The Riemann zeta function \(\zeta(s)=\sum_{n\ge1}n^{-s}\), defined for \(\Re(s)>1\), extends to a meromorphic function on \(\CC\) whose only singularity is a simple pole at \(s=1\). The extension satisfies the functional equation \[ \zeta(s) = \pi^{s-1/2}\,\frac{\Gamma\qty{\frac{1-s}{2}}}{\Gamma\qty{\frac{s}{2}}}\,\zeta(1-s). \]
For \(t>0\) let \(\theta(t)\coloneqq\sum_{n\in\ZZ}e^{-\pi n^2t}\). By Poisson summation \(\theta(1/t)=t^{1/2}\theta(t)\), and \(\abs{\theta(t)-1}\le ce^{-\pi t}\) for \(t\ge1\). For \(\Re(s)>1\) put \[ \xi(s)\coloneqq\frac12\int_0^\infty r^{s/2-1}\qty{\theta(r)-1}\,dr. \] Integrating term by term, \(\int_0^\infty r^{s/2-1}e^{-\pi n^2r}\,dr=\pi^{-s/2}n^{-s}\Gamma(s/2)\), so \(\xi(s)=\pi^{-s/2}\Gamma(s/2)\zeta(s)\) for \(\Re(s)>1\). Splitting the integral at \(r=1\) and applying \(\theta(1/r)=r^{1/2}\theta(r)\) on \((0,1)\) gives \[ \xi(s)=\frac{1}{s-1}-\frac1s+\frac12\int_1^\infty\qty{r^{s/2-1}+r^{-s/2-1/2}}\qty{\theta(r)-1}\,dr, \] where the integral is entire in \(s\) by the exponential decay of \(\theta-1\). So \(\xi\) extends meromorphically to \(\CC\) with simple poles only at \(0\) and \(1\), and the right side is unchanged under \(s\mapsto1-s\), so \(\xi(s)=\xi(1-s)\). Since \(1/\Gamma\) is entire with a simple zero at \(s=0\), \(\zeta(s)=\pi^{s/2}\xi(s)/\Gamma(s/2)\) is meromorphic on \(\CC\); the pole of \(\xi\) at \(0\) cancels against the zero of \(1/\Gamma(s/2)\), leaving only the simple pole at \(s=1\). Writing \(\xi(s)=\xi(1-s)\) in terms of \(\zeta\) gives the functional equation.