text:
\begin{document}
\markboth
{Y. Hashimoto}
{Selberg's zeta functions for congruence subgroups}
\pagestyle{myheadings}

\maketitle
\renewcommand{\thefootnote}{}
\footnote{MSC: primary: 11M36; secondary:11E41}

\begin{abstract}
It is known that the Selberg zeta function for the modular group has an expression
in terms of the class numbers and the fundamental units
of the indefinite binary quadratic forms.
In the present paper, we generalize such a expression
to any congruence subgroup of the modular groups in $\sr$ and $\slc$.
\end{abstract}

\section{Introduction}

Let $\bH$ be the upper half plane with the hyperbolic metric and
$\Gam$ a discrete subgroup of $\sr$
such that the volume of $\Gam\bsla\bH$ is finite.
We denote by $\Prim(\Gam)$ the set of primitive hyperbolic conjugacy classes of $\Gam$
and $N(\gam)$ the square of the larger eigenvalue of $\gam\in\Prim(\Gam)$.
The Selberg zeta functions for $\Gam$ are defined as follows.
\begin{align*}
Z_{\Gam}(s):=&\prod_{\gam\in\Prim(\Gam)}\prod_{n=0}^{\infty}(1-N(\gam)^{-s-n})\quad\Re{s}>1.
\end{align*}
It is known that $Z_{\Gam}(s)$ is analytically continued to the whole complex plane
as a meromorphic function of order $2$.
By virtue of the analytic properties of $Z_{\Gam}(s)$,
we can obtain the following asymptotic formula called by the prime geodesic theorem (see, e.g., \cite{He}).
\begin{align*}
\pi_{\Gam}(x):=\#\{\gam\in\Prim(\Gam)\divset N(\gam)<x\}=\li(x)+O(x^{\delta})\as x\tinf,
\end{align*}
where $\li(x):=\int_2^x(\log{t})^{-1}dt$ and the constant $\delta$ $(0<\delta<1)$ depends on $\Gam$.

When $\Gam=\sz$, since there is a one-to-one correspondence
between elements of $\Prim(\Gam)$ and equivalence classes
of primitive indefinite binary quadratic forms,
and $N(\gam)$ for $\gam\in\Prim(\Gam)$ coincides
the square of the fundamental unit of the quadratic form
corresponding to $\gam$ (see e.g. \cite{G}),
we can express the Selberg zeta function and the prime geodesic theorem
for $\Gam=\sz$ as follows (see \cite{Sa1}).
\begin{align}
Z_{\sz}(s)=&\prod_{D\in\D}\prod_{n=0}^{\infty}(1-\epsilon(D)^{-2(s+n)})^{h(D)}\quad \Re{s}>1,\label{arithselb}\\
\pi_{\sz}(x)=&\sum_{\begin{subarray}{c}D\in\D\\ \epsilon(D)<x\end{subarray}}h(D)
\sim \li(x^2)\as x\tinf,\label{arithgeod}
\end{align}
where $\D$ is the set of integers $D>0$ such that $D$ is not square and $D\equiv 1,0$ mod $4$,
$h(D)$ and $\epsilon(D)$ are respectively the class number
and the fundamental unit of the discriminant $D$ in the narrow sense.

For a three dimensional hyperbolic manifolds whose fundamental group is $\slt(\cO)$
where $\cO$ is the integer ring of the imaginary quadratic field,
similar expressions of the Selberg zeta function and the prime geodesic theorem
were obtained in \cite{Sa2}.

Such arithmetic expressions of the Selberg zeta functions essentially give
those of the length spectra on the corresponding manifolds.
Though it is not easy in general to study the distributions of the length spectra
and its multiplicities in detail, by virtue of the expressions by $h(D)$ and $\epsilon(D)$,
the distributions of them for the cases of the modular group and the congruence subgroups
could be studied by the classical tools in the analytic
number theory (see \cite{BLS}, \cite{Pe} and \cite{H2}).
Moreover, by writing down the prime geodesic theorem for the modular group
with $h(D)$ and $\epsilon(D)$,
one can get the density formula of the sum of the class numbers
with a different form to that conjectured by Gauss
and proven by Siegel \cite{Si}
(see \cite{Sa1}, \cite{H1} and \cite{H3}).

The aim of the present paper is to generalize
such arithmetic expressions of the Selberg zeta functions
for congruence subgroups of $\sz$ and $\slt(\cO)$.
For the groups $\Gam_0(N)$, $\Gam_1(N)$ and $\Gam(N)$ of $\sz$,
the author \cite{H1} already gave explicit expressions of the logarithmic derivatives of
the Selberg zeta functions in terms of the class numbers and the fundamental units.
In the present paper, we show that the Selberg zeta functions
for any congruence subgroup of $\sz$ and $\mathrm{SL}_2(\cO_K)$ are expressed
in terms of the class numbers and the fundamental units such like \eqref{arithselb}.

\section{Quadratic forms and modular groups}

In this section, we give notations of the quadratic forms,
explain the relations between the quadratic forms and
elements in the modular group
and study the difference between the conjugation in $\slt(\cO)$
and that in $\slt(\cO/\n)$ for an integer ring $\cO$ of an algebraic number fields
and an ideal $\n$ in $\cO$.

Let $K$ be an algebraic number field over $\bQ$ such that $[K:\bQ]<\infty$
and $\cO$ the ring of integers of $K$.
We denote by
\begin{align*}
Q(x,y)=[a,b,c]:=ax^2+bxy+cy^2
\end{align*}
a binary quadratic form over $\cO$ where $a,b,c\in\cO$.
We write $Q\sim Q'$ if there exists $g\in\slt(\cO)$ such that $Q(x,y)=Q'\big(g.(x,y)\big)$.
Let $D:=b^2-4ac$ be the discriminant of $[a,b,c]$ and
$P(D)$ the set of solutions $(t,u)\in\cO^2$ of the Pell equation $t^2-Du^2=4$.
The set $P(D)$ is an abelian group with the identity $(2,0)$ and the product
\begin{align*}
(t_1,u_1)*(t_2,u_2)=\Big(\frac{t_1t_2+u_1u_2D}{2},\frac{t_1u_2+t_2u_1}{2}\Big).
\end{align*}
It is known that $P(d)\simeq (\bZ/l\bZ)\times \bZ$ where $l$ is
the number of roots of $1$ in $\cO_{K(\sqrt{D})}$,
$r_1$ is the number of embeddings $K(\sqrt{D})$ into $\bR$
and $2r_2$ is the number of embeddings $K(\sqrt{D})$ into $\bC$ with $\Im(K(\sqrt{D}))\not\subset \bR$.
Note that $r_1+2r_2=[K(\sqrt{D}):\bQ]$.
For a quadratic form $Q=[a,b,c]$ and a solution $(t,u)\in P(D)$, we put
\begin{align}\label{1to1}
\gam\big(Q,(t,u)\big):=\begin{pmatrix}\disp\frac{t+bu}{2}&-cu\\ au& \disp\frac{t-bu}{2}\end{pmatrix}\in\slt(\cO).
\end{align}
Conversely, for $\gam=(\gam_{i,j})_{1\leq i,j\leq 2}\in \mr{SL}_2(\cO)$, we put
\begin{align*}
&t_{\gam}:=\gam_{11}+\gam_{22},\quad u_{\gam}:=\gcd{(\gam_{21},\gam_{11}-\gam_{22},-\gam_{12})},\\
&a_{\gam}:=\gam_{21}/u_{\gam},\quad b_{\gam}:=(\gam_{11}-\gam_{22})/u_{\gam},
\quad c_{\gam}:=-\gam_{12}/u_{\gam},\\
&Q_{\gam}:=[a_{\gam},b_{\gam},c_{\gam}],\quad
D_{\gam}:=\frac{t_{\gam}^2-4}{u_{\gam}^2}=b_{\gam}^2-4a_{\gam}c_{\gam}.
\end{align*}
We note the following elementary facts without proofs.
\begin{fact}\label{factquad}
Suppose that $\gam,\gam_1,\gam_2\in\slt(\cO)$ are not in the center of $\slt(\cO)$. Then we have
\begin{enumerate}
\item $\gam(Q_g,(t_g,u_g))=g$ for any $g\in\slt(\cO)$.
\item $D(Q)=D_{\gam(Q,(t,u))}$ for any $(t,u)\in P(D)$ and $D_{\gam}=D(Q_{\gam})$.
\item $(t_{\gam(Q,(t,u))},u_{\gam(Q,(t,u))})=(t,u)$ and
$(t_{\gam},u_{\gam})\in P(d_{\gam})$.
\item If $Q_{\gam_1}=Q_{\gam_2}$ then $Q_{\gam_1\gam_2}=Q_{\gam_1}=Q_{\gam_2}$
and $(t_{\gam_1\gam_2},u_{\gam_1\gam_2})=(t_{\gam_1},u_{\gam_1})*(t_{\gam_2},u_{\gam_2})$.
\item $Q_{g^{-1}\gam g}(x,y)=Q_{\gam}\big(g.(x,y)\big)$ for any $g\in\mr{SL}_2(\cO)$.
\end{enumerate}
\end{fact}

For a fixed discriminant $D\in\cO$, let $I(D)$ be the set of equivalence classes of quadratic forms $[a,b,c]$
with $b^2-4ac=D$.
It is known that $h(D):=\#I(D)$ is finite.
According to Fact \ref{factquad}, we see that
\begin{align}
h(d)=&\#\{\gam\in\Conj(\mr{SL}_2(\cO))\divset D_{\gam}=D\}/P(D)\nt\\
=&\#\{\gam\in\Conj(\mr{SL}_2(\cO))\divset t_{\gam}=t,u_{\gam}=u\},\label{hd}
\end{align}
where $(t,u)$ is a non-trivial solution of $t^2-Du^2=4$.

For quadratic forms $[a_1,b_1,c_1],[a_2,b_2,c_2]\in I(d)$,
we give the following product.
\begin{align}\label{composition}
[a_1,b_1,c_1]*[a_2,b_2,c_2]=\Big[\frac{a_1a_2}{\beta^2},
v_1\frac{a_1}{\beta}b_2+v_2\frac{a_2}{\beta}b_1+w\frac{b_1b_2+D}{2\beta},C\Big],
\end{align}
where $C\in\cO$ satisfies that the discriminant of the quadratic form of the right hand side is $d$,
$\beta:=\gcd(a_1,a_2,(b_1+b_2)/2)$ and $v_1,v_2,w\in\cO$ are chosen as
$v_1a_1+v_2a2+w(b_1+b_2)/2=\beta$.
It is easy to see that $I(D)$ is a finite abelian group with the identity $[1,\delta,(D-\delta^2)/4]$
and the product above, where $\delta\in\cO$ is taken as $4\mid d-\delta^2$.

For an ideal $\Da\subset\cO$, let
\begin{align*}
\cO_{\Da}^{(2)}:=&\{\alpha\in\cO\divset \alpha^2\equiv 1\bmod \Da\},\\
\hat{\Gam}(\Da):=&\Ker\big(\mr{SL}_2(\cO)\stackrel{\text{proj.}}{\rightarrow}
\mr{PSL}_2(\cO/\Da)\big),\\
=&\{\gam\in \mr{SL}_2(\cO)\divset \gam\equiv \alpha I\bmod{\Da}, \alpha\in \cO_{\Da}^{(2)}\}.
\end{align*}
Note that if the factorization of $\Da$ in $\cO$ is $\Da=\prod_{\p}\p^r$
and $\#\cO_K/\p=q$ (a power of a rational prime) then
\begin{align*}
\hGam(\Da)=&\bigcap_{\p}\hGam(\p^r), \quad
[\slt(\cO):\hGam(\Da)]=\#\mr{PSL}_2(\cO/\Da)
=\frac{\prod_{\p}q^{3r-2}(q^2-1)}{\#\cO_{\Da}^{(2)}}.
\end{align*}

We now study the conjugation in $\mr{PSL}_2(\cO/\Da)$.
\begin{lem}\label{conj01}
Let $\Da$ be an ideal in $\cO$.
For $a,b,c\in\cO/\Da$, $d=b^2-4ac$ and $t,u\in\cO/\Da$ satisfying $t^2-Du^2=4$
without $(t,u)=(\pm2,0)$,
we denote by
\begin{align*}
\gam:=\begin{pmatrix}\disp\frac{1}{2}(t+bu)&-cu\\au&\disp\frac{1}{2}(t-bu)\end{pmatrix}, \quad
\gam_{\nu}:=\begin{pmatrix}\disp\frac{t+\delta u}{2}&\disp\frac{D-\delta^2}{4}\nu^{-1}u\\
\nu u&\disp\frac{t-\delta u}{2}\end{pmatrix},
\end{align*}
where $\nu\in(\cO/\Da)^{*}$ and $\delta$ is taken as $2\mid b+\delta$ when $\gcd(\Da,2)\neq1$
and is taken by $0$ when $\gcd(\Da,2)=1$.
Then we have $\gam\sim\gam_{\nu}$ in $\mr{PSL}_2(\cO/\Da)$ for some $\nu\in \cO_{\Da}^{(2)}$.
\end{lem}

\begin{proof}
When $\gcd(a,\Da)\neq1$, it is easy to see that there exists $g\in\mr{PSL}_2(\cO/\Da)$ such that
$\gcd\big((g^{-1}\gam g)_{21}/u,\Da\big)=1$.
Then it is sufficiently to consider only the case for $\gcd(a,\Da)=1$.

When $a=\nu\alpha^2$ for $\nu\in\cO_{\Da}^{(2)}$ and $\alpha\in\cO/\Da$,
we have ${g'}^{-1}\gam g'=\gam_{\nu}$ where
\begin{align*}
g':=\begin{pmatrix}\alpha^{-1}&\disp(\nu\alpha)^{-1}(b+\delta)/2\\
0&\alpha\end{pmatrix}.
\end{align*}
This completes the proof.
\end{proof}

\begin{lem}\label{classdiv}
Let $\Da$ be an ideal in $\cO$.
For $(t,u)\in\cO^2-\{(\pm2,0)\}$,
we denote by $\cI(t,u):=\{\gam\in\Conj(\slt(\cO))\divset t_{\gam}=t,u_{\gam}=u\}$
and $\cJ_{\nu}(t,u):=\{\gam\in\cI(t,u)\divset \text{$\gam\sim\gam_{\nu}$ in $\psl(\cO/\Da)$}\}$.
For $\nu,\nu'\in\cO_{\Da}^{(2)}$, define the equivalence $\nu\sim\nu'$ such that
 $\gam_{\nu}$ is conjugate to $\gam_{\nu'}$ in $\psl(\cO/\Da)$.
Put $\cL_{\Da}(t,u):=\{\nu\in\cO_{\Da}^{(2)}/\sim\divset J_{\nu}(t,u)\neq\phi\}$
and $\mu_{\Da}(t,u):=\#\cL_{\Da}(t,u)$.
Number the elements of $\cL_{\Da}(t,u)$
by $\nu_1(=1),\nu_2,\dots,\nu_{\mu}$.
For simplicity, rewrite $\cJ_{\nu_i}$ by $\cJ_i$.
Then we have $\#\cJ_i=h(D)/\mu$ for any $1\leq i\leq \mu$.
\end{lem}

\begin{proof}
For $\gam_1,\gam_2\in\cI(t,u)$, denote by $\gam_1*\gam_2:=\gam(Q_{\gam_1}*Q_{\gam_2},(t,u))$,
where the product $*$ in the right hand side is given by \eqref{composition}.

Fix $\gam'\in\cJ_i$.
It is easy to see that $\gam*\gam'\in\cJ_i$ for any $\gam\in\cJ_1$.
When $\gam_1,\gam_2\in\cJ_i$ are not conjugate in $\mr{SL}_2(\cO)$,
it is trivial that $\gam_1*\gam'\not\sim \gam_2*\gam'$ in $\mr{PSL}_2(\cO)$.
Then we have $\cJ_i\supset \gam'*\cJ_1$.
Conversely, $\hat{\gam}*\gam'\in\cJ_1$ for any $\hat{\gam}\in\cJ_i$.
Then we can get $\cJ_1=\gam'*\cJ_i$ similarly.
This yields that $\#\cJ_1=\#\cJ_i$ for any $1\leq i\leq \mu$.
Since $\sum_{i=1}^{\mu}\#\cJ_i=h(d)$,
we conclude that $\#\cJ_i=h(d)/\mu$ for any $1\leq i\leq \mu$.
\end{proof}

Consider the case of $\cO=\bZ$. Note that, when $\cO=\bZ$.
\begin{align*}
\bZ_{p^r}^{(2)}=\begin{cases}\{1\},&p=2,r=1,\\
\{1,3\},&p=2,r=2,\\ \{1,3,5,7\},&p=2,r\geq3,\\
\{1,\eta\}, &p\geq3,
\end{cases}
\end{align*}
where $\eta\in(\bZ/p\bZ)^{\times}$ is a non-quadratic residue of $p$.

\begin{lem}\label{nonquad}
Let $p\geq5$ be a prime number.
If $p\nmid\alpha$ then there exists $l\in\bZ/p\bZ$
such that $1+\alpha l^2$ is a non-quadratic residue.
\end{lem}

\begin{proof}
When $(\alpha/p)=1$, it is easy to see that $\{\alpha l^2\divset l\in(\bZ/p\bZ)^*\}$
coincides $\{\beta\in(\bZ/p\bZ)^*\divset (\beta/p)=1\}$.
Since $\#\{\beta\in(\bZ/p\bZ)^*\divset (\beta/p)=1\}=1/2\#(\bZ/p\bZ)^*$,
the claim of the lemma is trivial.

When $(\alpha/p)=-1$, then $\{\alpha l^2\divset l\in(\bZ/p\bZ)^*\}
=\{\beta\in(\bZ/p\bZ)^*\divset (\beta/p)=-1\}$.
Assume that $1+\alpha l^2$ is quadratic residue for any $l\in(\bZ/p\bZ)^*$,
namely $\{1+\alpha l^2\divset l\in(\bZ/p\bZ)^*\}=\{\beta\in(\bZ/p\bZ)^*\divset (\beta/p)=1\}$.
Since $\#\{\beta\in(\bZ/p\bZ)^*\divset (\beta/p)=1\}=\#\{\beta\in(\bZ/p\bZ)^*\divset (\beta/p)=-1\}$,
it should hold that $(1/p)=(3/p)=\cdots=1$, $(2/p)=(4/p)=\cdots=-1$
or $(1/p)=(3/p)=\cdots=-1$, $(2/p)=(4/p)=\cdots=1$.
However $(1/p)=(4/p)=1$. This contradicts to the assumption.
Therefore the claim holds.
\end{proof}

\begin{lem}\label{conj2}
(i) When $p=2$ we have
\begin{align*}
&\text{if $d\equiv1$ mod $4$ then $\gam_1\sim\gam_{\nu}$ in $\mr{PSL}_2(\bZ/2^r\bZ)$
for $\nu=3,5,7$},\\
&\text{if $d\equiv3$ mod $4$ then $\gam_1\sim\gam_{5}$, $\gam_3\sim\gam_7$, $\gam_1^3\sim\gam_3$
and $\gam_3^3\sim\gam_1$ in $\mr{PSL}_2(\bZ/2^r\bZ)$},\\
&\text{if $d\equiv2$ mod $8$ then $\gam_1\sim\gam_7$, $\gam_3\sim\gam_5$, $\gam_1^3\sim\gam_3$
and $\gam_3^3\sim\gam_1$ in $\mr{PSL}_2(\bZ/2^r\bZ)$},\\
&\text{if $d\equiv6$ mod $8$ then $\gam_1\sim\gam_3$, $\gam_5\sim\gam_7$, $\gam_1^5\sim\gam_5$
and $\gam_5^5\sim\gam_1$ in $\mr{PSL}_2(\bZ/2^r\bZ)$},\\
&\text{if $d\equiv4$ mod $8$ then $\gam_1\sim\gam_5$, $\gam_3\sim\gam_7$, $\gam_1^3\sim\gam_3$
and $\gam_3^3\sim\gam_1$ in $\mr{PSL}_2(\bZ/2^r\bZ)$},\\
&\text{if $d\equiv0$ mod $8$ then $\gam_1^{\nu}\sim\gam_{\nu}$ and $\gam_{\nu}^{\nu}\sim\gam_1$
in $\mr{PSL}_2(\bZ/2^r\bZ)$ for $\nu=3,5,7$},
\end{align*}
where $d=D$ if the square free factor of $D$ is equivalent to $1$ modulo $4$
and $d=D/4$ otherwise.

\noindent(ii) When $p\geq3$ we have
\begin{align*}
\text{if $p\nmid D$}\quad &\text{then $\gam_1\sim\gam_{\eta}$ in $\mr{PSL}_2(\bZ/p^r\bZ)$},\\
\text{if $p\mid D$}\quad &\text{then $\gam_1^{\eta}\sim\gam_{\eta}$
and $\gam_{\eta}^{\mu}\sim\gam_1$ $(\eta\mu\equiv1\bmod{p^r})$ in $\mr{PSL}_2(\bZ/p^r\bZ)$}.
\end{align*}
\end{lem}

\begin{proof}
When $p=2$, let
\begin{align*}
g_i:=\begin{pmatrix}1&0\\i&1\end{pmatrix}.
\end{align*}
We have
\begin{align*}
(g_1^{-1}\gam_1 g_1)_{21}/u\equiv&\begin{cases}6&d\equiv2,\\5&d\equiv4,\\3&d\equiv6,\end{cases}&
(g_2^{-1}\gam_1 g_2)_{21}/u\equiv&\begin{cases}3&d\equiv1,\\5&d\equiv3,\\7&d\equiv5,\end{cases}\\
(g_4^{-1}\gam_1 g_4)_{21}/u\equiv&\begin{cases}3&d\equiv5,\\5&d\equiv1,\end{cases}&
(g_6^{-1}\gam_1 g_6)_{21}/u\equiv&\begin{cases}5&d\equiv5,\\7&d\equiv1.\end{cases}
\end{align*}
Then the relations $\gam_1\sim\gam_{\nu}$ in (i) of the lemma are true.

When $p=3$ and $3\nmid D_{\gam}$, we have $(g^{-1}\gam_1 g)_{21}\equiv 2u_{\gam}$ mod $3^r$,
where
\begin{align*}
g:=\begin{cases}\disp\begin{pmatrix}0&-(-D/8)^{-1/2}\\(-8/D)^{1/2}&0\end{pmatrix},
&D_{\gam}\equiv1\bmod{3},\\
\disp\begin{pmatrix}(2+D/4)^{1/2}&(2+D/4)^{1/2}-1\\ 1&1 \end{pmatrix},&D_{\gam}\equiv2\bmod{3}.
\end{cases}
\end{align*}
Then we see that $\gam_1\sim\gam_2$ in $\mr{PSL}_2(\bZ/3^r\bZ)$ if $3\nmid D$.

When $p\geq5$, according to Lemma \ref{nonquad}, we see that if $p\nmid \alpha$ the equation
$x^2-\alpha y^2\equiv \eta$ mod $p^r$ has a solution such that $p\nmid x$ or $p\nmid y$.
For $p\nmid D_{\gam}$, let $(l,m)\in(\bZ/p^r\bZ)^2$ be the solution of $x^2-4^{-1}Dy^2\equiv\eta$
and
\begin{align*}
g:=\begin{pmatrix}l&*\\m&*\end{pmatrix}.
\end{align*}
Then we have $(g^{-1}\gam_1 g)_{21}/u\equiv \eta$ mod $p^r$.
This means that $\gam_1$ is conjugate to $\gam_{\eta}$ in $\mr{PSL}_2(\bZ/p^r\bZ)$.

By virtue of 4. of Fact \ref{factquad}, we have
\begin{align*}
\frac{(\gam_1^{\nu})_{21}}{u}=\frac{u_{\gam^{\nu}}}{u}
=&\frac{\big((t+u \sqrt{D})/2\big)^{\nu}-\big((t-u \sqrt{D})/2\big)^{\nu}}{2u\sqrt{d}}\\
=&\sum_{l=0}^{[(\nu-1)/2]}\binom{\nu}{2l+1}\Big(\frac{t}{2}\Big)^{\nu-1}
\Big(\frac{Du^2}{Du^2+4}\Big)^l.
\end{align*}
When $\nu=3$, we have
\begin{align*}
\frac{(\gam_1^{\nu})_{21}}{u}
=&\Big(\frac{t}{2}\Big)^{2}\Big(\nu+\frac{Du^2}{Du^2+4}\Big)\equiv3\bmod{8}
\end{align*}
for $d\equiv0,2,3,4,7$ mod $8$.
Then in these cases $\nu_1^{3}\sim\nu_3$.
Similarly we can see that $\nu_1^5\sim\nu_5$ for $d\equiv6$ mod $8$.
When $8\mid d$ or $p\mid d$, we have
\begin{align*}
\frac{(\gam_1^{\nu})_{21}}{u}
=&\nu\Big(\frac{t}{2}\Big)^{\nu-1}
+\sum_{l=1}^{[(\nu-1)/2]}\binom{\nu}{2l+1}\Big(\frac{t}{2}\Big)^{\nu-1}
\Big(\frac{Du^2}{Du^2+4}\Big)^l
\equiv \nu\bmod{\text{$8$ or $p$}}.
\end{align*}
Then $\gam_1^{\nu}\sim\gam_{\nu}$ holds for $p=2$, $\nu=3,5,7$ and $p\geq3$, $\nu=\eta$.
\end{proof}

\section{Generalized Venkov-Zograf's formula}

In this section, we give a generalization of Venkov-Zograf's formula \cite{VZ}
used in the proof of the main theorem.

Let $G$ be a connected non-compact semi-simple Lie group with finite center,
and $K$ a maximal compact subgroup of $G$.
We put $d=\dim{(G/K)}$.
We denote by $\g$, $\Dk$ the Lie algebras of $G$, $K$ respectively
and $\g=\Dk+\p$ a Cartan decomposition with respect to the Cartan involution $\theta$.
Let $\Da_{\p}$ be a maximal abelian subspace of $\p$.
Throughout this paper we assume that $\rank(G/K)=1$, namely $\dim{\Da_{\p}}=1$.
We extend  $\Da_{\p}$ to a $\theta$-stable maximal abelian subalgebra $\Da$ of $\g$,
so that $\Da=\Da_{\p}+\Da_{\Dk}$, where $\Da_{\p}=\Da\cap\p$ and $\Da_{\Dk}=\Da\cap\Dk$.
We put $A=\exp{\Da}$, $A_{\p}=\exp{\Da_{\p}}$ and $A_{\Dk}=\exp{\Da_{\Dk}}$.

We denote by $\g^{\bC}$, $\Da^{\bC}$ the complexification of $\g$, $\Da$ respectively.
Let $\Phi$ be the set of roots of $(\g^{\bC},\Da^{\bC})$,
$\Phi^+$ the set of positive roots in $\Phi$,
$P_+=\{\alpha\in\Phi^+ | \alpha\not\equiv 0$ on $\Da_{\p}\}$, and $P_-=\Phi^+-P_+$.
We put $\rho=1/2\sum_{\alpha\in P_+}\alpha$.
For $h\in A$ and linear form $\lambda$ on $\Da$,
we denote by $\xi_{\lambda}$ the character of $\Da$ given by
$\xi_{\lambda}(h)=\exp{\lambda(\log{h})}$.
Let $\Sigma$ be the set of restrictions to $\Da_{\p}$ of the elements of $P_+$.
Then the set $\Sigma$ is either of the form $\{\beta\}$ or $\{\beta,2\beta\}$
with some element $\beta\in\Sigma$.
We fix an element $H_0\in\Da_{\p}$ such that $\beta(H_0)=1$, and put $\rho_0=\rho(H_0)$.
Note that
\begin{align*}
2\rho_0=\begin{cases} m-1,\\ 2m,\\ 4m+2,\\ 22,\\ \end{cases}\quad
d=\begin{cases}m,& \quad (G=\mr{SO}(m,1),m\geq2),\\
2m, & \quad (G=\mr{SU}(m,1),m\geq2),\\ 4m, & \quad (G=\mr{SP}(m,1),m\geq1),\\
16, & \quad (G=F_4).\end{cases}
\end{align*}

Let $\Gamma$ be a discrete subgroup of $G$
such that the volume of $X_{\Gam}=\Gam\bsla G/K$ is finite.
We denote by $\Prim{(\Gamma)}$ the set of primitive hyperbolic conjugacy classes of $\Gamma$,
and $Z(\Gamma)$ the center of $\Gamma$.
For $\gamma\in \Prim(\Gamma)$,
we denote by $h(\gamma)$ an element of $A$ which is conjugate to $\gamma$,
and $h_{\p}(\gamma)$, $h_{\Dk}(\gamma)$ the elements of $A_{\p}$, $A_{\Dk}$ respectively
such that $h(\gamma)=h_{\p}(\gamma)h_{\Dk}(\gamma)$.
Let $N(\gamma)$ be the norm of $\gamma$ given by $N(\gamma)=\exp{(\beta(\log{(h_{\p}(\gamma))}))}$.

For a finite dimensional representation $\chi$ of $\Gam$,
we define the Selberg zeta function as follows.
\begin{align*}
Z_{\Gam}(s,\chi):=&\prod_{\gam\in\Prim(\Gam)}\prod_{\lambda\in L}
\det(\Id-\chi(\gam)\xi_{\lambda}(h(\gam))^{-1}N(\gam)^{-s})^{m_{\lambda}} \quad \Re{s}>2\rho_0,
\end{align*}
where $L$ is the semi-lattice of linear forms on $\Da$ given by
$L:=\{\sum_{i=1}^{l}m_i \alpha_i\divset \alpha_i\in P_+,m_i\in\bZ_{\geq0}\}$
and $m_{\lambda}$ is the number of distinct $(m_1,\cdots,m_l)\in(\bZ_{\geq0})^l$
such that $\lambda=\sum_{i=1}^{l}m_i \alpha_i\in L$.
We note that the logarithmic derivative of the above is written as follows.
\begin{align}
\frac{Z'_{\Gamma}(s,\chi)}{Z_{\Gamma}(s,\chi)}&=\sum_{\gamma\in\Prim(\Gam),k\geq1}
\tr\chi(\gam^k)\log{N(\gam)}D(\gamma^k)^{-1}N(\gamma)^{-ks} \quad \Re{s}>2\rho_0,
\label{log}
\end{align}
where $D(\gamma):=\prod_{\alpha\in P_{+}}|1-\xi_{\alpha}(h(\gamma))^{-1}|$.

We now generalize the Venkov-Zograf formula \cite{VZ} for two dimensional case as follows.

\begin{lem}\label{venkov}
Let $\Gam$ be a discrete subgroup of $G$ such that the volume of $\Gam\bsla G/K$ is finite
and $\Gam'$ a subgroup of $\Gam$ with the finite index.
Then, for a finite dimensional representation $\chi$ of $\Gam'$, we have
\begin{align*}
Z_{\Gam'}(s,\chi)=Z_{\Gam}(s,\Ind_{\Gam'}^{\Gam}\chi),
\end{align*}
where $\Ind_{\Gam'}^{\Gam}\chi$ is a representation of $\Gam$ induced by $\chi$.
\end{lem}

\begin{proof}
It is easy to see that, for $\gam\in\Gam$, there exist positive integers $m_1,\cdots,m_k$
and a complete system of representatives $\{a_j^(i)\}$ of $\Gam/\Gam'$
such that $m_1\geq\cdots \geq m_k$, $\sum_{i=1}^{k}m_i=[\Gam:\Gam']$
and $(a_j^{(i)})^{-1}\gam a_{j+1}^{(i)}$ for $m_i\geq2$, $1\leq j\leq m_i-1$
and $(a_{m_i}^{(i)})^{-1}\gam a_{1}^{(i)}$ are in $\Gam'$.
Then by the definition of the induced representation, we see that
\begin{align}\label{ind0}
(\Ind_{\Gam'}^{\Gam}\chi)(\gamma)\sim\left(\begin{array}{ccc}
S_{m_1}&\cdots&0\\
\vdots&\ddots&\vdots\\
0&\cdots&S_{m_k}
\end{array}\right),
\end{align}
where $S_{m_i}$ is given by
\begin{align*}
S_{m_i}=\begin{cases}\chi({a_1^{(i)}}^{-1}\gam a_1^{(i)})&(m_i=1),\\
\left(\begin{array}{cccccc}
0&\chi({a_1^{(i)}}^{-1}\gam a_2^{(i)})&\cdots&0\\
\vdots&\vdots&\ddots&\vdots\\
0&0&\cdots&\chi({a_{m_i-1}^{(i)}}^{-1}\gam a_{m_i}^{(i)})\\
\chi({a_{m_i}^{(i)}}^{-1}\gam a_1^{(i)})&0&\cdots&0
\end{array}\right)&(m_i\geq2).\end{cases}
\end{align*}
Then we have
\begin{align*}
Z_{\Gam}(s,\Ind_{\Gam'}^{\Gam}\chi)
=&\prod_{\gam\in\Prim(\Gam)}\prod_{\lambda\in L}
\det(\Id-\Ind_{\Gam'}^{\Gam}\chi(\gam)\xi_{\lambda}(h(\gam))^{-1}N(\gam)^{-s})^{m_{\lambda}}\\
=&\prod_{\gam\in\Prim(\Gam)}\prod_{\lambda\in L}\prod_{i=1}^{k}
\det(\Id-\chi({a_1^{(i)}}^{-1}\gam^{m_i}a_1^{(i)})\xi_{\lambda}(h(\gam^{m_i}))^{-1}N(\gam^{m_i})^{-s})^{m_{\lambda}}.
\end{align*}
According to Lemma 2.3 of \cite{HW}, we see that
$(\Ind_{\Gam'}^{\Gam}\chi)(\gam)$ is expressed as \eqref{ind0}
if and only if ${a_1^{(i)}}^{-1}\gam^{m_i}a_1^{(i)}$ $(1\leq i\leq k)$ are primitive and
are not conjugate to each other in $\Gam'$.
This means that
\begin{align*}
\{{a_1^{(i)}}^{-1}\gam^{m_i}a_1^{(i)}\divset \gam\in\Prim(\Gam), 1\leq i\leq k\}=\Prim(\Gam').
\end{align*}
Then we have
\begin{align*}
Z_{\Gam}(s,\Ind_{\Gam'}^{\Gam}\chi)=&\prod_{\gam'\in\Prim(\Gam')}\prod_{\lambda\in L}
\det(\Id-\chi(\gam')\xi_{\lambda}(h(\gam'))^{-1}N(\gam')^{-s})^{m_{\lambda}}
=Z_{\Gam'}(s,\chi).
\end{align*}
This completes the proof.
\end{proof}

We furthermore prepare the following two lemmas.
\begin{lem}\label{prod} (see \cite{HW})
Let $\Gam,\Gam_1,\Gam_2$ be discrete subgroups of $G$ such that $\vol(\Gam\bsla G/K)<\infty$,
$\Gam_1,\Gam_2\subset \Gam$, $\Gam_1\Gam_2=\Gam$ and $[\Gam:\Gam_1\cap\Gam_2]<\infty$.
Then, for any $\gam\in\Gam$, we have
\begin{align*}
\big(\Ind_{\Gam_1\cap\Gam_2}^{\Gam}1\big)(\gam)
=\big(\Ind_{\Gam_1}^{\Gam}1\big)(\gam)\otimes \big(\Ind_{\Gam_2}^{\Gam}1\big)(\gam).
\end{align*}
\end{lem}

\begin{lem}\label{div}
Let $\Gam$ be a congruence subgroup of level $N=\prod_{p\mid N}p^r$.
Then we have $\Gam=\cap_{p\mid N}\Gam_{p^r}$ for some $\Gam_{p^r}\supset \hGam(p^r)$.
\end{lem}
\begin{proof}
Put $\Gam_{p^r}=\Gam\hGam(p^r)$.
It is easy to check that $\cap_{p\mid N}\Gam_{p^r}=\Gam$.
\end{proof}

\section{Main theorem and its proof}

In the case of $G=\mr{SO}(2,1)\simeq\sr$ and $G=\mr{SO}(3,1)\simeq\slc$,
the Selberg zeta function is written as follows.
\begin{align*}
Z_{\Gam}(s)=&\prod_{\gam\in\Prim(\Gam)}\prod_{n=0}^{\infty}(1-N(\gam)^{-s-n}),
&&\Re{s}>1,\quad G=\sr,\\
Z_{\Gam}(s)=&\prod_{\gam\in\Prim(\Gam)}
\prod_{\begin{subarray}{c}m,n\geq0\\ m\equiv n\bmod{m(\gam)}\end{subarray}}
(1-a(\gam)^{-2m}\overline{a(\gam)}^{-2n}N(\gam)^{-s}),&&\Re{s}>2,\quad G=\slc,
\end{align*}
where $a(\gam)$ is the eigenvalue of $\gam\in\Prim(\Gam)$ such that $|a(\gam)|>1$
and $m(\gam)$ is the order of the torsion of the centralizer of $\gam$.
Note that $|a(\gam)|^2=N(\gam)$.

It is known that $\Gam=\sz$ and $\slt(\cO)$ where $\cO$ is the ring of integers of $\bQ(\sqrt{-M})$
and $M>0$ is not square are discrete subgroups of $G=\sr$ and $\slt(\bC)$ respectively such that
the volume of $\Gam\bsla G/K$ is finite.

Let
\begin{align*}
\D_K=\D:=\begin{cases}\{D\in\bZ_{>0}\divset \text{$D$ is not a square and $D\equiv0,1$ mod $4$}\}&K=\bQ,\\
\{D\in\cO\divset\text{$D$ is not a square and $d\equiv \exists\alpha^2$ mod $4$}\}&K=\bQ(\sqrt{-M}).
\end{cases}
\end{align*}
For $D\in\D_K$, we see that $P(D)\simeq (\bZ/l\bZ)\times \bZ$.
We call that $\epsilon(D):=(t_1+u_1\sqrt{D})/2$ by the fundamental unit of $D$.
Then the primitive element $\gam$ of $\slt(\cO)$ is written as \eqref{1to1} for $(t,u)=(t_1,u_1)$.
For such a $\gam$, we see that $a(\gam)=\epsilon(D)$ and $N(\gam)=|\epsilon(D)|^2$.
Since $m(\gam)=l$, by using $(t,u)\in P(D)$ with $0\leq \arg\epsilon(D)\leq \pi/2$,
we see that the Selberg zeta function for $\Gam=\slt(\cO)$ as follows (see \cite{Sa1} and \cite{Sa2}).
\begin{align*}
Z_{\sz}(s)&=\prod_{d\in\D}\prod_{n=0}^{\infty}(1-\epsilon(D)^{-2(s+n)})^{h(D)}&\Re{s}>1,\\
Z_{\slt(\cO)}(s)&=\prod_{D\in\D}\prod_{m,n=0}^{\infty}
(1-\epsilon(D)^{-2m}\overline{\epsilon(D)}^{-2n}|\epsilon(D)|^{-2s})^{h(D)}&\Re{s}>2.
\end{align*}

Let $\Da$ be an ideal in $\cO$.
We call that $\Gam\subset \slt(\cO)$ is a congruence subgroup of $\slt(\cO)$ with the level $\Da$
if $\Gam\supset\hGam(\Da)$ and $\Gam\not\supset\hGam(\Da')$ for any $\Da'\mid\Da$.
The main result of the present paper is as follows.

\begin{thm}\label{thm}
Let $\cO$ be the integer ring of $\bQ$ or its imaginary quadratic extension,
$\Da$ an ideal in $\cO$
and $\Gam$ a congruence subgroup of $\slt(\cO)$ of level $\Da$.
Then the following formulas hold: \\
(i) The case of $\cO=\bZ$,
\begin{align*}
&Z_{\Gam}(s)=\prod_{D\in\D}\prod_{n=0}^{\infty}
\det(I-\chi_{\Gam}(\gam_1)\epsilon(D)^{-2(s+n)})^{h(D)},\\
&\frac{Z'_{\Gam}(s)}{Z_{\Gam}(s)}=\sum_{D\in\D}\sum_{j\geq1}
\tr\chi_{\Gam}(\gam_1)h(D)\frac{2\log{\epsilon(D)}}{1-\epsilon(D)^{-2j}}\epsilon(D)^{-2js}.
\end{align*}
\noindent (ii) The case of $\cO\subset\bQ(\sqrt{-M})$,
\begin{align*}
&Z_{\Gam}(s)=\prod_{D\in\D}\prod_{n,m=0}^{\infty}\prod_{\nu\in\cL_{\Da}(t_1(D),u_1(D))}
\det(I-\chi_{\Gam}(\gam_{\nu})\epsilon(D)^{-2m}\overline{\epsilon(D)}^{-2n}|\epsilon(D)|^{-2s})^{h(D)/\mu_{\Da}(D)},\\
&\frac{Z'_{\Gam}(s)}{Z_{\Gam}(s)}=\sum_{D\in\D}\sum_{j\geq1}\sum_{\nu\in\cL_{\Da}(t_1(D),u_1(D))}
\tr\chi_{\Gam}(\gam_\nu)\frac{h(D)}{\mu_{\Da}(D)}
\frac{2\log{\epsilon(D)}}{|1-\epsilon(D)^{-2j}|^2}|\epsilon(D)|^{-2js}.
\end{align*}
where $\gam_{\nu}$ and $\cL_{\Da}(t,u)$ are respectively defined in Lemma \ref{conj01} and \ref{classdiv}.
\end{thm}

\begin{proof}
Since the expressions of the logarithmic derivatives of $Z_{\Gam}(s)$ can be obtained easily
from those of $Z_{\Gam}(s)$, we treat only $Z_{\Gam}(s)$.

According to Lemma \ref{venkov}, we have
\begin{align*}
Z_{\Gam}(s)=&Z_{\slt(\cO)}(s,\chi_{\Gam})\\
=&\prod_{\gam\in\Prim(\slt(\cO))}\prod_{\lambda\in L}
\det(I-\chi_{\Gam}(\gam)\xi_{\lambda}(h(\gam))^{-1}N(\gam)^{-s})^{m_{\lambda}}\\
=&\prod_{D\in\D}\prod_{\lambda\in L}\prod_{\gam\in\cI(t_1(D),u_1(D))}
\det(I-\chi_{\Gam}(\gam)\xi_{\lambda}(h(\gam))^{-1}N(\gam)^{-s})^{m_{\lambda}}\\
=&\prod_{d\in\D}\prod_{\lambda\in L}\prod_{i=1}^{\mu_{\Da}(D)}\prod_{\gam\in\cJ_i(t_1(D),u_1(D))}
\det(I-\chi_{\Gam}(\gam)\xi_{\lambda}(h(\gam))^{-1}N(\gam)^{-s})^{m_{\lambda}}.
\end{align*}
where $\chi_{\Gam}:=\Ind_{\Gam}^{\sz}1$.
Since $\gam\sim\gam_{\nu_i}$ in $\mr{PSL}_2(\cO/\Da)$ and
$\chi_{\Gam}\sim \Ind_{\Gam/\hGam(\Da)}^{\slt(\cO/\Da)}1$,
we see that $\chi_{\Gam}(\gam)\sim\chi_{\Gam}(\gam_{\nu_i})$.
Then, from Lemma \ref{classdiv}, we have
\begin{align*}
&Z_{\Gam}(s)\\ &=\begin{cases}\disp \prod_{D\in\D}\prod_{n=0}^{\infty}\prod_{i=1}^{\mu_{\Da}(D)}
\det(I-\chi_{\Gam}(\gam_{\nu_i})\epsilon(D)^{-2(s+n)})^{h(D)/\mu_{\Da}(D)},&\cO=\bZ,\\
\\ \disp \prod_{D\in\D}\prod_{m,n=0}^{\infty}\prod_{i=1}^{\mu_{\Da}(D)}
\det(I-\chi_{\Gam}(\gam_{\nu_i})\epsilon(D)^{-m}\overline{\epsilon(D)}^{-n}|\epsilon(D)|^{-2(s+n)})
^{h(D)/\mu_{\Da}(D)},&\cO\subset\bQ(\sqrt{-M}).
\end{cases}
\end{align*}
This completes the proof for imaginary quadratic $\cO$.

Consider the case of $\cO=\bZ$.
When $\Gam\supset \hGam(p^r)$, by virtue of Lemma \ref{conj2},
we see that $\gam_1^{l_1}\sim\gam_{\nu_i}$ and $\gam_{\nu_i}^{m_i}\sim\gam_1$
in $\mr{PSL}_2(\bZ/p^r\bZ)$ for some integers $l_i,m_i\geq1$.
Then we can get $\tr\chi_{\Gam}(\gam_1)=\tr\chi_{\Gam}(\gam_{\nu_i})$.
Similarly we have $\tr\chi_{\Gam}(\gam_1^k)=\tr\chi_{\Gam}(\gam_{\nu_i}^k)$ for any $k$.
This means that $\chi_{\Gam}(\gam_1)$ and $\chi_{\Gam}(\gam_{\nu_i})$
give the same type of permutations.
Then we have $\chi_{\Gam}(\gam_1)\sim\chi_{\Gam}(\gam_{\nu_i})$.
Furthermore when $\Gam\supset\hGam(N)$ with $N=\prod_{p\mid N}p^r$,
we see that $\chi_{\Gam}(\gam)\sim\otimes_{p\mid N} \chi_{\Gam_{p^r}}(\gam)$
from Lemma \ref{prod} and \ref{div}.
Then $\chi_{\Gam}(\gam_1)\sim\chi_{\Gam}(\gam_{\nu_i})$ holds for any congruence subgroup of $\sz$.
Therefore we get
\begin{align*}
Z_{\Gam}(s)=\disp \prod_{D\in\D}\prod_{n=0}^{\infty}
\det(I-\chi_{\Gam}(\gam_{1})\epsilon(D)^{-2(s+n)})^{h(D)}.
\end{align*}
This completes the proof of the main theorem for $Z_{\Gam}(s)$.
Getting the expressions for $Z'_{\Gam}(s)/Z_{\Gam}(s)$ is not difficult.
\end{proof}

\section{Examples}
In this section, we give two examples.
For convenience, we put
\begin{align*}
H\big(x;(m_1^{n_1},\cdots,m_l^{n_l})\big):=\prod_{i=1}^l(1-x^{m_i})^{n_i},
\end{align*}
where $m_1,\dots, m_l$ and $n_1,\dots,n_l$ are positive integers.
Then Theorem \ref{thm} and some elementary calculations give
the following expressions of the Selberg zeta functions.

\noindent {\bf Example 1.} In the case of $\Gam=\Gam_0(p):=\{\gam\in\sz\divset p\mid\gam_{21}\}$
with prime $p\geq3$, we have
\begin{align*}
Z_{\Gam_0(p)}(s)=&\prod_{\begin{subarray}{c}D\in\D\\ p\mid u_1 \end{subarray}}\prod_{n=0}^{\infty}
H\big(\epsilon(D)^{-2(s+n)};(1^{p+1})\big)^{h(D)}\\
&\times\prod_{(1<)l_1\mid (p-1)/2}
\prod_{\begin{subarray}{c}D\in\D\\ p\nmid u_1,p\mid u_{l_1}\\ (D/p)=1\end{subarray}}\prod_{n=0}^{\infty}
H\big(\epsilon(D)^{-2(s+n)};(1^2,l_1^{(p-1)/l_1})\big)^{h(D)}\\
&\times\prod_{\begin{subarray}{c}D\in\D\\ p\nmid u_1,p\mid D\end{subarray}}\prod_{n=0}^{\infty}
H\big(\epsilon(D)^{-2(s+n)};(1,p)\big)^{h(D)}\\
&\times \prod_{(1<)l_2\mid (p+1)/2}
\prod_{\begin{subarray}{c}D\in\D\\ p\nmid u_1,p\mid u_{l_2}\\ (D/p)=-1\end{subarray}}\prod_{n=0}^{\infty}
H\big(\epsilon(D)^{-2(s+n)};(l_2^{(p+1)/l_2})\big)^{h(D)},
\end{align*}
\begin{align*}
\frac{Z'_{\Gam_0(p)}(s)}{Z_{\Gam_0(p)}(s)}=&
(p+1)\sum_{\begin{subarray}{c}D\in\D,j\geq1 \\ p\mid u_j\end{subarray}}
h(D)\frac{2\log{\epsilon(D)}}{1-\epsilon(D)^{-2j}}\epsilon(D)^{-2js}\\
&+2\sum_{\begin{subarray}{c}D\in\D,j\geq1 \\ p\nmid u_j, (D/p)=1\end{subarray}}\\
h(D)\frac{2\log{\epsilon(D)}}{1-\epsilon(D)^{-2j}}\epsilon(D)^{-2js}
&+\sum_{\begin{subarray}{c}D\in\D,j\geq1 \\ p\nmid u_j, p\mid D \end{subarray}}
h(D)\frac{2\log{\epsilon(D)}}{1-\epsilon(D)^{-2j}}\epsilon(D)^{-2js}.
\end{align*}

\vspace{12pt}

\noindent {\bf Example 2.} In the case of $\Gam=\Gam_0(\p):=\{\gam\in\slt(\cO)\divset \gam_{21}\equiv0\bmod{\p}\}$
with prime ideal $\p$ such that $\p$ does not divide $2$ and the discriminant of $\cO$, we have
\begin{align*}
Z_{\Gam_0(\p)}(s)=&\prod_{\begin{subarray}{c}D\in\D\\ \p\mid u_1 \end{subarray}}\prod_{n,m=0}^{\infty}
H\big(\epsilon(D)^{-2m}\overline{\epsilon(D)}^{-2n}|\epsilon(D)|^{-2s};(1^{p^2+1})\big)^{h(D)}\\
&\times\prod_{(1<)l_1\mid (p^2-1)/2}
\prod_{\begin{subarray}{c}D\in\D\\ \p\nmid u_1,\p\mid u_{l_1}\\ (D/\p)=1\end{subarray}}\prod_{n,m=0}^{\infty}
H\big(\epsilon(D)^{-2m}\overline{\epsilon(D)}^{-2n}|\epsilon(D)|^{-2s};(1^2,l_1^{(p^2-1)/l_1})\big)^{h(D)}\\
&\times\prod_{\begin{subarray}{c}D\in\D\\ \p\nmid u_1,\p\mid D\end{subarray}}\prod_{n,m=0}^{\infty}
H\big(\epsilon(D)^{-2m}\overline{\epsilon(D)}^{-2n}|\epsilon(D)|^{-2s};(1,p^p)\big)^{h(D)}\\
&\times \prod_{(1<)l_2\mid (p^2+1)/2}
\prod_{\begin{subarray}{c}D\in\D\\ \p\nmid u_1,\p\mid u_{l_2}\\ (D/\p)=-1\end{subarray}}\prod_{n,m=0}^{\infty}
H\big(\epsilon(D)^{-2m}\overline{\epsilon(D)}^{-2n}|\epsilon(D)|^{-2s};(l_2^{(p^2+1)/l_2})\big)^{h(D)},\\
\frac{Z'_{\Gam_0(\p)}(s)}{Z_{\Gam_0(\p)}(s)}
=&(p^2+1)\sum_{\begin{subarray}{c}D\in\D,j\geq1\\ \p\mid u_j\end{subarray}}
h(D)\frac{2\log{\epsilon(D)}}{|1-\epsilon(D)^{-2j}|^2}|\epsilon(D)|^{-2js},\\
&+2\sum_{\begin{subarray}{c}D\in\D,j\geq1\\ \p\nmid u_j, (D/\p)=1\end{subarray}}
h(D)\frac{2\log{\epsilon(D)}}{|1-\epsilon(D)^{-2j}|^2}|\epsilon(D)|^{-2js},\\
&+\sum_{\begin{subarray}{c}D\in\D,j\geq1\\ \p\nmid u_j, \p\mid D\end{subarray}}
h(D)\frac{2\log{\epsilon(D)}}{|1-\epsilon(D)^{-2j}|^2}|\epsilon(D)|^{-2js}.
\end{align*}


TITLE: Double integral question involving substitution
QUESTION [1 upvotes]: $$  \iint_D\frac{x-y+1}{x+y}\,d(x,y)$$
$$D: 1 \leq x+y \leq 2x-4y \leq 3$$
I was trying to solve this integral using uv substitution.
I called: $$u = x+y $$
And:
$$v = 2x-4y$$
I can find the Jacobian and from there the determinant, however, I get stuck on the actual substitution. I do know that the denominator can be put to $u,$ but what do I change the numerator to?
Or do I have the wrong substitution for $u$ and $v$?

REPLY [1 votes]: If we take $$u=x+y\\v=2x-4y$$ then we get $$x=\frac{4u+v}{6} \\ y=\frac{2u-v}{6}$$ Note that $\frac{\partial(x,y)}{\partial(u,v)}=-\frac{1}{6}$ and the image of your region $D$ under $(x,y)\longrightarrow (x+y,2x-4y)$ is the region $\tilde{D}$ defined by $$\tilde{D}:1\leq u \leq v \leq 3$$ So we get $$\iint_{D}\frac{x-y+1}{x+y}\mathrm{d}y\mathrm{d}x=\frac{1}{6}\int_1^3 \int_u^3\Bigg[\frac{v}{3u}+\frac{1}{u}+\frac{1}{3}\Bigg]\mathrm{d}v\mathrm{d}u$$ Can you finish?


TITLE: integration F(x)
QUESTION [0 upvotes]: $$F(x)=\int\limits_{\ln2}^{x}\tanh(t) \mathrm{d}t$$
would this become $$F(x)=\tanh(t) \mathrm{d}t$$
or i have to integrate it
$$F(x)=\int \tanh(t) \mathrm{d}t$$
$$F(x)=\int \frac{e^{t}-e^{-t}}{e^{t}+e^{-t}} \mathrm{d}t$$
$$F(x)= \ln(e^{t}+e^{-t})$$ and sub $x$ and $\ln 2$ to find the solution?

REPLY [1 votes]: So the last expression can be expressed as $$F(x)=ln(e^t+e^{-t})|_{ln(2)}^x,$$ from there you just plug in the top bound $x$, and subtract it by plugging in the lower boun $ln(2)$.