I would like to compute the integral $\int_0^{\infty} e^{-(1-i \xi) y^2} dy$. Instead of doing a crazy complex contour integral, I want to use a complex continuation method.
Let $f(z) = \int_0^{\infty} e^{-z y^2} dy$ and $g(z) = \frac 1 2 \sqrt{\frac{\pi} z}$. Both of these functions are defined for $z \in \mathbb{C}$ such that $Re(z) > 0$.
I know that $g$ is holomorphic in the complex half plane, but I don't know how to prove that $f$ is as well.
Since they are holomorphic and agree on positive real values, they agree everywhere in the complex half plane (where the real component is positive). In particular, they agree for $z = 1 - i \xi$. Therefore, $$\int_0^{\infty} e^{-(1-i \xi) y^2} dy = \frac 1 2 \sqrt{\frac{\pi}{1 - i\xi}}.$$
Any help would be appreciated in proving that the function $f$ is holomorphic.