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  "body": "\n## Statement\n\nLet $\\zeta$ be the [[Riemann zeta function|1XWBY4N0VT]] and $\\Gamma$ the [[gamma function|0H5SYBSED9]].\nThen, as an identity of [[meromorphic functions|0ZZ7ZYDR8W]] on $\\C$,\n$$\\zeta(s) = 2^s \\pi^{s-1} \\sin\\!\\left(\\frac{\\pi s}{2}\\right) \\Gamma(1 - s) \\, \\zeta(1 - s).$$\nEquivalently, the completed zeta function $\\xi_0(s) = \\pi^{-s/2} \\, \\Gamma(s/2) \\, \\zeta(s)$ satisfies $\\xi_0(s) = \\xi_0(1 - s)$ for all $s \\in \\C \\setminus \\set{0, 1}$.\n\n## Remarks\n\nOne proof uses the theta function $\\theta(t) = \\sum_{n \\in \\Z} e^{-\\pi n^2 t}$ for $t \u003e 0$.\nPoisson summation gives $\\theta(1/t) = \\sqrt{t} \\, \\theta(t)$.\nA Mellin transform expresses $\\xi_0(s)$ as an integral of $(\\theta(t) - 1)/2$ against $t^{s/2 - 1}$.\nSplitting the integral at $t = 1$ and using the symmetry of $\\theta$ gives a formula that is visibly invariant under $s \\mapsto 1 - s$.\nThe same formula gives the meromorphic continuation of $\\zeta$ to $\\C$.\nExample: at $s = -1$ the right side is $\\frac{1}{2} \\cdot \\frac{1}{\\pi^2} \\cdot (-1) \\cdot \\Gamma(2) \\cdot \\zeta(2) = -\\frac{1}{12}$, using $\\zeta(2) = \\pi^2/6$, so $\\zeta(-1) = -1/12$.\nAt $s = -2, -4, -6, \\dots$ the sine factor vanishes while $\\Gamma(1 - s) \\zeta(1 - s)$ is finite and nonzero, so these are zeros of $\\zeta$, the trivial zeros.\n\n## Explanations\n\n### L1 human\n\nThe zeta function is first defined only for $\\operatorname{Re} s \u003e 1$.\nThe functional equation links its value at $s$ to its value at $1 - s$, a reflection in the point $1/2$.\nSo knowing $\\zeta$ on the right half of the plane determines it on the left half.\n\n### L2 human\n\nThe equation shows that the zeros of $\\zeta$ with $\\operatorname{Re} s \u003c 0$ are the trivial zeros $-2, -4, \\dots$.\nThe other zeros lie in the critical strip $0 \\le \\operatorname{Re} s \\le 1$, placed symmetrically about the line $\\operatorname{Re} s = 1/2$.\nThe [[Riemann hypothesis|VCGF9YR36X]] says that they lie on this line.\n[[Dirichlet L-functions|9D2Z543EJQ]] of primitive characters and many other L-functions satisfy functional equations of the same shape.\n",
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