Laplace Transform of tan(t)

gAr · 2011-08-12 06:03:54

Yes, that's what I observed too...

Anyway, the output is not useful.

bobbym · 2011-08-12 06:16:03

Also there is a Inverse Laplace Transform that could be tried on that answer. 5 to 1 says it does not return tan(t).

gAr · 2011-08-12 06:25:37

Yes, bur I'll not try that!
I checked numerical results for s=1, they didn't match.

zetafunc. · 2011-08-12 06:27:33

alexfloo, Physics Forums wrote:

Let me start by saying I've never seen a Cauchy principal value integral. All my comments are from a few minutes of research.
The Cauchy principal value integral is essentially a method of assigning mathematically useful values to divergent improper integrals. Wikipedia, for instance, defines the Laplace transform in terms of a Lebesgue integral. The key point really is that the Lebesgue and the Cauchy principle value are *not* the same, so in the traditional sense, what Wolfram|Alpha computed is not actually the Laplace transform of the tangent, which doesn't exist, because the Lebesgue integral diverges. However, (it is likely that - again, I've never seen this integral before) the function Wolfram computed has many of the useful properties of the Laplace Transform, and is therefore a reasonable substitute.

Posted here if you might find it useful. If this is true then the Inverse Laplace transform of that function won't return tan(t)...

bobbym · 2011-08-12 06:34:46

That is almost like Borel summation. Some Sums that diverge ( Ramanujan sums ) converge in a Borel sense.

zetafunc. · 2011-08-12 06:40:29

What I want to know though is how you can change the integral so that it does not diverge...

gAr · 2011-08-12 06:41:23

Hmmm, okay.

It's my EOD, see you both later.
Good discussion.

bobbym · 2011-08-12 06:41:37

Simplest way might be by truncating it. But why? It would not be the Laplace Transform.

See you later gAr!

zetafunc. · 2011-08-12 06:45:08

You mean a Laplace transform such that the transform is valid for some interval [a,b] provided that the interval does not contain any singularities? That might work but I agree it would not be the Laplace transform. Is there a way you can do it without including any singularities?

bobbym · 2011-08-12 06:48:20

Hi zetafunc.;

Not for tan(t) which has an infinite amount of them. You would have to go from 0 to pi / 2. Then it might converge, but what does it represent?

zetafunc. · 2011-08-12 06:48:37

I have a whiteboard at home and on it still has written the result of performing integration by parts twice on tan(t) defined in terms of e and i.

What I wanted to ask you earlier was if i can be treated the same way as a real constant can? In other words, can you say that ∫if(t)dt = i∫f(t)dt?

bobbym · 2011-08-12 06:49:52

Yes, move the i right through it is a constant.

zetafunc. · 2011-08-12 06:51:34

Do you know anything about Cauchy principal values? As you said it is similar to Borel sums where you can turn a diverging summation into a converging one that can hold true for all values.

zetafunc. · 2011-08-12 06:54:43

Thanks, I can get on with performing another IBP now now that I know factoring out the i is okay...

Sorry, I am just a bit new to Laplace transforms, I feel bad because I don't really have a conceptual understanding of what I'm doing so to speak; all I know how to do is use it to solve ordinary differential equations and to turn a function of t into a function of s. I've read that it can change it from the time-domain into the frequency-domain but I don't know what that means or how that makes sense.

bobbym · 2011-08-12 06:55:03

Hi;

I am sorry but I do not. Borel sumation is something physicists use more, I think?

zetafunc. · 2011-08-12 06:56:58

Borel summations are used by physicists? Hmm... what would they use them for? (sorry for going off-topic)

zetafunc. · 2011-08-12 07:01:24

Never mind, found it:

Wikipedia wrote:

Borel summation finds application in perturbation expansions in quantum field theory. In particular in 2-dimensional Euclidean field theory the Schwinger functions can often be recovered from their perturbation series using Borel summation (Glimm & Jaffe 1987, p. 461). Some of the singularities of the Borel transform are related to instantons and renormalons in quantum field theory (Weinberg 2005, 20.7).

bobbym · 2011-08-12 07:02:02

As near as I understand it was convenient for certain series that diverged to be convergent for quantum mechanics. I do not know any more about that. Except that using Borel summation a sequence of positive numbers when added, can sum to a negative number!

zetafunc. · 2011-08-12 07:04:43

Thanks for the reply -- are you saying that Borel summation can mean a sequence of positive numbers can sum to a negative number?

zetafunc. · 2011-08-12 07:05:48

Maybe I should post the problem on lots of other forums, no one seems to be responding on PF. Maybe someone knows how to apply Cauchy principal values here...

bobbym · 2011-08-12 07:14:42

You could do that but remember you can not force this one to have the answer you want. It will always come up the same, divergent.

bobbym · 2011-08-12 07:15:34

-- are you saying that Borel summation can mean a sequence of positive numbers can sum to a negative number?

Ramanujan did a couple of sums like that.

zetafunc. · 2011-08-13 03:37:52

Still no luck with converting it to a Cauchy principal value integral... I'm not sure how to set it up because tan(t) is periodic with singularities every interval of π. Do you think I could work backwards from W|A's result and see where that gets if I calculate the inverse Laplace transform of that by hand? I know I shouldn't completely trust Wolfram Alpha's result but the direction that this problem has taken interests me as I didn't think you could find a way to define the Laplace transform for a non-piecewise continuous periodic function...

bobbym · 2011-08-13 03:47:03

Hi;

You could try that but I think the Alpha answer is gibberish. What if I am right?

zetafunc. · 2011-08-13 03:54:39

Well, I'll never know until I try... I don't know how to use a computer to do it either, and taking the inverse Laplace transform of all those digamma functions looks sticky.

I just tried integration by parts 4 times using the definition of tan in terms of e and i and it seems to follow a similar pattern to just taking the improper integral of e[sup]-st[/sup]tan(t)dt. That would make sense. Unless my IBP is wrong...

Math Is Fun Forum

#26 2011-08-12 06:03:54

Re: Laplace Transform of tan(t)

#27 2011-08-12 06:16:03

Re: Laplace Transform of tan(t)

#28 2011-08-12 06:25:37

Re: Laplace Transform of tan(t)

#29 2011-08-12 06:27:33

Re: Laplace Transform of tan(t)

#30 2011-08-12 06:34:46

Re: Laplace Transform of tan(t)

#31 2011-08-12 06:40:29

Re: Laplace Transform of tan(t)

#32 2011-08-12 06:41:23

Re: Laplace Transform of tan(t)

#33 2011-08-12 06:41:37

Re: Laplace Transform of tan(t)

#34 2011-08-12 06:45:08

Re: Laplace Transform of tan(t)

#35 2011-08-12 06:48:20

Re: Laplace Transform of tan(t)

#36 2011-08-12 06:48:37

Re: Laplace Transform of tan(t)

#37 2011-08-12 06:49:52

Re: Laplace Transform of tan(t)

#38 2011-08-12 06:51:34

Re: Laplace Transform of tan(t)

#39 2011-08-12 06:54:43

Re: Laplace Transform of tan(t)

#40 2011-08-12 06:55:03

Re: Laplace Transform of tan(t)

#41 2011-08-12 06:56:58

Re: Laplace Transform of tan(t)

#42 2011-08-12 07:01:24

Re: Laplace Transform of tan(t)

#43 2011-08-12 07:02:02

Re: Laplace Transform of tan(t)

#44 2011-08-12 07:04:43

Re: Laplace Transform of tan(t)

#45 2011-08-12 07:05:48

Re: Laplace Transform of tan(t)

#46 2011-08-12 07:14:42

Re: Laplace Transform of tan(t)

#47 2011-08-12 07:15:34

Re: Laplace Transform of tan(t)

#48 2011-08-13 03:37:52

Re: Laplace Transform of tan(t)

#49 2011-08-13 03:47:03

Re: Laplace Transform of tan(t)

#50 2011-08-13 03:54:39

Re: Laplace Transform of tan(t)

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