Can a Double Integral be Simplified Using a Substitution of Polar 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.