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Double integral.

  1. Jul 3, 2007 #1
    If we wish to calculate the integral.

    [tex] \int_{0}^{\infty}dx \int_{0}^{\infty}dy e^{i(x^{2}-y^{2}} [/tex]

    which under the simmetry [tex] (x,y) \rightarrow (y,x) [/tex] it gives you the complex conjugate counterpart.

    my idea is to make the substitution (as an analogy of Laplace method)

    [tex] x=rcosh(u) [/tex] , [tex] y=rsinh(u) [/tex] (or viceversa) using the fact that the square of cosh minus square of sinh is equal to one and expressing the integral as:

    [tex] \int_{0}^{\infty}dr re^{ir^{2}}d\Omega [/tex]

    the integral Omega is over the angle variable u, if we knew tha exact value of Omega the the radial part is just the imaginary number "i" if i am not wrong.
  2. jcsd
  3. Jul 3, 2007 #2
    Here is a way do to this without any substitutions.

    We can write:
    [tex]\int_0^{\infty} e^{-iy^2} \ dy \cdot \int_0^{\infty} e^{ix^2} \ dx[/tex]

    Use Euler's formula:
    [tex]\left( \int_0^{\infty} \cos y^2 - i\sin y^2 \ dy \right) + \left( \int_0^{\infty} \cos x^2 + i \sin x^2 \ dx \right)[/tex]

    Which can be combined into:
    [tex] 2 \int_0^{\infty} \cos x^2 \ dx = \sqrt{ \frac{\pi}{2}}[/tex]
  4. Jul 3, 2007 #3
    Thanks kummer, i know that the integra proposed can be made by simple analytic method however i would like to know how to perform the integral with my transformation of coordinates, so if we can map the rectangle.

    [tex] R= (0, \infty) x(0,\infty) [/tex] Cartesian coordinates

    to [tex] V= (0, \infty) x(-\pi,\pi) [/tex] Polar (??) coordinates

    [tex] x=r cosh(u) [/tex] and the same for y with the hyperbolic sine.
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