Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

**gAr****Member**- Registered: 2011-01-09
- Posts: 3,478

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

Anyway, the output is not useful.

"Believe nothing, no matter where you read it, or who said it, no matter if I have said it, unless it agrees with your own reason and your own common sense" - Buddha?

"Data! Data! Data!" he cried impatiently. "I can't make bricks without clay."

Offline

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 84,454

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

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

Offline

**gAr****Member**- Registered: 2011-01-09
- Posts: 3,478

Yes, bur I'll not try that!

I checked numerical results for s=1, they didn't match.

"Believe nothing, no matter where you read it, or who said it, no matter if I have said it, unless it agrees with your own reason and your own common sense" - Buddha?

"Data! Data! Data!" he cried impatiently. "I can't make bricks without clay."

Offline

**zetafunc.****Guest**

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****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 84,454

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

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

Offline

**zetafunc.****Guest**

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

**gAr****Member**- Registered: 2011-01-09
- Posts: 3,478

Hmmm, okay.

It's my EOD, see you both later.

Good discussion.

"Believe nothing, no matter where you read it, or who said it, no matter if I have said it, unless it agrees with your own reason and your own common sense" - Buddha?

"Data! Data! Data!" he cried impatiently. "I can't make bricks without clay."

Offline

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 84,454

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

See you later gAr!

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

Offline

**zetafunc.****Guest**

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****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 84,454

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?

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

Offline

**zetafunc.****Guest**

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****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 84,454

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

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

Offline

**zetafunc.****Guest**

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.****Guest**

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****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 84,454

Hi;

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

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

Offline

**zetafunc.****Guest**

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

**zetafunc.****Guest**

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****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 84,454

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!

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

Offline

**zetafunc.****Guest**

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

**zetafunc.****Guest**

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****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 84,454

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.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

Offline

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 84,454

-- 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.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

Offline

**zetafunc.****Guest**

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****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 84,454

Hi;

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

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

Offline

**zetafunc.****Guest**

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...