= Solution
Use the <Fourier transform> normalization $\widehat f(\xi)=\int_{\mathbb R}f(x)e^{-2\pi ix\xi}\,dx$. A precise version of the <Poisson summation formula> is that, for every <Schwartz function> $f\in\mathcal S(\mathbb R)$,
$$
\boxed{\sum_{n\in\mathbb Z}f(n)=\sum_{m\in\mathbb Z}\widehat f(m).}
$$
Both series have <absolute convergence>. Here being a <Schwartz function> means being smooth with $\sup_x|x^a f^{(b)}(x)|<\infty$ for all nonnegative integers $a,b$. This hypothesis makes the sums and the termwise operations below legitimate; the formula is not being asserted for arbitrary integrable <functions>.
Form the <periodization of a Schwartz function> $P_f(x)=\sum_n f(x+n)$. Every differentiated series has <uniform convergence> on $[0,1]$, by rapid decay, so $P_f$ is a smooth <periodic function> of period one. Its $m$th <Fourier coefficient> is
$$
\int_0^1P_f(x)e^{-2\pi imx}\,dx
=\sum_n\int_n^{n+1}f(u)e^{-2\pi imu}\,du=\widehat f(m).
$$
The interchange follows from <absolute convergence> uniformly on this interval. <Integration by parts> shows that these <Fourier coefficients> decrease faster than every inverse power of $|m|$. Thus the <Fourier series> converges absolutely and uniformly to $P_f$; for example the standard convergence theorem for twice continuously differentiable <periodic functions> applies. Evaluating at zero proves the <Poisson summation formula>.
For $t>0$, scaling the given Gaussian <Fourier transform> gives
$$
\widehat{e^{-\pi t x^2}}(\xi)=t^{-1/2}e^{-\pi\xi^2/t}.
$$
Applying the <Poisson summation formula> gives the real-parameter <Jacobi theta function> transformation
$$
\boxed{\theta(t)=\sum_{n\in\mathbb Z}e^{-\pi n^2t}=t^{-1/2}\theta(1/t).}
$$
For $\operatorname{Re}s>1$, termwise application of the <Mellin transform> is justified by integrating absolute values and gives
$$
\Lambda(s):=\pi^{-s/2}\Gamma(s/2)\zeta(s)
=\frac12\int_0^\infty(\theta(t)-1)t^{s/2-1}\,dt.
$$
Each positive $n$ contributes $\Gamma(s/2)(\pi n^2)^{-s/2}$, and the factor one half removes the equal positive and negative terms. This is the <Mellin representation of the completed Riemann zeta function>.
Split the integral at one. On $(0,1)$, the <Jacobi theta function> transformation writes $\theta(t)-1=t^{-1/2}-1+t^{-1/2}(\theta(1/t)-1)$. Integrating the first two terms and substituting $u=1/t$ in the third yields the <pole-subtracted theta integral for the completed zeta function>:
$$
\Lambda(s)=\frac1{s-1}-\frac1s+
\frac12\int_1^\infty(\theta(t)-1)
\left(t^{s/2-1}+t^{(1-s)/2-1}\right)\,dt.
$$
The remaining integral is an <entire function> of $s$: $\theta(t)-1=O(e^{-\pi t})$, and on each compact set of $s$ this exponential dominates all powers of $t$ and all factors arising from differentiation. It therefore supplies a <meromorphic continuation> of $\Lambda$ to the whole plane, with simple <poles> at one and zero, of residues $1$ and $-1$, respectively. Its expression is unchanged by $s\mapsto1-s$.
The reciprocal <Gamma function> is entire, with simple zeros at its nonpositive integer arguments. Consequently
$$
\zeta(s)=\pi^{s/2}\Gamma(s/2)^{-1}\Lambda(s)
$$
is holomorphic everywhere except for a simple <pole> at $s=1$, of residue one. The apparent <pole> at $s=0$ cancels; in fact $\Gamma(s/2)^{-1}\sim s/2$ gives $\zeta(0)=-1/2$. This proves the <analytic continuation> of the <Riemann zeta function>. The completed <functional equation of the Riemann zeta function> is
$$
\boxed{\pi^{-s/2}\Gamma(s/2)\zeta(s)
=\pi^{-(1-s)/2}\Gamma((1-s)/2)\zeta(1-s).}
$$
These are equalities of <meromorphic functions>; at apparent singularities they are interpreted by continuation. Equivalently, $\xi(s)=\tfrac12s(s-1)\Lambda(s)$ is an <entire function> with $\xi(s)=\xi(1-s)$.
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