Evaluating Fresnel Integral: sin(x^2) from 0 to $\infty$

[Forny]

Somebody could please tell me how to evaluate the integral:

integral(sin(x^2)) from o to infinity

 

[TD]

You can find a strategy http://en.wikipedia.org/wiki/Fresnel_integral#Error_function".

 

[Forny]

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.

 

[WigneRacah]

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.

Since $e^{iz^2}$ is an analytic function on $\mathbb{C}$ we have:

$$\int_{\Gamma_R} e^{iz^2} d z = 0$$ where $\Gamma_R$ is the "pizza-slice" countour given in the link.
Now
$$\int_{\Gamma_R} e^{iz^2} d z = \int_0^R e^{i t^2} d t + \int_0^{\frac{\pi}{4}} e^{i R^2 e^{i 2 \theta}} i R e^{i \theta} d \theta + \int_R^0 e^{- t^2} e^{i \frac{\pi}{4}} d t$$.

By taking the limit $R \rightarrow \infty$ and observing that

$$\lim_{R\rightarrow \infty} \int_0^{\frac{\pi}{4}} e^{i R^2 e^{i 2 \theta}} i R e^{i\theta} d \theta = 0$$

we obtain

$$\int_0^\infty e^{i t^2} d t = e^{i \frac{\pi}{4}} \int_0^\infty e^{- t^2} d t$$.

By Euler's formula and gaussian integral we get:

$$\int_0^\infty \sin(t^2) dt = \int_0^\infty \cos(t^2) dt = \sqrt{\frac{\pi}{8}}$$.

 

[dextercioby]

Somebody could please tell me how to evaluate the integral:

integral(sin(x^2)) from o to infinity

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).


Daniel.

 

[bmanbs2]

Reviving this tread, but why does

$$\lim_{R\rightarrow \infty} \int_0^{\frac{\pi}{4}} e^{i R^2 e^{i 2 \theta}} i R e^{i\theta} d \theta = 0$$

?

 

[ZZappaZZappa]

This is why:

$$\left\arrowvert\int_0^{\pi/4}e^{iR^2e^{i2\theta}}iRe^{i\theta}d\theta \right\arrowvert = \left\arrowvert\int_0^{\pi/4}exp\{iR^2(\cos(2\theta)+i \sin(2\theta))\}iRe^{i\theta}d\theta \right\arrowvert$$

Evaluating the absolute value, this equals

$$\int_0^{\pi/4}exp\{-R^2 \sin(2\theta)\}R d\theta$$

Now find the maximum of $$g(\theta) = exp\{-R^2 \sin(2\theta)\}$$ by differentiation (check it's a max by second derivative test).
This occurs at $$\theta = \pi/4$$.
Then the above is less than or equal to

$$R \cdot exp\{-R^2\} \frac{\pi}{4}$$

Take the limit as $$R\to \infty$$ to get 0.
Absolute value to zero, original to zero.

 

[ikaveh]

This is why:

$$\left\arrowvert\int_0^{\pi/4}e^{iR^2e^{i2\theta}}iRe^{i\theta}d\theta \right\arrowvert = \left\arrowvert\int_0^{\pi/4}exp\{iR^2(\cos(2\theta)+i \sin(2\theta))\}iRe^{i\theta}d\theta \right\arrowvert$$

Evaluating the absolute value, this equals

$$\int_0^{\pi/4}exp\{-R^2 \sin(2\theta)\}R d\theta$$

Now find the maximum of $$g(\theta) = exp\{-R^2 \sin(2\theta)\}$$ by
differentiation (check it's a max by second derivative test).

This occurs at $$\theta = \pi/4$$.
Then the above is less than or equal to

$$R \cdot exp\{-R^2\} \frac{\pi}{4}$$

Take the limit as $$R\to \infty$$ to get 0.
Absolute value to zero, original to zero.

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.!
Any other proof that is right?
this is my homework pls

 

[ikaveh]

here is the right proof:
http://planetmath.org/encyclopedia/FresnelIntegralsAtInfinity.html