I do not realy understand your question. If rR then the equation is the same than forr r=0 and R=Infinty.

The task is to calculate explicitly the function f(x,y), which is  defined as the sum.

I do not understand why this should be the same for r=0 and R=\infty? If r=0 and R=\infty then you sum over the whole plane Z^2, while for 0

Why do you think that you have an explicit solution when k^2+I^2 is bounded on a ring?

There is no reason for it if there is no other conditions. If you do not know how large it can  k^2 + I^2 be, what would be the condition that would simplify the expression?

r=0.0001 R=100,000,000 ? How can this simplifay the expression?

As I said above both  r  and R are  large.  And r=0.0001 is not really large, or?

And as I also said above:

I would be happy with r=aR for some fixed a \in(0,1) and  large R

In that case the function f(x,y) is large in an O(1/R) neighborhood around (0,0) and due to averaging effects it is comparably small for (x,y) much bigger than O(1/R). But I was wondering it there is a similar trick as in one dimension, where one can explicitly calculate the sum for all r and R.

It is very relative what is large. 0.0001 light year is very large, dont you agree, and 100,000,000 pm is very small. So, maby you meant that (R-r)/R

The only unit that is available in the real numbers is 1. There are no units like pm or lightyears in the question. So I do not understand the remark.

Let me just remark that a\in(0,1), ie by definition the interval is open and 0 < a < 1. So a can not be zero and for the limit R\to\infty the lower bound on k,l goes to infinity ...

No, I did not mean (R-r)/R

My remark about units was only to explain to you, that large is relative. Something can be large only when yo compere it to something else. For instance earth is large compared to size of human beeing and very small compared to galaksy.

if you have only one real number youvcannot say if it is parge or small. Did you mean that R>>1 ? But although you did, it cannot help.

Your condition 0

Do not worry about typos. I am sure I have many without using a phone.

Actually, the problem for r=0 and R =\infty is solvable, as f(x,y)dxdy is then a multiple of the Dirac measure in the point (0,0) --  Ok, plus the obvious 2\pi-periodic extension. So  f(x,y)dxdy is the counting measure of the lattice  2\pi \N^2.

I hoped that fixing 01 would give some structure to the problem, as the resulting f is a sum of many terms with wave-number of the same order of magnitude.

As I have already mentioned 0 1 is irlevant to the problem as  K2 + I2 can be arbitrarly small.

Sorry, i couldnt help.

But k^2 and l^2 cannot be arbitrary small, as they are bounded below by aR^2 >> 1