\(\DeclareMathOperator{\KR}{K_{\mathbb{R}}}\) \(\DeclareMathOperator{\diffx}{\, d^{\times}}\) \(\DeclareMathOperator{\diff}{\, d}\)
The following is an integration formula due to Hecke who used it to prove the analytic continuation of Dedekind zeta function.
Let \(E^*(\Lambda,z)\) be the completed Epstein zeta function of a lattice \(\Lambda \subseteq \mathbb{R}^{n}\).
Let \(\zeta_K(s)\) be the Dedekind zeta function of a number field \(K\) of degree \(n\). Then, the completed Dedekind zeta function is defined as \[\begin{equation} \zeta_{K}^{*}(s) = \pi^{-\frac{s}{2}} \Gamma(\tfrac{s}{2})^{r_1} \left( (2 \pi)^{-s} \Gamma(s)\right)^{r_2} |\Delta_K|^{\frac{s}{2}} \zeta_K(s), \end{equation}\] where \(\Delta_K\) is the discriminant of the number field \(K\) and \(r_1\) is the number of real embeddings and \(r_2\) is the number of pairs of complex embeddings.
Then, the following theorem is true.
(Hecke)
Identify \(K\otimes \mathbb{R} = \mathbb{R}^{n} = \mathbb{R}^{r_1} \oplus \mathbb{C}^{r_2}\).
Consider the Arakelov class group \(\widetilde{\mathop{\mathrm{Cl}}}(K)\) and identify it as a collection of lattices in \(\mathop{\mathrm{SL}}_{n}(\mathbb{R})/\mathop{\mathrm{SL}}_{n}(\mathbb{Z})\). Then, the Hecke integration formula basically says that \[\begin{equation} \int_{\widetilde{\mathop{\mathrm{Cl}}}(K)} E^{*}(\Lambda,ns) d\Lambda = \frac{\omega_K}{2^{r_1} n R_K} \zeta_K^{*}(s) \end{equation}\] Here \(\omega_K\) is the number of roots of unity in \(K\) and \(R_K\) is the regulator of \(K\). The measure \(d\Lambda\) is with respect to a Haar measure on \(\widetilde{\mathop{\mathrm{Cl}}}(K)\) that makes every connected component of unit covolume.
The same theorem can also be done with class group L functions. See [1]
Let us outline the proof of this theorem.
Let us reestablish the so-called “Hecke’s trick” [1]. This is the crucial step that connects diagonal orbits to norms.
{For \(\Re(s)>1\)} we have that \[\begin{align} \int_{a \in \KR^{(1)}} \|av\|^{-sd} \diffx a & = |\mathop{\mathrm{N}}(v)|^{-s} \cdot { \frac{(2 \pi )^{r_2} \Gamma(\tfrac{s}{2})^{r_1} \Gamma(s)^{r_2} }{ 2^{sr_2} \cdot \frac{d}{2} \Gamma(\frac{sd}{2}) } }. \end{align}\]
Let us prove it for \(v=1\). Consider the integral \[\begin{equation} \label{eq:first_one} \int_{x \in \KR} e^{-\|x\|^{2}} |\mathop{\mathrm{N}}(x)|^{s-1} \diff x = \int_{x \in \KR^{\times}} e^{-\|x\|^{2}} |\mathop{\mathrm{N}}(x)|^{s} |\mathop{\mathrm{N}}(x)|^{-1}\diff x. \end{equation}\] We can evaluate it two ways. One way is to write the Haar measure on \(|\mathop{\mathrm{N}}(x)|^{-1} \diff x\) on \(\KR^{\times}\) by writing \[\begin{align} \KR^{\times} & \simeq \mathbb{R}_{>0} \times \KR^{(1)} \\ x & = t \cdot a \\ |\mathop{\mathrm{N}}(x)|^{-1} \diff x & = \deg K \cdot \left( \tfrac{1}{t}\diff t \right)\cdot \diffx a. \end{align}\] Observe then that for \(d=\deg K\) we have \(\mathop{\mathrm{N}}(at) =t^{d}\mathop{\mathrm{N}}(a)\). Hence, \[\begin{align} \int_{x \in \KR} e^{-\|x\|^{2}} |\mathop{\mathrm{N}}(x)|^{s-1} \diff x & = \int_{t \in \mathbb{R}_{>0}}\int_{a \in \KR^{(1)}} e^{-t^2\|a\|^{2}} d t^{sd-1} \diffx a \diff t .\\ & = \int_{t \in \mathbb{R}_{>0}}e^{-t^2} d t^{sd-1} \diff t \int_{a \in \KR^{(1)}} \|a\|^{-sd} \diffx a \\ & = \tfrac{d}{2}\Gamma(\tfrac{sd}{2}) \int_{a \in \KR^{(1)}} \|a\|^{-sd}\diffx a. \end{align}\] On the other hand, one can also write \(x=(x_1,\dots,x_{r_1+r_2}) \in \mathbb{R}^{r_1} \times \mathbb{C}^{r_2}\). \[\begin{align} \int_{x \in \KR} & e^{-\|x\|^{2}} |\mathop{\mathrm{N}}(x)|^{s-1} \diff x \\ & = \int_{x \in \KR} e^{-\left( \sum_{i=1}^{r_1}|x_i|^{2} +2 \sum_{i=r_1+1}^{r_2}|x_i|^{2} \right)} |x_1 x_2 \dots x_{r_1}|^{s-1} |x_{r_1+1} \dots x_{r_1+r_2}|^{2( s-1 )} \diff x\\ & = \prod_{i=1}^{r_1}\left( 2\int_{x \in {\mathbb{R}_{> 0}} } e^{- x^{2}} x^{s-1} \diff x \right) \prod_{i=r_1+1}^{r_1+r_2}\left( 2\cdot 2 \pi \int_{x \in {\mathbb{R}_{> 0}} } e^{- 2 x^{2}} x^{2(s-1)} x\diff x \right)\\ & = (2\pi)^{r_2} 2^{-sr_2} \Gamma(\tfrac{s}{2})^{r_1} \Gamma(s)^{r_2}. \end{align}\] The factor of \(2\) appearing before \(2\pi\) is a bit subtle.
For \(v \neq 1\), replace \(x\) with \(xv\) in the integral in the first equation and follow the same process.
This page was updated on October 8, 2025.
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