# 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$(1)$? – Zophikel Feb 25 '18 at 22:59 • See Fresnel Integral. – Felix Marin Feb 25 '18 at 23:01 • @FelixMarin thanks for the hint in due time i'll be able to answer my own question – Zophikel Feb 25 '18 at 23:13 • You're welcome !!!. – Felix Marin Feb 25 '18 at 23:27 • Also is my current progress on the problem accurate ? – Zophikel Feb 25 '18 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{2}}{2}(1+i)dx = 0i + \frac{\sqrt{\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. – Zophikel Jun 12 '18 at 18:23