Integration of function in a region

[phonic]

Dear All,

I want to calculate some convolution like integrations:
g_1(k,l)=\int\int_A \cos(k(x+y))\cos(ly) f(x,y)dx dy
g_2(k,l)=\int\int_A \cos(k(x+y))\sin(ly) f(x,y)dx dy
g_3(k,l)=\int\int_A \sin(k(x+y))\cos(ly) f(x,y)dx dy
g_4(k,l)=\int\int_A \sin(k(x+y))\sin(ly) f(x,y)dx dyf(x,y) =\cos(k(x-x_0)) \cos(l(y-y_0)) - b

|x-x_0|< \pi/k, |y-y_0|< \pi/l

k,l are integers, and x_0,y_0,b are constant real numbers, 0<b<1. Region A is the area where f(x,y)\geq 0.

I thought about transforming the coordinate into curvilinear coordinate, so that the two base vectors are tangent and normal to the line f(x,y). But I don't know how to derive.

Can anyone provide some help and guidance? Thanks!

 

[aesir]

It is much simpler in terms of exponentials, for example write cos(x)=(e^{ix}+e^{-ix})/2

 

[phonic]

Using your idea, define u=(e^{ik(x-x_0)+il(y-y0)}+e^{-ik(x-x_0)-il(y-y0)})/2 and v=(e^{-ik(x-x_0)-il(y-y0)}+e^{-ik(x-x_0)+il(y-y0)})/2, then the intergration region A looks simpler: A: u+v>b , but then \cos(k(x+y))\cos(ly) is not possible to write as a function of u and v.Is there other transform help to have a better shape of region A?
The main difficulty to directly calculate the double integration is that when I break the double integration into two 1-d integration, first x then y, then the integration for x is over a y dependent range, and this in turn makes the integration for y very complicated.

It is much simpler in terms of exponentials, for example write cos(x)=(e^{ix}+e^{-ix})/2