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\begin{document}

\title{Lecture {\rm 34}}

\author{Mustafa Arslan}

\address{Louisiana State University \\
Department of Mathematics \\
Baton Rouge, LA 70803, USA}

\date{\today}

\email{arslan@math.lsu.edu}

\maketitle

Last time we were proving the following theorem:
\begin{thm}[Riemann]\label{ARS4:Riemann}
The Riemann $\zeta$-function, defined as
\[\zeta(s) =\sum_{n=1}^\infty\frac{1}{n^s}\]
for $\re s$ extends to an entire meromorphic function of $s$
having a simple pole at $s=1$ wit residue $1$, and having no other
poles. Moreover we have the functional equation
\begin{equation}\label{ARS4eqn1}
Z(s) = Z(1-s)
\end{equation}
where $Z(s) = \pi^{-s/2}\Gamma(s/2)\zeta(s)$.

\end{thm}
We proved that the function
\[\varphi(s) = \int_1^\infty t^{s/2}[\theta(t)-1]\frac{dt}{t} +
\int_0^1t^{s/2}\left[\theta(t)-\frac{1}{\sqrt{t}}\right]\frac{dt}{t}\]
is an entire (holomorphic) function of $s$ where
\[\theta(t)=\sum_{n\in\Z}e^{-\pi n^2t}.\]
\begin{lem}
For some constant $c>0$
\[\left|\theta(t)-\frac{1}{\sqrt{t}}\right|<e^{-c/t}\]
when $t>0$ is small enough.
\end{lem}
\begin{proof}
By the functional equation
\[\theta(t) =\frac{1}{\sqrt{t}}\theta\left(\frac{1}{t}\right)\]
of the theta series we have
\begin{eqnarray}
\nonumber\theta(t)-\frac{1}{\sqrt{t}}&=&\frac{1}{\sqrt{t}}
\left(\theta\left(\frac{1}{t}\right)-1\right)\\
\nonumber&=&\frac{1}{\sqrt{t}}\left(2\sum_{n=1}^\infty e^{-\pi
n^2/t}\right)\\
&=&\frac{2}{\sqrt{t}}(e^{\pi/t}+e^{-4\pi/t}+\cdots).\label{ARS4eqn2}
\end{eqnarray}
If choose $t$ small enough so that $\sqrt{t}>4e^{-1/t}$ and
$e^{-3\pi/t}<\frac{1}{2}$ then we will have
\begin{eqnarray*}
\frac{2}{\sqrt{t}}(e^{\pi/t}+e^{-4\pi/t}+\cdots)
&<&\frac{1}{2}e^{1/t}(e^{\pi/t}+e^{-4\pi/t}+\cdots)\\
&\leq&\frac{1}{2}e^{-(\pi-1)/t}\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}\cdots\right)\\
&\leq&\frac{1}{2}e^{-2(\pi-1)/t}
\end{eqnarray*}
this proves the lemma.
\end{proof}
Now for $\re s>1$,
\begin{equation}
\begin{split}
\varphi(s) &= 2\sum_{n=1}^\infty \int_1^\infty e^{-\pi
n^2t}t^{s/2}\frac{dt}{t}+\int_1^\infty t^{s/2}\frac{dt}{t}\\ &
\quad\quad\quad +2\sum_{n=1}^\infty\int_0^1e^{\pi
n^2t}t^{s/2}\frac{dt}{t}-\int_0^1\frac{1}{\sqrt{t}}t^{s/2}\frac{dt}{t}
\end{split}
\end{equation}
and with simple calculations we have
\[\int_0^1t^{s/2}\frac{dt}{t}=\frac{2}{s},\]
\[\int_0^1\frac{1}{\sqrt{t}}t^{s/2}\frac{dt}{t}=\frac{2}{s-1}\]
and hence
\begin{eqnarray}
\nonumber\varphi(s) &=&\frac{2}{s}+\frac{2}{1-s}+
2\sum_{n=1}^\infty\int_0^\infty e^{-\pi n^2t}t^{s/2}\frac{dt}{t}\\
&=&\frac{2}{s}
+\frac{2}{1-s}+2\pi^{-s/2}\Gamma\left(\frac{s}{2}\right)\zeta(s)\label{ARS4eqn3}
\end{eqnarray}
The relation \ref{ARS4eqn3} says that $\Gamma(s/2)\zeta(s)$ admits
a meromorphic continuation, which has poles at most at $s=0$ or
$s=1$. Applying the change of variable $u = 1/t$ and using the
functional equation
\begin{equation}\label{ARS4eqn4}
\theta\left(\frac{1}{u}\right)=\sqrt u\theta(u)
\end{equation}
 we obtain

\begin{eqnarray*}
\varphi(s) &=& \int_1^\infty t^{s/2}[\theta(t)-1]\frac{dt}{t}
+\int_0^1t^{s/2}\left[\theta(t)-\frac{1}{\sqrt{t}}\right]\frac{dt}{t}\\
&=&-\int_1^0u^{-s/2}\left[\theta\left(\frac{1}{u}\right)-1\right]\frac{du}{u}
-\int_\infty^1u^{-s/2}\left[\theta\left(\frac{1}{u}\right)-\sqrt{u}\right]\frac{du}{u}\\
&=&\int_0^1u^{s/2}[u^{1/2}\theta(u) -1]\frac{du}{u}+\int_1^\infty
u^{-s/2}[u^{1/2}\theta(u)-u^{1/2}]\frac{du}{u}\\
&=&\int_0^1u^{(1-s)/2}\left[\theta(u)-\frac{1}{\sqrt
u}\right]\frac{du}{u}
+\int_1^\infty u^{(1-s)/2}[\theta(u)-1]\frac{du}{u}\\
&=&\varphi(1-s)
\end{eqnarray*}
This proves relation \ref{ARS4eqn1} of the theorem
\ref{ARS4:Riemann}.

Now, consider
\begin{equation}\label{ARS4eqn5}
\zeta(s) =\frac{\pi^{s/2}}{\Gamma(s/2)}\left(\frac{1}{2}\varphi(s)
-\frac{1}{s}-\frac{1}{1-s}\right).
\end{equation}
>From the functional equation for $\Gamma(s)$
\[\Gamma(s+1) =
s\Gamma(s)\]we get
\[\Gamma\left(\frac{s}{2}\right)=\frac{2}{s}\Gamma\left(\frac{s}{2}
+1\right).\] Hence with this relation and \ref{ARS4eqn5} one can
easily deduce that the function $\zeta$ has a pole at $s=1$ with
residue $1$.
\begin{defn}
The {\em Mellin transform} of a function $f(t)$ is the function
$M[f](s)$ of $s$ defined by
\begin{equation}\label{ARS4:Mellin}
M[f](s)=\int_0^\infty f(t)t^s\frac{dt}{t}.
\end{equation}
\end{defn}
As an example the Mellin transform of $e^{-t}$ is $\Gamma(s)$.
Another example is
\[\zeta(s) = M[\theta(t)](s).\]
Riemann's argument is that we can obtain the functional equation
for $\zeta(s)$ from the functional equation of $\theta(t)$.

For convergence of \ref{ARS4:Mellin} in some half plane $\re s> C$
one needs a bound of the form $|f(t)|\leq Ke^{-C't}$ for some
constants $K$ and $C'$.
\begin{rem}
Mellin transform of a modular form is a Dirichlet series with an
analytic continuation and a functional equation.
\end{rem}
Let \[\chi:\left(\Z\big/N\right)^\times\rightarrow\C^\times\] be a
homomorphism (``Dirichlet character"). Define the function
$L(s,\chi)$ for $\re s>1$ as
\[L(s,\chi)=\sum_{n=1}^\infty\frac{\chi(n)}{n^s}.\]
with the convention $\chi(n) = 0$ if $[n]$ is not invertible in
$\Z/N$. Note that if $N=1$, and $\chi=1$ then $L=\zeta$.
\begin{thm}
Define
\[\Lambda(s,\chi)
=\left(\frac{f}{\pi}\right)^{s/2}\Gamma(\frac{s+\delta}{2}L(s,\chi)\]
where $f\in \Z_+$ (``conductor"), and
\[\delta =\left\{\begin{array}{ccc}0&& \mbox{if }\chi(-1) =1\\
&& \\ 1&& \mbox{if }\chi(-1)=-1
\end{array}\right.\]
The function $\Lambda$ has an analytic continuation to the entire
$s$-plane with the relation
\[\Lambda(s,\chi)=W_\chi\cdot\Lambda(1-s,\bar\chi)\]where
\[W_\chi =\frac{\sum_{\alpha
=1}^f\chi(\alpha)e^{2\pi\alpha/f}}{\sqrt f i^s}\]
\end{thm}
Note that the sum
\[\tau(\chi)=\sum_{\alpha
=1}^f\chi(\alpha)e^{2\pi\alpha/f}\] is called the ``Gauss sum".







\end{document}