# Trying to invert an expression

1. May 26, 2015

### Morberticus

I've come across an expression that looks like

n(k) = ∫cos(x-y)f(x,y)dxdy

Is there a name for this transform? I would like to invert it to obtain f(x,y) but I'm not used to the 2D integral on the RHS. I tried to turn it into a fourier transform:

n(k) = 1/2 ( ∫eixe-iyf(x,y)dxdy + ∫e-ixeiyf(x,y)dxdy)

but got stuck. Any help would be appreciated.

Thanks

2. May 27, 2015

### Ssnow

Hi, there is something of strange, where is $k$ in the integral?, you say $n(k)$ but in the right side I don't see $k$, and what is the domain of integration ?

3. May 27, 2015

### Morberticus

You are right! Sorry, I forgot about the k when writing down the expression(s). The correct expression is

n(k) = ∫ cos( k(x-y) ) f(x,y) dxdy

Thanks

4. May 27, 2015

### Ssnow

I think your transform is analogous to the ''cosine transform'', I hope you can find something useful here

http://dsp-book.narod.ru/TAH/ch03.pdf

but I think (as you wrote...) it is possible to use the inverse Fourier transform...