$\int_{-\infty}^{\infty}e^{-\pi x^2}\cdot e^{-2\pi ix\xi}dx = e^{\pi\xi^2}$ 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?
 A: 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}
A: Here is a solution avoiding the residu theorem or any change of variables over the complex plane.
Let $g(\xi)$ be the left hand side :$$g(\xi)=\int_{\mathbb R} e^{-\pi x^2}e^{-2i\pi x\xi}dx.$$
One can easily check that $f:\mathbb R\times \mathbb C\rightarrow \mathbb C$ given by $f:(x,\xi)\mapsto e^{-\pi x^2}e^{-2i\pi x\xi}$ satisfies :


*

*$f$ is measruable,

*$\xi\mapsto f(x,\xi)$ is holomorphic for every $x\in \mathbb R$,

*for any $R>0$, $$|f(x,z)|\leq e^{-\pi x^2}e^{2\pi x R}$$ where the right hand side belongs to $L^1(\mathbb R)$. 


Hence, $g$ is a holomorphic function over $\mathbb C$ and one has $$ g'(\xi) = \int_{\mathbb R} \dfrac{\partial f}{\partial \xi}(x,\xi)dx.$$
However, using integration by parts, we get :
$$\begin{array}{rcl} g'(\xi) & = & \displaystyle \int_{\mathbb R} \dfrac{\partial f}{\partial \xi}(x,\xi)dx\\
& = & \displaystyle  i\int_{\mathbb R} -2x\pi e^{-\pi x^2}e^{-2i\pi x\xi}dx \\
& = & -2\pi \xi g(\xi).\end{array}$$
Since $g(0)=1$ (this is a Gaussian integral), $g$ satisfies $$\left\{\begin{array}{l} y'(\xi)=-2\pi y(\xi) \\ y(0)=1\end{array}\right.$$ and so does $\xi \mapsto \exp(-\pi \xi ^2)$.
Thus, $$g(\xi)=\int_{\mathbb R} e^{-\pi x^2}e^{-2i\pi x\xi}dx=\exp(\color{red}-\pi \xi ^2).$$
A: Hint: 
$$-\pi x^2 - 2\pi i x \xi = -\pi(x^2+2ix\xi) = -\pi(x+i\xi)^2+(?)$$
Try to figure out what $(?)$ should be, and you should get something resembling a known integral. You might have to do some substitutions and/or contour deformations.
