Show that

$$\lim_{R\rightarrow \infty} \int_{-R}^Re^{iz^2}dz = \sqrt \pi e^{i\pi/4} + \mathcal O \bigg(\frac 1R \bigg)$$

by using contour integration


First I observed that

$$\int_{-R}^Re^{iz^2}dz =2\int_0^Re^{iz^2}dz$$

So I chose the contour $\Gamma_R = \gamma_1 * \gamma_2 * \gamma_3$ where

\begin{alignat}{2} \gamma_1(t) & = Re^{it} \qquad & t \in \Big[0,\frac \pi 2\Big] \\ \gamma_2(t) & = (R-t)e^{i\pi/4} \qquad & t \in [0,R] \\ \gamma_3(t) & = t \qquad & t \in [0,R] \end{alignat}

which is a sector of radius $R$ and angle $\pi /4$.

The integrand $e^{iz^2}$ is an entire function, so by Cauchy's Theorem we have

$$\int_{\Gamma_R}e^{iz^2}dz \equiv 0$$

for all $R$. Moreover, I was able to estimate the integral along the arc $\gamma_1$ as

$$\bigg|\int_{\gamma_1} e^{iz^2}dz \bigg| \leq \mathcal O\bigg(\frac 1R \bigg) \qquad \text{as } R\rightarrow \infty$$

However, I am stuck with the integral along $\gamma_2$:

\begin{align} \int_{\gamma_2}e^{iz^2}dz & = \int_0^R \exp \Big(i{\underbrace{\big[(R-t)e^{i\pi/4}\big]}_{\gamma_2(t)}}^2\Big) \cdot -e^{i\pi/4}dt \\ & = -e^{i\pi/4}\int_0^R\exp\big(-(R-t)^2\big)dt \\ & = -e^{i\pi/4}\int_0^R\exp\big(-t^2\big)dt \\ & = ? \end{align}

How should I proceed? (I feel like I have already gone wrong somewhere)

  • $\begingroup$ Probably you have misprinted in definition of $\gamma_1$. $\endgroup$
    – user
    Feb 14, 2019 at 22:07

1 Answer 1


In fact you have already done it as $$ \int_0^\infty \exp (-t^2)dt=\frac {\sqrt\pi}2. $$ Quite surely the value of this integral was assumed to be known.


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