Question:
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
Attempt:
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)
