For a full-span lattice \(\Lambda \subseteq \mathbb{R}^{n}\), we define the Epstein zeta function as \[\begin{equation} E(\Lambda,s) = \tfrac{1}{2} |\mathop{\mathrm{vol}}(\mathbb{R}^{n}/\Lambda)|^{\frac{1}{n}s}\sum_{v \in \Lambda \subseteq 0} \frac{1}{\| v\|^{s}} \end{equation}\]
The completed Epstein zeta function is defined as \[\begin{equation} E^{*}(\Lambda,s) = {\Gamma\left(\tfrac{s}{2}\right)}{\pi^{-\frac{s}{2}}} E(\Lambda,s). \end{equation}\]
The right hand side is absolutely convergent for \(\Re( s ) > n\).
Due to Poisson summation formula, the completed Epstein zeta function satisfies
\[\begin{equation} E^{*}(\Lambda,s) = E^{*}(\Lambda^{*},n-s). \end{equation}\]
By taking writing a theta function of a lattice as the inverse Mellin transform of the Epstein zeta function of \(\Lambda\), we can show the following proposition by doing some contour shifting.
\[\begin{equation} E^{*}(\Lambda,s) = -\tfrac{1}{s} - \tfrac{1}{n-s} + \frac{1}{2} \sum_{v \in \Lambda\setminus \{0\}} f\left(s, \tfrac{\|v\|}{\mathop{\mathrm{vol}}(\mathbb{R}^{n}/\Lambda)^{\frac{1}{n}}}\right) + \frac{1}{2}\sum_{v \in \Lambda^{*}\setminus \{0\}} f\left(s, \tfrac{\|v\|}{\mathop{\mathrm{vol}}(\mathbb{R}^{n}/\Lambda^{*})^{\frac{1}{n}}}\right), \end{equation}\] where \[\begin{equation} f(s,a) = \tfrac{1}{( \pi a^{2} )^{\frac{s}{2}}} \Gamma(\tfrac{s}{2}, \pi a^{2}) = \int_{1}^{\infty} t^{\frac{s}{2}-1} \exp( - \pi t a^{2}) dt. \end{equation}\]
The proof of this expression also implies thee existence of a meromorphic continuation of the Epstein zeta function.
This page was updated on November 13, 2023.
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