Double integral with polar-like substitution using hyperbolic coordinates

[Klaus_Hoffmann]

If we wish to calculate the integral.

$$\int_{0}^{\infty}dx \int_{0}^{\infty}dy e^{i(x^{2}-y^{2}}$$

which under the symmetry $$(x,y) \rightarrow (y,x)$$ it gives you the complex conjugate counterpart.

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

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

$$\int_{0}^{\infty}dr re^{ir^{2}}d\Omega$$

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.

 

[Kummer]

Here is a way do to this without any substitutions.

We can write:
$$\int_0^{\infty} e^{-iy^2} \ dy \cdot \int_0^{\infty} e^{ix^2} \ dx$$

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

Which can be combined into:
$$2 \int_0^{\infty} \cos x^2 \ dx = \sqrt{ \frac{\pi}{2}}$$

 

[Klaus_Hoffmann]

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.

$$R= (0, \infty) x(0,\infty)$$ Cartesian coordinates

to $$V= (0, \infty) x(-\pi,\pi)$$ Polar (??) coordinates

$$x=r cosh(u)$$ and the same for y with the hyperbolic sine.