# Integration of function in a region

## Main Question or Discussion Point

Dear All,

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

$f(x,y) =\cos(k(x-x_0)) \cos(l(y-y_0)) - b$

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

$k,l$ 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!

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

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 $cos(x)=(e^{ix}+e^{-ix})/2$