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We wish to evaluate the integral $I=\int_{0}^{\infty}e^{ix^{2}}dx$. 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 Semicircular 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 $$ \lim_{R \rightarrow \infty}\bigg | \oint_{\gamma_{R}^{2}}e^{iz^{2}} dz \bigg| \rightarrow 0 $$ $\text{Lemma (1.1)}$ $\text{Cauchy Integral Theorem}$ $(1.1.3)$ 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 $(1.1.9)$ $(1.1.9)$ $$I=\int_{0}^{\infty}e^{ix^{2}}dx = \int_{0}^{\infty}(cos(x^{2}) + isin(x^{2}))dx = e^{i\pi/4} \sqrt(\pi)/2$$
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Tue, 12 Jun 2018 18:22 GMT