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

You are not logged in.

- Topics: Active | Unanswered

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 107,810

That is very true. So if we can not show analytically that it converges and we can not even show it empirically then how?

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

Offline

It may be possible analytically with a particular change of co-ordinates. I'm going to talk to my supervisor in a couple of hours to find out how to do it.

**LearnMathsFree: Videos on various topics.New: Integration Problem | Adding FractionsPopular: Continued Fractions | Metric Spaces | Duality**

Offline

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 107,810

Okay, good luck. Let me know what he says.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

Offline

He said my method was wrong but somehow I ended up with the right integral, except the limits are R^d rather than [0,2pi)^d.

**LearnMathsFree: Videos on various topics.New: Integration Problem | Adding FractionsPopular: Continued Fractions | Metric Spaces | Duality**

Offline

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 107,810

R^d

I am sorry, I am not following that. What does that mean for Mathematica?

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

Offline

So if d = 2 it would be a double integral such as the one in post #1 but without the sum.

**LearnMathsFree: Videos on various topics.New: Integration Problem | Adding FractionsPopular: Continued Fractions | Metric Spaces | Duality**

Offline

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 107,810

That might be a lot easier to do. Anyway, did he give some hint as to what the value of the integral will be. Smaller than 1?

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

Offline

He has no idea. I think I did try this integral once but I can't remember what the results were, it might be useful to investigate for smaller dimensions.

New: Integration Problem | Adding Fractions

Popular: Continued Fractions | Metric Spaces | Duality

Offline

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 107,810

Okay, post your code if you get some.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

Offline

Managed to show the integral converges in all dimensions remains unknown. That is, the integral

converges in those dimensions.

New: Integration Problem | Adding Fractions

Popular: Continued Fractions | Metric Spaces | Duality

Offline

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 107,810

The integral from post #3?

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

Offline

Yeah, but without the sum in front. (And we integrate over all of R^d instead of [1,infinity)^d.)

I suspect that the integral in the case d = 2 does not converge, though I've yet to prove that concretely.

*Last edited by zetafunc (2016-11-01 00:17:13)*

New: Integration Problem | Adding Fractions

Popular: Continued Fractions | Metric Spaces | Duality

Offline

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 107,810

Why do you think that?

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

Offline

I can't say anything concrete, but I think I might be able to show that the integral is bounded below by something which is divergent.

New: Integration Problem | Adding Fractions

Popular: Continued Fractions | Metric Spaces | Duality

Offline

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 107,810

That would be a good way to do it.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

Offline

I've posted a related question on MO. Received 1 upvote so far, but no responses. My supervisor will not be able to help for the next fortnight or so because he is going to the US.

The question can be found **here**.

Here is a simplified version of the same question on MathSE: http://math.stackexchange.com/questions … -variables

*Last edited by zetafunc (2016-11-01 09:33:48)*

New: Integration Problem | Adding Fractions

Popular: Continued Fractions | Metric Spaces | Duality

Offline

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 107,810

I upvoted it too but I do not see any responses yet.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

Offline

Thanks. The lack of responses on MO is not a good sign -- it means that the only one who I can get help from is my supervisor.

New: Integration Problem | Adding Fractions

Popular: Continued Fractions | Metric Spaces | Duality

Offline

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 107,810

Are you sure he can answer the question?

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

Offline

Yes, because he tried to show me a few weeks ago but he didn't go into detail. Essentially I don't understand how to take the Cauchy product of series whose indices vary over any arbitrary rational lattice. A simplified version of what I need to know is asked here, but I also have received no responses: http://math.stackexchange.com/questions … -variables

New: Integration Problem | Adding Fractions

Popular: Continued Fractions | Metric Spaces | Duality

Offline

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 107,810

But is that double integral correct?

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

Offline

Which double integral?

New: Integration Problem | Adding Fractions

Popular: Continued Fractions | Metric Spaces | Duality

Offline

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 107,810

The one in the other thread?! Excuse me, but I thought both threads were essentially talking about the same or almost the same one?!

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

Offline

The integrands are the same, but the integral in this thread ranges over [0,2pi)^2, whilst the other one ranges over all of R^2.

New: Integration Problem | Adding Fractions

Popular: Continued Fractions | Metric Spaces | Duality

Offline

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 107,810

I remember the code I was looking at:

```
Sum[Abs[Integrate[((x^2 + y^2)^(-1/2))*(((b - x)^2 + (c - y)^2)^(-1/2))*
BesselJ[1, k*Sqrt[x^2 + y^2]]*
BesselJ[1, k*Sqrt[(b - x)^2 + (y - c)^2]], {x, 0, 2*Pi}, {y, 0, 2*Pi}]^2], {b, 1, Infinity}, {c, 1, Infinity}]
```

I remember that it did not seem to be converging. I asked if your supervisor knew what it converged to. I do not remember your answer...

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

Offline