# Showing that $\lim_{R \rightarrow \infty} \bigg(\int_{0}^{R}e^{ix^{2}}dx-e^{i \pi/4}\int_{0}^{R}e^{-r^{2}}dr \bigg)=0$?

In the text "Complex Variables Introduction and Applications Second Edition", I'm having trouble proving the proposition in $$(1)$$, could this be done through Cauchy's Theorem ?

We wish to evaluate the integral $$I=\int_{0}^{\infty}e^{ix^{2}}$$. Consider the contour $$I=\oint_{\gamma_{(R_{1})}}e^{iz^{2}}$$, where $$\gamma_{(R)}$$ is the closed circular sector in the upper half plane with boundary points $$0,R$$ and $$Re^{i\pi/4}$$. Show that $$I_{R}=0$$ and that $$\lim_{R \rightarrow \infty} \oint_{\gamma_{(R)}}e^{iz^{2}}dz=0$$, where $$\gamma_{{(R_{2})}}$$ is the line integral along the circular sector from $$R$$ to $$Re^{i \pi/4}$$. Then, breaking up the contour $$\gamma_{(R)}$$ into three component parts, deduce in $$(1)$$

$$(1)$$\lim_{R \rightarrow \infty} \bigg(\int_{0}^{R}e^{ix^{2}}dx-e^{i \pi/4}\int_{0}^{R}e^{-r^{2}}dr \bigg)=0.$$and from the well-known result of real integration, \int_{0}^{\infty}e^{-x^{2}}dx= \sqrt(\pi)/2, deduce that I=e^{i\pi/4}\sqrt(\pi)/2 $$\text{Lemma (0.0)}$$ Since our function has no poles, one can pick a Contour $$\gamma_{R}$$ such that: $$\gamma_{R}^{1}(t) = 0 \, \, \text{if} \, \, R \leq t \leq R$$ $$\gamma_{R}^{2}(t) = Re^{i \pi/4} \, \, \text{if} \, \, \, \, 0\leq t \leq \pi$$ $$\text{Lemma (1.0)}$$ In order to show that $$\lim_{R \rightarrow \infty} \oint_{\gamma_{2(R)}} e^{iz^{2}}dz = 0$$, one must rely on the ML-Estimates, as formally discussed in $$(1.1.2)$$. $$\text{Estimation Lemma}$$ $$(1.1.2)$$ Let $$U \subset \mathbb{C}$$ be open and $$f \in C^{0}(U)$$. If $$\gamma :[a,b] \rightarrow U$$ is a $$C^{1}$$ curve, then in $$(1.1.3)$$ $$(1.1.3)$$ $$\bigg | \oint_{\gamma}f(z)dz \bigg | \leq \bigg( \sup_{t \in [a,b]}|f(\gamma(t))| \bigg) \cdot \int_{b}^{a} \bigg | D_{t}\gamma(t) \bigg |dt.$$ Utilizing $$(1.1.3)$$ one can achieve the upper bound for $$\gamma_{R}^{2}$$ in $$(1.1.4)$$ $$(1.1.4)$$ $$\bigg |\oint_{\gamma_{R}^{2}}e^{iz^{2}} dz \bigg | \leq \big\{\text{length}(\gamma_{R}^{2}) \big\} \cdot \sup_{\gamma_{R}^{2}}|e^{iz}|\leq \pi R(e^{R}-0)$$ From $$(1.1.4)$$ thus we have $$\lim_{R \rightarrow \infty}\bigg | \oint_{\gamma_{R}^{2}}e^{iz^{2}} dz \bigg| \rightarrow 0$$ • Why does the author does the author in his hint in (1) subtract the integrals over the given contour it one can via Cauchy's Theorem in (*)$$ (*) \, \, \, \ \, \, \, \, \lim_{R \rightarrow \infty}\oint_{\gamma_{R}^{1}}f(x)dx + \oint_{\gamma_{R}^{2}}f(z)dz \, \, \, \, ?$$From all the examples of doing Integrals over a given semicircular contour one add's up the integrals via Cauchy's Theorem and takes the limit so I'm not sure particularly why the author subtracts the integrals in (1) ? Feb 25, 2018 at 22:59 • See Fresnel Integral. Feb 25, 2018 at 23:01 • @FelixMarin thanks for the hint in due time i'll be able to answer my own question Feb 25, 2018 at 23:13 • You're welcome !!!. Feb 25, 2018 at 23:27 • Also is my current progress on the problem accurate ? Feb 25, 2018 at 23:53 ## 1 Answer From Felix Martin's comment/hint and taking inspiration from Josh Keneda, I managed to get a solution everything pertaining to the solution can be seen in \text{Lemma (1.1)} \text{Lemma (1.1)} \text{Cauchy Integral Theorem} (1.1.3) Let U be an open subset of C which is simply connected, let f : U → C be a holomorphic function, and let {\displaystyle \!\,\Gamma } \!\, be a rectifiable path in U whose start point is equal to its end point. Then in (1.1.4) (1.1.4)$$\oint_{\Gamma}f(z)dz = 0.$$In view of (1.1.4), we can make the following conclusions in (1.1.5) (1.1.5)$$\oint_{\gamma_{R}^{2}}e^{iz^{2}} dz = \bigg( \oint_{0}^{R}e^{iz^{2}}dz + \oint_{0}^{\pi / 4} e^{iz^{2}}dz + \oint_{R}^{0}e^{iz^{2}}dz \bigg) = 0.$$Clearly, it's trival to see that z=x and in view of \text{Lemma (0.0})  we have the following developments in (1.1.6) (1.1.6)$$\oint_{\gamma_{R}^{2}}e^{iz^{2}} dz = \bigg( \oint_{0}^{R}e^{iz^{2}}dz + \oint_{0}^{\pi / 4}e^{(iRe^{i \theta})^{2}}Rie^{i \theta}d \theta + \oint_{R}^{0}e^{(iRe^{\pi i /4})^{2}}e^{\pi i /4} dr \bigg) = 0.$$From (1.1.6), one can make the observation$$\bigg |\int_{0}^{\pi /4}e^{iR^{2}}e^{2i\theta}Rie^{i\theta}d \theta \bigg | \leq R \int_{0}^{\pi /4} |e^{[iR^{2}(cos2 \theta + i\sin 2 \theta)]}|d\theta  \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \leq \int_{0}^{\pi / 4} e^{-R^{2} \sin 2\theta} d \theta \leq r \int_{0}^{\pi / 4 }e^{-R^{2}(4 \theta / \pi)}d \theta\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, = \frac{\pi}{4} \frac{1-e^{-R^{2}}}{R} \rightarrow 0 \, \text{as}\, R \rightarrow \infty.whereas we have the following developments in (1.1.7) (1.1.7) \begin{align*}\lim_{R \rightarrow \infty} R e^{i\pi/4} \int_R^0 e^{-u^2} \frac{-1}{R} du &= \lim_{R\rightarrow \infty} R e^{i\pi/4} \int_0^R e^{-u^2} \frac{1}{R} du\\ &= \lim_{R \rightarrow \infty} e^{i \pi/4} \int_0^R e^{-u^2} du\\ &= e^{i\pi/4} \int_0^\infty e^{-u^2} du\\&=e^{i\pi/4}\frac{\sqrt{\pi}}{2}.\end{align*} Combining (1.1.6) - (1.1.7) one can arrive at the following conclusions in (1.1.8)

$(1.1.8)$

$$\lim_{R \rightarrow \infty}\oint_{R}^{0}e^{ix^{2}} = \int_{0}^{\infty}(cos(x^{2}) + isin(x^{2}) \frac{\sqrt[1]{2}}{2}(1+i)dx = 0i + \frac{\sqrt[1]{\pi}}{2}$$

Comparing the real and imagery parts of $(1.1.8)$ gives in $(1.1.9)$

$(1.1.9)$

$$I=\int_{0}^{\infty}e^{ix^{2}}dx = \int_{0}^{\infty}(cos(x^{2}) + isin(x^{2}))dx = e^{i\pi/4} \sqrt(\pi)/2.$$

• Also a full corresponding mathb.in post has been made. Jun 12, 2018 at 18:23