Contour integration of $\int_{-\infty}^{\infty}e^{iax^2}dx$

Consider the following integral:

$$\int_{-\infty}^{\infty}e^{iax^2}dx$$

Here I believe we have to consider the two cases when $a<0$ and $a>0$, as they need different contours. For $a>0$ I've been using the upper half circle, radius $R$. I shall call the whole semi circle $\Gamma$ and the curved section $\Gamma_1$

As the function is holomorphic we have:

$$\int_\Gamma e^{iaz^2} dz = \int_{-R}^R e^{iax^2} dx + \int_{\Gamma_1} e^{iaz^2} dz =0$$

$$\Rightarrow \int_{-R}^R e^{iax^2} dx = - \int_{\Gamma_1} e^{iaz^2} dz = -\int_{0}^\pi iRe^{i\theta} e^{iaR^2e^{2i\theta}} d\theta$$

It's this point i'm stuck on, any help would be greatly appreciated!

EDIT: So from the comments, I've seen my method was incorrect, if anyone could outline this other contour it would be great, I have not seen it before.

• For $a > 0$, you should not use the upper half circle, you should use two 1/8 circles for $\theta \in [0,\frac{\pi}{4}]$ and $[\pi,\frac{5\pi}{4}]$ and the line from $[-\infty e^{i\frac{\pi}{4}}, +\infty e^{i^\frac{\pi}{4}}]$. – achille hui Mar 17 '13 at 13:08
• @achillehui Thanks for the hint, I will try that now – Freeman Mar 17 '13 at 13:13
• please note that you don't need to evaluate the integrals on the two arcs, you need to bound them and show they converge to $0$ as $R \to \infty$. – achille hui Mar 17 '13 at 13:19
• The integral along the line is a regular gaussian integral. If you don't want to quote the result, you can use the trick to evaluate its square in polar coordinate: $$\left(\int_{-\infty}^{\infty} e^{-\frac{x^2}{2}} dx\right)^2 = \iint e^{-\frac{r^2}{2}} r drd\theta = 2\pi$$ – achille hui Mar 17 '13 at 13:28
• For $a>0$, Jordan's inequality is not applicable unless the angle the sector makes with the positive real axis is $\frac{\pi}{4}$ or less. A half circle makes an angle of $\pi$ with the positive real axis. For $f(z) = e^{iaz^{3}}$, the angle would need to be $\frac{\pi}{6}$ or less. In general for $b>1$, the angle would need to be $\frac{\pi}{2b}$ or less. – Random Variable Mar 17 '13 at 16:15

If $a<0$ we have $b:=-a>0$, and $$J(a):=\int_{-\infty}^\infty e^{iax^2}\,dx=\int_{-\infty}^\infty e^{-ibx^2}\,dx=\overline{J(b)}=\overline{J(-a)}.$$ Therefore, it is enough to evaluate $J(a)$ for $a>0$. So from now on we suppose that $a>0$.

Notice that $$J(a)=\int_{-\infty}^\infty e^{iax^2}\,dx=\frac{1}{\sqrt{a}}\int_{-\infty}^\infty e^{ix^2}\,dx=\frac{1}{\sqrt{a}}J(1).$$ Hence, we only have to compute $J(1)$.

Given $r>0$ we denote by $\Gamma_R$ the boundary of $$\Omega_r=\left\{z \in \mathbb{C}:\ |z|\le r,\ 0\le \arg z\le \frac{\pi}{4}\right\}.$$ Consider the parametrization $\gamma_r:[0,1] \to \mathbb{C}$ of $\Gamma_r$ given by $$\gamma_r(t)=\begin{cases} \gamma_1(3t) & \text{ for } 0 \le t \le \frac13\\ \gamma_2(3t-1) & \text{ for } \frac13 < t \le \frac23\\ \gamma_3(3t-2) & \text{ for } \frac23 < t \le 1 \end{cases},$$ where $$\gamma_1(t)= rt,\ \gamma_2(t)= r\exp\left(i\frac{\pi t}{4}\right),\ \gamma_3(t)= r(1-t)\exp\left(i\frac{\pi}{4}\right).$$ Since $$f: \mathbb{C} \to \mathbb{C},\ f(z):=e^{iz^2}$$ is holomorphic and $\Omega_r$ is simply connected, we have thanks to the Cauchy Integral formula: $$0=:\int_{\Gamma_r}f(z)\,dz=\sum_{k=1}^3I_k(r),$$ with $$I_k(r):=\int_{\frac{k-1}{3}}^{\frac{k}{3}}\gamma_R'(t)f(\gamma_R(t))\,dt, \ k=1,2,3.$$ I claim that $$\lim_{r\to \infty}I_2(r)=0.$$ In fact, since $\sin \theta \ge \frac{\theta}{2}$ for $\theta \le \theta \le \frac{\pi}{2}$, we have: \begin{eqnarray} |I_2(r)|&=&\left|\int_0^13\gamma_2'(3t-1)f(\gamma_2(3t-1))\,dt\right|=\left|\int_0^1\gamma_2'(t)f(\gamma_2(t))\,dt\right|\\ &\le& \int_0^1\frac{\pi r}{4}\left|\exp\left\{ir^2\cos\left(\frac{\pi t}{2}\right)-r^2\sin\left(\frac{\pi t}{2}\right)\right\}\right|\,dt\\ &=&\int_0^1\frac{r\pi}{4}\exp\left[-r^2\sin\left(\frac{\pi t}{2}\right)\right]\,dt\\ &\le&\int_0^1\frac{r\pi}{4}\exp\left(-\frac{r^2\pi t}{4}\right)\,dt =\frac{1}{r}\left[1-\exp\left(-\frac{\pi r^2}{4}\right)\right]. \end{eqnarray} Thus $\lim_{r \to \infty}I_2(r)=0$ as claimed.

Clearly \begin{eqnarray} I_3(r)&=&\int_{\frac23}^13\gamma_3'(3t-2)f(\gamma_3(3t-2))\,dt=\int_0^1\gamma_3'(t)f(\gamma_3(t))\,dt\\ &=&-r\exp\left(i\frac{\pi}{4}\right)\int_0^1\exp\left(-r^2t^2\right)\,dt=-\exp\left(i\frac{\pi}{4}\right)\int_0^{r}\exp\left(-x^2\right)\,dx\\ I_1(r)&=&\int_0^{\frac13}3\gamma_1'(3t)f(\gamma_1(3t))\,dt=r\int_0^1\exp\left(ir^2t^2\right)\,dt=\int_0^{r}\exp\left(ix^2\right)\,dx. \end{eqnarray} So we have $$0=\sum_{k=1}^3I_k(r)=\int_0^{r} e^{ix^2}\,dx-\exp\left(i\frac{\pi}{4}\right)\int_0^{r} e^{-x^2}\,dx+I_2(r).$$ This implies that \begin{eqnarray} J(1)&=&2\int_0^\infty e^{ix^2}\,dx=2\exp\left(i\frac{\pi}{4}\right)\int_0^\infty e^{-x^2}\,dx=\exp\left(i\frac{\pi}{4}\right)\int_{-\infty}^\infty e^{-x^2}\,dx\\ &=&\sqrt{\pi}\exp\left(i\frac{\pi}{4}\right)=(1+i)\sqrt{\frac{\pi}{2}}. \end{eqnarray} Thus we have $$J(a)=\begin{cases} (1+i)\sqrt{\frac{\pi}{2a}} & \text{ for } a>0\\ (1-i)\sqrt{\frac{\pi}{-2a}} & \text{ for } a<0 \end{cases}.$$

• Note that your $J(a)$ exists for no real number $a$. – Did Jun 30 '13 at 13:35
• @Did yes it does! this is a well-known integral, see e.g. en.wikipedia.org/wiki/Fresnel_integral – Mercy King Jun 30 '13 at 13:38
• Sorry but no. On the very page you link to: "The limits of these functions as x goes to infinity are known..." So you might define $J(a)$ as the limit when $R\to\infty$ of an integral from $-R$ to $+R$. But the integral on $(-\infty,\infty)$ does not exist since $e^{iax^2}$ does not converge to zero when $|x|\to\infty$. – Did Jun 30 '13 at 13:45
• @Did An integral over $(-\infty,\infty)$ is always defined as the limit of an integral over some $(r,R)$ when $r\to -\infty$ and $R\to \infty$, or you could say the sum of two limits: $\lim_{r \to -\infty}\int_r^0(\ldots)+\lim_{R\to \infty}\int_0^R(\ldots)$! – albmiz-mth Jul 6 '13 at 6:42
• @albmiz-mth No it is not, not always (grab a textbook). – Did Jul 6 '13 at 7:15