Fresnel Integrals, Contour Integration

[Vale132]

Homework Statement

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, \int_0^u sin (u^2)\,du and \int_0^u cos (u^2)\,du, are important in optics. For the case of infinite upper limits, evaluate these integrals as follows: Make the change of variable x = u^2; to evaluate the resulting integrals, find \oint e^{iz}z^{-1/2}\,dz around the contour shown. Let r \rightarrow 0 and R \rightarrow \infty 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 r) cut out around the origin to avoid the singularity there.)

Homework Equations

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

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:

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

and

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


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

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

The second integral tending to zero as R \rightarrow \infty makes less sense. Doesn't the numerator tend to infinity, and the denominator stay a constant 1, as R \rightarrow \infty? And substituting z = Re^{iθ} 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!

 

[lurflurf]

The numerator goes to 0
You forgot to change a z
$$\int_0^{\pi/2} e^{i R e^{iθ}}Rie^{iθ}/(R^{1/2}e^{iθ/2})\,dθ$$

 

[Vale132]

How does that work when R \rightarrow \infty ?

 

[lurflurf]

$$e^{i R e^{iθ}}\rightarrow 0$$
very fast
compare to e^-x

 

[Dick]

$$e^{i R e^{iθ}}\rightarrow 0$$
very fast
compare to e^-x

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.