is confusing

Hi, I just wanted to note, that the notation u(t) for the unit step is not very well chosen, because u(t) is also often used in electronics and control engineering for the input signal, which is often Laplace-transformed itself. I would suggest to use another name for the unit step, for example σ(t) or H(t) for the Heaviside-function. Your opinions?

I have to disagree. In my technical experience, I have encountered the unit step function as u(t) most of the time. I agree that this is troubling notation because it can be used for the input signal in a control systems setting, but in context I usually think it is clear. At most, I think we probably ought to clarify that u(t) is the unit step function in this context. I'm open to suggestion, however. -SocratesJedi | Talk 08:56, 29 January 2007 (UTC)

I agree with the first chap. As a mathematician, I have never seen the Heaviside function denoted by anything other than H(t). —Preceding unsigned comment added by 62.31.164.38 (talk) 21:41, 14 May 2011 (UTC)

In systems theory it is usually denoted by 1(t) which was probably Heaviside's notation.JFB80 (talk) 19:36, 15 May 2011 (UTC)

The u(t) notation is used in engineering. The notation, however, usually carries a (power of s) subscript to denote doublet, impulse, step, ramp, etc. IIRC, u0(t) is an impluse; u-1(t) is a step. That some people have never seen it says little. Glrx (talk) 19:48, 15 May 2011 (UTC)

Why not check up on a few standard books, e.g. Schaum's Outline. JFB80 (talk) 12:59, 16 May 2011 (UTC)

## erfc

Hello, in the table the last transform we have is:

${\displaystyle \mathrm {erf} (t)\cdot u(t)\leftrightarrow {e^{s^{2}/4}\operatorname {erfc} \left(s/2\right) \over s}}$

Where I'm fairly sure that since ${\displaystyle \operatorname {erfc} (z)=1-\operatorname {erf} (z)}$ we could express instead as:

${\displaystyle \mathrm {erf} (t)\cdot u(t)\leftrightarrow {e^{s^{2}/4}\left(1-\operatorname {erf} \left(s/2\right)\right) \over s}}$

I think this would be a better way of expressing this because it is already clear what erf(z) means in the original context of the table, but it is not clear what erfc(z) is unless the Error function article has already been read (and it seems problematic to link directly there lest erf(z) and erfc(z) be confused).

I will make this edit in a few days if no one objects, but I wanted to see if I an consensus existed as well. Please let me know your thoughts. Thanks. -SocratesJedi | Talk 08:53, 29 January 2007 (UTC)

I am boldly making this edit. Feel free to revert and discuss if you feel it is inappropriate. -SocratesJedi | Talk 22:01, 31 January 2007 (UTC)

## N-tilde

Under Differential equation example 1 we have:

${\displaystyle {\frac {dN}{dt}}+\lambda N=0}$

Next, we take the Laplace transform of both sides of the equation:

${\displaystyle \left(s{\tilde {N}}(s)-N_{o}\right)+\lambda {\tilde {N}}(s)\ =\ 0}$

where

${\displaystyle {\tilde {N}}(s)={\mathcal {L}}\{N(t)\}}$

The ${\displaystyle {\tilde {N}}(s)}$ notation was removed in favor of just keeping N(s) where N(t) represented time domain and N(s) represented s-domain signals to keep the notation simple. I reverted this to reinclude the tilde N because I think it's necessary to make it clear that N(t) and N(s) are not the same function with t --> s, but are entirely different (but related through the Laplace transform) functions. I think anyone familiar with the field would understand the N(s) notation, but I also think it would be very confusing to someone who didn't understand it thinking N(s) = N(t) where t --> s. Rather, I think

Usually I've seen this in books where they might write x(t) [lower case] and X(s) [upper case] to distinguish the functions. I recommend we stay with the tilde notation or switch to the n(t) N(s) type of notation (lower-case time/upper case laplace] for clarify and simplicity.

Could I get comments on this? I'd like to build consensus, if possible. -SocratesJedi | Talk 18:40, 2 February 2007 (UTC)

## Transformation Table

I think there should be an Re() everywhere around the s, in the ROC column.

I agree. It just doesnt make any sense to say s > a for a complex number s and a real number a. It is not possible to define a "consistent" ordering on the complex numbers. Anybody listening here? —Preceding unsigned comment added by 62.214.248.128 (talk) 15:25, 3 October 2007 (UTC)

Lantanov, your edit summary indicates that the text you have inserted into the intro is the intro from the Korn & Korn reference. Perhaps I misunderstand what you mean by is but if you've copied the intro verbatim here, that's a no-no. Alfred Centauri 14:32, 30 August 2007 (UTC)

It is not verbatim, I re-wrote it and shortened it, and you can check this. Besides, I have the K&K book only in Russian translation, so I had to re-translate the text back from Russian to English. And in any case: how inventive and original can one be when he quotes a mathematical definition in which every deviation has the risk of being erroneous? For instance, you mistook my user name by not checking the original.:) --Lantonov 15:32, 30 August 2007 (UTC)

No worries mate, just checking. Alfred Centauri 16:59, 30 August 2007 (UTC)

It's ok. BTW, I slept over another idea for reshuffling intro. See if it pans out. Have a good day editing.--Lantonov 05:08, 31 August 2007 (UTC)

## Confusion at bottom of table

The explanatory notes at the bottom of the table states that "In general, the ROC for causal systems is not the same as the ROC for anticausal systems." Can someone explain what ROC stands for? —Preceding unsigned comment added by JonathonReinhart (talkcontribs) 16:03, 3 December 2007 (UTC)

The article explains it! Laplace transform#Region of convergence Oli Filth(talk) 19:19, 3 December 2007 (UTC)

## fourier

I removed a statement to the effect of laplace x-form is in the branch of mathematics known as fourier analysis; I looked up the history, it was originally functional analysis. —Preceding unsigned comment added by Blablablob (talkcontribs) 19:07, 21 June 2008 (UTC)

what is bi-laterial laplace transforms step function?80.191.172.10 (talk) 13:00, 29 April 2009 (UTC)

using laplace transformation,show that integral of sint/2*dt from 0 to infinity=pie/2 —Preceding unsigned comment added by 175.40.52.234 (talk) 15:05, 17 March 2010 (UTC)

## Emphasis on Positive Infinity?

                 Pray believe me dear friend(s) -
J.A.G.
(Physics, Physical Sciences, Sciences, and Mathematics).  —Preceding unsigned comment added by 108.10.51.252 (talk) 19:34, 14 April 2010 (UTC)


Reply. WP:TLDR, but I've changed the article to use ${\displaystyle \infty }$ for ${\displaystyle +\infty }$ consistently throughout. I believe this is what you wanted. Sławomir Biały (talk) 22:48, 14 April 2010 (UTC)

## Moments

My attempts at a compromise wording to bring the relationship with moments again to the lead was reverted. Now, obviously, there is a relationship between the Laplace transform and moments, either by an exponential change of variables, or by means of the generating function. The issue therefore is how to address this issue constructively in a manner appropriate to the general discussion in the lead of the article. I am open to suggestions about how to proceed, but I disagree with outright removal of content from the general discussion on the grounds that it isn't sufficiently precise. Try to improve the wording if it isn't to your liking. Sławomir Biały (talk) 22:04, 21 July 2010 (UTC)

My issue with this statement is that it doesn't belong in the introduction, as it's confusing for someone who is not already intimately familiar with the Laplace transform; should probably be in the section about moments. Let me try to move it there with more precise wording. To say that the Laplace transform "resolves a function in its moments" is just so outrageously imprecise as to be misleading. There is a relationship, but a very tenuous one, and it doesn't belong at the top of the article. Danpovey (talk) 22:38, 21 July 2010 (UTC)

Actually, after trying to figure out how to make this statement precise I was unable to because the statement is just plain wrong. But I won't continue an edit war. I will try to persuade you instead as you are obviously a person of some mathematical sophistication who can appreciate my point. The Laplace transform does not resolve a function into its moments because for no value of s does L(s) correspond to a moment; I think the person who orginally posted that statement must have thought that was the case (e.g. they thought it was the Mellin transform). There are further transformations you can do that would reveal the moments, but that would define a different integral transform that is different from the Laplace transform (in fact, for any integral transform we can apply its inverse and then the Mellin transform and get the moments, so this isn't unique to the Laplace transform). The exponential change of variables is not relevant because then we are not talking about the moments of the original function but the transformed version. I think there should be some place for removing content when the content is inaccurate, rather than bending over backwards to interpret in such a way that it's correct. Danpovey (talk) 22:49, 21 July 2010 (UTC)

I'm not absolutely married to the idea of mentioning moments in the lead. However, I do feel as though some effort should be made to communicate what the Laplace transform "is" rather than succumb into the trap that most elementary treatments fall into: treating it as a purely operational gadget. I also feel that it is important to draw an intuitive distinction between the Fourier transform and the Laplace transform. Referring to Widder's treatment in the 1945 Notices article "What is the Laplace transform?", he appears to favor the view that the Laplace transform generalizes the classical power series (more directly the Dirichlet series) by replacing the summation with integration. The Fourier transform, by contrast, is the "continuous spectrum" analog of a trigonometric (Fourier) series. Perhaps this is worth trying to work into the lead here. At any rate, in that same article, Widder does go on to make the connection with moments very precise (using the exponential change of variables). It does seem to me worth trying to get this in the lead, because it seems likely to me at least that thinking about the Laplace transform in terms of moments is more intuitive than thinking about it as a generalization of Dirichlet's series. Sławomir Biały (talk) 16:36, 22 July 2010 (UTC)
I agree that it would be good to give some kind of intuition. I originally tried to think of something but was not able to come up with something that is both brief and precise. One way is to view it as a generalization of the Fourier transform where the coefficient of x in the exponential is a general complex number, not purely imaginary. Perhaps you would be better qualified than me to think of an alternative explanation. Danpovey (talk) 20:20, 22 July 2010 (UTC)

I thought I had a handle on Laplace transforms, but it took some reading to figure out that 'resolves a function into it's moments' statement. But I Absolutely agree with Sławomir Biały, on having a "what it is" in the intro.
Here is the explanation I've retained : It is a leaky bucket (or any other low pass filter). Given a time history, f(t), the Laplace transform tells you how much is in the bucket now, as a function of the leakage rate (s).
Similarly, it can be put in finance terms: given a (continuous time) future income stream, f(t), the Laplace transform tells you the present value of the income stream, as a function of the interest rate (s).
Sukisuki (talk) 10:28, 30 July 2010 (UTC), fixed interest rate

## Remove howto box on examples section

It's well known in certain fields of instruction that the reader can not confidently understand the factual presentation without working some pertinent examples. Witness volumes from Knuth with more space devoted to exercises and answers to exercises than expository text.

The problem with the examples section is that it seems to invite expansion beyond serving the purpose of permitting the reader to double check his/her factual comprehension.

What is needed here instead of deprecating necessary corroborative material is a notice that the examples serve a corroborative purpose and that this section is not there to grow into a giant cheat sheet.

The reason I removed the howto box is that it deflates the spirit of collaboration and draws more attention to policy than the effectiveness of the article at presenting the material. Much of the original contribution at Wikipedia was fueled by the joy of escaping this kind of pettiness. I can't see how reworking the exposition here would make the section less "how to". The only logical outcome is to remove the example section completely, which would damage the article's intelligibility. Deprecating labels should be bandied with extreme care.

Rules aren't much use if no one remains to follow them. — MaxEnt 21:46, 12 September 2010 (UTC)

I agree with the removal. I'm not sure the alternate disclaimer is needed. These examples are not in the style of a howto, and provide a good practical illustration of the subject matter in a manner appropriate to an encyclopedia article. Sławomir Biały (talk) 00:15, 13 September 2010 (UTC)

## Laplace transform of f(t)/t^n , where n is a positive integer start from 2

It is known that for positive integer ${\displaystyle n>1}$,

${\displaystyle {\mathcal {L}}\left\{{\frac {f(t)}{t^{n}}}\right\}=\int _{s}^{\infty }\int _{\sigma _{1}}^{\infty }\cdots \int _{\sigma _{n-1}}^{\infty }F(\sigma _{n})\,d\sigma _{n}\cdots \,d\sigma _{2}\,d\sigma _{1}}$

Can we borrow the idea from Cauchy formula for repeated integration and have this simplification?

${\displaystyle {\mathcal {L}}\left\{{\frac {f(t)}{t^{n}}}\right\}}$
${\displaystyle =\int _{s}^{\infty }\int _{\sigma _{1}}^{\infty }\cdots \int _{\sigma _{n-1}}^{\infty }F(\sigma _{n})\,d\sigma _{n}\cdots \,d\sigma _{2}\,d\sigma _{1}}$
${\displaystyle =(-1)^{n}\int _{\infty }^{s}\int _{\infty }^{\sigma _{1}}\cdots \int _{\infty }^{\sigma _{n-1}}F(\sigma _{n})\,d\sigma _{n}\cdots \,d\sigma _{2}\,d\sigma _{1}}$
${\displaystyle ={\frac {(-1)^{n}}{(n-1)!}}\int _{\infty }^{s}\left(s-\sigma \right)^{n-1}F(\sigma )\,d\sigma }$
${\displaystyle ={\frac {(-1)^{n-1}}{(n-1)!}}\int _{s}^{\infty }\left(s-\sigma \right)^{n-1}F(\sigma )\,d\sigma }$

Doraemonpaul (talk) 00:51, 27 September 2010 (UTC)

## Laplace transform of f(t)g(t)

In fact ${\displaystyle {\mathcal {L}}\left\{f(t)g(t)\right\}={\frac {1}{2\pi i}}\int _{c-i\infty }^{c+i\infty }F(\sigma )G(s-\sigma )\,d\sigma \ }$.

Proof:

${\displaystyle {\mathcal {L}}\left\{f(t)g(t)\right\}}$
${\displaystyle =\int _{0}^{\infty }e^{-st}f(t)g(t)\,dt}$
${\displaystyle =\int _{0}^{\infty }e^{-st}{\frac {1}{2\pi i}}\int _{c-i\infty }^{c+i\infty }e^{\sigma t}F(\sigma )\,d\sigma \,g(t)\,dt}$
${\displaystyle ={\frac {1}{2\pi i}}\int _{0}^{\infty }\int _{c-i\infty }^{c+i\infty }e^{\sigma t-st}F(\sigma )g(t)\,d\sigma \,dt}$
${\displaystyle ={\frac {1}{2\pi i}}\int _{c-i\infty }^{c+i\infty }\int _{0}^{\infty }e^{(\sigma -s)t}F(\sigma )g(t)\,dt\,d\sigma \ }$
${\displaystyle ={\frac {1}{2\pi i}}\int _{c-i\infty }^{c+i\infty }F(\sigma )\int _{0}^{\infty }e^{-(s-\sigma )t}g(t)\,dt\,d\sigma \ }$
${\displaystyle ={\frac {1}{2\pi i}}\int _{c-i\infty }^{c+i\infty }F(\sigma )G(s-\sigma )\,d\sigma \ }$

Doraemonpaul (talk) 00:58, 6 October 2010 (UTC)

## New entry to table

The new entry to the table is redundant with the entry on the delayed nth power frequency shift, which I think is clearer. Also, the new entry tries to do too much, by bringing in partial fractions and indicating this by a perhaps not so easily recognizable limit formula. This doesn't seem to be the sort of information that is suitable for a table entry. There is an example that discusses partial fractions already, and the article partial fractions goes into much more detail about how to find partial fraction expansions. Someone who is unfamiliar with partial fractions is likely to be totally mystified by this table entry. Someone already familiar with them is likely to have better ways to compute the decomposition than by using the limit formula. Sławomir Biały (talk) 10:53, 5 November 2011 (UTC)

## Complex domain of integration in multiplication identity

I've seen it a million times before, but now that I actually need to calculate one I can't find any reference to this particular meaning of complex integration. Nor does Wolfram Alpha help me out any. For one, why does the transform use the limit as T goes to infinity instead of simply integrating over c-i∞ to c+i∞? And does that mean I can use the fundamental theorem of calculus to evaluate the functions? For example, I have a term ${\displaystyle \int _{c-i\infty }^{c+i\infty }\sigma ^{-3/2}d\sigma }$. Can I evaluate it as: ${\displaystyle \lim _{T\rightarrow \infty }((c+iT)^{-1/2}-(c-iT)^{-1/2})=0}$? ᛭ LokiClock (talk) 02:53, 13 December 2011 (UTC)

That's right. The integral ${\displaystyle \int _{c-i\infty }^{c+i\infty }\sigma ^{-3/2}d\sigma \,}$ is a contour integral. As long as the contour contour lies entirely in the half-plane ${\displaystyle \Re (\sigma )>0\,}$, it does not cross a branch cut of the square root. The fundamental theorem of calculus (actually, a version of the fundamental theorem of line integrals) can be applied as you have done, since the primitive ${\displaystyle -2\sigma ^{-1/2}}$ (with the principal value of the square root) is valid along the contour. The only situation where you would need to worry is if two different primitives would be needed, depending on what part of the contour you are on. This happens precisely when you cross a branch cut of a multiple-valued function such as the square root. Sławomir Biały (talk) 12:23, 14 December 2011 (UTC)
So for a lot of these integrals, I end up with a function of c, the real part, besides s, the frequency. How do I get rid of c? And if I do end up crossing over a branch cut, what do I do with result? I think I did this, because I have solutions with ${\displaystyle {\sqrt[{4}]{-1}}}$ factors and things like ${\displaystyle {\sqrt {p}}{\sqrt {i/p}}}$. ᛭ LokiClock (talk) 20:28, 14 December 2011 (UTC)
See Bromwich integral for first Q. Modify the contour to go around the branch cut rather than cross it. Glrx (talk) 00:49, 16 December 2011 (UTC)

## Confusing boxes (frames)

Those entries in the table in the Properties and theorems section that contain integral symbols are incorrectly formatted by default to display the expressions using boxes or frames, as viewed in Firefox 8.0. I haven't learned the language used to represent such expressions, so I can't fix this problem. Integral expressions appearing outside of this table are displayed correctly, and in a nice, distinctive, bold italic font. Probably there is an entry in WP-space that describes how to use this language. Can anyone help? David Spector (talk) 21:42, 29 December 2011 (UTC)

I'm not sure I understand what you mean. Do you know how to take a screenshot? The equations should be rendered by default as embedded images, so it doesn't make any sense why they would look good in one place and not another. Sławomir Biały (talk) 22:25, 29 December 2011 (UTC)

I can understand your confusion if you don't see the problem. It happens because a different sublanguage is used at the two places (look at the wiki source). I've created a temporary screenshot where you can see the boxes. This might be a Firefox bug. David Spector (talk) 11:38, 30 December 2011 (UTC)

Now I see. I was able to reproduce this issue by going into My preferences and selecting under the Math header HTML if possible or else PNG. You can fix this issue by selecting instead HTML if very simple or else PNG (which I thought was the default, but maybe that's changed). Anyway, obviously there is an issue with the Wikipedia software (or possibly a limitation inherent to CSS.) If necessary, these formulas can be fixed for all users by adding to the end of each of them (inside the [itex] tags) a "\,\!" (without the quotes) Sławomir Biały (talk) 11:57, 30 December 2011 (UTC)

Thanks for the excellent analysis. Since this seems to be a real bug, I am submitting it as a bug. Will report status here. David Spector (talk) 15:06, 30 December 2011 (UTC)