Author Topic: Q4 TUT 0202  (Read 800 times)

Victor Ivrii

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Q4 TUT 0202
« on: October 26, 2018, 05:52:17 PM »
Integrate $e^{iz^2}$ around the contour y shown in Figure below to obtain the Fresnel
integrals:
$$
\int_0^\infty \cos(x^2)\,dx = \int_0^\infty \sin(x^2)\,dx=\frac{\sqrt{2\pi}}{4}.
$$
(Use that $\int_0^\infty e^{-x^2}\,dx =\frac{\sqrt{\pi}}{2}$).


$z=x$, $0\le x\le R$;  $z=Re^{i\theta}$, $0\le \theta\le \frac{\pi}{4}$; $z=te^{i\pi/4}$, $R\ge t\ge 0$; $R\to \infty$.

Yuechen Huang

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Re: Q4 TUT 0202
« Reply #1 on: October 26, 2018, 06:53:54 PM »
Answer is attached.

Victor Ivrii

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Re: Q4 TUT 0202
« Reply #2 on: October 26, 2018, 09:23:10 PM »
Very difficult to read

Jeffery Mcbride

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Re: Q4 TUT 0202
« Reply #3 on: November 07, 2018, 06:24:36 PM »
By Cauchy's theorem, we get:
\begin{equation*}
0\ =\ \int _{\gamma } e^{iz^{2}} dz\ =\int ^{R}_{0} e^{ix^{2}} dx\ +\int ^{\pi /4}_{0} ie^{i\left( Re^{i\theta }\right)^{2}} Re^{i\theta } d\theta \ +\ \int ^{0}_{R} e^{ir^{2} \ e^{i\pi /2}} e^{\pi /4} \ dr\\
\\
=\int ^{R}_{0} e^{ix^{2}} dx\ +\int ^{\pi /4}_{0} e^{iR^{2}( cos2\theta \ +\ isin( 2\theta ) \ } iRe^{i\theta } d\theta \ -\ e^{i\pi /4} \ \int ^{R}_{0} e^{ir^{2}( cos\pi /2\ _{\ } +\ isin\pi /2)} dr\\
\\
=\int ^{R}_{0} e^{ix^{2}} dx\ +\int ^{\pi /4}_{0} e^{iR^{2}( cos2\theta \ +\ isin( 2\theta ) \ } iRe^{i\theta } d\theta \ -\ e^{i\pi /4} \ \int ^{R}_{0} e^{-r^{2}} dr\\
\\
\\
\end{equation*}

Now, just handling the middle integral, we get:


\begin{equation*}
\int ^{\pi /4}_{0} e^{iR^{2}( cos2\theta \ +\ isin( 2\theta ) \ } iRe^{i\theta } d\theta \\
=R\int ^{\pi /4}_{0} e^{iR^{2} cos2\theta \ -\ R^{2} sin( 2\theta ) \ } ie^{i\theta } d\theta \\
This\ is\ estimated\ above\ by:\\
\\
R\int ^{\pi /4}_{0} e^{-R^{2} sin( 2\theta ) \ } d\theta \ \leq R\int ^{\pi /4}_{0} e^{-2R^{2}\frac{\theta }{\pi } \ } d\theta \ < \ \frac{\pi }{2R}\\
\end{equation*}


\begin{equation*}
So\ the\ limit\ as\ R\ \rightarrow \ \infty ,\ this\ will\ be\ 0,\ and\ we\ will\ get:\\
0=\int ^{R}_{0} e^{ix^{2}} dx\ -\ e^{i\pi /4} \ \int ^{R}_{0} e^{-r^{2}} dr\\
0=\int ^{\infty }_{0}\left( cos\left( x^{2}\right) +\ isin\left( x^{2}\right) dx\right) \ -\ \frac{1+i}{\sqrt{2}} \ \int ^{\infty }_{0} e^{-r^{2}} dr\\
\frac{1+i}{\sqrt{2}} \ \int ^{\infty }_{0} e^{-r^{2}} dr\ =\ \int ^{\infty }_{0} cos\left( x^{2}\right) dx+\ i\int ^{\infty }_{0} sin\left( x^{2}\right) dx\\
\frac{1+i}{\sqrt{2}} \ \frac{\sqrt{\pi }}{2} \ =\ \int ^{\infty }_{0} cos\left( x^{2}\right) dx+\ i\int ^{\infty }_{0} sin\left( x^{2}\right) dx\\
\frac{\sqrt{2\pi }}{4} \ +\ i\ \frac{\sqrt{2\pi }}{4} \ =\ \int ^{\infty }_{0} cos\left( x^{2}\right) dx+\ i\int ^{\infty }_{0} sin\left( x^{2}\right) dx\\
\\
\end{equation*}

So, we separate imaginary and real parts and we get

\begin{equation*}
\int ^{\infty }_{0} cos\left( x^{2}\right) dx\ =\ \int ^{\infty }_{0} sin\left( x^{2}\right) dx\ =\frac{\sqrt{2\pi }}{4} \
\end{equation*}

« Last Edit: November 09, 2018, 01:42:45 PM by Jeff Mcbride »

Jeffery Mcbride

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Re: Q4 TUT 0202
« Reply #4 on: December 02, 2018, 01:45:54 PM »
Never got any feedback on this answer, is it correct?

Victor Ivrii

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Re: Q4 TUT 0202
« Reply #5 on: December 02, 2018, 03:48:23 PM »
Indeed, correct