Since [itex]e^{iz^2}[/itex] is an analytic function on [itex]\mathbb{C}[/itex] we have:Forny said:well, I had already checked the link, but I don't undestand the strategy, could somebody please explain it to me?, or if you have a text where I can find some explanation about that kind of process, I would really apreciate if you help me.
Because i'm lazy to write all that much, see pages 98 & 99 of Max Planck's "Theory of Light", Macmillan, 1932 (a.k.a. the IV-the volume of the "Introduction to Theoretical Physics" set).Forny said:Somebody could please tell me how to evaluate the integral:
integral(sin(x^2)) from o to infinity
Buddy you are wrong the maximum of exp(-R^2sin(2thta)) is not at theta=pi/4 its maximum is at theta=0 ! So The maximum of g(theta) = exp{-R^2 *sin(2*theta)} is 1, not exp(-R^2), and this makes you proof wrong.!This is why:
[tex]
\left\arrowvert\int_0^{\pi/4}e^{iR^2e^{i2\theta}}iRe^{i\theta}d\theta \right\arrowvert = \left\arrowvert\int_0^{\pi/4}exp\{iR^2(\cos(2\theta)+i \sin(2\theta))\}iRe^{i\theta}d\theta \right\arrowvert
[/tex]
Evaluating the absolute value, this equals
[tex]
\int_0^{\pi/4}exp\{-R^2 \sin(2\theta)\}R d\theta
[/tex]
Now find the maximum of [tex] g(\theta) = exp\{-R^2 \sin(2\theta)\} [/tex] by
differentiation (check it's a max by second derivative test).
This occurs at [tex] \theta = \pi/4 [/tex].
Then the above is less than or equal to
[tex]
R \cdot exp\{-R^2\} \frac{\pi}{4}
[/tex]
Take the limit as [tex] R\to \infty [/tex] to get 0.
Absolute value to zero, original to zero.