= Solution
Use the <Fourier transform> convention $\widehat g(\xi)=\int_{\mathbb R}g(x)e^{-2\pi ix\xi}\,dx$. For $t>0$, completing the square in the <Gaussian integral> gives
$$
g_t(x)=e^{-\pi tx^2},\qquad \widehat g_t(\xi)=t^{-1/2}e^{-\pi\xi^2/t}.
$$
Both functions are rapidly decreasing, so the <Poisson summation formula> applies:
$$
\sum_{n\in\mathbb Z}e^{-\pi tn^2}=t^{-1/2}\sum_{n\in\mathbb Z}e^{-\pi n^2/t}.
$$
In terms of the <Jacobi theta function>, this is $\vartheta(i/t)=\sqrt t\,\vartheta(it)$. The theta series converges locally uniformly in the <complex upper half-plane>, defining a <holomorphic function>. The branch of $\sqrt{-i\tau}$ with positive value at $\tau=it$ is holomorphic there, because $-i\tau$ lies in the right half-plane. Both sides of the proposed inversion identity are holomorphic; agreement on the positive imaginary axis and the <identity theorem> give
$$
\boxed{\vartheta(-1/\tau)=\sqrt{-i\tau}\,\vartheta(\tau).}
$$
Equivalently, the complex Gaussian has transform $(-i\tau)^{-1/2}\exp(-\pi i\xi^2/\tau)$, which produces the same identity directly by <Poisson summation>.
Put $\psi(t)=(\vartheta(it)-1)/2=\sum_{n\geq1}e^{-\pi n^2t}$. For $\operatorname{Re}s>1$, integrate term by term in its <Mellin transform>:
$$
Z(s):=\int_0^\infty\psi(t)t^{s/2-1}\,dt
=\pi^{-s/2}\Gamma(s/2)\zeta(s).
$$
The <theta function> inversion gives
$$
\psi(t)=t^{-1/2}\psi(1/t)+\frac{t^{-1/2}-1}{2}.
$$
Split the integral at one. In the term containing $\psi(1/t)$ put $t=1/u$. The elementary remainder integral is
$$
\frac12\int_0^1\left(t^{(s-1)/2-1}-t^{s/2-1}\right)dt
=\frac1{s-1}-\frac1s.
$$
Thus the <pole-subtracted theta integral for the completed zeta function> is
$$
\boxed{Z(s)=\int_1^\infty\psi(t)\left(t^{s/2-1}+t^{(1-s)/2-1}\right)dt+\frac1{s-1}-\frac1s.}
$$
The integral is entire by exponential decay. This supplies a <meromorphic continuation> with simple poles at zero and one, of residues $-1$ and $1$. The expression is unchanged by $s\mapsto1-s$, proving the <functional equation of the Riemann zeta function>
$$
\boxed{\pi^{-s/2}\Gamma(s/2)\zeta(s)=\pi^{-(1-s)/2}\Gamma((1-s)/2)\zeta(1-s).}
$$
The reflection and duplication identities $\Gamma(z)\Gamma(1-z)=\pi/\sin\pi z$ and $\Gamma(z)\Gamma(z+1/2)=2^{1-2z}\sqrt\pi\Gamma(2z)$ convert it to
$$
\boxed{\zeta(s)=2^s\pi^{s-1}\sin(\pi s/2)\Gamma(1-s)\zeta(1-s).}
$$
Multiplying $Z$ by $s(s-1)/2$ removes its two poles and yields the <Riemann xi function>, which is an <entire function>, satisfying $\xi(s)=\xi(1-s)$.
Back to article page
