jordans lemma application 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?
 A: 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
$$0=\int_0^\infty\cos x^2\,dx=\int_0^\infty \sin x^2\,dx=\frac{\sqrt \pi}{2\sqrt 2}$$
