Sum of an integral involving Bessel functions

zetafunc · 2016-10-09 01:26:13

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 ?

bobbym · 2016-10-09 03:06:33

Hi;

Can you please show me your M code?

zetafunc · 2016-10-09 03:14:50

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}]

but that will not give an answer, I don't think. I'm currently trying to use NSum/NIntegrate to try to find some partial sums given parameters .

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

bobbym · 2016-10-09 03:22:39

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

zetafunc · 2016-10-09 03:24:40

Sorry, I fixed it. There was a ] missing.

bobbym · 2016-10-09 03:39:09

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

zetafunc · 2016-10-09 03:42:35

I do not necessarily need to know the exact sum. I do know that it should converge, but I am really trying to find a bound for it in terms of .

bobbym · 2016-10-09 03:46:33

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

zetafunc · 2016-10-09 03:50:15

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)

bobbym · 2016-10-09 03:51:22

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

zetafunc · 2016-10-09 03:53:53

Sorry, the k should be a . I just entered k because it was easier than entering into M.

Last edited by zetafunc (2016-10-09 03:54:07)

bobbym · 2016-10-09 03:56:12

So, you want k to be another free variable or I am hoping we can at least bound it...

zetafunc · 2016-10-09 04:07:13

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)

bobbym · 2016-10-09 04:10:25

Can you write that up in Mathematica speak?

zetafunc · 2016-10-09 04:15:41

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)

bobbym · 2016-10-09 04:21:01

Bracket missing.

zetafunc · 2016-10-09 04:23:15

Sorry, fixed it again.

bobbym · 2016-10-09 04:25:15

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

zetafunc · 2016-10-09 04:28:08

OK, thanks. I am currently waiting to see what Mathematica does with the original sum if b and c vary from 1 to 100.

bobbym · 2016-10-09 04:29:25

Are there any singularities in that integral?

zetafunc · 2016-10-09 04:31:40

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)

bobbym · 2016-10-09 04:37:57

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.

zetafunc · 2016-10-09 04:39:58

Hmm, I don't think so. The limit as (x,y) tends to (0,0) blows up to infinity. That is a big problem. I may need to talk to my supervisor about this.

bobbym · 2016-10-09 04:42:09

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.

zetafunc · 2016-10-09 04:44:22

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.

Math Is Fun Forum

#1 2016-10-09 01:26:13

Sum of an integral involving Bessel functions

#2 2016-10-09 03:06:33

Re: Sum of an integral involving Bessel functions

#3 2016-10-09 03:14:50

Re: Sum of an integral involving Bessel functions

#4 2016-10-09 03:22:39

Re: Sum of an integral involving Bessel functions

#5 2016-10-09 03:24:40

Re: Sum of an integral involving Bessel functions

#6 2016-10-09 03:39:09

Re: Sum of an integral involving Bessel functions

#7 2016-10-09 03:42:35

Re: Sum of an integral involving Bessel functions

#8 2016-10-09 03:46:33

Re: Sum of an integral involving Bessel functions

#9 2016-10-09 03:50:15

Re: Sum of an integral involving Bessel functions

#10 2016-10-09 03:51:22

Re: Sum of an integral involving Bessel functions

#11 2016-10-09 03:53:53

Re: Sum of an integral involving Bessel functions

#12 2016-10-09 03:56:12

Re: Sum of an integral involving Bessel functions

#13 2016-10-09 04:07:13

Re: Sum of an integral involving Bessel functions

#14 2016-10-09 04:10:25

Re: Sum of an integral involving Bessel functions

#15 2016-10-09 04:15:41

Re: Sum of an integral involving Bessel functions

#16 2016-10-09 04:21:01

Re: Sum of an integral involving Bessel functions

#17 2016-10-09 04:23:15

Re: Sum of an integral involving Bessel functions

#18 2016-10-09 04:25:15

Re: Sum of an integral involving Bessel functions

#19 2016-10-09 04:28:08

Re: Sum of an integral involving Bessel functions

#20 2016-10-09 04:29:25

Re: Sum of an integral involving Bessel functions

#21 2016-10-09 04:31:40

Re: Sum of an integral involving Bessel functions

#22 2016-10-09 04:37:57

Re: Sum of an integral involving Bessel functions

#23 2016-10-09 04:39:58

Re: Sum of an integral involving Bessel functions

#24 2016-10-09 04:42:09

Re: Sum of an integral involving Bessel functions

#25 2016-10-09 04:44:22

Re: Sum of an integral involving Bessel functions

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