On a Question in Fourier Transform

Someone asked about the second last line below:

Where $A,\omega\in \R$. And I want to record the explanation:

Fact. For any $y\in \R$, we define $g(y):\R\to \R$ by $$g(y) = \int_{-\infty}^\infty e^{-A(x+iy)^2}\,dx,$$ then $g'(y)=0$, and therefore $$\int_{-\infty}^\infty e^{-A(x+iy)^2}\,dx = g(y) = g(0) = \int_{-\infty}^\infty e^{-Ax^2}\,dx ,$$ and the last one is well-known that can be converted to the standard normal density function.

Proof. Just note that $g(-y)=g(y)$ and therefore $g'(-y)(-1) = g'(y)$, next by the formula of $g'(y)$ we can show that $g'(-y)=g'(y)$, therefore $$g'(y) = -g'(-y) = -g'(y) \implies g'(y)=0.\qed$$