If we wish to calculate the integral.(adsbygoogle = window.adsbygoogle || []).push({});

[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.

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# Double integral.

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