<s>[INST] $\int_{-\infty}^{\infty}e^{-\pi x^2}\cdot e^{-2\pi ix\xi}dx = e^{\pi\xi^2}$ Prove that for all $\xi \in \mathbb{C}$,
$$\int_{-\infty}^{\infty}e^{-\pi x^2}\cdot e^{-2\pi ix\xi}dx = e^{\pi\xi^2}$$
I don't really know how to compute this integral. Can you please help me?
 [/INST] By noticing that 
$$a x^{2} + b x = \left( \sqrt{a} x + \frac{b}{2 \sqrt{a}} \right)^{2} - \frac{b^{2}}{4 a}$$
then
\begin{align}
I &= \int_{-\infty}^{\infty} e^{-a x^{2} - b x} \, dx \\
&= e^{\frac{b^{2}}{4 a}} \, \int_{-\infty}^{\infty} e^{- \left( \sqrt{a} x + \frac{b}{2 \sqrt{a}} \right)^{2}} \, dx 
\end{align}
Making the change $t = \sqrt{a} x + \frac{b}{2 \sqrt{a}}$ leads to
\begin{align}
I &= \frac{1}{\sqrt{a}} \, e^{\frac{b^{2}}{4 a}} \, \int_{-\infty}^{\infty} e^{-t^{2}} \, dt \\
&= \frac{2}{\sqrt{a}} \, e^{\frac{b^{2}}{4 a}} \, \int_{0}^{\infty} e^{-t^{2}} \, dt \\
&= \frac{1}{\sqrt{a}} \, e^{\frac{b^{2}}{4 a}} \, \int_{0}^{\infty} e^{- u} \, u^{-1/2} \, du \hspace{5mm} \mbox{ where } t = \sqrt{u} \\
&= \sqrt{\frac{\pi}{a}} \, e^{\frac{b^{2}}{4 a}}.
\end{align}
Hence 
\begin{align}
\int_{-\infty}^{\infty} e^{-a x^{2} - b x} \, dx = \sqrt{\frac{\pi}{a}} \, e^{\frac{b^{2}}{4 a}}.
\end{align}
For the case $a = \pi$ and $b = 2 \pi i \, \eta$ the result becomes
\begin{align}
\int_{-\infty}^{\infty} e^{- \pi x^{2} - 2 \pi i \eta x} \, dx =  e^{- \pi \eta^{2}}.
\end{align}
</s>