# Math Help - Double integral problem

1. ## 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!

2. I think you will need to apply Euler's Forumla here...

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

Here $\displaystyle \theta = -u\,x - v\,y$.

3. Beautiful, that's probably the key.

I'll get on it, thanks!

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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!