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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

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.**

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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: 109,606

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.**

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

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.**

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

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.**

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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
- Posts: 109,606

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.**

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

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.**

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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
- Posts: 109,606

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.**

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

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.**

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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 is a simplified version of the same question on MathSE:

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

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

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.**

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

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.**

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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:

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

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.**

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

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.**

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

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.**

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