The functional equation is
$$
\boxed{\pi^{-s/2}\Gamma(s/2)\zeta(s)
=\pi^{-(1-s)/2}\Gamma((1-s)/2)\zeta(1-s)}.
$$
Equivalently,
$$
\zeta(s)=2^s\pi^{s-1}\sin(\pi s/2)\Gamma(1-s)\zeta(1-s).
$$

For a proof, let
$$
\Theta(u)=\sum_{n\in\mathbb Z}e^{-\pi n^2u}.
$$
The <Poisson summation formula> applied to a <Gaussian function> gives the theta transformation
$$
\Theta(u)=u^{-1/2}\Theta(1/u).
$$
The standard <Gamma function> integral and termwise integration initially give, for $\Re s>1$,
$$
\Lambda(s):=\pi^{-s/2}\Gamma(s/2)\zeta(s)
=\frac12\int_0^\infty(\Theta(u)-1)u^{s/2}\frac{du}{u}.
$$
Split the integral at one, substitute $u\mapsto1/u$ in the lower half, and use the theta transformation. The result is
$$
\Lambda(s)=\frac1{s(s-1)}
+\frac12\int_1^\infty(\Theta(u)-1)
\left(u^{s/2}+u^{(1-s)/2}\right)\frac{du}{u}.
$$
The integral is an <entire function> of $s$ because $\Theta(u)-1$ decays exponentially. The right side is visibly invariant under $s\mapsto1-s$, proving both the <analytic continuation> and the <Functional equation of the Riemann zeta function>. This is the <Mellin representation of the completed Riemann zeta function>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 150 / 2 / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/iii/paper-150/2/b/solution