Evaluating Fresnel Integrals Using Euler Formula

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The discussion focuses on evaluating Fresnel integrals C and S, defined as C = ∫0 cos(x²) dx and S = ∫0 sin(x²) dx, using the Euler formula. The solution reveals that both integrals equal sqrt(π/8). The approach involves squaring the integral F = C + iS, transforming it into polar coordinates, and applying a convergence factor. The final result confirms that C and S are equal, leading to the conclusion that C = S = sqrt(π/8).

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Homework Statement


Evaluate the following integrals C = 0inf∫cos(x2) dx and S = 0inf∫sin(x2) dx

Homework Equations


[/B]
Hint: use Euler formula to write the integral for F = C + iS. Square the integral and evaluate it in polar coordinates. Temporary add a convergence factor.

Answer: C = S = sqrt(pi/8)

The Attempt at a Solution



(Abbreviated form)

F2 = 1/4 0inf∫ eix^2dx 0inf∫eiy^2dy

F2 = pi/2 0inf∫ eir^2 r dr

F2 = pi/4 0inf∫eiudu

Now I think the convergence factor comes in here, but I am not entirely sure how that works.

Thanks!
 
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iggyonphysics said:

Homework Statement


Evaluate the following integrals C = 0inf∫cos(x2) dx and S = 0inf∫sin(x2) dx

Homework Equations


[/B]
Hint: use Euler formula to write the integral for F = C + iS. Square the integral and evaluate it in polar coordinates. Temporary add a convergence factor.

Answer: C = S = sqrt(pi/8)

The Attempt at a Solution



(Abbreviated form)

F2 = 1/4 0inf∫ eix^2dx 0inf∫eiy^2dy

F2 = pi/2 0inf∫ eir^2 r dr

F2 = pi/4 0inf∫eiudu

Now I think the convergence factor comes in here, but I am not entirely sure how that works.

I imagine I wind up with something like F2= i*pi/4 and F = sqrt(i*pi/4) and since S and C are equal, they must each equal sqrt(pi/8).

Thanks!

You can look at
J_r = \int_0^{\infty} e^{-ru} e^{iu} \, du,
where ##r > 0##. Then take the limit as ##r \to 0##.
 
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Great, so that just comes to i, correct?

So,

F^2=π/4 eiπ/2 --> F = ei/4sqrt(π/4)

How do I get to S and C from here? (I know if I evaluate the product of sqrt(π/4) and sin(π/4)2 it spits out sqrt(π/8). Why does this work?)
 

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