Translated complex gaussian-type integral: $\int_0^{\infty} \exp(i(t-\alpha)^2) dt$

Question

It's fairly straight forward to show that

$$ \int_0^{\infty} \exp(it^2) dt = \frac{\sqrt{\pi}}{2}\exp\left(i\frac{\pi}{4}\right) $$

via complex contour integration over a contour shaped like a piece of pie:

However, I am trying to determine the value of a similar integral with similar integrand where $\alpha \in \mathbb{C}$ and $\text{Im}(\alpha) < 0 $:

$$ \int_0^{\infty} \exp(i(t-\alpha)^2) dt $$

The above integral works nicely because along the top left curve, $t=r\exp(i\pi/4)$, so that $it^2 = -r^2$. I thought maybe I could set up the equation: $$ i(r^2\exp(i2\theta)-\alpha)^2 = -r^2 $$ to determine some fixed $\theta$ (perhaps dependent on $\alpha$) along which to integrate, but this didn't really get me anywhere. Any ideas?

Answer 1

I'll summarize the Dr. MV's comments above as an answer:

If $\alpha = a+bi$, the change of variables $t \to t-\alpha$, changes the path of integration to the ray in the complex plane, $\left\{ -a+iy : y \in [-b,\infty) \right\}$. Since the integrand is holomorphic, the integral is path independent, and thus equivalent to integrating form $\alpha$ to the origin, and from the origin to $(-a,\infty)$. This requires the use of the imaginary error function which has no closed-form solution.

Answer 2

By change of variables, the integral you are trying to evaluate is simply $$ \int_{\alpha}^{\alpha+\infty}e^{it^2}\ dt, $$ which means "integrate along the horizontal ray starting at $\alpha$".

While you haven't provided your argument handling the case $\alpha=0$, you can adapt this argument to handle other $\alpha$ as well; the contour will shift, but should still wrap around the correct singularity (provided $\alpha$ belongs to a certain domain).