# Fresnel Integrals, Contour Integration

1. Feb 15, 2014

### Vale132

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, $\int_0^u sin (u^2)\,du$ 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.)

2. Relevant equations

$\oint f(z)\,dz = 0$ 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:

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

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!

2. Feb 15, 2014

### lurflurf

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

3. Feb 16, 2014

### Vale132

How does that work when $R \rightarrow \infty$ ?

4. Feb 16, 2014

### lurflurf

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

5. Feb 16, 2014

### Dick

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