Integrating interpolated function in polar coordinates

Question

I have a set of data that I wish to express in polar coordinates.

The data represents the function $f(k_x,k_y)$ in the integral below
$ \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} dk_x dk_y f(k_x,k_y) = \int_0^{2\pi} d \theta \int_0^\infty dk \, n(k) $.

I am looking to find the corresponding function $n(k)$ in polar coordinates, where $k = \sqrt{k_x^2 + k_y^2}$ and it's possible to write $k_x = k \cos \theta$, $k_y = k \sin \theta$.

With a analytic function, mathematica can easily do this: if the function is f, then the following

Integrate[f[k Cos[\[Theta]], k Sin[\[Theta]]], {\[Theta], 0, 2 \[Pi]}]

easily gives me the right function. However, instead of an analytic function I tried to make a function by interpolating my dataset with

f=ListInterpolate[data]

Now, I can no longer integrate out the $\theta$ dependence. I have looked at directly transforming the data as a list, but could not find an appropriate transform. Is there a way to get the interpolated function to behave, or perhaps a much more obvious method of doing this that I'm missing?

(The data I am interpolating is of the form

{{ z@k_x1@k_y1, z@k_x2@k_y1 ...},{z@k_x1@k_y2,z@k_x2@k_y2 }...}

i.e no explicit $k_x$ and $k_y$ coordinates to work with.)

Answer 1

It is exactly the same way as in the case of the function given in the analytical form. This might be the end of my message, but I think, it may help to have a look at an example.

Since you did not give any data to play with, I will generate a simple example myself. Here is a list of an exponential:

lst1 = Table[Exp[-5 (x^2 + y^2)], {x, -1, 1, 0.1}, {y, -1, 1, 0.1}];

Here is the interpolation of this list:

f2 = ListInterpolation[lst1, {{-1, 1}, {-1, 1}}]

This is the check that we obtained something reasonable:

Plot3D[f2[x, y], {x, -1, 1}, {y, -1, 1}]

Now let us integrate

NIntegrate[f[k*Cos[\[CurlyPhi]], k*Sin[\[CurlyPhi]]]*k, {k, 0,0.5},{\[CurlyPhi], 0, 2 \[Pi]}]

(*  0.0000147   *)

Have fun!