Fourier transform of $e^{-4\pi ^2 x^2}$ How do you prove $$\int_{-\infty}^{\infty}e^{-(2\pi x + i\xi/2)^2}dx=\int_{-\infty}^{\infty}e^{-(2\pi x)^2}dx$$ for $\xi \in \mathbb{R}$.
The Question arises from calculating the Fourier Transform for the function $e^{-4\pi ^2 x^2}$
Substituting $2\pi x$ with $x$ I suppose it is true that 
$$\int_{-\infty}^{\infty}e^{-(x + i\xi)^2}dx=\int_{-\infty}^{\infty}e^{-x^2}dx$$ 
 A: The integrand is entire so we can use Cauchy's theorem.
Consider the curve $\gamma_M$ that joins the points
$-M, M, M+iy, -M+iy, -M$ (where $y = {\xi \over 4 \pi}$).
Show that the portions $M,M+iy$ and $-M+iy, -M$ are bounded by an expression of the form $B e^{-(2 \pi M)^2}$ (for $M$ sufficiently large). Then switch the
limits of integration on the 'upper' integral.
A: First note by absorbing some constants into $\xi$ that you just need to prove
$$\int_{\mathbb{R}}e^{-4\pi^2x^2}dx =  \int_{\mathbb{R}} e^{-4\pi^2(x+\xi i)^2}dx  $$
for any real $\xi$.
To prove this, do a contour integral around the rectangle with corners $\pm R$ and $\pm R + \xi$ (we will send $R$ to infinity). The integral is zero as the integrand is holomorphic. So you just need to show that the integral along the two 'vertical' parts of the contour go to zero with $R$.
One of these contributions will be 
$$ \int_0^{\xi} i e^{-4\pi^2(R+it)^2} dt . $$
Note that 
$$ \bigg{\vert} \int_0^{\xi} i e^{-4\pi^2(R+it)^2} dt \bigg{\vert}  \leq \int_0^{\xi} e^{-4\pi^2(R^2-t^2)} dt \rightarrow 0    $$
as $R \rightarrow \infty$ (by the bounded convergence theorem). The other 'vertical' contribution is similar. 
A: $$
    \int_{-\infty}^{\infty}e^{-(2\pi x+i\xi/2)^{2}}dx = \int_{-\infty}^{\infty}e^{-4\pi^{2}x^{2}}e^{-2\pi i \xi x}dx\; e^{\xi^{2}/4}.
$$
If you know the Fourier transform of $e^{-4\pi^{2}x^{2}}$, you can evaluate at $\xi$, and you're done.
