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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. – Siminore Aug 1 '12 at 9:27

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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Thanks so far! I don't understand how you worked out the integral II though? Why is it at most this and that, and why is the max of e^iz^2 = 1/e^R^2 ? – Boudewijn Aug 1 '12 at 20:45
Integral II's estimation is just...yes, the [estimation lemma][1], and $$iz^2=i(R^2e^{2i\theta})=i(R^2[\cos 2\theta+i\sin theta)]=-R^2\sin 2\theta + iR^2\cos 2\theta$$ and using now that for $\,x\in\Bbb R\,\,,\,|e^{ix}|=1\,$ we get $$|e^{iz^2}|=e^{-R^2\sin 2\theta}\xrightarrow [R\to\infty]{} 0$$You're right, no inequality's needed there as $\,\sin 2\theta\geq 0\,\,,\text{for}\,\,0\leq\theta\leq \pi/4$ [1] – DonAntonio Aug 1 '12 at 23:59
what if $\theta = 0$ ? – Epsilon May 9 '13 at 14:41
Right @André , then define $\,\gamma_r\,$ only for $\,|z|=R\;,\;\text{Im}(z)>0\,$ ...:) Thanks. – DonAntonio May 9 '13 at 20:14
Is $\Gamma$ then still closed ? I think so but I am not sure :D – Epsilon May 10 '13 at 8:48

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