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Can anyone determine whether or not the following integral converges?

where is large (but independent of and ), and denotes the standard Euclidean norm on i.e. and denotes the Bessel function of the first kind.I would like for this integral to be at least Equivalently, this integral can be written (less compactly) as:**LearnMathsFree: Videos on various topics.New: Integration Problem | Adding FractionsPopular: Continued Fractions | Metric Spaces | Duality**

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
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Is this symmetric around either axis?

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No, it shouldn't be.

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**bobbym****bumpkin**- From: Bumpkinland
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You say p>>0, any idea about the size of it?

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in this case represents the radius in the generalised circle problem -- we take it to approach infinity, i.e. it is arbitrarily large.

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**bobbym****bumpkin**- From: Bumpkinland
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When we encountered this integral before we were concerned about the many singularities. It is possible that this double integral does not exist because of that. What if that is the case?

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There is a singularity at (0,0) and another at (b,c), yes. But in dimensions 3 and higher, it can be shown that the integral converges (the terms have slightly different powers, which depend on d, but the locations of the singularities are the same). To prove convergence for d > 2, we can cut R^d into overlapping pieces -- but we can't do that for d = 2 because we encounter a 1-dimensional integral of r^[(d-3)/2] through r = 0. (And that integral converges iff d > 3.) I can write up the details, but the case d = 2 may be more delicate.

If the double integral diverges, then we will be forced not to estimate the sum with an integral and try to bound the sum in a different way.

*Last edited by zetafunc (2016-11-02 09:42:22)*

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**bobbym****bumpkin**- From: Bumpkinland
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Since I am not getting anywhere fast with that integral perhaps the sum might be better.

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The sum was:

where (\substack does not seem to work on here.)We could also use the asymptotic bound (That was the bound I used for d > 2.) Doing that gives us the integral:or, as a double integral,

*Last edited by zetafunc (2016-11-02 23:02:16)*

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**bobbym****bumpkin**- From: Bumpkinland
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Hi;

b and c will be a problem for M as before.

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I agree. It may be more fruitful to integrate around small neighbourhoods of b,c and shrink those neighbourhoods around the singularities.

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**bobbym****bumpkin**- From: Bumpkinland
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What are b and c again? Reals? Integers?

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Technically they are elements of some arbitrary rational lattice, but we can treat them as integers here without loss of generality.

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**bobbym****bumpkin**- From: Bumpkinland
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So that means an infinite number of singularities.

I need to eat something and that means I have to cook it up first. Sorry, to break off but I am starving see you later.

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No -- when we estimate the sum with an integral, we can treat b,c as being fixed.

OK, enjoy your meal, and see you later!

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**bobbym****bumpkin**- From: Bumpkinland
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The double integral you posted in post #9,

that one does not converge.

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Interesting -- how do you know?

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**bobbym****bumpkin**- From: Bumpkinland
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I do not know but M does not think so.

```
ClearAll[b, c];
Integrate[Integrate[1/((x^2 + y^2)^(-3/4) ((b - x)^2 + (c - y)^2)^(-3/4)), {x, -\[Infinity], \[Infinity]}], {y, -\[Infinity], \[Infinity]}]
```

If that is the correct integral then it is possible that M might be wrong but it is at least 10 to 1 against.

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**anonimnystefy****Real Member**- From: Harlan's World
- Registered: 2011-05-23
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bobbym wrote:

The double integral you posted in post #9,

that one does not converge.

There shouldn't minuses in those exponents.

*Last edited by anonimnystefy (2016-11-02 20:32:40)*

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anonimnystefy wrote:

There shouldn't minuses in those exponents.

Thanks for pointing that out -- I've removed them from my post.

bobbym wrote:

I do not know but M does not think so.

`ClearAll[b, c]; Integrate[Integrate[1/((x^2 + y^2)^(-3/4) ((b - x)^2 + (c - y)^2)^(-3/4)), {x, -\[Infinity], \[Infinity]}], {y, -\[Infinity], \[Infinity]}]`

If that is the correct integral then it is possible that M might be wrong but it is at least 10 to 1 against.

I am running it now in M with the negative signs removed.

*Last edited by zetafunc (2016-11-02 20:50:09)*

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**bobbym****bumpkin**- From: Bumpkinland
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That will make a big difference. Howdy anonimnystefy!

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If we get something which looks to be divergent, all is not lost: it just means that we lose too much information by bounding the Bessel functions in modulus. In which case, it may be necessary to use the asymptotics of the Bessel functions, which can get very messy. (But it might be necessary in order to take advantage of the positive-negative cancellation that occurs.)

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**bobbym****bumpkin**- From: Bumpkinland
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Hi;

The problem is that just like Mathematica, we find that if an expression is too complicated it is hard to do anything analytical to it.

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We may need to take the integral at the bottom of post #1, set rho = 1000, say, and sum over all b,c such that b,c are non-zero and b+c=0. If we can find roughly what that should be for some ranges of b,c then we might be able to guess what the answer should be by varying rho.

*Last edited by zetafunc (2016-11-20 22:02:23)*

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