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I am interested in the following sum:

where is a constant, and denotes the Bessel function of the first kind. This is a special case of a more general sum I'd like to consider, with :where denotes the standard Euclidean norm on , i.e. . I've tried computing this in a few different ways using Mathematica. The first way to get rid of the Bessel functions is to use the bound for some constant depending on . However, this may be dangerous, since by taking absolute values of the Bessel functions, we lose the ability to take advantage of any positive-negative cancellation that occurs. Mathematica doesn't seem to be able to compute the integral for , although one can get a numerical result by replacing with numbers instead (but since I want to sum over , this is not entirely helpful). We can also use the asymptotic expansions of , and in the case of , the Bessel function has a very simple closed form:and in general, there are also finite sum expansions for half-integer values of Bessel functions. But I haven't managed to get any semblance of a result, using any of these methods.

Can anyone find a way of computing this sum for , or perhaps a better method for general ?Offline

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

Hi;

Can you please show me your M code?

**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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The input for that sum for d = 2 (restricted to positive integers) is

```
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}]
```

*Last edited by zetafunc (2016-10-09 03:24:16)*

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

I am getting a syntax error out of that. Please check your brackets.

**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 is a very tough problem and may not have a closed form. What kind of answer are you looking for?

**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

Perhaps I can get something out of this. First, I would like to test empirically your assertion that it converges.

**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 is possible it may diverge, but if that is the case, it means that I have done something wrong (it should be true that the above sum is actually ).

*Last edited by zetafunc (2016-10-09 03:50:57)*

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

One problem is that I do not know what k is. Can you say something about k?

**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

So, you want k to be another free variable or I am hoping we can at least bound 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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We can use the bound mentioned in post #1 but it may cause the integral to either converge or diverge. If d = 2 then, after bounding the Bessel functions, one gets:(I left out the as neither the sum nor integral depend on it now, if we choose to bound the Bessel functions in this way.)

*Last edited by zetafunc (2016-10-09 04:08:16)*

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

Can you write that up in Mathematica speak?

**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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Here it is:

```
Sum[(Integrate[((x^2 + y^2)^(-3/4))*(((b - x)^2 + (c - y)^2)^(-3/
4)), {x, 0, 2*Pi}, {y, 0, 2*Pi}]^2), {b, 1, Infinity}, {c, 1,
Infinity}]
```

If it is possible to show this converges then we are done.

*Last edited by zetafunc (2016-10-09 04:23:06)*

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

Bracket missing.

**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

Let me see what can be done with that now. Please hold on.

**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 there any singularities in that integral?

**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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You mean the one in post #13? There is definitely a singularity at (x,y) = (0,0). Others may occur too if at any point (b,c) = (x,y). For the integral by itself though, there should only be a singularity at (0,0).

*Last edited by zetafunc (2016-10-09 04:32:01)*

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

A singularity at (0,0) is one of the endpoints of the integral can be a big problem. Is it a removable singularity?

There will of course be chances for more singularities at (b,c) = (x,y) as you point out.

**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

It is not impossible but usually singularities unless they are removable cause integrals to equal infinity, in other words they do not converge and therefore do not exist.

**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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The only way out of this that I can see would be to try to use the asymptotic expansions of the Bessel functions so that we end up with something with positive powers rather than negative ones.

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