# Fresnel Integrals, Contour Integration

• Vale132
In summary, the conversation is discussing the evaluation of Fresnel integrals using a change of variable and a contour integral. The integrals along the y-axis and x-axis are found to be related to the original integrals, and the conversation then focuses on proving that the integrals along the two quarter-circles tend to zero. The conversation mentions substituting values and using Euler's formula, but there are concerns about the behavior of the exponential as the variables tend to infinity. Ultimately, it is concluded that the integrals do tend to zero, but the process is not as simple as initially thought.
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!

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

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

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

lurflurf said:
$$e^{i R e^{iθ}}\rightarrow 0$$
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.

Last edited:

## 1. What are Fresnel Integrals?

Fresnel Integrals are a pair of mathematical functions that are used in the study of optics and diffraction. They are named after French physicist Augustin-Jean Fresnel, who first introduced them in 1818.

## 2. How are Fresnel Integrals used in optics?

Fresnel Integrals are used to calculate the diffraction pattern of light when it passes through a small aperture or is reflected off a curved surface. They are also used in the study of electromagnetic waves and their behavior.

## 3. What is Contour Integration?

Contour Integration is a technique used in complex analysis to evaluate integrals of complex-valued functions along a given path in the complex plane. It is a powerful tool in solving various mathematical problems, including those involving Fresnel Integrals.

## 4. How is Contour Integration used to solve problems involving Fresnel Integrals?

Contour Integration allows us to evaluate integrals involving Fresnel Integrals by choosing a suitable contour in the complex plane and applying the residue theorem. This makes the calculation of these integrals much easier and more efficient.

## 5. Are there any applications of Fresnel Integrals and Contour Integration in other fields?

Yes, both Fresnel Integrals and Contour Integration have various applications in physics, engineering, and mathematics. They are used in signal processing, optics, acoustics, and many other fields to solve a wide range of problems involving complex-valued functions.

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