= Solution
Use the <Jacobi theta function> $\Theta(t)=\sum_{n\in\mathbb Z}e^{-\pi n^2t}$, $t>0$. Its transformation law follows directly from the <Gaussian Fourier transform>. With the <Fourier transform> convention $\widehat f(\xi)=\int_{\mathbb R}f(u)e^{-2\pi i\xi u}\,du$,
$$
\widehat{e^{-\pi t u^2}}(\xi)=t^{-1/2}e^{-\pi\xi^2/t}.
$$
For example, differentiating the transform and using <integration by parts> gives $\widehat f'(\xi)=-(2\pi\xi/t)\widehat f(\xi)$; its value at zero is the <Gaussian integral> $t^{-1/2}$. Periodize the Gaussian. Its <Fourier coefficients> are these transforms at <integer> frequencies, and both series converge absolutely, so evaluating its <Fourier series> at zero proves the <Poisson summation formula> here:
$$
\Theta(t)=t^{-1/2}\Theta(1/t).
$$
For $\Re s>1$, integrating the exponentially convergent series termwise gives the <Mellin representation of the completed Riemann zeta function>
$$
\Lambda(s):=\pi^{-s/2}\Gamma(s/2)\zeta(s)
=\frac12\int_0^\infty(\Theta(t)-1)t^{s/2-1}\,dt.
$$
Split at one and substitute $u=1/t$ in the lower part, using the transformation just proved. The contribution of $t^{-1/2}-1$ is $1/(s-1)-1/s$. We obtain 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 integral is <entire>: $\Theta(t)-1$ decays exponentially, uniformly dominating the integrand and all its $s$ derivatives on <compact sets>. Both the rational term and the integral are invariant under $s\mapsto1-s$. This proves the meromorphic identity
$$
\boxed{\pi^{-s/2}\Gamma(s/2)\zeta(s)
=\pi^{-(1-s)/2}\Gamma((1-s)/2)\zeta(1-s).}
$$
Equivalently the <Riemann xi function>
$$
\xi(s)=\tfrac12s(s-1)\Lambda(s)
$$
is <entire> and satisfies $\xi(s)=\xi(1-s)$. The reflection and duplication formulas for the <Gamma function> turn the same identity into
$$
\boxed{\zeta(s)=2^s\pi^{s-1}\sin(\pi s/2)\Gamma(1-s)\zeta(1-s),}
$$
understood by <meromorphic continuation> at removable exceptional points. This establishes the <functional equation of the Riemann zeta function>, rather than assuming it.
If $N(T)$ counts zeros $\rho=\beta+i\gamma$ in the <critical strip> with $0<\gamma\le T$, \b[including <zero multiplicity>], the <Riemann–von Mangoldt formula> is
$$
\boxed{N(T)=\frac{T}{2\pi}\log\frac{T}{2\pi}-\frac{T}{2\pi}+O(\log T).}
$$
At a zero ordinate one may instead use the conventional half-weight boundary count; that only alters the stated error. In particular $N(T)\sim T\log T/(2\pi)$.
To invert without assuming simple zeros, fix $0<\varepsilon<1$ and set $T_n^\pm=(1\pm\varepsilon)2\pi n/\log n$. Then $\log T_n^\pm/\log n\to1$, so
$$
\frac{N(T_n^\pm)}n\longrightarrow1\pm\varepsilon.
$$
For large $n$, $N(T_n^-)<n<N(T_n^+)$, and hence $T_n^-<\gamma_n\le T_n^+$. Let $\varepsilon\downarrow0$. The <asymptotic inversion of the zeta zero count> gives
$$
\boxed{\gamma_n\sim\frac{2\pi n}{\log n}.}
$$
As in the zero-counting formula, the sequence lists ordinates with <zero multiplicity>; “increasing” means nondecreasing here. The argument does not require the <Riemann hypothesis> or simplicity of zeros.
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