I wonder if the integral $\int_{-\infty}^{{\infty}}e^{-\alpha x^2}=\sqrt{\frac{\pi}{\alpha}}$, for $\alpha\neq 0$, how could the integral $\int_{-\infty}^{\infty} e^{-(2\pi x +i\omega)^2}dx$ be simplified as follows: $\omega \in \mathbb{R}$:
$$\int_{-\infty}^{\infty} e^{-(2\pi x +i\omega)^2}dx=\int_{-\infty}^{\infty} e^{-(2\pi x)^2}dx=\frac{1}{2\pi}\int_{-\infty}^{\infty} e^{-x^2}dx=\frac{1}{2\sqrt{\pi}}.$$ I understand how to pass from the second equality to last one via scaling $x\rightarrow \frac{x}{2\pi}$, however to get the immediate second part from the first needs some complex integration identity. What sort of identity allows us to do that? I count on your answer.
$$ \eqalign{ & I(n) = \int_{x = - n}^{\;n} {e^{\, - \left( {2\pi x + iw} \right)^{\,2} } dx} = e^{\,w^{\,2} } \int_{x = - n}^{\;n} {e^{\, - 4\pi ^2 x^2 } e^{\, - 4iw\pi x} dx} = \cr & = e^{\,w^{\,2} } \left( {\int_{x = - n}^{\;n} {e^{\, - 4\pi ^2 x^2 } \cos \left( {\,4w\pi x} \right)dx} - i\int_{x = - n}^{\;n} {e^{\, - 4\pi ^2 x^2 } \sin \left( {\,4w\pi x} \right)dx} } \right) = \cr & = e^{\,w^{\,2} } 2\int_{x = 0}^{\;n} {e^{\, - 4\pi ^2 x^2 } \cos \left( {\,4w\pi x} \right)dx} = \cr & = {{2e^{\,w^{\,2} } } \over {4w\pi }}\int_{x = 0}^{\;n} {e^{\, - \left( {{{2\pi } \over {4w\pi }}} \right)^2 \left( {4w\pi x} \right)^2 } \cos \left( {4w\pi x} \right)d\left( {4w\pi x} \right)} = \cr & = {{2e^{\,w^{\,2} } } \over {4w\pi }}\int_{y = 0}^{\;4w\pi n} {e^{\, - \left( {{1 \over {2w}}} \right)^2 y^2 } \cos y\,dy} \cr} $$ and $$ \mathop {\lim }\limits_{n\; \to \,\infty } I(n) = {{2e^{\,w^{\,2} } } \over {4w\pi }}{{\sqrt \pi e^{\, - w^{\,2} } } \over {2{1 \over {2w}}}} = {1 \over {2\sqrt \pi }} $$