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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

$$ \lim_{R \rightarrow \infty}\bigg | \oint_{\gamma_{R}^{2}}e^{iz^{2}} dz \bigg| \rightarrow 0 $$

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  • $\begingroup$ 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 $(1)$ ? $\endgroup$
    – Zophikel
    Feb 25, 2018 at 22:59
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    $\begingroup$ See Fresnel Integral. $\endgroup$ Feb 25, 2018 at 23:01
  • $\begingroup$ @FelixMarin thanks for the hint in due time i'll be able to answer my own question $\endgroup$
    – Zophikel
    Feb 25, 2018 at 23:13
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    $\begingroup$ You're welcome !!!. $\endgroup$ Feb 25, 2018 at 23:27
  • $\begingroup$ Also is my current progress on the problem accurate ? $\endgroup$
    – Zophikel
    Feb 25, 2018 at 23:53

1 Answer 1

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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 $(1.1.9)$

$(1.1.9)$

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

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  • $\begingroup$ Also a full corresponding mathb.in post has been made. $\endgroup$
    – Zophikel
    Jun 12, 2018 at 18:23

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