Cauchy's Integral with quadratic exponential term As I was studying the Cauchy's integral formula, I tried to do the integral:
\begin{equation}
I = \int\limits_{-\infty}^{\infty} \frac{1}{x - a} e^{(i A x^2 + i B x)} dx
\end{equation}
with $A>0, B>0$ and $a > 0$.
Consider an integral on a complex plan:
\begin{equation}
J = \int\limits_{C + C_R}  \frac{1}{z - a} e^{(i A z^2 + i B z)} dz
\end{equation}
where $C$ is along the real axis $-\infty \rightarrow +\infty$ and $C_R$ is the upper half circle $z = Re^{i\theta}$ with $R \rightarrow \infty$ and $\theta \in [0, \pi]$.
Naively, I would expect $C_R$ part of the integral gives zero and $C$ part of the integral gives $I$, then the $I$ can be derived by Cauchy's integral formula.
However, as I tried to check the $C_R$ part of the integral, I found that ($z = Re^{i\theta}$):
$$
\begin{split}
I_R &= \int\limits_0^{\pi} d\theta \frac{iRe^{i\theta}}{Re^{i\theta} - a} \exp\big(iAR^2e^{2i\theta}+iBRe^{i\theta}\big) \\
|I_R| &\leq \int\limits_0^{\pi} d\theta\left |\frac{iRe^{i\theta}}{Re^{i\theta} - a}\right| \Big|\exp\big(iAR^2e^{2i\theta}+iBRe^{i\theta}\big)\Big|
\end{split}
$$
where the first term
\begin{equation}
\left|\frac{iRe^{i\theta}}{Re^{i\theta} - a}\right| \leq \frac{R}{R-a} \rightarrow 1 \ as\ R \rightarrow \infty
\end{equation}
and the second term
\begin{equation}
\left|\exp(iAR^2e^{2i\theta}+iBRe^{i\theta})\right| \leq e^{-AR^2\sin(2\theta) - BR\sin(\theta)}
\end{equation}
will not approach to zero because of $e^{-AR^2\sin(2\theta)}$.
Is there anything wrong in my approach? And is there any other way I can perform this integral $I$?
Thanks a million for advises!
 A: Let me first remove the $Bx$ term by completing the square,
$$I=\int\limits_{-\infty}^{\infty} \frac{e^{i A x^2+iBx}}{x - a}\,dx=e^{-iB^2/4A}\int\limits_{-\infty}^{\infty} \frac{e^{i A x^2}}{x - a-B/2A}\,dx.$$
Mathematica evaluates the Cauchy principal value of the integral in terms of Meijer G-functions,
$$I=-\tfrac{1}{8} \pi ^{-5/2} e^{-iB^2/4A}\biggl\{G_{3,5}^{5,3}\left(\alpha\,\biggl|
\begin{array}{c}
 0,\frac{1}{4},\frac{3}{4} \\
 0,0,\frac{1}{4},\frac{1}{2},\frac{3}{4} \\
\end{array}
\right)+8 \pi ^4 G_{7,9}^{5,3}\left(\alpha\,\biggl|
\begin{array}{c}
 0,\frac{1}{4},\frac{3}{4},-\frac{1}{8},\frac{1}{8},\frac{3}{8},\frac{5}{8} \\
 0,0,\frac{1}{4},\frac{1}{2},\frac{3}{4},-\frac{1}{8},\frac{1}{8},\frac{3}{8},\frac{5}{8} \\
\end{array}
\right)+i G_{3,5}^{5,3}\left(\alpha\,\biggl|
\begin{array}{c}
 \frac{1}{4},\frac{1}{2},\frac{3}{4} \\
 0,\frac{1}{4},\frac{1}{2},\frac{1}{2},\frac{3}{4} \\
\end{array}
\right)+8 \pi ^4 i G_{7,9}^{5,3}\left(\alpha\,\biggl|
\begin{array}{c}
 \frac{1}{4},\frac{1}{2},\frac{3}{4},-\frac{1}{8},\frac{1}{8},\frac{3}{8},\frac{5}{8} \\
 0,\frac{1}{4},\frac{1}{2},\frac{1}{2},\frac{3}{4},-\frac{1}{8},\frac{1}{8},\frac{3}{8},\frac{5}{8} \\
\end{array}
\right)\biggr\},$$
with 
$$\alpha=\left(a+\frac{B}{2A}\right)^4\frac{A^2}{4}.$$
