1

I was reading the following: Integrating Fresnel Integrals with Cauchy Theorem?

And I do not understand how the two inequalities below follow:

$$\left |\int_0^{\pi/4} e^{i(Re^{i\theta})^2}(iRe^{i\theta})d\theta \right |\le \int_0^{\pi/4} Re^{-R^2\sin(\theta)}d\theta \le \int_0^{\pi/4}Re^{-R^2\left (\frac{2}{\pi}\theta\right)}d\theta$$

I can get at least a partial result on the first one:

$\le \int_0^{\pi/4} |e^{i(Re^{i\theta})^2}||R||i||e^{i\theta}|d\theta = \int_0^{\pi/4} Re^{iR^2e^{2i\theta}}d\theta$. I'm thinking that I need to use the fact that $e^{ix}=\cos(x)+i\sin(x)$ but I'm not entirely sure how it is used.

user351797
  • 1,005

1 Answers1

1

We have $$ e^{i(Re^{i\theta})^2} = e^{iR^2e^{2i\theta}}, $$ and $ \lvert e^{z} \rvert = e^{\Re(z)} $, so $$ \lvert e^{iR^2e^{2i\theta}} \rvert = e^{-R^2\sin{2\theta}} $$ (and so there is a typo in the image in the quoted question). This, along with $\lvert iRe^{i\theta} \rvert = R$ and the triangle inequality $ \lvert \int_C f(z) dz \rvert \leq \int_C \lvert f(z) \rvert \lvert dz \rvert $ is what is used in the first inequality.

The second is easiest to understand by drawing a graph: $\sin{x}$ is concave on $(0,\pi/2]$, so it lies below its tangent and above its chord, which gives $$ \frac{2}{\pi}x \leq \sin{x} \leq x \qquad (0<x<\pi/2), $$ and the former is used to bound $-\sin{2\theta}$ above on $(0,\pi/4]$.

Chappers
  • 67,606