# Help with double integral of exp(ixy)

1. Dec 27, 2006

### lemma28

$$\int\limits_{ - \infty }^\infty {\int\limits_{ - \infty }^\infty {e^{ixy} dxdy = 2\pi}}$$

(x,y, real)

It came out analyzing the relation between DiracDelta and the Fourier Transform formula. (it's the reason why insert the constant 1/sqrt(2pi) in the fourier transform formula to be consistent with the diracdelta definition).
I know that it's value is 2pi. But I'd like to see how to actually calculate it. (Possibly in some elegant way...)

There must be some tricky "magic" based on symmetry consideration to reduce the double integral to the length of a unit circle. But I can't find it.

Thanks

2. Dec 27, 2006

### jpr0

You could try converting the integral to polar coordinates,

$$\int_{0}^{2\pi}d\varphi \int_0^{\infty}e^{i\frac{r^2}{2}\sin 2\varphi}rdr$$

then let $z = r^2/2$ and $\theta = 2\varphi$. Your angular integral should then look like a Bessel J function. http://www.math.sfu.ca/~cbm/aands/page_360.htm look at relation 9.1.21.

Last edited: Dec 27, 2006
3. Dec 27, 2006

### StatusX

If you change the bounds to some finite range, say -a<x<a and -b<y<b, then it shouldn't be too hard to show that the integral reduces to:

$$2 \int_{-ab}^{ab} \frac{\sin(u)}{u} du$$

Then you can take the limit as a,b-> infinity. The improper integral of sin(x)/x is known to be pi, and the easiest way to derive this is probably using a laplace transform.

4. Dec 27, 2006

### lemma28

Thanks. I got it!