Approximation of error function-type integral

[RedSonja]

Hi! How do I approximate the integral
\begin{equation} \int_0^{\infty} dt \:e^{-iA(t-B)^2} \end{equation}
with A, B real, A > 0, and B=b \cos\theta where 0 \leq \theta < 2\pi?
I guess for B\ll 0 the lower limit may be extended to - \infty to yield a full complex gaussian integral, but what about B \geq 0? And what happens for A \gg 1 and A \ll 1 respectively?
Thanks for your help!

 

[HallsofIvy]

If you let x= \sqrt{A}(t- B) then the integral becomes
\frac{1}{\sqrt{A}}\int_{-\sqrt{A}B}^\infty e^{-x^2}dx

 

[RedSonja]

Ok, so for

\begin{equation}
\frac{1}{\sqrt{A}} \int_{-\sqrt{A}B}^{\infty} dx \: e^{-ix^2}
\end{equation}

with \sqrt{A}B large and positive we may extend the limit to -\infty and obtain \sqrt{\frac{\pi}{A}} e^{-i\frac{\pi}{4}}, and for \sqrt{A}B\approx 0 we get half of that, but what happens for \sqrt{A}B negative? I suppose we get 0 for large, negative \sqrt{A}B (?), but I don't know how to handle the approximation for "intermediate" negative \sqrt{A}B.

Can anyone help? Thanks!