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Doing complex analysis, I encountered a problem that I do not know how to solve. I am to prove the Fresnel integrals from $x = 0$ to infinity, using a contour integral of $e^{i x^2}$. The hint said to use Jordan's lemma, but that pertains to a function $e^{ix}$ times a function $G(x)$, but as far as I can tell there is no way to pick $G$ so that the total ends up as $e^{i x^2}$.

Does anyone know what to do?

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    Inside an integral, $x$ is $x^2$. Just change variables, but be careful.2012-08-01

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If you're using the "usual" contour to do these integrals you don't even need Jordan's lemma:

$\Gamma:=[0,R]\cup\gamma_R\cup\{z\in\Bbb C\,:\,z=te^{\pi i/4}\,,\,0\leq t\leq R\}\,,\,R\in\Bbb R^+ $ with $\,\gamma_R:=\{z\in \Bbb C\,:\,z=Re^{i\theta}\,,\,0\leq \theta\leq \pi/4\}\,$ . Then, as $\,f(z):=e^{iz^2}\,$ analytic everywhere, we get $0=\oint_\Gamma f(z)\,dz=\stackrel{I}{\int_0^R e^{ix^2}dx}+\stackrel{II}{\int_{\gamma_R}e^{iz^2}dz}-\stackrel{III}{\int_0^Re^{iz^2}dz}$ (the minus sign before III is due to the fact that we "walk" the contour in the positive direction) , and we have:

$I\,\longrightarrow\,\int_0^Re^{ix^2}dx\xrightarrow[R\to\infty]{} \int_0^\infty\cos x^2\,dx+i\int_0^R\sin x^2\,dx$

$II\,\longrightarrow\,\left|\int_{\gamma_R}e^{iz^2}dz\right|\leq\max_{z\in\gamma_R}\left|e^{iz^2}\right|\cdot\frac{R\pi}{3}=\frac{R\pi}{3\,e^{R^2}}\xrightarrow [R\to\infty]{} 0$

$III\,\longrightarrow\,z=te^{\pi i/4}\Longrightarrow dz=e^{\pi i/4}dt\,,\,z^2=it^2\Longrightarrow \int_0^Re^{iz^2}dz=e^{\pi i/4}\int_0^Re^{i^2t^2e^{\pi i/2}}dt=$

$=e^{\pi i/4}\int_0^Re^{-t^2}dt\xrightarrow [R\to\infty]{}\frac{1}{\sqrt 2}(1+i)\sqrt{\frac{\pi}{2}}=\frac{\sqrt \pi}{2\sqrt 2}+i\frac{\sqrt \pi}{2\sqrt 2}$ From the above, letting $\,R\to\infty\,$ and comparing real and imaginary parts, we finally get

$0=\int_0^\infty\cos x^2\,dx=\int_0^\infty \sin x^2\,dx=\frac{\sqrt \pi}{2\sqrt 2}$

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    @André, no it isn't, but who cares? The estimation theorem only talks of a smooth (piecewise, even) path with finite length.2013-05-10