Integral of Complex Gaussian: $\int_{-\infty}^{\infty} e^{-(2\pi x +i\omega)^2}dx$. 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.
 A: $$
\eqalign{
  & I(n) = \int_{x =  - n}^{\;n} {e^{\, - \left( {2\pi x + iw} \right)^{\,2} } dx}  = e^{\,w^{\,2} } \int_{x =  - n}^{\;n} {e^{\, - 4\pi ^2 x^2 } e^{\, - 4iw\pi x} dx}  =   \cr 
  &  = e^{\,w^{\,2} } \left( {\int_{x =  - n}^{\;n} {e^{\, - 4\pi ^2 x^2 } \cos \left( {\,4w\pi x} \right)dx} 
 - i\int_{x =  - n}^{\;n} {e^{\, - 4\pi ^2 x^2 } \sin \left( {\,4w\pi x} \right)dx} } \right) =   \cr 
  &  = e^{\,w^{\,2} } 2\int_{x = 0}^{\;n} {e^{\, - 4\pi ^2 x^2 } \cos \left( {\,4w\pi x} \right)dx}  =   \cr 
  &  = {{2e^{\,w^{\,2} } } \over {4w\pi }}\int_{x = 0}^{\;n} {e^{\, - \left( {{{2\pi } \over {4w\pi }}} \right)^2 \left( {4w\pi x} \right)^2 }
 \cos \left( {4w\pi x} \right)d\left( {4w\pi x} \right)}  =   \cr 
  &  = {{2e^{\,w^{\,2} } } \over {4w\pi }}\int_{y = 0}^{\;4w\pi n} {e^{\, - \left( {{1 \over {2w}}} \right)^2 y^2 } \cos y\,dy}  \cr} 
$$
and
$$
\mathop {\lim }\limits_{n\; \to \,\infty } I(n) = {{2e^{\,w^{\,2} } } \over {4w\pi }}{{\sqrt \pi  e^{\, - w^{\,2} } } \over {2{1 \over {2w}}}} = {1 \over {2\sqrt \pi  }}
$$
