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!
