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\title{Math 678, Lecture 3}
\author{Notes by Bryden Cais}
\date{Sept. 9, 2002}
\begin{document}
\maketitle

\section{Fourier Expansions}

The main goal today will be to compute the fourier expansions of the modular forms $G_k(z)$.
Let $f(w)$ be holomorphic on $\H$ with period 1.  Recall that we can then write
$$f(z)=\sum_{n\in\Z}a(n)e(nz),$$
where
$$a(n)=\int_{0+iy}^{1+iy}f(z)e(-nz)\,\mathrm{d}z,$$
for any $y$.  Observe also that
\begin{align*}
G_k(z)&=\sum_{n\in\Z\setminus 0}n^{-k}+\sum_{m\in\Z\setminus 0}\sum_{n\in\Z}\frac{1}{(mz+n)^k}\\
&=2\z(k)+2\sum_{m=1}^{\infty}\sum_{n\in\Z}\frac{1}{(mz+n)^k}.
\end{align*}
So in order to determine the fourier expansion of $G_k(z),$ it suffices to determine that of
$$f(w)=\sum_{n\in\Z}\frac{1}{(w+n)^k}.$$
Write
$$f(w)=\sum_{n\in\Z}a(n)e(nw),$$
where as usual,
\begin{align*}
a(n)&=\int_{0+iv}^{1+iv}f(w)e(-nw)\,\mathrm{d}w\\
&=\sum_{k=-\infty}^{\infty}\int_{k+iv}^{k+1+iv}\frac{e(-nw)}{w^k}\,\mathrm{d}w\\
&=\int_{-\infty+iv}^{\infty+iv}\frac{e(-nw)}{w^k}\,\mathrm{d}w.
\end{align*}
We now split into three cases:
\begin{enumerate}
    \item $n=0$: Letting $v\rightarrow\infty$ we see that $$\int_{-\infty+iv}^{\infty+iv}\frac{\mathrm{d}w}{w^k}\rightarrow 0$$
    since the minimum value of the norm of $w$ tends to $\infty$ and the contour does not include $0$.

    \item $n<0$: In this case we have $|e(-nw)|=|e(-2\pi i n(u+iv))|=e^{2\pi n v}$.  Since $n<0$, we can again take the limit
    as $v\rightarrow\infty$ and we see that the integrand again tends to $0$ and that the contour does not contain $0$, so that
    the integral is $0$ also.

    \item $n>0$: This is the only tricky case.  We find that
    $$\int_{-\infty+iv}^{\infty+iv}\frac{e(-nw)}{w^k}\,\mathrm{d}w=\int_C\frac{e(-nw)}{w^k}\,\mathrm{d}w,$$
    where $C$ is the contour with vertices at $-\infty+iv,\infty+iv,-\infty-it,\infty-it$ for $v,t>0$ oriented clockwise.  Observe
    that the integral along the vertical sides is $0$.  We now consider the limit of this integral as $t\rightarrow\infty$.
    As before, the integral along the bottom horizontal segment of $C$ tends to $0$ as $t\rightarrow\infty.$  Cauchy's Theorem
    then tells us that
    $$\int_{-\infty+iv}^{\infty+iv}\frac{e(-nw)}{w^k}\,\mathrm{d}w=-2\pi i\res_0\left(\frac{e(-nw)}{w^k}\right)=\frac{(-2\pi i)^k n^{k-1}}{(k-1)!}$$
\end{enumerate}

We have therefore shown that
$$f(w)=\sum_{n\in\Z}\frac{1}{(n+w)^k}=\sum_{n=1}^{\infty}\frac{(-2\pi i)^k n^{k-1}}{(k-1)!}e(nw).$$

\begin{exer}
    Use the above formula to show that
    $$\z(k)=-\frac{(2\pi i)^k B_k}{2k!},$$
    where $B_k$ is defined by
    $$\frac{x}{e^x-1}=\sum_{k=0}^{\infty}B_k\frac{x^k}{k!}.$$
\end{exer}

We then have

\begin{align*}
G_k(z)&=2\z(k)+\frac{2(-2\pi i)^k}{(k-1)!}\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}n^{k-1}e(mnz)\\
&=2\z(k)+\frac{2(-2\pi i)^k}{(k-1)!}\sum_{r=1}^{\infty}\sigma_{k-1}(r)e(rz).
\end{align*}

\begin{exer}
    Show that $\sigma_{k-1}$ is multiplicative; more generally, show that
    $$\sigma_{k-1}(m)\sigma_{k-1}(n)=\sum_{d|(m,n)}d^{k-1}\sigma_{k-1}(\frac{mn}{d^2}).$$
\end{exer}

\begin{exer}
    Show that
    $$n^{k-1}<\sigma_{k-1}(n)<\zeta(k-1)n^{k-1}$$ and that these are the sharpest asymptotics one can hope for in the sense that
    \begin{align*}
        \liminf \frac{\sigma_{k-1}(n)}{n^{k-1}}&=1\\
        \intertext{and}
        \limsup \frac{\sigma_{k-1}(n)}{n^{k-1}}&=\z(k-1).
    \end{align*}
\end{exer}

Now let $l$ be any integer.  Then we claim that for $k$ even, almost all $\sigma_{k-1}(n)$ are divisible by $l$ in the sense that
$$\lim_{x\rightarrow\infty}\frac{\#\left\{n\leq x: l|\sigma_{k-1}(n)\right\}}{x}=1.$$
To see this, observe that if $p\equiv -1\mod l$ and $p|n$ but $p^2\not|n$ then
$$\sigma_{k-1}(n)=\sigma_{k-1}(p)\sigma_{k-1}(n/p),$$
and $l|\sigma_{k-1}(p)$ since $\sigma_(k-1)(p)=p^{k-1}+1\equiv 0\mod l.$  Now notice that for large $x$ one has
$$\#\left\{p\leq x: p|x,p^2\not|x\right\}\simeq x\left(\frac{1}{p}-\frac{1}{p^2}\right),$$
and therefore the probability that an arbitrary integer $n$ has no prime factors congruent to $-1\mod l$ that divide $n$
exactly once is
$$\prod_{p\equiv -1\mod l}\left(1-\frac{1}{p}+\frac{1}{p^2}\right)=0$$
by Dirichlet's Theorem (which tells us that $\#\left\{p\leq x:p\equiv a\mod q\right\}\simeq x/(\varphi(q)\log(x))$ if $(p,a)=1$).
\end{document}