= Solution
Use $e(t)=e^{2\pi it}$ and the <Fourier transform> convention $\widehat f(\xi)=\int_{\mathbb R}f(x)e(-x\xi)\,dx$, initially for an <integrable function>. For the interval <indicator function>, direct integration gives
$$
\widehat{1_{[0,1]}}(\xi)=\begin{cases}(1-e(-\xi))/(2\pi i\xi),&\xi\ne0,\\1,&\xi=0.\end{cases}
$$
The integral bound is one, and the displayed numerator has modulus at most two. By the <convolution theorem>, the <Fourier transform> of the $m$-fold <convolution> is its $m$th power. Thus the <Fourier decay of repeated interval convolutions> gives
$$
\boxed{|\widehat f(\xi)|\le\min(1,(\pi|\xi|)^{-m})\ll\min(1,|\xi|^{-m}).}
$$
At zero, interpret the second term of the minimum as infinity.
The <Schwartz space> $\mathcal S(\mathbb R)$ consists of <smooth functions> for which every <Schwartz seminorm> $\sup_x|x^a f^{(b)}(x)|$ is finite, for nonnegative integers $a,b$. <Differentiation under the integral sign> and <integration by parts> give
$$
\widehat f^{(b)}(\xi)=\widehat{(-2\pi ix)^bf}(\xi),\qquad
(2\pi i\xi)^a\widehat f^{(b)}(\xi)=\widehat{\frac{d^a}{dx^a}\bigl((-2\pi ix)^bf(x)\bigr)}(\xi).
$$
All differentiated integrands are <integrable>, and their boundary terms vanish because of rapid decay. The last transform is bounded by its integrand's $L^1$ <norm>. The same argument applies at every derivative order, so $\widehat f\in\mathcal S(\mathbb R)$.
For $f\in\mathcal S(\mathbb R)$, the <Poisson summation formula> is
$$
\boxed{\sum_{n\in\mathbb Z}f(n)=\sum_{k\in\mathbb Z}\widehat f(k).}
$$
To prove it, form the <periodization of a Schwartz function> $F(x)=\sum_n f(x+n)$. Rapid decay gives <uniform convergence> of this series and of every differentiated series on $[0,1]$, so $F$ is a smooth <periodic function>. Its <Fourier coefficients> are
$$
\int_0^1F(x)e(-kx)\,dx=\sum_n\int_n^{n+1}f(u)e(-ku)\,du=\widehat f(k).
$$
The series $\sum_k\widehat f(k)e(kx)$ converges absolutely and uniformly. Its <Fourier coefficients> equal those of $F$. The uniqueness theorem for <Fourier series> says that two <integrable functions> with the same coefficients agree almost everywhere; as both functions here are continuous, they agree everywhere. Evaluating at zero proves <Poisson summation>.
The <Gaussian Fourier transform> of $e^{-\pi tx^2}$, $t>0$, is $t^{-1/2}e^{-\pi\xi^2/t}$. Apply <Poisson summation> to obtain the transformation of the <Jacobi theta function>
$$
\vartheta(t)=\sum_{n\in\mathbb Z}e^{-\pi n^2t},\qquad \vartheta(t)=t^{-1/2}\vartheta(1/t).
$$
For $\Re s>1$, termwise integration in the <Mellin transform> gives the <Mellin representation of the completed Riemann zeta function>
$$
\Lambda(s):=\pi^{-s/2}\Gamma(s/2)\zeta(s)=\frac12\int_0^\infty(\vartheta(t)-1)t^{s/2-1}\,dt.
$$
Split at one and substitute $t=1/u$ in the lower integral. The theta transformation gives
$$
\Lambda(s)=\frac1{s-1}-\frac1s+\frac12\int_1^\infty(\vartheta(t)-1)\bigl(t^{s/2-1}+t^{(1-s)/2-1}\bigr)\,dt.
$$
The integral is an <entire function> of $s$, since $\vartheta(t)-1$ decays exponentially. This <pole-subtracted theta integral for the completed zeta function> supplies a <meromorphic continuation> with simple <poles> at zero and one. Both the rational term and the integral are unchanged by $s\mapsto1-s$, proving the <functional equation of the Riemann zeta function>
$$
\boxed{\pi^{-s/2}\Gamma(s/2)\zeta(s)=\pi^{-(1-s)/2}\Gamma((1-s)/2)\zeta(1-s).}
$$
Equivalently the <Riemann xi function> $\xi(s)=\frac12s(s-1)\Lambda(s)$ is <entire> and satisfies $\xi(s)=\xi(1-s)$. The reflection and duplication identities for the <Gamma function> also give $\zeta(s)=2^s\pi^{s-1}\sin(\pi s/2)\Gamma(1-s)\zeta(1-s)$ by <meromorphic continuation>.
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