As I was studying the Cauchy’s integral formula, I tried to do the integral:

begin{equation}

I(x) = int_{-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(z) = int_{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 $z = Re^{itheta}$ with $R rightarrow infty$ and $theta in (0, 2pi)$.

Naively, I would expect $C_R$ part of the integral gives zero and $C$ part of the integral gives $I(x)$, then the $I(x)$ can be derived by Cauchy’s integral formula.

However, as I tried to check the $C_R$ part of the integral, I found that:

begin{array}

$I_R(z = Re^{itheta}) &=& int_0^{pi} dtheta frac{iRe^{itheta}}{Re^{itheta} – a} exp(iAR^2e^{2itheta}+iBRe^{itheta})

end{array}

begin{array}

$|I_R| &leq & int_0^{pi} dtheta |frac{iRe^{itheta}}{Re^{itheta} – a}| |exp(iAR^2e^{2itheta}+iBRe^{itheta})|

end{array}

where the first term

begin{equation}

|frac{iRe^{itheta}}{Re^{itheta} – a}| leq frac{R}{R-a} rightarrow 1 as R rightarrow infty

end{equation}

and the second term

begin{equation}

|exp(iAR^2e^{2itheta}+iBRe^{itheta})| leq e^{-AR^2sin(2theta) – BRsin(theta)}

end{equation}

will not approach to zero because of $e^{-AR^2sin(2theta)}$.

Is there anything wrong in my approach? And is there any other way I can perform this integral I(x)?

Thanks a million for advises!