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

Jordan Bell
Department of Mathematics, University of Toronto
January 22, 2022

## 1 Nonholomorphic Eisenstein series

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

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

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

 $|m\tau+n|^{2}=\left|\frac{m}{\sqrt{m^{2}+n^{2}}}\tau+\frac{n}{\sqrt{m^{2}+n^{2% }}}\right|^{2}(m^{2}+n^{2})\geq C_{K}(m^{2}+n^{2}),$

and hence

 $\left|\frac{y^{s}}{|m\tau+n|^{2s}}\right|=\frac{y^{\sigma}}{|m\tau+n|^{2\sigma% }}\leq\frac{y^{\sigma}}{(C_{K}(m^{2}+n^{2}))^{\sigma}}.$

Because $\sigma>1$,

 $\sum_{(0,0)\neq(m,n)\in\mathbb{Z}^{2}}\frac{1}{(m^{2}+n^{2})^{\sigma}}<\infty.$

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

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

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

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

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

Then

 $\displaystyle G(\tau,s)$ $\displaystyle=\frac{1}{2}\sum_{\nu\geq 1}\sum_{(m,n)\in\mathbb{Z}^{2},\gcd(m,n% )=\nu}\frac{y^{s}}{|m\tau+n|^{2s}}$ $\displaystyle=\frac{1}{2}\sum_{\nu\geq 1}\sum_{(c,d)\in\mathbb{Z}^{2},\gcd(c,d% )=1}\frac{y^{s}}{|\nu c\tau+\nu d|^{2s}}$ $\displaystyle=\frac{1}{2}\sum_{(c,d)\in\mathbb{Z}^{2},\gcd(c,d)=1}\frac{y^{s}}% {|c\tau+d|^{2s}}\sum_{\nu\geq 0}\nu^{-2s}$ $\displaystyle=\zeta(2s)E(\tau,s).$

## 2 Modular functions

###### Theorem 2.

For $\begin{pmatrix}a&b\\ c&d\end{pmatrix}\in\mathrm{SL}_{2}(\mathbb{Z})$, $\tau\in\mathbb{H}$, and $\mathrm{Re}\,s>1$,

 $G\left(\frac{a\tau+b}{c\tau+d},s\right)=G(\tau,s).$
###### 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 $ad-bc=1$,

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

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,

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

But $\begin{pmatrix}a&b\\ c&d\end{pmatrix}\in\mathrm{SL}_{2}(\mathbb{Z})$ 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

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

and thus we get

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

completing the proof. ∎

## 3 Fourier expansion

We now derive the Fourier series of $G(\cdot,s)$.11 1 Henri Cohen, Number Theory, vol. II: Analytic and Modern Tools, p. 211, Theorem 10.4.3. $K_{s-\frac{1}{2}}$ denotes the Bessel function.

###### Theorem 3.

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

 $\displaystyle G(\tau,s)$ $\displaystyle=\zeta(2s)y^{s}+\pi^{\frac{1}{2}}\frac{\Gamma\left(s-\frac{1}{2}% \right)}{\Gamma(s)}\zeta(2s-1)y^{-s+1}$ $\displaystyle+2\frac{\pi^{s}}{\Gamma(s)}\sum_{n=1}^{\infty}n^{s-1}\sigma_{-2s+% 1}(n)F_{s-\frac{1}{2}}(2\pi ny)\cos(2\pi nx),$

where

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

Define

 $S(z,s)=\sum_{n\in\mathbb{Z}}\frac{|y|^{s}}{|z+n|^{2s}},\qquad z=x+iy,y\neq 0,% \quad\mathrm{Re}\,s>1.$

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

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

The Poisson summation formula22 2 Henri Cohen, Number Theory, vol. I: Tools and Diophantine Equations, p. 46, Corollary 2.2.17. states that if $f:\mathbb{R}\to\mathbb{C}$ is continuous and of locally bounded variation, then for all $x\in\mathbb{R}$,

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

where

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

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

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

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

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

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

 $\displaystyle\widehat{f}_{y}(k)$ $\displaystyle=\int_{\mathbb{R}}e^{-2\pi ikyu}(y^{2}u^{2}+y^{2})^{-s}|y|du$ $\displaystyle=|y|^{-2s+1}\int_{\mathbb{R}}e^{-2\pi ikyu}(u^{2}+1)^{-s}du$ $\displaystyle=2|y|^{-2s+1}\int_{0}^{\infty}\cos(2\pi kyu)(u^{2}+1)^{-s}du;$

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

We use the following identity:33 3 Henri Cohen, Number Theory, vol. II: Analytic and Modern Tools, p. 117, Theorem 9.8.9. for $a>0$ and $\mathrm{Re}\,s>\frac{1}{2}$,

 $\int_{0}^{\infty}\cos(au)(u^{2}+1)^{-s}du=\pi^{1/2}\cdot\left(\frac{a}{2}% \right)^{s-\frac{1}{2}}\cdot\frac{1}{\Gamma(s)}\cdot K_{s-\frac{1}{2}}(a).$

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

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

Therefore (1) becomes

 $\displaystyle S(z,s)$ $\displaystyle=|y|^{s}\widehat{f}_{y}(0)+|y|^{s}\sum_{k\neq 0}2|y|^{-s+\frac{1}% {2}}\pi^{s}|k|^{s-\frac{1}{2}}\cdot\frac{1}{\Gamma(s)}\cdot K_{s-\frac{1}{2}}(% 2\pi|ky|)\cdot e^{2\pi ikx}$ $\displaystyle=|y|^{s}\widehat{f}_{y}(0)+2|y|^{\frac{1}{2}}\pi^{s}\cdot\frac{1}% {\Gamma(s)}\sum_{k\neq 0}|k|^{s-\frac{1}{2}}\cdot K_{s-\frac{1}{2}}(2\pi|ky|)% \cdot e^{2\pi ikx}$ $\displaystyle=|y|^{s}\widehat{f}_{y}(0)+4|y|^{\frac{1}{2}}\pi^{s}\cdot\frac{1}% {\Gamma(s)}\sum_{k=1}^{\infty}k^{s-\frac{1}{2}}\cdot K_{s-\frac{1}{2}}(2\pi k|% y|)\cdot\cos(2\pi kx).$

We use the following identity for the beta function:44 4 Henri Cohen, Number Theory, vol. II: Analytic and Modern Tools, p. 93, Corollary 9.6.40. For $\mathrm{Re}\,b>\frac{1}{2}\mathrm{Re}\,a>0$,

 $\int_{0}^{\infty}u^{a-1}(u^{2}+1)^{-b}du=\frac{1}{2}B\left(\frac{a}{2},b-\frac% {a}{2}\right)=\frac{\Gamma\left(\frac{a}{2}\right)\Gamma\left(b-\frac{a}{2}% \right)}{2\Gamma(b)}.$

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

 $\widehat{f}_{y}(0)=2|y|^{-2s+1}\int_{0}^{\infty}(u^{2}+1)^{-s}du=2|y|^{-2s+1}% \frac{\pi^{\frac{1}{2}}\Gamma\left(s-\frac{1}{2}\right)}{2\Gamma(s)}.$

Therefore

 $\displaystyle S(z,s)$ $\displaystyle=\pi^{\frac{1}{2}}\cdot|y|^{-s+1}\cdot\frac{\Gamma\left(s-\frac{1% }{2}\right)}{\Gamma(s)}$ $\displaystyle+4|y|^{\frac{1}{2}}\pi^{s}\cdot\frac{1}{\Gamma(s)}\sum_{k=1}^{% \infty}k^{s-\frac{1}{2}}\cdot K_{s-\frac{1}{2}}(2\pi k|y|)\cdot\cos(2\pi kx)$ $\displaystyle=\pi^{\frac{1}{2}}\cdot|y|^{-s+1}\cdot\frac{\Gamma\left(s-\frac{1% }{2}\right)}{\Gamma(s)}$ $\displaystyle+2\frac{\pi^{s}}{\Gamma(s)}\sum_{k=1}^{\infty}k^{s-1}F_{s-\frac{1% }{2}}(2\pi k|y|)\cdot\cos(2\pi kx).$

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

 $\displaystyle G(\tau,s)$ $\displaystyle=y^{s}\zeta(2s)+\frac{1}{2}\sum_{m\neq 0}\frac{S(m\tau,s)}{|m|^{s}}$ $\displaystyle=y^{s}\zeta(2s)+\sum_{m=1}^{\infty}\frac{S(m\tau,s)}{m^{s}}$ $\displaystyle=y^{s}\zeta(2s)+\pi^{\frac{1}{2}}\frac{\Gamma\left(s-\frac{1}{2}% \right)}{\Gamma(s)}\sum_{m=1}^{\infty}\frac{(my)^{-s+1}}{m^{s}}$ $\displaystyle+2\frac{\pi^{s}}{\Gamma(s)}\sum_{m=1}^{\infty}\frac{1}{m^{s}}\sum% _{k=1}^{\infty}k^{s-1}F_{s-\frac{1}{2}}(2\pi kmy)\cdot\cos(2\pi kmx)$ $\displaystyle=y^{s}\zeta(2s)+\pi^{\frac{1}{2}}\frac{\Gamma\left(s-\frac{1}{2}% \right)}{\Gamma(s)}y^{-s+1}\zeta(2s-1)$ $\displaystyle+2\frac{\pi^{s}}{\Gamma(s)}\sum_{k,m\geq 1}\frac{k^{s-1}}{m^{s}}F% _{s-\frac{1}{2}}(2\pi kmy)\cdot\cos(2\pi kmx).$

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

 $\displaystyle G(\tau,s)$ $\displaystyle=y^{s}\zeta(2s)+\pi^{\frac{1}{2}}\frac{\Gamma\left(s-\frac{1}{2}% \right)}{\Gamma(s)}y^{-s+1}\zeta(2s-1)$ $\displaystyle+2\frac{\pi^{s}}{\Gamma(s)}\sum_{N=1}^{\infty}N^{s-1}\sigma_{-2s+% 1}(N)F_{s-\frac{1}{2}}(2\pi Ny)\cos(2\pi Nx),$

completing the proof. ∎

We use the above Fourier expansion to establish that for all $t\in\mathbb{H}$, $G(\tau,s)$ has a meromorphic continuation to $\mathbb{C}$ and satisfies a certain functional equation.55 5 Henri Cohen, Number Theory, vol. II: Analytic and Modern Tools, p. 212, Corollary 10.4.4. The meromorphic continuation and functional equation of $G(\tau,s)$ can also be obtained without using its Fourier expansion.66 6 Paul Garrett, The simplest Eisenstein series, http://www.math.umn.edu/~garrett/m/mfms/notes_c/simplest_eis.pdf

###### Theorem 4.

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

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

satisfies the functional equation

 $\mathcal{G}(\tau,1-s)=\mathcal{G}(\tau,s).$
###### Proof.

For $\nu\in\mathbb{C}$ and for $w>0$ we have77 7 Henri Cohen, Number Theory, vol. II: Analytic and Modern Tools, p. 113, Proposition 9.8.6.

 $K_{\nu}(w)=\int_{0}^{\infty}e^{-w\cosh t}\cosh(\nu t)dt$

and88 8 Henri Cohen, Number Theory, vol. II: Analytic and Modern Tools, p. 115, Proposition 9.8.7.

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

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 $s\in\mathbb{C}$, the function $w\mapsto K_{s-\frac{1}{2}}(w)$ belongs to $C^{\infty}(\mathbb{R}_{>0})$. We have a fortiori that for any $s\in\mathbb{C}$,

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

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

 $\Lambda(1-s)=\Lambda(s).$

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

 $\displaystyle\mathcal{G}(\tau,s)$ $\displaystyle=\pi^{-s}\Gamma(s)G(\tau,s)$ $\displaystyle=\Lambda(2s)y^{s}+\Lambda(2s-1)y^{-s+1}$ $\displaystyle+2\sum_{n=1}^{\infty}n^{s-1}\sigma_{-2s+1}(n)F_{s-\frac{1}{2}}(2% \pi ny)\cos(2\pi nx).$

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

For $s\in\mathbb{C}$,

 $\displaystyle n^{(1-s)-1}\sigma_{-2(1-s)+1}(n)$ $\displaystyle=n^{-s}\sigma_{2s-1}(n)$ $\displaystyle=\sum_{ef=n}(ef)^{-s}e^{2s-1}$ $\displaystyle=\sum_{ef=n}e^{s-1}f^{-s}$ $\displaystyle=\sum_{ef=n}(ef)^{s-1}f^{-2s+1}$ $\displaystyle=n^{s-1}\sigma_{-2s+1}(n).$

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 $\mathcal{G}(\tau,s)$ is unchanged if $s$ is replaced with $1-s$, and together with $\Lambda(1-w)=\Lambda(w)$ this yields

 $\displaystyle\mathcal{G}(\tau,1-s)$ $\displaystyle=\Lambda(2-2s)y^{1-s}+\Lambda(2-2s-1)y^{-(1-s)+1}$ $\displaystyle+2\sum_{n=1}^{\infty}n^{s-1}\sigma_{-2s+1}(n)F_{s-\frac{1}{2}}(2% \pi ny)\cos(2\pi nx)$ $\displaystyle=\Lambda(1-(2s-1))y^{1-s}+\Lambda(1-2s)y^{s}$ $\displaystyle+2\sum_{n=1}^{\infty}n^{s-1}\sigma_{-2s+1}(n)F_{s-\frac{1}{2}}(2% \pi ny)\cos(2\pi nx)$ $\displaystyle=\Lambda(2s-1)y^{1-s}+\Lambda(2s)y^{s}$ $\displaystyle+2\sum_{n=1}^{\infty}n^{s-1}\sigma_{-2s+1}(n)F_{s-\frac{1}{2}}(2% \pi ny)\cos(2\pi nx)$ $\displaystyle=\mathcal{G}(\tau,s).$

## 4 Kronecker limit formula

For $\tau\in\mathbb{H}$, 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 $s=1$,

 $G(\tau,s)=\frac{\pi}{2}\cdot\frac{1}{s-1}+a_{0}(\tau)+a_{1}(\tau)\cdot(s-1)+\cdots,$

and so defining $\frac{\pi}{2}C(\tau)=a_{0}(\tau)$,

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

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

 $\eta(\tau)=e^{\frac{\pi i\tau}{12}}\prod_{n=1}^{\infty}(1-q^{n}),\qquad\tau\in% \mathbb{H},$

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,99 9 Henri Cohen, Number Theory, vol. II: Analytic and Modern Tools, p. 213, Theorem 10.4.6. which expresses $C(\tau)$ in terms of the Dedekind eta function.

###### Theorem 5.

For $\tau=x+iy\in\mathbb{H}$,

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

with

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

Define

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

Then

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

We use the asymptotic formula

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

and with

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

this gives, as $s\to 1$,

 $\displaystyle\log\zeta(s)$ $\displaystyle=\log\left(\frac{1}{s-1}+\gamma+O(|s-1|)\right)$ $\displaystyle=-\log(s-1)+\log(1+\gamma(s-1)+O(|s-1|^{2}))$ $\displaystyle=-\log(s-1)+\gamma(s-1)+O(|s-1|^{2}),$

and hence

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

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

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

for $|z-\frac{1}{2}|<\frac{1}{2}$, and the Taylor series of $\log\Gamma(1+z)$ about $z=0$ is

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

for $|z|<1$. Using these we have

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

and

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

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

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

Taking the exponential and using

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

as $s\to 1$ we have

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

Using this and the fact that

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

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

 $\displaystyle G(\tau,s)$ $\displaystyle=\frac{\pi^{2}}{6}y+\frac{\pi}{2}\cdot\frac{1}{s-1}+\frac{\pi}{2}% \cdot(2\gamma-2\log 2-\log y)+O(|s-1|)$ $\displaystyle+2\frac{\pi^{s}}{\Gamma(s)}\sum_{n=1}^{\infty}n^{s-1}\sigma_{-2s+% 1}(n)F_{s-\frac{1}{2}}(2\pi ny)\cos(2\pi nx).$

We have

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

As well,

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

and

 $\sigma_{-2s+1}(n)=\sum_{d|n}d^{-2s+1}=\sum_{d|n}(d^{-1}+O(|s-1|))=\sigma_{-1}(% n)+O(|s-1|),\qquad s\to 1.$

Finally, we use the fact that that for all $t>0$,1010 10 Henri Cohen, Number Theory, vol. II: Analytic and Modern Tools, p. 112, Theorem 9.8.5.

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

giving

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

and hence

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

Therefore, as $s\to 1$,

 $\displaystyle G(\tau,s)$ $\displaystyle=\frac{\pi^{2}}{6}y+\frac{\pi}{2}\cdot\frac{1}{s-1}+\frac{\pi}{2}% \cdot(2\gamma-2\log 2-\log y)$ $\displaystyle+2\pi\sum_{n=1}^{\infty}1\cdot\sigma_{-1}(n)e^{-2\pi ny}\cos(2\pi nx% )+O(|s-1|).$

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

 $a_{0}(\tau)=\frac{\pi^{2}}{6}y+\frac{\pi}{2}\cdot(2\gamma-2\log 2-\log y)+2\pi% \sum_{n=1}^{\infty}\sigma_{-1}(n)e^{-2\pi ny}\cos(2\pi nx).$

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

 $\displaystyle\mathrm{Re}\,\left(\sum_{n=1}^{\infty}\sigma_{-1}(n)q^{n}\right)$ $\displaystyle=\sum_{n=1}^{\infty}\sigma_{-1}(n)\mathrm{Re}\,(e^{2\pi in\tau})$ $\displaystyle=\sum_{n=1}^{\infty}\sigma_{-1}(n)\mathrm{Re}\,(e^{-2\pi ny}e^{2% \pi inx})$ $\displaystyle=\sum_{n=1}^{\infty}\sigma_{-1}(n)e^{-2\pi ny}\cos(2\pi nx),$

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

 $S(\tau)=\mathrm{Re}\,\left(\sum_{n=1}^{\infty}\sigma_{-1}(n)q^{n}\right).$

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

 $\displaystyle\log\prod_{n=1}^{\infty}(1-q^{n})$ $\displaystyle=\sum_{n=1}^{\infty}\log(1-q^{n})$ $\displaystyle=-\sum_{n=1}^{\infty}\sum_{m=1}^{\infty}\frac{q^{nm}}{m}$ $\displaystyle=-\sum_{N=1}^{\infty}\sum_{d|N}\frac{q^{N}}{d}$ $\displaystyle=-\sum_{N=1}^{\infty}\sigma_{-1}(N)q^{N},$

so

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

Then, because

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

and because

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

this becomes

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

Thus

 $\displaystyle a_{0}(\tau)$ $\displaystyle=\frac{\pi^{2}}{6}y+\frac{\pi}{2}\cdot(2\gamma-2\log 2-\log y)-% \frac{\pi^{2}}{6}y-2\pi\log|\eta(\tau)|$ $\displaystyle=\frac{\pi}{2}\cdot(2\gamma-2\log 2-\log y)-2\pi\log|\eta(\tau)|,$

so

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

completing the proof. ∎

## 5 Hyperbolic Laplacian

For $f\in C^{2}(\mathbb{H})$, we define $\Delta_{\mathbb{H}}f:\mathbb{H}\to\mathbb{C}$ by

 $(\Delta_{\mathbb{H}}f)(\tau)=-y^{2}(\partial_{x}^{2}f+\partial_{y}^{2}f)(\tau)% ,\qquad\tau=x+iy\in\mathbb{H}.$

For more on $\Delta_{\mathbb{H}}$ see the below references.1111 11 Daniel Bump, Spectral Theory and the Trace Formula, http://sporadic.stanford.edu/bump/match/trace.pdf; Fredrik Strömberg, Spectral theory and Maass waveforms for modular groups– from a computational point of view, http://www.cams.aub.edu.lb/events/confs/modular2012/files/lecture_notes_spectral_theory.pdf; cf. Anton Deitmar, Automorphic Forms, p. 54, Lemma 2.7.3.

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

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

and

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

from which we have

 $\displaystyle(\partial_{x}^{2}f)(x,y)$ $\displaystyle=s(s+1)m^{2}(mx+n+imy)^{-2}f(x,y)$ $\displaystyle+2s^{2}m^{2}|m\tau+n|^{-2}f(x,y)$ $\displaystyle+s(s+1)m^{2}(mx+n-imy)^{-2}f(x,y).$

We also calculate

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

Now,

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

and

 $\displaystyle(\partial_{y}^{2}f)(x,y)$ $\displaystyle=s(s-1)y^{s-2}g(x,y)+2sy^{s-1}(\partial_{y}g)(x,y)$ $\displaystyle+y^{s}(\partial_{y}^{2}g)(x,y),$

from which we have

 $\displaystyle(\partial_{y}^{2}f)(x,y)$ $\displaystyle=s(s-1)y^{s-2}g(x,y)$ $\displaystyle-2s^{2}imy^{s-1}(mx+n+imy)^{-1}g(x,y)$ $\displaystyle+2s^{2}imy^{s-1}(mx+n-imy)^{-1}g(x,y)$ $\displaystyle-s(s+1)m^{2}y^{s}(mx+n+imy)^{-2}g(x,y)$ $\displaystyle+2s^{2}m^{2}y^{s}(mx+n+imy)^{-1}(mx+n-imy)^{-1}g(x,y)$ $\displaystyle-s(s+1)m^{2}y^{s}(mx+n-imy)^{-2}g(x,y)$ $\displaystyle=s(s-1)y^{-2}f(x,y)$ $\displaystyle-2s^{2}imy^{-1}(mx+n+imy)^{-1}f(x,y)$ $\displaystyle+2s^{2}imy^{-1}(mx+n-imy)^{-1}f(x,y)$ $\displaystyle-s(s+1)m^{2}(mx+n+imy)^{-2}f(x,y)$ $\displaystyle+2s^{2}m^{2}|m\tau+n|^{-2}f(x,y)$ $\displaystyle-s(s+1)m^{2}(mx+n-imy)^{-2}f(x,y).$

Combining the above expressions we get

 $\displaystyle(\partial_{x}^{2}f+\partial_{y}^{2})(x,y)$ $\displaystyle=m^{2}f(x,y)\cdot\bigg{(}2s^{2}|m\tau+n|^{-2}+2s^{2}|m\tau+n|^{-2% }\bigg{)}$ $\displaystyle+s(s-1)y^{-2}f(x,y)$ $\displaystyle-2s^{2}imy^{-1}(mx+n+imy)^{-1}f(x,y)$ $\displaystyle+2s^{2}imy^{-1}(mx+n-imy)^{-1}f(x,y)$ $\displaystyle=m^{2}f(x,y)\cdot\bigg{(}2s^{2}|m\tau+n|^{-2}+2s^{2}|m\tau+n|^{-2% }\bigg{)}$ $\displaystyle+s(s-1)y^{-2}f(x,y)-4s^{2}m^{2}|m\tau+n|^{-2}f(x,y)$ $\displaystyle=s(s-1)y^{-2}f(x,y).$

Thus

 $(\Delta_{\mathbb{H}}f)(x,y)=s(s-1)f(x,y),$

i.e.,

 $\Delta_{\mathbb{H}}f=s(s-1)f.$

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

 $\displaystyle\Delta_{\mathbb{H}}G(\cdot,s)=s(s-1)G(\cdot,s).$

Because the coefficients of the differential operator $L=\Delta_{\mathbb{H}}-s(s-1)$ are real analytic, a function $f:\mathbb{H}\to\mathbb{C}$ satisfying $Lf=0$ is real analytic.1212 12 Lipman Bers and Martin Schechter, Elliptic Equations, in Lipman Bers, Fritz John, and Martin Schechter, eds., Partial Diferential Equations, pp. 207–210, Chapter 4, Appendix. Therefore, for $\mathrm{Re}\,s>1$, $G(\cdot,s)$ is real analytic.