For all $\xi \in \mathbb{C}$ we have $e^{-\pi\xi^2}=\int_{-\infty}^\infty \! e^{-\pi x^2}e^{2\pi ix\xi}\ \mathrm{d}x.$ This is Exercise 2.4 in Stein & Shakarchi's Complex Analysis. Prove that for all $\xi \in \mathbb{C}$
$$e^{-\pi\xi^2}=\int_{-\infty}^\infty \! e^{-\pi x^2}e^{2\pi ix\xi}\ \mathrm{d}x.$$
They prove it for the real case, so I assume that I'm supposed to use that. All I can think to do is write $\xi=a+ib$, which gives a term $e^{2\pi i xa}$ in the integral which becomes 1, since the power is a real multiple of $2\pi$ and $i$. That leaves
$$e^{-\pi\xi^2}=\int_{-\infty}^\infty \! e^{-\pi x^2}e^{-2\pi xb}\ \mathrm{d}x$$
but I don't have any idea where to go from there, or if this is even a fruitful direction. Any hints would be appreciated.
 A: Write $\xi=a+bi$. The exponent inside the integrand looks like
$$\begin{array}{ll} -\pi x^2+2\pi ix\xi & =-\pi x^2+2\pi ix(a+bi) \\ & =-\pi x^2+2\pi iax-2\pi bx \\ & = -\pi (x^2+2bx)+2\pi iax \end{array}$$
The real part has both an $x$ and an $x^2$ term, but we may complete the square:
$$ \begin{array}{l} = -\pi(x^2+2bx+b^2)+\pi b^2+2\pi iax \\  =-\pi(x+b)^2+\pi b^2+2\pi iax. \end{array} $$
With the substitution $u=x+b$ ($\Leftrightarrow x=u-b$) this becomes
$$ \begin{array}{l} =-\pi u^2+\pi b^2+2\pi ia(u-b) \\ =-\pi u^2+2\pi iau+[\pi b^2-2\pi iab]. \end{array}$$
Therefore
$$ \begin{array}{ll} \displaystyle \int_{-\infty}^{\infty} \exp(-\pi x^2+2\pi ix\xi)\,\mathrm{d}x & \displaystyle =\exp(\pi b^2-2\pi iab)\int_{-\infty}^{\infty} \exp(-\pi u^2+2\pi iau)\,\mathrm{d}u \\ & \displaystyle =\exp(\pi b^2-2\pi iab)\exp(-\pi a^2) \\ & =\exp(-\pi(a+bi)^2) \\ & =\exp(-\pi \xi^2). \end{array} $$
A: Note that the integral
$$
\int_{-\infty}^{\infty} e^{-\pi x^2}e^{-2\pi i x\xi}\,\mathrm{d}x
$$
is the Fourier transform of $e^{-\pi x^2}$. Now simplify
$$
e^{-\pi x^2}e^{-2\pi ix\xi} = e^{-\pi(x^2+2\pi i x)}
$$
and complete the square of $x^2 + 2\pi i x$ to deduce that
$$
x^2 + 2\pi i x = x^2 + 2\pi i x -\xi^2 + \xi^2 = (x+i\xi) + \xi^2.
$$
Substituting, this gives
$$
e^{-\pi(x^2 + 2\pi i x)} = e^{-\pi(x+i\xi)^2-\pi\xi^2} = e^{-\pi(x+i\xi)^2} e^{- \pi\xi^2}
$$
and hence
$$
\int_{-\infty}^{\infty} e^{-\pi(x^2+2\pi ix\xi)}\,\mathrm{d}x= e^{-\pi\xi^2}\int_{-\infty}^{\infty} e^{-\pi(x+i\xi)^2}\,\mathrm{d}x.
$$
Making the substitution $u = x+i\xi$ so that $\mathrm{d}x = \mathrm{d}u$ then yields
$$
e^{-\pi\xi^2}\int_{-\infty}^{\infty} e^{-\pi(x+i\xi)^2}\,\mathrm{d}x = e^{-\pi\xi^2}\int_{-\infty}^{\infty}e^{-\pi u^2}\,\mathrm{d}u= e^{-\pi\xi^2}\frac{\sqrt{\pi}}{\sqrt{\pi}} = e^{-\pi\xi^2},
$$
as was to be shown. Note that this shows that
$$
\mathscr{F}(e^{-\pi x^2}) = e^{-\pi \xi^2},
$$
where $\mathscr{F}$ is the Fourier transform on $L^2(\mathbb{R})$.
A: This is, in my opinion, the most conventional solution that uses Cauchy's theorem. Hope this helps others like myself :)
Let $\gamma$ be a closed parallelogram, starting from the origin, consisting of 4 segments $\gamma_1, \gamma_2, \gamma_3, \gamma_4$ in clockwise orientation sequentially.
plotting of the curve $\gamma$
As a complex function $f(z) = e^{\pi z^2}$ is holomorphic in this simply connected region, we can apply Cauchy's theorem to get
\begin{align*}
    0
    = \int_\gamma f(z)\, dz
    = \underset{(1)}{\underbrace{\int_{\gamma_1} e^{\pi z^2}\, dz}} +
      \underset{(2)}{\underbrace{\int_{\gamma_2} e^{\pi z^2}\, dz}} +
      \underset{(3)}{\underbrace{\int_{\gamma_3} e^{\pi z^2}\, dz}} +
      \underset{(4)}{\underbrace{\int_{\gamma_4} e^{\pi z^2}\, dz}}
\end{align*}
For $(1)$,
\begin{align*}
 \lim_{R \rightarrow \infty} (1)
    = \lim_{R \rightarrow \infty} \int_{-R}^R e^{\pi(ix)^2} i\, dx
    = i \int_{-\infty}^\infty e^{-\pi x^2} i\, dx
    = i
\end{align*}
Assume $\xi = a + ib$, $(2)$ is bounded absolutely by
\begin{align*}
 |(2)|
    &= \left| \int_{\gamma_2} e^{\pi z^2}\, dz \right| \\
    &\leq \sup_{0 \leq t \leq 1} \left| exp(\pi (\xi t + iR)^2)\, dz \right| \cdot 1 \\
    &=\sup_{0 \leq t \leq 1} exp(\pi ((ta)^2-(tb+R)^2))\, dz \\
\end{align*}
which goes to $0$ as $R \rightarrow \infty$, so that $\lim_{R \rightarrow \infty} (2) = 0$.
Similarly, $\lim_{R \rightarrow \infty} (4) = 0$.
For $(3)$, it can be written as
\begin{align*}
 (3)
    = \int_R^{-R} exp(\pi(\xi + iy)^2) i\, dy
    = -i\int_{-R}^R exp(\pi\xi^2 -\pi y^2 + 2\pi i\xi y)\, dy
    = - i e^{\pi\xi^2} \int_{-R}^R e^{-\pi y^2} e^{2 \pi i \xi y}\, dy
\end{align*}
Assembling $(1), (2), (3)$ gives,
\begin{align*}
 i = 
 i e^{\pi\xi^2} \int_{-R}^R e^{-\pi y^2} e^{2 \pi i \xi y}\, dy
\end{align*}
Dividing both sides with ${i e^{\pi\xi^2}}$ gives the desired answer.
