How to evaluate $\lim_{\epsilon\rightarrow 0^+}\int_{-\infty\cdot e^{i\epsilon}}^{\infty\cdot e^{i\epsilon}}\, dz\,e^{-iz^2}$? How to evaluate 
$$\lim_{\epsilon\rightarrow 0^+}\int_{-\infty\cdot e^{i\epsilon}}^{\infty\cdot e^{i\epsilon}}\, dz\,e^{-iz^2}?$$
Can we evaluate this integral using the stationary phase approximation?
 A: 
The integral $\displaystyle \lim_{L\to \infty}\int_{-Le^{i\epsilon}}^{Le^{i\epsilon}}e^{-iz^2}\,dz$ diverges for all $0<\epsilon<\pi/2$.
To see this, we enforce the substitution $\displaystyle z=e^{i\epsilon}t$ to find 
$$\begin{align}\int_{-Le^{i\epsilon}}^{Le^{i\epsilon}}e^{-iz^2}\,dz&=e^{i\epsilon}\int_{-L}^L e^{-ie^{i2\epsilon}t^2}\,dt\\\\
&=e^{i\epsilon}\int_{-L}^L e^{\sin(2\epsilon)t^2}e^{-i\cos(2\epsilon)t^2}\,dt\tag 1
\end{align}$$
For $0<\epsilon<\pi/2$, the exponential growth of the integrand on the right-hand side of $(1)$ renders the integral divergent as $L\to \infty$.  If $-\pi/2<\epsilon<0$, however, then the integral converges as $L\to \infty$.  We now proceed to evaluate the limit of the integral on the left-hand side of $(1)$ when $-\pi/2<\epsilon<0$.

Let $L>0$ and $-\pi/2<\epsilon<0$ be fixed and $C_\pm$ be the closed contour comprised of the straight line segments (i) from $\pm L$ to $\pm Le^{i\epsilon}$, (ii) from $\pm Le^{i\epsilon}$ to $0$, (iii)  from $0$ to $\pm L$.
Since $e^{-iz^2}$ is entire, Cauchy's Integral Theorem guarantees that 
$$\begin{align}
0&=\oint_{C_\pm } e^{-iz^2}\,dz\\\\
&=\int_{\pm L}^{\pm Le^{i\epsilon}}e^{-iz^2}\,dz+\int_{\pm Le^{i\epsilon}}^{0}e^{-iz^2}\,dz+\int_{0}^{\pm L}e^{-iz^2}\,dz \tag 2
\end{align}$$
Using $(2)$ reveals
$$\int_{-Le^{i\epsilon}}^{Le^{i\epsilon}}e^{-iz^2}\,dz=\int_{-L}^L e^{iz^2}\,dz +\int_{L}^{Le^{i\epsilon}}e^{-iz^2}\,dz-\int_{-L}^{-Le^{i\epsilon}}e^{-iz^2}\,dz \tag 3$$
As $L \to \infty$, the second and third integrals on the right-hand side of $(3)$ approach $0$.  Hence, for $-\pi/2<\epsilon<0$,
$$\lim_{L \to \infty}\int_{-Le^{i\epsilon}}^{Le^{i\epsilon}}e^{-iz^2}\,dz=\lim_{L\to \infty}\int_{-L}^L e^{-iz^2}\,dz =(1-i)\sqrt{\frac\pi2}$$
