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Integration of function in a region

  1. Jan 10, 2012 #1
    Dear All,

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

    [itex] f(x,y) =\cos(k(x-x_0)) \cos(l(y-y_0)) - b [/itex]

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

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

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

    Can anyone provide some help and guidance? Thanks!
    Last edited: Jan 10, 2012
  2. jcsd
  3. Jan 10, 2012 #2
    It is much simpler in terms of exponentials, for example write [itex]cos(x)=(e^{ix}+e^{-ix})/2[/itex]
  4. Jan 10, 2012 #3
    Using your idea, define [itex]u=(e^{ik(x-x_0)+il(y-y0)}+e^{-ik(x-x_0)-il(y-y0)})/2[/itex] and [itex]v=(e^{-ik(x-x_0)-il(y-y0)}+e^{-ik(x-x_0)+il(y-y0)})/2[/itex], then the intergration region A looks simpler: [itex] A: u+v>b[/itex] , but then [itex]\cos(k(x+y))\cos(ly)[/itex] 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.

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