linguistic:paraphrasing
Riemann's functional equation

This zeta function satisfies the functional equation ζ (s) = 2 s π s − 1 sin (π s 2) Γ (1 − s) ζ (1 − s), {\displaystyle \zeta (s)=2^{s}\pi ^{s-1}\ \sin \left({\frac {\pi s}{2}}\right)\ \Gamma (1-s)\ \zeta (1-s),} where Γ(s) is the gamma function. This is an equality of meromorphic functions valid on the whole complex plane. The equation relates values of the Riemann zeta function at the points s and 1 − s, in particular relating even positive integers with odd negative integers. Owing to the zeros of the sine function, the functional equation implies that ζ (s) has a simple zero at each even negative integer s = −2 n, known as the trivial zeros of ζ (s). When s is an even positive integer, the product sin(πs / 2)Γ(1 − s) on the right is non-zero because Γ(1 − s) has a simple pole, which cancels the simple zero of the sine factor.

A proof of the functional equation proceeds as follows:We observe that if σ > 0 {\displaystyle \sigma >0}, then ∫ 0 ∞ x 1 2 s − 1 e − n 2 π x d x = Γ (s 2) n s π s 2. {\displaystyle \int _{0}^{\infty }x^{{1 \over 2}{s}-1}e^{-n^{2}\pi x}\,dx={\Gamma \left({s \over 2}\right) \over {n^{s}\pi ^{s \over 2}}}.}

As a result, if σ > 1 {\displaystyle \sigma >1} then Γ (s 2) ζ (s) π s / 2 = ∑ n = 1 ∞ ∫ 0 ∞ x s 2 − 1 e − n 2 π x d x = ∫ 0 ∞ x s 2 − 1 ∑ n = 1 ∞ e − n 2 π x d x, {\displaystyle {\frac {\Gamma \left({\frac {s}{2}}\right)\zeta (s)}{\pi ^{s/2}}}=\sum _{n=1}^{\infty }\int _{0}^{\infty }x^{{s \over 2}-1}e^{-n^{2}\pi x}\,dx=\int _{0}^{\infty }x^{{s \over 2}-1}\sum _{n=1}^{\infty }e^{-n^{2}\pi x}\,dx,} with the inversion of the limiting processes justified by absolute convergence (hence the stricter requirement on σ {\displaystyle \sigma }).

For convenience, let ψ (x):= ∑ n = 1 ∞ e − n 2 π x {\displaystyle \psi (x):=\sum _{n=1}^{\infty }e^{-n^{2}\pi x}}

which is a special case of the theta function. Then ζ (s) = π s 2 Γ (s 2) ∫ 0 ∞ x 1 2 s − 1 ψ (x) d x. {\displaystyle \zeta (s)={\pi ^{s \over 2} \over \Gamma ({s \over 2})}\int _{0}^{\infty }x^{{1 \over 2}{s}-1}\psi (x)\,dx.}

By the Poisson summation formula we have ∑ n = − ∞ ∞ e − n 2 π x = 1 x ∑ n = − ∞ ∞ e − n 2 π x, {\displaystyle \sum _{n=-\infty }^{\infty }{e^{-n^{2}\pi x}}={1 \over {\sqrt {x}}}\sum _{n=-\infty }^{\infty }{e^{-n^{2}\pi \over x}},}

so that 2 ψ (x) + 1 = 1 x { 2 ψ (1 x) + 1 }. {\displaystyle 2\psi (x)+1={1 \over {\sqrt {x}}}\left\{2\psi \left({1 \over x}\right)+1\right\}.}

Hence π − s 2 Γ (s 2) ζ (s) = ∫ 0 1 x s 2 − 1 ψ (x) d x + ∫ 1 ∞ x s 2 − 1 ψ (x) d x. {\displaystyle \pi ^{-{s \over 2}}\Gamma \left({s \over 2}\right)\zeta (s)=\int _{0}^{1}x^{{s \over 2}-1}\psi (x)\,dx+\int _{1}^{\infty }x^{{s \over 2}-1}\psi (x)\,dx.}

This is equivalent to ∫ 0 1 x s 2 − 1 { 1 x ψ (1 x) + 1 2 x − 1 2 } d x + ∫ 1 ∞ x s 2 − 1 ψ (x) d x {\displaystyle \int _{0}^{1}x^{{s \over 2}-1}\left\{{1 \over {\sqrt {x}}}\psi \left({1 \over x}\right)+{1 \over 2{\sqrt {x}}}-{1 \over 2}\right\}\,dx+\int _{1}^{\infty }x^{{s \over 2}-1}\psi (x)\,dx} or 1 s − 1 − 1 s + ∫ 0 1 x s 2 − 3 2 ψ (1 x) d x + ∫ 1 ∞ x s 2 − 1 ψ (x) d x. {\displaystyle {1 \over {s-1}}-{1 \over s}+\int _{0}^{1}x^{{s \over 2}-{3 \over 2}}\psi \left({1 \over x}\right)\,dx+\int _{1}^{\infty }x^{{s \over 2}-1}\psi (x)\,dx.}

So π − s 2 Γ (s 2) ζ (s) = 1 s (s − 1) + ∫ 1 ∞ (x − s 2 − 1 2 + x s 2 − 1) ψ (x) d x {\displaystyle \pi ^{-{s \over 2}}\Gamma \left({s \over 2}\right)\zeta (s)={1 \over {s({s-1})}}+\int _{1}^{\infty }\left({x^{-{{s} \over 2}-{1 \over 2}}+x^{{{s} \over 2}-1}}\right)\psi (x)\,dx}

which is convergent for all s, so holds by analytic continuation. Furthermore, the RHS is unchanged if s is changed to 1 − s. Hence π − s 2 Γ (s 2) ζ (s) = π − 1 2 + s 2 Γ (1 2 − s 2) ζ (1 − s) {\displaystyle \pi ^{-{s \over 2}}\Gamma \left({s \over 2}\right)\zeta (s)=\pi ^{-{1 \over 2}+{s \over 2}}\Gamma \left({1 \over 2}-{s \over 2}\right)\zeta (1-s)} which is the functional equation. Attributed to Bernhard Riemann.

The functional equation was established by Riemann in his 1859 paper " On the Number of Primes Less Than a Given Magnitude " and used to construct the analytic continuation in the first place.