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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,467

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

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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.

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,467

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

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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.

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,467

R^d

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

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

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
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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?

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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.

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
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Okay, post your code if you get some.

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

converges in those dimensions.

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
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The integral from post #3?

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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)*

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
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Why do you think that?

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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.

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,467

That would be a good way to do it.

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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)*

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
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I upvoted it too but I do not see any responses yet.

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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.

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
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Are you sure he can answer the question?

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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

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
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But is that double integral correct?

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Which double integral?

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
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The one in the other thread?! Excuse me, but I thought both threads were essentially talking about the same or almost the same one?!

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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.

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
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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...

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