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