# Approximation of error function-type integral

1. Oct 24, 2012

### RedSonja

Hi! How do I approximate the integral
$$\int_0^{\infty} dt \:e^{-iA(t-B)^2}$$
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?

Last edited: Oct 24, 2012
2. Oct 24, 2012

### HallsofIvy

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

3. Oct 24, 2012

### RedSonja

Ok, so for

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

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!

Last edited: Oct 24, 2012