Nonholomorphic Eisenstein series, the Kronecker limit formula, and the hyperbolic Laplacian

1 Nonholomorphic Eisenstein series

Let $ℍ=\{x+iy\in ℂ:y>0\}$ For $\tau =x+iy\in ℍ$ and $s=\sigma +it,\sigma >1$, we define the nonholomorphic Eisenstein series

$$G(\tau ,s)=\frac{1}{2}\sum _{(0,0)\ne (m,n)\in {\mathbb{Z}}^2}\frac{y^s}{{|m\tau +n|}^{2s}}.$$

The function $(\tau ,a,b)\mapsto a\tau +b$ is continuous $ℍ\times S^1\to ℂ$, and for all $\tau \in ℍ$ and $(a,b)\in S^1$ we have $a\tau +b\ne 0$. It follows that if $K$ is a compact subset of $ℍ$ then there is some $C_K>0$ such that $|a\tau +b|\ge C_K$ for all $\tau \in K$, $(a,b)\in S^1$. Then, for all $\tau \in K$ and for all $(0,0)\ne (m,n)\in {\mathbb{Z}}^2$,

$${|m\tau +n|}^2={\left|\frac{m}{\sqrt{m^2+n^2}}\tau +\frac{n}{\sqrt{m^2+n^2}}\right|}^2(m^2+n^2)\ge C_K(m^2+n^2),$$

and hence

$$\left|\frac{y^s}{{|m\tau +n|}^{2s}}\right|=\frac{y^{\sigma }}{{|m\tau +n|}^{2\sigma }}\le \frac{y^{\sigma }}{{(C_K(m^2+n^2))}^{\sigma }}.$$

Because $\sigma >1$,

It follows that for any $s=\sigma +it$ with $\sigma >1$, the function $\tau \mapsto G(\tau ,s)$ is continuous $ℍ\to ℂ$.

It is sometimes useful to write $G$ in another way. For $\tau =x+iy\in ℍ$ and $\mathrm{Re}s>1$, define

$$E(\tau ,s)=\frac{1}{2}\sum _{(c,d)\in {\mathbb{Z}}^2,\gcd (c,d)=1}\frac{y^s}{{|c\tau +d|}^{2s}}.$$

Theorem 1.

For all $\tau \in ℍ$ and $\mathrm{Re}s>1$,

$$G(\tau ,s)=\zeta (2s)E(\tau ,s).$$

Proof.

First we remark that for $0\ne a\in \mathbb{Z}$, $\gcd (a,0)=|a|$. For $(0,0)\ne (m,n)\in {\mathbb{Z}}^2$, with $\nu =\gcd (m,n)$,

$$\gcd (\frac{m}{\nu },\frac{n}{\nu })=1.$$

Then

$G(\tau ,s)$ $={\displaystyle \frac{1}{2}}{\displaystyle \sum _{\nu \ge 1}}{\displaystyle \sum _{(m,n)\in {\mathbb{Z}}^2,\gcd (m,n)=\nu }}{\displaystyle \frac{y^s}{{|m\tau +n|}^{2s}}}$

$={\displaystyle \frac{1}{2}}{\displaystyle \sum _{\nu \ge 1}}{\displaystyle \sum _{(c,d)\in {\mathbb{Z}}^2,\gcd (c,d)=1}}{\displaystyle \frac{y^s}{{|\nu c\tau +\nu d|}^{2s}}}$

$={\displaystyle \frac{1}{2}}{\displaystyle \sum _{(c,d)\in {\mathbb{Z}}^2,\gcd (c,d)=1}}{\displaystyle \frac{y^s}{{|c\tau +d|}^{2s}}}{\displaystyle \sum _{\nu \ge 0}}{\nu }^{-2s}$

$=\zeta (2s)E(\tau ,s).$

∎

2 Modular functions

Theorem 2.

For , $\tau \in ℍ$, and $\mathrm{Re}s>1$,

$$G(\frac{a\tau +b}{c\tau +d},s)=G(\tau ,s).$$

Proof.

$$\frac{a\tau +b}{c\tau +d}=\frac{a\tau +b}{c\tau +d}\cdot \frac{c\overline{\tau }+d}{c\overline{\tau }+d}=\frac{ac{|\tau |}^2+ad\tau +bc\overline{\tau }+bd}{{|c\tau +d|}^2},$$

so, for $\tau =x+iy$ and $\frac{a\tau +b}{c\tau +d}=u+iv$, using that $ad-bc=1$,

$$u=\frac{ac{|\tau |}^2+bd+x(ad+bc)}{{|c\tau +d|}^2},v=\frac{y}{{|c\tau +d|}^2};$$

we shall only use the expression for $v$. Also, for $(m,n)\in {\mathbb{Z}}^2$,

$$m\left(\frac{a\tau +b}{c\tau +d}\right)+n=\frac{(ma+nc)\tau +mb+nd}{c\tau +d}.$$

Then,

$G({\displaystyle \frac{a\tau +b}{c\tau +d}},s)$ $={\displaystyle \frac{1}{2}}{\displaystyle \sum _{(0,0)\ne (m,n)\in {\mathbb{Z}}^2}}{\left({\displaystyle \frac{y}{{|c\tau +d|}^2}}\right)}^s{\left|{\displaystyle \frac{(ma+nc)\tau +mb+nd}{c\tau +d}}\right|}^{-2s}$

$={\displaystyle \frac{1}{2}}{\displaystyle \sum _{(0,0)\ne (m,n)\in {\mathbb{Z}}^2}}{\displaystyle \frac{y^s}{{|(ma+nb)\tau +mb+nd|}^{2s}}}.$

But implies that

$$(m,n)\mapsto (ma+nc,mb+nd)$$

is a bijection ${\mathbb{Z}}^2\setminus \{(0,0)\}\to {\mathbb{Z}}^2\setminus \{(0,0)\}$, so

$$\sum _{(0,0)\ne (m,n)\in {\mathbb{Z}}^2}\frac{1}{{|(ma+nb)\tau +mb+nd|}^{2s}}=\sum _{(0,0)\ne (\mu ,\nu )\in {\mathbb{Z}}^2}\frac{1}{{|\mu \tau +\nu |}^{2s}},$$

and thus we get

$$G(\frac{a\tau +b}{c\tau +d},s)=G(\tau ,s),$$

completing the proof. ∎

3 Fourier expansion

We now derive the Fourier series of $G(\cdot ,s)$. $K_{s-\frac{1}{2}}$ denotes the Bessel function.

Theorem 3.

If $\tau \in ℍ$ and $\mathrm{Re}s>1$, then

$G(\tau ,s)$ $=\zeta (2s)y^s+{\pi }^{\frac{1}{2}}{\displaystyle \frac{\mathrm{\Gamma }\left(s-\frac{1}{2}\right)}{\mathrm{\Gamma }(s)}}\zeta (2s-1)y^{-s+1}$

$+2{\displaystyle \frac{{\pi }^s}{\mathrm{\Gamma }(s)}}{\displaystyle \sum _{n=1}^{\mathrm{\infty }}}{n}^{s-1}{\sigma }_{-2s+1}(n)F_{s-\frac{1}{2}}(2\pi ny)\cos (2\pi nx),$

where

$$F_{s-\frac{1}{2}}(w)={\left(\frac{2w}{\pi }\right)}^{1/2}{K}_{s-\frac{1}{2}}(w).$$

Proof.

Define

$$S(z,s)=\sum _{n\in \mathbb{Z}}\frac{{|y|}^s}{{|z+n|}^{2s}},z=x+iy,y\ne 0,\mathrm{Re}s>1.$$

We can write $G(\tau ,s)$ using this as

$$G(\tau ,s)=\frac{1}{2}\sum _{n\ne 0}\frac{y^s}{{|n|}^{2s}}+\frac{1}{2}\sum _{m\ne 0}\sum _{n\in \mathbb{Z}}\frac{y^s}{{|m\tau +n|}^{2s}}=y^s\zeta (2s)+\frac{1}{2}\sum _{m\ne 0}\frac{S(m\tau ,s)}{{|m|}^s}.$$

The Poisson summation formula states that if $f:ℝ\to ℂ$ is continuous and of locally bounded variation, then for all $x\in ℝ$,

$$\sum _{n\in \mathbb{Z}}f(x+n)=\sum _{k\in \mathbb{Z}}\widehat{f}(k)e^{2\pi ikx},$$

where

$$\widehat{f}(\xi )={\int }_{ℝ}{e}^{-2\pi i\xi t}f(t)𝑑t,\xi \in ℝ.$$

Let $z=x+iy,y\ne 0$, let $\mathrm{Re}s>1$, and define $f_y:ℝ\to ℂ$ by

$$f_y(t)={|t+iy|}^{-2s}={((t+iy)(t-iy))}^{-s}={(t^2+y^2)}^{-s},t\in ℝ.$$

Applying the Poisson summation formula we get

$$\sum _{n\in \mathbb{Z}}{|x+n+iy|}^{-2s}=\sum _{k\in \mathbb{Z}}{\widehat{f}}_y(k)e^{2\pi ikx},$$

i.e.,

$$S(z,s)={|y|}^s\sum _{k\in \mathbb{Z}}{\widehat{f}}_y(k)e^{2\pi ikx},$$ (1)

with

$${\widehat{f}}_y(k)={\int }_{ℝ}{e}^{-2\pi ikt}{(t^2+y^2)}^{-s}𝑑t.$$

As $y\ne 0$, doing the change of variable $t=yu$ we get

${\widehat{f}}_y(k)$ $={\displaystyle {\int }_{ℝ}}{e}^{-2\pi ikyu}{(y^2{u}^2+y^2)}^{-s}|y|𝑑u$

$={|y|}^{-2s+1}{\displaystyle {\int }_{ℝ}}{e}^{-2\pi ikyu}{(u^2+1)}^{-s}𝑑u$

$=2{|y|}^{-2s+1}{\displaystyle {\int }_0^{\mathrm{\infty }}}\cos (2\pi kyu){(u^2+1)}^{-s}𝑑u;$

the final equality is because the function $u\mapsto {(u^2+1)}^{-s}$ is even.

We use the following identity: for $a>0$ and $\mathrm{Re}s>\frac{1}{2}$,

$${\int }_0^{\mathrm{\infty }}\cos (au){(u^2+1)}^{-s}𝑑u={\pi }^{1/2}\cdot {\left(\frac{a}{2}\right)}^{s-\frac{1}{2}}\cdot \frac{1}{\mathrm{\Gamma }(s)}\cdot K_{s-\frac{1}{2}}(a).$$

For $k\in \mathbb{Z}\setminus \{0\}$, using this with $a=2\pi |ky|>0$ gives

${\widehat{f}}_y(k)$ $=2{|y|}^{-2s+1}\cdot {\pi }^{1/2}\cdot {(\pi |ky|)}^{s-\frac{1}{2}}\cdot {\displaystyle \frac{1}{\mathrm{\Gamma }(s)}}\cdot K_{s-\frac{1}{2}}(2\pi |ky|)$

$=2{|y|}^{-s+\frac{1}{2}}{\pi }^s{|k|}^{s-\frac{1}{2}}\cdot {\displaystyle \frac{1}{\mathrm{\Gamma }(s)}}\cdot K_{s-\frac{1}{2}}(2\pi |ky|).$

Therefore (1) becomes

$S(z,s)$ $={|y|}^s{\widehat{f}}_y(0)+{|y|}^s{\displaystyle \sum _{k\ne 0}}2{|y|}^{-s+\frac{1}{2}}{\pi }^s{|k|}^{s-\frac{1}{2}}\cdot {\displaystyle \frac{1}{\mathrm{\Gamma }(s)}}\cdot K_{s-\frac{1}{2}}(2\pi |ky|)\cdot e^{2\pi ikx}$

$={|y|}^s{\widehat{f}}_y(0)+2{|y|}^{\frac{1}{2}}{\pi }^s\cdot {\displaystyle \frac{1}{\mathrm{\Gamma }(s)}}{\displaystyle \sum _{k\ne 0}}{|k|}^{s-\frac{1}{2}}\cdot K_{s-\frac{1}{2}}(2\pi |ky|)\cdot e^{2\pi ikx}$

$={|y|}^s{\widehat{f}}_y(0)+4{|y|}^{\frac{1}{2}}{\pi }^s\cdot {\displaystyle \frac{1}{\mathrm{\Gamma }(s)}}{\displaystyle \sum _{k=1}^{\mathrm{\infty }}}{k}^{s-\frac{1}{2}}\cdot K_{s-\frac{1}{2}}(2\pi k|y|)\cdot \cos (2\pi kx).$

We use the following identity for the beta function: For $\mathrm{Re}b>\frac{1}{2}\mathrm{Re}a>0$,

$${\int }_0^{\mathrm{\infty }}{u}^{a-1}{(u^2+1)}^{-b}𝑑u=\frac{1}{2}B(\frac{a}{2},b-\frac{a}{2})=\frac{\mathrm{\Gamma }\left(\frac{a}{2}\right)\mathrm{\Gamma }\left(b-\frac{a}{2}\right)}{2\mathrm{\Gamma }(b)}.$$

Using this with $a=1$ and $b=s$, and since $\mathrm{\Gamma }\left(\frac{1}{2}\right)={\pi }^{\frac{1}{2}}$,

$${\widehat{f}}_y(0)=2{|y|}^{-2s+1}{\int }_0^{\mathrm{\infty }}{(u^2+1)}^{-s}𝑑u=2{|y|}^{-2s+1}\frac{{\pi }^{\frac{1}{2}}\mathrm{\Gamma }\left(s-\frac{1}{2}\right)}{2\mathrm{\Gamma }(s)}.$$

Therefore

$S(z,s)$ $={\pi }^{\frac{1}{2}}\cdot {|y|}^{-s+1}\cdot {\displaystyle \frac{\mathrm{\Gamma }\left(s-\frac{1}{2}\right)}{\mathrm{\Gamma }(s)}}$

$+4{|y|}^{\frac{1}{2}}{\pi }^s\cdot {\displaystyle \frac{1}{\mathrm{\Gamma }(s)}}{\displaystyle \sum _{k=1}^{\mathrm{\infty }}}{k}^{s-\frac{1}{2}}\cdot K_{s-\frac{1}{2}}(2\pi k|y|)\cdot \cos (2\pi kx)$

$={\pi }^{\frac{1}{2}}\cdot {|y|}^{-s+1}\cdot {\displaystyle \frac{\mathrm{\Gamma }\left(s-\frac{1}{2}\right)}{\mathrm{\Gamma }(s)}}$

$+2{\displaystyle \frac{{\pi }^s}{\mathrm{\Gamma }(s)}}{\displaystyle \sum _{k=1}^{\mathrm{\infty }}}{k}^{s-1}{F}_{s-\frac{1}{2}}(2\pi k|y|)\cdot \cos (2\pi kx).$

We now express $G(\tau ,s)$ using this formula for $S(z,s)$. For $\tau \in ℍ$ and $\mathrm{Re}s>1$, since $S(z,s)=S(-z,s)$,

$G(\tau ,s)$ $=y^s\zeta (2s)+{\displaystyle \frac{1}{2}}{\displaystyle \sum _{m\ne 0}}{\displaystyle \frac{S(m\tau ,s)}{{|m|}^s}}$

$=y^s\zeta (2s)+{\displaystyle \sum _{m=1}^{\mathrm{\infty }}}{\displaystyle \frac{S(m\tau ,s)}{m^s}}$

$=y^s\zeta (2s)+{\pi }^{\frac{1}{2}}{\displaystyle \frac{\mathrm{\Gamma }\left(s-\frac{1}{2}\right)}{\mathrm{\Gamma }(s)}}{\displaystyle \sum _{m=1}^{\mathrm{\infty }}}{\displaystyle \frac{{(my)}^{-s+1}}{m^s}}$

$+2{\displaystyle \frac{{\pi }^s}{\mathrm{\Gamma }(s)}}{\displaystyle \sum _{m=1}^{\mathrm{\infty }}}{\displaystyle \frac{1}{m^s}}{\displaystyle \sum _{k=1}^{\mathrm{\infty }}}{k}^{s-1}{F}_{s-\frac{1}{2}}(2\pi kmy)\cdot \cos (2\pi kmx)$

$=y^s\zeta (2s)+{\pi }^{\frac{1}{2}}{\displaystyle \frac{\mathrm{\Gamma }\left(s-\frac{1}{2}\right)}{\mathrm{\Gamma }(s)}}{y}^{-s+1}\zeta (2s-1)$

$+2{\displaystyle \frac{{\pi }^s}{\mathrm{\Gamma }(s)}}{\displaystyle \sum _{k,m\ge 1}}{\displaystyle \frac{k^{s-1}}{m^s}}{F}_{s-\frac{1}{2}}(2\pi kmy)\cdot \cos (2\pi kmx).$

As

$$\sum _{km=N}\frac{k^{s-1}}{m^s}=\sum _{km=N}\frac{{(km)}^{s-1}}{m^{2s-1}}=N^{s-1}\sum _{km=N}{m}^{-2s+1}=N^{s-1}{\sigma }_{-2s+1}(N),$$

this can be written as

$G(\tau ,s)$ $=y^s\zeta (2s)+{\pi }^{\frac{1}{2}}{\displaystyle \frac{\mathrm{\Gamma }\left(s-\frac{1}{2}\right)}{\mathrm{\Gamma }(s)}}{y}^{-s+1}\zeta (2s-1)$

$+2{\displaystyle \frac{{\pi }^s}{\mathrm{\Gamma }(s)}}{\displaystyle \sum _{N=1}^{\mathrm{\infty }}}{N}^{s-1}{\sigma }_{-2s+1}(N)F_{s-\frac{1}{2}}(2\pi Ny)\cos (2\pi Nx),$

completing the proof. ∎

We use the above Fourier expansion to establish that for all $t\in ℍ$, $G(\tau ,s)$ has a meromorphic continuation to $ℂ$ and satisfies a certain functional equation. The meromorphic continuation and functional equation of $G(\tau ,s)$ can also be obtained without using its Fourier expansion.

Theorem 4.

For any $\tau \in ℍ$, $G(\tau ,s)$ has a meromorphic continuation to $ℂ$ whose only pole is at $s=1$, which is a simple pole with residue $\frac{\pi }{2}$. The function

$$𝒢(\tau ,s)={\pi }^{-s}\mathrm{\Gamma }(s)G(\tau ,s)$$

satisfies the functional equation

$$𝒢(\tau ,1-s)=𝒢(\tau ,s).$$

Proof.

For $\nu \in ℂ$ and for $w>0$ we have

$$K_{\nu }(w)={\int }_0^{\mathrm{\infty }}{e}^{-w\cosh t}\cosh (\nu t)𝑑t$$

and

$$K_{\nu }(w)\sim {\left(\frac{2w}{\pi }\right)}^{-\frac{1}{2}}{e}^{-w},w\to +\mathrm{\infty }.$$

Using the above identity, one checks that for $w>0$, the function $s\mapsto K_{s-\frac{1}{2}}(w)$ is entire, and that for any $s\in ℂ$, the function $w\mapsto K_{s-\frac{1}{2}}(w)$ belongs to $C^{\mathrm{\infty }}({ℝ}_{>0})$. We have a fortiori that for any $s\in ℂ$,

$$K_{s-\frac{1}{2}}(w)=O(e^{-w}),w\to +\mathrm{\infty }.$$

Let $\mathrm{\Lambda }(s)={\pi }^{-\frac{s}{2}}\mathrm{\Gamma }\left(\frac{s}{2}\right)\zeta (s)$. The functional equation for the Riemann zeta function states that $\mathrm{\Lambda }$ has a meromorphic continuation to $ℂ$ whose only poles are at $s=0$ and $s=1$, which are simple poles, and satisfies, for all $s\ne 0,1$,

$$\mathrm{\Lambda }(1-s)=\mathrm{\Lambda }(s).$$

Using the Fourier series for $G(\cdot ,s)$, we have that for $\tau \in ℍ$ and $\mathrm{Re}s>1$,

$𝒢(\tau ,s)$ $={\pi }^{-s}\mathrm{\Gamma }(s)G(\tau ,s)$

$=\mathrm{\Lambda }(2s)y^s+\mathrm{\Lambda }(2s-1)y^{-s+1}$

$+2{\displaystyle \sum _{n=1}^{\mathrm{\infty }}}{n}^{s-1}{\sigma }_{-2s+1}(n)F_{s-\frac{1}{2}}(2\pi ny)\cos (2\pi nx).$

The residue of $\mathrm{\Lambda }(2s)$ at $s=0$ is $-\frac{1}{2}$; the residue of $\mathrm{\Lambda }(2s)$ at $s=\frac{1}{2}$ is $\frac{1}{2}$; the residue of $\mathrm{\Lambda }(2s-1)$ at $s=\frac{1}{2}$ is $-\frac{1}{2}$; and the residue of $\mathrm{\Lambda }(2s-1)$ at $s=1$ is $\frac{1}{2}$. It follows that the residue of $𝒢(\tau ,s)$ at $s=0$ is $-\frac{1}{2}$; the residue of $𝒢(\tau ,s)$ at $s=\frac{1}{2}$ is $\frac{1}{2}{y}^{1/2}-\frac{1}{2}{y}^{1/2}=0$; the residue of $𝒢(\tau ,s)$ at $s=1$ is $\frac{1}{2}$; and these are no other poles of $𝒢(\tau ,s)$. Because $\mathrm{\Gamma }(s)$ has a simple pole at $s=0$, $G(\tau ,s)=\frac{{\pi }^s}{\mathrm{\Gamma }(s)}𝒢(\tau ,s)$ does not have a pole at $s=0$. The residue of $G(\tau ,s)$ at $s=1$ is $\frac{\pi }{\mathrm{\Gamma }(1)}\cdot \frac{1}{2}=\frac{\pi }{2}$, and this is the only pole of $G(\tau ,s)$.

For $s\in ℂ$,

$n^{(1-s)-1}{\sigma }_{-2(1-s)+1}(n)$ $=n^{-s}{\sigma }_{2s-1}(n)$

$={\displaystyle \sum _{ef=n}}{(ef)}^{-s}{e}^{2s-1}$

$={\displaystyle \sum _{ef=n}}{e}^{s-1}{f}^{-s}$

$={\displaystyle \sum _{ef=n}}{(ef)}^{s-1}{f}^{-2s+1}$

$=n^{s-1}{\sigma }_{-2s+1}(n).$

Generally, $K_{\nu }=K_{-\nu }$, so $F_{s-\frac{1}{2}}=F_{(1-s)-\frac{1}{2}}$. Thus each term in the series in the above formula for $𝒢(\tau ,s)$ is unchanged if $s$ is replaced with $1-s$, and together with $\mathrm{\Lambda }(1-w)=\mathrm{\Lambda }(w)$ this yields

$𝒢(\tau ,1-s)$ $=\mathrm{\Lambda }(2-2s)y^{1-s}+\mathrm{\Lambda }(2-2s-1)y^{-(1-s)+1}$

$+2{\displaystyle \sum _{n=1}^{\mathrm{\infty }}}{n}^{s-1}{\sigma }_{-2s+1}(n)F_{s-\frac{1}{2}}(2\pi ny)\cos (2\pi nx)$

$=\mathrm{\Lambda }(1-(2s-1))y^{1-s}+\mathrm{\Lambda }(1-2s)y^s$

$+2{\displaystyle \sum _{n=1}^{\mathrm{\infty }}}{n}^{s-1}{\sigma }_{-2s+1}(n)F_{s-\frac{1}{2}}(2\pi ny)\cos (2\pi nx)$

$=\mathrm{\Lambda }(2s-1)y^{1-s}+\mathrm{\Lambda }(2s)y^s$

$+2{\displaystyle \sum _{n=1}^{\mathrm{\infty }}}{n}^{s-1}{\sigma }_{-2s+1}(n)F_{s-\frac{1}{2}}(2\pi ny)\cos (2\pi nx)$

$=𝒢(\tau ,s).$

∎

4 Kronecker limit formula

For $\tau \in ℍ$, Theorem 4 shows that $G(\tau ,s)$ is meromorphic and that its only pole is at $s=1$, which is a simple pole with residue $\frac{\pi }{2}$. It follows that $G(\tau ,s)$ has the Laurent expansion about $s=1$,

$$G(\tau ,s)=\frac{\pi }{2}\cdot \frac{1}{s-1}+a_0(\tau )+a_1(\tau )\cdot (s-1)+\mathrm{\cdots },$$

and so defining $\frac{\pi }{2}C(\tau )=a_0(\tau )$,

$$G(\tau ,s)=\frac{\pi }{2}\left(\frac{1}{s-1}+C(\tau )+O(|s-1|)\right),s\to 1.$$

We define the Dedekind eta function $\eta :ℍ\to ℂ$ by

$$\eta (\tau )=e^{\frac{\pi i\tau }{12}}\prod _{n=1}^{\mathrm{\infty }}(1-q^n),\tau \in ℍ,$$

where $q=e^{2\pi i\tau }=e^{-2\pi y}{e}^{2\pi ix}$, for $\tau =x+iy$. We now prove the Kronecker limit formula, which expresses $C(\tau )$ in terms of the Dedekind eta function.

Theorem 5.

For $\tau =x+iy\in ℍ$,

$$G(\tau ,s)=\frac{\pi }{2}\left(\frac{1}{s-1}+C(\tau )+O(|s-1|)\right),s\to 1,$$

with

$$C(\tau )=2\gamma -2\log 2-\log y-4\log |\eta (\tau )|.$$

Proof.

Define

$$G(s)={\pi }^{\frac{1}{2}}\frac{\mathrm{\Gamma }\left(s-\frac{1}{2}\right)}{\mathrm{\Gamma }(s)}\zeta (2s-1)y^{-s+1}.$$

Then

$$\log G(s)=\frac{1}{2}\log \pi +\log \zeta (2s-1)+(-s+1)\log y+\log \left(\frac{\mathrm{\Gamma }\left(s-\frac{1}{2}\right)}{\mathrm{\Gamma }(s)}\right).$$ (2)

We use the asymptotic formula

$$\zeta (s)=\frac{1}{s-1}+\gamma +O(|s-1|),s\to 1,$$

and with

$$\log (1+w)=w+O({|w|}^2),w\to 1,$$

this gives, as $s\to 1$,

$\log \zeta (s)$ $=\log \left({\displaystyle \frac{1}{s-1}}+\gamma +O(|s-1|)\right)$

$=-\log (s-1)+\log (1+\gamma (s-1)+O({|s-1|}^2))$

$=-\log (s-1)+\gamma (s-1)+O({|s-1|}^2),$

and hence

$$\log \zeta (2s-1)=-\log (2s-2)+\gamma (2s-2)+O({|s-1|}^2),s\to 1.$$

The Taylor series for $\log \mathrm{\Gamma }(z)$ about $z=\frac{1}{2}$ is

$$\log \mathrm{\Gamma }(z)=\frac{1}{2}\log \pi -(2\log 2+\gamma )\left(z-\frac{1}{2}\right)+\sum _{k=2}^{\mathrm{\infty }}{(-1)}^k(2^k-1)\frac{\zeta (k)}{k}{\left(z-\frac{1}{2}\right)}^k,$$

for , and the Taylor series of $\log \mathrm{\Gamma }(1+z)$ about $z=0$ is

$$\log \mathrm{\Gamma }(1+z)=-\gamma z+\sum _{k=2}^{\mathrm{\infty }}{(-1)}^k\frac{\zeta (k)}{k}{z}^k,$$

for . Using these we have

$$\log \mathrm{\Gamma }\left(s-\frac{1}{2}\right)=\frac{1}{2}\log \pi -(2\log 2+\gamma )(s-1)+O({|s-1|}^2),s\to 1$$

and

$$\log \mathrm{\Gamma }(s)=-\gamma (s-1)+O({|s-1|}^2),s\to 1.$$

Applying these approximations with (2) we get, as $s\to 1$,

$\log G(s)$ $={\displaystyle \frac{1}{2}}\log \pi -\log (2s-2)+\gamma (2s-2)+O({|s-1|}^2)+(-s+1)\log y$

$+{\displaystyle \frac{1}{2}}\log \pi -(2\log 2+\gamma )(s-1)+O({|s-1|}^2)$

$+\gamma (s-1)+O({|s-1|}^2)$

$=\log \pi -\log 2-\log (s-1)+2\gamma (s-1)+(-s+1)\log y$

$-(2\log 2+\gamma )(s-1)+\gamma (s-1)+O({|s-1|}^2)$

$=\log {\displaystyle \frac{\pi }{2}}-\log (s-1)+(2\gamma -2\log 2-\log y)(s-1)+O({|s-1|}^2).$

Taking the exponential and using

$$e^w=1+w+O({|w|}^2),w\to 0,$$

as $s\to 1$ we have

$G(s)$ $={\displaystyle \frac{\pi }{2}}\cdot {\displaystyle \frac{1}{s-1}}\cdot \left(1+(2\gamma -2\log 2-\log y)(s-1)+O({|s-1|}^2)\right)$

$={\displaystyle \frac{\pi }{2}}\cdot {\displaystyle \frac{1}{s-1}}+{\displaystyle \frac{\pi }{2}}\cdot (2\gamma -2\log 2-\log y)+O(|s-1|).$

Using this and the fact that

$$\zeta (2s)y^s=\frac{{\pi }^2}{6}y+O(|s-1|),s\to 1,$$

Theorem 3 thus yields that as $s\to 1$,

$G(\tau ,s)$ $={\displaystyle \frac{{\pi }^2}{6}}y+{\displaystyle \frac{\pi }{2}}\cdot {\displaystyle \frac{1}{s-1}}+{\displaystyle \frac{\pi }{2}}\cdot (2\gamma -2\log 2-\log y)+O(|s-1|)$

$+2{\displaystyle \frac{{\pi }^s}{\mathrm{\Gamma }(s)}}{\displaystyle \sum _{n=1}^{\mathrm{\infty }}}{n}^{s-1}{\sigma }_{-2s+1}(n)F_{s-\frac{1}{2}}(2\pi ny)\cos (2\pi nx).$

We have

$$\frac{{\pi }^s}{\mathrm{\Gamma }(s)}=\pi +O(|s-1|),s\to 1.$$

As well,

$$n^{s-1}=1+O(|s-1|),s\to 1,$$

and

$${\sigma }_{-2s+1}(n)=\sum _{d|n}{d}^{-2s+1}=\sum _{d|n}(d^{-1}+O(|s-1|))={\sigma }_{-1}(n)+O(|s-1|),s\to 1.$$

Finally, we use the fact that that for all $t>0$,

$$K_{\frac{1}{2}}(t)=\sqrt{\frac{\pi }{2t}}{e}^{-t},$$

giving

$$F_{\frac{1}{2}}(t)={\left(\frac{2t}{\pi }\right)}^{\frac{1}{2}}{K}_{\frac{1}{2}}(t)=e^{-t},$$

and hence

$$F_{s-\frac{1}{2}}(2\pi ny)=e^{-2\pi ny}+O(|s-1|),s\to 1.$$

Therefore, as $s\to 1$,

$G(\tau ,s)$ $={\displaystyle \frac{{\pi }^2}{6}}y+{\displaystyle \frac{\pi }{2}}\cdot {\displaystyle \frac{1}{s-1}}+{\displaystyle \frac{\pi }{2}}\cdot (2\gamma -2\log 2-\log y)$

$+2\pi {\displaystyle \sum _{n=1}^{\mathrm{\infty }}}1\cdot {\sigma }_{-1}(n)e^{-2\pi ny}\cos (2\pi nx)+O(|s-1|).$

This implies that the constant term in the Laurent expansion of $G(\tau ,s)$ about $s=1$ is

$$a_0(\tau )=\frac{{\pi }^2}{6}y+\frac{\pi }{2}\cdot (2\gamma -2\log 2-\log y)+2\pi \sum _{n=1}^{\mathrm{\infty }}{\sigma }_{-1}(n)e^{-2\pi ny}\cos (2\pi nx).$$

But, with $q=e^{2\pi i\tau }$,

$\mathrm{Re}\left({\displaystyle \sum _{n=1}^{\mathrm{\infty }}}{\sigma }_{-1}(n)q^n\right)$ $={\displaystyle \sum _{n=1}^{\mathrm{\infty }}}{\sigma }_{-1}(n)\mathrm{Re}(e^{2\pi in\tau })$

$={\displaystyle \sum _{n=1}^{\mathrm{\infty }}}{\sigma }_{-1}(n)\mathrm{Re}(e^{-2\pi ny}{e}^{2\pi inx})$

$={\displaystyle \sum _{n=1}^{\mathrm{\infty }}}{\sigma }_{-1}(n)e^{-2\pi ny}\cos (2\pi nx),$

so

$$a_0(\tau )=\frac{{\pi }^2}{6}y+\frac{\pi }{2}\cdot (2\gamma -2\log 2-\log y)+2\pi S(\tau ),$$

where

$$S(\tau )=\mathrm{Re}\left(\sum _{n=1}^{\mathrm{\infty }}{\sigma }_{-1}(n)q^n\right).$$

Using the power series for $\log (1+z)$ about $z=0$,

$\log {\displaystyle \prod _{n=1}^{\mathrm{\infty }}}(1-q^n)$ $={\displaystyle \sum _{n=1}^{\mathrm{\infty }}}\log (1-q^n)$

$=-{\displaystyle \sum _{n=1}^{\mathrm{\infty }}}{\displaystyle \sum _{m=1}^{\mathrm{\infty }}}{\displaystyle \frac{q^{nm}}{m}}$

$=-{\displaystyle \sum _{N=1}^{\mathrm{\infty }}}{\displaystyle \sum _{d|N}}{\displaystyle \frac{q^N}{d}}$

$=-{\displaystyle \sum _{N=1}^{\mathrm{\infty }}}{\sigma }_{-1}(N)q^N,$

so

$$S(\tau )=-\mathrm{Re}\left(\log \prod _{n=1}^{\mathrm{\infty }}(1-q^n)\right).$$

Then, because

$$\mathrm{Re}\log z=\log |z|$$

and because

$$|\eta (\tau )|=\left|e^{\frac{\pi i\tau }{12}}\prod _{n=1}^{\mathrm{\infty }}(1-q^n)\right|=e^{-\frac{\pi y}{12}}\prod _{n=1}^{\mathrm{\infty }}|1-q^n|,$$

this becomes

$$S(\tau )=-\log \prod _{n=1}^{\mathrm{\infty }}|1-q^n|=-\frac{\pi y}{12}-\log |\eta (\tau )|.$$

Thus

$a_0(\tau )$ $={\displaystyle \frac{{\pi }^2}{6}}y+{\displaystyle \frac{\pi }{2}}\cdot (2\gamma -2\log 2-\log y)-{\displaystyle \frac{{\pi }^2}{6}}y-2\pi \log |\eta (\tau )|$

$={\displaystyle \frac{\pi }{2}}\cdot (2\gamma -2\log 2-\log y)-2\pi \log |\eta (\tau )|,$

so

$$C(\tau )=2\gamma -2\log 2-\log y-4\log |\eta (\tau )|,$$

completing the proof. ∎

5 Hyperbolic Laplacian

For $f\in C^2(ℍ)$, we define ${\mathrm{\Delta }}_{ℍ}f:ℍ\to ℂ$ by

$$({\mathrm{\Delta }}_{ℍ}f)(\tau )=-y^2({\partial }_x^{2}f+{\partial }_y^{2}f)(\tau ),\tau =x+iy\in ℍ.$$

For more on ${\mathrm{\Delta }}_{ℍ}$ see the below references.

Let $(0,0)\ne (m,n)\in {\mathbb{Z}}^2$ and $\mathrm{Re}s>1$, and define $f:ℍ\to ℂ$ by

$$f(x,y)=y^s{|mx+n+imy|}^{-2s}=y^s{(mx+n+imy)}^{-s}{(mx+n-imy)}^{-s}.$$

Write

$$g(x,y)={(mx+n+imy)}^{-s}{(mx+n-imy)}^{-s}.$$

We calculate

$({\partial }_{x}g)(x,y)$ $=-s{(mx+n+imy)}^{-s-1}m{(mx+n-imy)}^{-s}$

$-s{(mx+n+imy)}^{-s}{(mx+n-imy)}^{-s-1}m,$

and

$({\partial }_x^{2}g)(x,y)$ $=s(s+1){(mx+n+imy)}^{-s-2}{m}^2{(mx+n-imy)}^{-s}$

$+s^2{(mx+n+imy)}^{-s-1}{(mx+n-imy)}^{-s-1}{m}^2$

$+s^2{(mx+n+imy)}^{-s-1}{m}^2{(mx+n-imy)}^{-s-1}$

$+s(s+1){(mx+n+imy)}^{-s}{(mx+n-imy)}^{-s-2}{m}^2$

$=s(s+1)m^2{(mx+n+imy)}^{-2}g(x,y)$

$+2s^2{m}^2{(mx+n+imy)}^{-1}{(mx+n-imy)}^{-1}g(x,y)$

$+s(s+1)m^2{(mx+n-imy)}^{-2}g(x,y),$

from which we have

$({\partial }_x^{2}f)(x,y)$ $=s(s+1)m^2{(mx+n+imy)}^{-2}f(x,y)$

$+2s^2{m}^2{|m\tau +n|}^{-2}f(x,y)$

$+s(s+1)m^2{(mx+n-imy)}^{-2}f(x,y).$

We also calculate

$({\partial }_{y}g)(x,y)$ $=-s{(mx+n+imy)}^{-s-1}im{(mx+n-imy)}^{-s}$

$-s{(mx+n+imy)}^{-s}{(mx+n-imy)}^{-s-1}(-im),$

$({\partial }_y^{2}g)(x,y)$ $=s(s+1){(mx+n+imy)}^{-s-2}(-m^2){(mx+n-imy)}^{-s}$

$+s^2{(mx+n+imy)}^{-s-1}{(mx+n-imy)}^{-s-1}{m}^2$

$+s^2{(mx+n+imy)}^{-s-1}(im){(mx+n-imy)}^{-s-1}(-im)$

$+s(s+1){(mx+n+imy)}^{-s}{(mx+n-imy)}^{-s-2}{(-im)}^2$

$=-s(s+1)m^2{(mx+n+imy)}^{-2}g(x,y)$

$+2s^2{m}^2{(mx+n+imy)}^{-1}{(mx+n-imy)}^{-1}g(x,y)$

$-s(s+1)m^2{(mx+n-imy)}^{-2}g(x,y).$

Now,

$$({\partial }_{y}f)(x,y)=s{y}^{s-1}g(x,y)+y^s({\partial }_{y}g)(x,y)$$

and

$({\partial }_y^{2}f)(x,y)$ $=s(s-1)y^{s-2}g(x,y)+2s{y}^{s-1}({\partial }_{y}g)(x,y)$

$+y^s({\partial }_y^{2}g)(x,y),$

from which we have

$({\partial }_y^{2}f)(x,y)$ $=s(s-1)y^{s-2}g(x,y)$

$-2s^{2}im{y}^{s-1}{(mx+n+imy)}^{-1}g(x,y)$

$+2s^{2}im{y}^{s-1}{(mx+n-imy)}^{-1}g(x,y)$

$-s(s+1)m^2{y}^s{(mx+n+imy)}^{-2}g(x,y)$

$+2s^2{m}^2{y}^s{(mx+n+imy)}^{-1}{(mx+n-imy)}^{-1}g(x,y)$

$-s(s+1)m^2{y}^s{(mx+n-imy)}^{-2}g(x,y)$

$=s(s-1)y^{-2}f(x,y)$

$-2s^{2}im{y}^{-1}{(mx+n+imy)}^{-1}f(x,y)$

$+2s^{2}im{y}^{-1}{(mx+n-imy)}^{-1}f(x,y)$

$-s(s+1)m^2{(mx+n+imy)}^{-2}f(x,y)$

$+2s^2{m}^2{|m\tau +n|}^{-2}f(x,y)$

$-s(s+1)m^2{(mx+n-imy)}^{-2}f(x,y).$

Combining the above expressions we get

$({\partial }_x^{2}f+{\partial }_y^2)(x,y)$ $=m^{2}f(x,y)\cdot \left(2s^2{|m\tau +n|}^{-2}+2s^2{|m\tau +n|}^{-2}\right)$

$+s(s-1)y^{-2}f(x,y)$

$-2s^{2}im{y}^{-1}{(mx+n+imy)}^{-1}f(x,y)$

$+2s^{2}im{y}^{-1}{(mx+n-imy)}^{-1}f(x,y)$

$=m^{2}f(x,y)\cdot \left(2s^2{|m\tau +n|}^{-2}+2s^2{|m\tau +n|}^{-2}\right)$

$+s(s-1)y^{-2}f(x,y)-4s^2{m}^2{|m\tau +n|}^{-2}f(x,y)$

$=s(s-1)y^{-2}f(x,y).$

Thus

$$({\mathrm{\Delta }}_{ℍ}f)(x,y)=s(s-1)f(x,y),$$

i.e.,

$${\mathrm{\Delta }}_{ℍ}f=s(s-1)f.$$

Thus we immediately get that for $\mathrm{Re}s>1$,

${\mathrm{\Delta }}_{ℍ}G(\cdot ,s)=s(s-1)G(\cdot ,s).$

Because the coefficients of the differential operator $L={\mathrm{\Delta }}_{ℍ}-s(s-1)$ are real analytic, a function $f:ℍ\to ℂ$ satisfying $Lf=0$ is real analytic. Therefore, for $\mathrm{Re}s>1$, $G(\cdot ,s)$ is real analytic.