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

In summary, to evaluate the integral of sin(x^2) from o to infinity, we can use the strategy provided in the link to find the limit of the integral as R approaches infinity. This limit can be simplified using Euler's formula and the gaussian integral, giving us the final result of the integral as the square root of pi over 8. Further explanation and proof can be found in Max Planck's "Theory of Light" on pages 98 and 99 of the IV-volume of the "Introduction to Theoretical Physics" set. This proof can also be found on PlanetMath.org.
  • #1
Forny
2
0
Somebody could please tell me how to evaluate the integral:

integral(sin(x^2)) from o to infinity
 
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  • #3
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.
 
  • #4
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.

Since [itex]e^{iz^2}[/itex] is an analytic function on [itex]\mathbb{C}[/itex] we have:

[tex]\int_{\Gamma_R} e^{iz^2} d z = 0[/tex] where [itex]\Gamma_R[/itex] is the "pizza-slice" countour given in the link.
Now
[tex]\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 [/tex].

By taking the limit [itex]R \rightarrow \infty[/itex] and observing that

[tex]\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 [/tex]

we obtain

[tex]\int_0^\infty e^{i t^2} d t = e^{i \frac{\pi}{4}} \int_0^\infty e^{- t^2} d t [/tex].

By Euler's formula and gaussian integral we get:

[tex]\int_0^\infty \sin(t^2) dt = \int_0^\infty \cos(t^2) dt = \sqrt{\frac{\pi}{8}}[/tex].
 
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  • #5
Forny said:
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.
 
  • #6
Reviving this tread, but why does

[tex]\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 [/tex]

?
 
  • #7
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.
 
  • #8
ZZappaZZappa said:
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.

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
 
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  • #9
here is the right proof:
http://planetmath.org/encyclopedia/FresnelIntegralsAtInfinity.html [Broken]
 
Last edited by a moderator:

1. What is the purpose of evaluating the Fresnel Integral?

The Fresnel Integral is a mathematical tool used in physics and engineering to calculate the diffraction of waves, such as light or sound, when they pass through an aperture or around an obstacle.

2. How is the Fresnel Integral calculated?

To evaluate the Fresnel Integral, the function sin(x^2) from 0 to infinity, it is necessary to use numerical methods such as Simpson's rule, Gaussian quadrature, or Monte Carlo integration.

3. What are the applications of the Fresnel Integral?

The Fresnel Integral has numerous applications in optics, acoustics, and signal processing. It is commonly used in the design of lenses, antennas, and diffraction gratings.

4. Is there a closed-form solution for the Fresnel Integral?

No, there is no known closed-form solution for the Fresnel Integral. It can only be evaluated using numerical methods.

5. Can the Fresnel Integral be evaluated for values other than sin(x^2)?

Yes, the Fresnel Integral can be evaluated for other functions, such as cos(x^2), e^(-x^2), and many others. However, the method of evaluation may vary depending on the specific function.

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