= Solution
Use the <Fourier transform> convention $\widehat f(u)=\int_{\mathbb R}f(x)e^{-2\pi iux}\,dx$. A version of the <Poisson summation formula> sufficient here is
$$
\boxed{\sum_{n\in\mathbb Z}f(n)=\sum_{m\in\mathbb Z}\widehat f(m)
\qquad(f\in\mathcal S(\mathbb R)).}
$$
Both sums converge absolutely. To prove it, form the <periodization of a Schwartz function> $P(x)=\sum_{n\in\mathbb Z}f(x+n)$. The rapid decay of $f$ and all its derivatives makes this series and its differentiated series uniformly convergent on $[0,1]$, so $P$ is smooth and one-periodic. Its <Fourier coefficients> are
$$
\begin{aligned}
\int_0^1P(x)e^{-2\pi imx}\,dx
&=\sum_n\int_0^1f(x+n)e^{-2\pi imx}\,dx\\
&=\int_{\mathbb R}f(y)e^{-2\pi imy}\,dy=\widehat f(m).
\end{aligned}
$$
Repeated integration by parts makes these coefficients rapidly decreasing. Thus $\sum_m\widehat f(m)e^{2\pi imx}$ converges absolutely and uniformly and has the same coefficients as $P$. By uniqueness of <Fourier series> it equals $P$. Evaluation at zero proves the formula.
For $u>0$, the Gaussian $f(x)=e^{-\pi ux^2}$ has transform $u^{-1/2}e^{-\pi m^2/u}$. This follows from the <Gaussian integral> at frequency zero and integration by parts, which gives the differential equation $\widehat f'(v)=-(2\pi v/u)\widehat f(v)$. Hence the <Jacobi theta function>
$$
\Theta(u)=\sum_{n\in\mathbb Z}e^{-\pi n^2u}
$$
satisfies $\Theta(u)=u^{-1/2}\Theta(1/u)$.
Use the normalization of the <completed Riemann zeta function>
$$
\Xi(s)=\pi^{-s/2}\Gamma(s/2)\zeta(s).
$$
Initially, for $\Re s>1$, the <Gamma function> integral and <absolute convergence> give its <Mellin transform> representation
$$
\Xi(s)=\frac12\int_0^\infty(\Theta(u)-1)u^{s/2-1}\,du.
$$
Indeed, the two equal terms for $n$ and $-n$ cancel the factor $1/2$, and the $n$th integral is $\pi^{-s/2}\Gamma(s/2)n^{-s}$.
Split at $u=1$ and use the theta transformation in the integral over $(0,1)$. Its elementary part is
$$
\frac12\int_0^1(u^{-1/2}-1)u^{s/2-1}\,du
=\frac1{s-1}-\frac1s.
$$
In the remaining part substitute $v=1/u$. Define
$$
J(s)=\int_1^\infty(\Theta(u)-1)u^{s/2-1}\,du.
$$
The <pole-subtracted theta integral for the completed zeta function> is therefore
$$
\boxed{\Xi(s)=\frac1{s-1}-\frac1s+\frac12\bigl(J(s)+J(1-s)\bigr).}
$$
Since $\Theta(u)-1=O(e^{-\pi u})$ for $u\geq1$, both integrals and every derivative in $s$ converge uniformly on compact subsets of $\mathbb C$. Thus $J$ is an <entire function>. The boxed identity supplies the <meromorphic continuation> of $\Xi$ to the whole plane, with no other singularities. Its <poles> really are simple, because the residues of the displayed rational terms are nonzero:
$$
\boxed{\operatorname{Res}_{s=0}\Xi=-1,\qquad
\operatorname{Res}_{s=1}\Xi=1.}
$$
The identity also gives $\Xi(s)=\Xi(1-s)$, the completed <functional equation of the Riemann zeta function>.
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