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Fresnel Integrals, Contour Integration

  1. Feb 15, 2014 #1
    1. The problem statement, all variables and given/known data

    Please let me know if this kind of posting of exact problems from a textbook isn't allowed; if that's the case I'll delete it immediately.

    From Boas's Mathematical Methods in the Physical Sciences, Third Edition: The Fresnel integrals, [itex] \int_0^u sin (u^2)\,du [/itex] and [itex] \int_0^u cos (u^2)\,du [/itex], are important in optics. For the case of infinite upper limits, evaluate these integrals as follows: Make the change of variable [itex] x = u^2 [/itex]; to evaluate the resulting integrals, find [itex] \oint e^{iz}z^{-1/2}\,dz [/itex] around the contour shown. Let [itex] r \rightarrow 0 [/itex] and [itex] R \rightarrow \infty [/itex] and show that the integrals along these quarter-circles tend to zero. Recognize the integral along the y-axis as a gamma function and so evaluate it. Hence evaluate the integral along the x-axis; the real and imaginary parts of this integral are the integrals you are trying to find.

    (Note: I already solved the problem correctly, but I don't know how to prove the bold part above; see work below.)

    (The "contour shown" looks like a quarter circle in the first quadrant, except that there is a small quarter circle (of radius [itex]r[/itex]) cut out around the origin to avoid the singularity there.)

    2. Relevant equations

    [itex] \oint f(z)\,dz = 0 [/itex] when the function is analytic on and inside a simple closed curve.

    3. The attempt at a solution

    The contour integral is the sum of four integrals, and it is equal to zero since the singularity at the origin is outside the contour. I omit the integrals along the x and y-axes because I'm just wondering about the ones along the two semicircles:

    [itex] \int_0^{π/2} e^{iz}Rie^{iθ}/R^{1/2}e^{iθ/2}\,dθ [/itex]

    and

    [itex] \int_{π/2}^{0} e^{iz}rie^{iθ}/r^{1/2}e^{iθ/2}\,dθ [/itex]


    These were obtained by substituting [itex] z=Re^{iθ} [/itex] and [itex] z=re^{iθ} [/itex], respectively, into the contour integral given in the problem statement.

    I can see that in the second integral, the [itex]r[/itex] in the numerator cancels with the [itex]r^{1/2}[/itex] in the denominator, and since [itex]r[/itex] (and therefore [itex]z[/itex]) are tending to zero, the integral tends to zero. Is this sufficient?

    The second integral tending to zero as [itex] R \rightarrow \infty [/itex] makes less sense. Doesn't the numerator tend to infinity, and the denominator stay a constant 1, as [itex] R \rightarrow \infty [/itex]? And substituting [itex] z = Re^{iθ} [/itex] into the exponential only makes things worse. I considered rewriting the exponentials using Euler's formula and examining their behavior, but I don't see how that would help.

    Thanks!
     
  2. jcsd
  3. Feb 15, 2014 #2

    lurflurf

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    The numerator goes to 0
    You forgot to change a z
    $$\int_0^{\pi/2} e^{i R e^{iθ}}Rie^{iθ}/(R^{1/2}e^{iθ/2})\,dθ$$
     
  4. Feb 16, 2014 #3
    How does that work when [itex] R \rightarrow \infty [/itex] ?
     
  5. Feb 16, 2014 #4

    lurflurf

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    $$e^{i R e^{iθ}}\rightarrow 0$$
    very fast
    compare to e^-x
     
  6. Feb 16, 2014 #5

    Dick

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    It's not that simple. You wind up needing to estimate ##\int_0^{\pi/4} e^{-R\sin\theta} d\theta##. It doesn't even really go to zero all that quickly. But it does go to zero.
     
    Last edited: Feb 16, 2014
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