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Math Help - Double integral problem

  1. #1
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    Double integral problem

    I want to integrate the following function: e^(-i*(u*x+v*y)) with respect to x and y. Evaluated where x value is between -W and W, while y value is between -H and H. And i in the function is the imaginary number, while u and v are arbitrary constant variable.

    I was able to get the integral itself, which evaluates to: -e ^(-i*(u*x + v*y))/(u*v)

    However mathematica was able to simplify the final solution to: (4 Sin[H v] Sin[u W])/(u v)

    Whereas the furthest I was able to get was: 4 * e ^ ( i * (u*w - v *h))/(uv)

    Where did I go wrong? What am I missing?



    Any help is greatly appreciated.

    Thank you!
    Last edited by stclouds; December 18th 2010 at 10:58 PM. Reason: Added work done
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  2. #2
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    I think you will need to apply Euler's Forumla here...

    \displaystyle e^{i\theta} = \cos{\theta} + i\sin{\theta}.

    Here \displaystyle \theta = -u\,x - v\,y.
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  3. #3
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    Beautiful, that's probably the key.

    I'll get on it, thanks!

    _____________________________________

    Sorry that's still a no go... Please take a look at the scribbles above. Even with the Euler's formula I wasn't able to drop the terms sufficiently to leave only the Sine functions like the Mathematica's equation.

    Equation: Integrate[E^(-I *(u*x + v * y)), {x, -W, W}, {y, -H, H}]
    Result: (4 Sin[H v] Sin[u W])/(u v)
    ---------------------------------------------------------

    Finally was able to get the right result, as it turned out I'm forgetting my trig identities, mainly:

    Cos(-x) = Cos (x)

    also

    Cos(x+y) = Cos(x)Cos(y) - Sin(x) Sin(y)

    >.< doh! In both instances of the identities I was confusing the plus (+) for the minus (-) sign.

    At any rate here's the corrected simplification:


    p.s. I should have carried over the (uv) denominator, but I just put that back at the final step.

    Thank you all!
    Last edited by stclouds; December 19th 2010 at 07:59 AM.
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