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\newtheorem{exer}{Exercise}
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\title{Modular Forms, PS 5}
\author{Bryden Cais}
\begin{document}
\maketitle

\begin{exer}
Recall the functional equation for the Riemann zeta function:
$$\pi^{-s/2}\G\left(\frac{s}{2}\right)\z(s)=\pi^{(s-1)/2}\G\left(\frac{1-s}{2}\right)\z(1-s).$$
Supposing that $k$ is any nonnegative even integer and replacing $s$ by $s-k+1$ above yields
$$\pi^{(k-s-1)/2}\G\left(\frac{s-k+1}{2}\right)\z(s-k+1)=\pi^{(s-k)/2}\G\left(\frac{k-s}{2}\right)\z(k-s).$$
Multiplying these two equalities together gives
\begin{eqnarray}
&&\pi^{(k-2s-1)/2}\G\left(\frac{s}{2}\right)\G\left(\frac{s-k+1}{2}\right)\z(s-k+1)\z(s)\nonumber\\
&=&\pi^{(2s-k-1)/2}\G\left(\frac{k-s}{2}\right)\G\left(\frac{1-s}{2}\right)\z(1-s)\z(k-s).\label{zeta1}
\end{eqnarray}
Since $k$ is even, $k/2=l$ is an integer.  Therefore, using the recurrence
$$\G(1+z)=z\G(z)$$ we arrive at
\begin{align*}
    \G\left(\frac{s+1}{2}\right)&=\left(\frac{s+1}{2}-l\right)\left(\frac{s+1}{2}-l+1\right)\cdots \left(\frac{s+1}{2}-1\right)\G\left(\frac{s+1}{2}-l\right)\\
    \G\left(l-\frac{s}{2}\right)&=\left(-\frac{s}{2}+l-1\right)\left(-\frac{s}{2}+l-2\right)\cdots \left(-\frac{s}{2}+1\right)\G\left(1-\frac{s}{2}\right).
\end{align*}
Now recall the \textit{duplication} formula for the $\G$ function:
$$\G(2x)=\frac{2^{2x-1}}{\sqrt{\pi}}\G(x)\G\left(x+\frac{1}{2}\right).$$
We thus obtain
\begin{align*}
\G\left(\frac{s}{2}\right)\G\left(\frac{s-k+1}{2}\right)&=\frac{\G\left(\frac{s}{2}\right)\G\left(\frac{s+1}{2}\right)}
{\left(\frac{s+1}{2}-l\right)\left(\frac{s+1}{2}-l+1\right)\cdots \left(\frac{s+1}{2}-1\right)}\\
&=\frac{\pi^{1/2}\G(s)}{2^{s-1}}\frac{1}{\left(\frac{s+1}{2}-l\right)\left(\frac{s+1}{2}-l+1\right)\cdots \left(\frac{s+1}{2}-1\right)}\\
\intertext{and}
\G\left(\frac{k-s}{2}\right)\G\left(\frac{1-s}{2}\right)&=
\left(-\frac{s}{2}+l-1\right)\left(-\frac{s}{2}+l-2\right)\cdots \left(-\frac{s}{2}+1\right)\G\left(1-\frac{s}{2}\right)\G\left(\frac{1-s}{2}\right)\\
&=\left(-\frac{s}{2}+l-1\right)\left(-\frac{s}{2}+l-2\right)\cdots \left(-\frac{s}{2}+1\right)\frac{\pi^{1/2}\G(1-s)}{2^{-s}}.\\
\end{align*}
Substituting these expressions in to (\ref{zeta1}) yields
\begin{eqnarray*}
&&\pi^{(k-2s-1)/2}\frac{\pi^{1/2}\G(s)}{2^{s-1}}\frac{1}{\left(\frac{s+1}{2}-l\right)\left(\frac{s+1}{2}-l+1\right)\cdots \left(\frac{s+1}{2}-1\right)}\z(s-k+1)\z(s)\\
&=&\pi^{(2s-k-1)/2}\left(-\frac{s}{2}+l-1\right)\left(-\frac{s}{2}+l-2\right)\cdots \left(-\frac{s}{2}+1\right)\frac{\pi^{1/2}\G(1-s)}{2^{-s}}\z(1-s)\z(k-s)
\end{eqnarray*}
Now observe that
\begin{align*}
&\left(\frac{s+1}{2}-l\right)\left(\frac{s+1}{2}-l+1\right)\cdots \left(\frac{s+1}{2}-1\right)\left(-\frac{s}{2}+l-1\right)\left(-\frac{s}{2}+l-2\right)\cdots \left(-\frac{s}{2}+1\right)\G(1-s)\\
&=(i)^k\left(\frac{k-1-s}{2}\right)\left(\frac{k-2-s}{2}\right)\cdots \left(\frac{1-s}{2}\right)\G(1-s)\\
&=\frac{(i)^k}{2^{k-1}}\G(k-s),
\end{align*}
so that we have
\begin{eqnarray*}
\pi^{(k-2s-1)/2}\frac{\G(s)}{2^{s}}\z(s-k+1)\z(s)
&=&(i)^k\pi^{(2s-k-1)/2}\frac{\G(k-s)}{2^{k-s}}\z(1-s)\z(k-s),
\end{eqnarray*}
which upon simplification gives
\begin{eqnarray*}
(2\pi)^{-s}\G(s)\z(s-k+1)\z(s)
&=&(i)^k(2\pi)^{s-k}\G(k-s)\z(1-s)\z(k-s),
\end{eqnarray*}
as required.
\end{exer}


\begin{exer}
    Observe that the above functional equation applies for $k=2$: that is, if we set
    $$\Lambda(s)=(2\pi)^{-s}\G(s)\z(s)\z(s-1)$$
    then we have
    $$\Lambda(s)=-\Lambda(2-s).$$  Let us apply Heck's argument for proving the converse theorem to see what
    object gives rise to $\Lambda(s)$.
    Observe that $\Lambda(s)$ has simple poles at $s=0,1,2$ (the poles at $0,2$ are clear since $\z$ has a simple pole at $s=1$,
    and $\G$ has a simple pole at $s=0$, while the pole at $1$ arises since $\z(0)\neq 0$).  Now let $c\in \R$ be large.
    Then
    \begin{align*}
    \frac{1}{2\pi i}\int_{c-i\infty}^{c-i\infty}\Lambda(s)y^{-s}\,\mathrm{d}s&=
    \frac{1}{2\pi i}\int_{c-i\infty}^{c-i\infty}\sum_{n=1}^{\infty}\sum_{m=1}^{\infty}\frac{m}{mn^s}\G(s)(2\pi y)^{-s}\,\mathrm{d}s\\
     &=\frac{1}{2\pi i}\int_{c-i\infty}^{c-i\infty}\sum_{n=1}^{\infty}\frac{\s_1(n)}{n^s}\G(s)(2\pi y)^{-s}\,\mathrm{d}s\\
    &=\frac{1}{2\pi i}\sum_{n=1}^{\infty}\s_1(n)\int_{c-i\infty}^{c-i\infty}\G(s)(2\pi n y)^{-s}\,\mathrm{d}s,
    \end{align*}
    where the change in order of summation is justified since we have chosen $c$ large enough
    that the sum converges absolutely and since for any $s$ with large imaginary part, $\G(s)$ decays exponentially as $\Im(s)\rightarrow \infty$
    so that the integral also converges absolutely.  Employing the fact that
    $$\frac{1}{2\pi i}\int_{c-i\infty}^{c-i\infty}\G(s)x^{-s}\,\mathrm{d}s=e^{-x},$$
    we therefore have
\begin{align*}
    f(iy)-a(0):=\frac{1}{2\pi i}\int_{c-i\infty}^{c-i\infty}\Lambda(s)y^{-s}\,\mathrm{d}s&=\sum_{n=1}^{\infty}\sigma_1(n)e^{-2\pi n y}.
\end{align*}
    We now evaluate the LHS by considering
    $$\int_C \Lambda(s) y^{-s}\,\mathrm{d}s,$$
    where $C$ is the rectangle with vertical sides passing through $c$ and $2-c$, and horizontal sides passing through $\pm iN$ for large $N$.
    Indeed, we will consider this integral as $N\rightarrow\infty$: it is not hard to see that the integrals along the horizontal sides
    tend rapidly to $0$---this follows from the behavior of $\G(s)$ for $s$ with large imaginary part.  Therefore, we have
\begin{align}
    \frac{1}{2\pi i}\int_{c-i\infty}^{c-i\infty}\Lambda(s)y^{-s}\,\mathrm{d}s-
    \frac{1}{2\pi i}\int_{2-c-i\infty}^{2-c-i\infty}\Lambda(s)y^{-s}\,\mathrm{d}s&=\sum_{\rho} \res{\Lambda}{\rho}, \label{zeta2}
\end{align}
where the sum is over all poles $\rho$ of $\Lambda$ inside $C$ and $\res{\Lambda}{\rho}$ denotes the residue of $\Lambda$ at $\rho$.
As mentioned above, the only poles inside $C$ are simple poles at $s=0,1,2$.  The determination of the residues is routine:
\begin{align*}
    \res{0}{\Lambda}=\z(0)\z(-1)&=\frac{1}{24}\\
    \res{1}{\Lambda}=(2\pi y)^{-1}\z(0)&=\frac{1}{4\pi y}\\
    \res{2}{\Lambda}=(2\pi y)^{-2}\z(2)&=\frac{1}{24 y^2}.
\end{align*}
We now use the functional equation $\Lambda(s)=-\Lambda(2-s)$ to rewrite the integral in the RHS of (\ref{zeta2}) as
\begin{align}
    \frac{1}{2\pi i}\int_{c-i\infty}^{c-i\infty}\Lambda(s)y^{-s}\,\mathrm{d}s+
    \frac{1}{2\pi i}\int_{c-i\infty}^{c-i\infty}\Lambda(2-s)y^{s-2}\,\mathrm{d}s&=\frac{1}{24}+\frac{1}{24 y^2}+\frac{1}{4\pi y}, \label{zeta3}
\end{align}
so that by our definition of $f(iy)$ we have
\begin{align}
    (f(iy)-a(0))+\frac{1}{y^2}(f(i/y)-a(0))&=\frac{1}{24}+\frac{1}{24 y^2}+\frac{1}{4\pi y}.
\end{align}
Therefore, with $a(0)=1/24$, or equivalently, with
$$F(iy):=1-24\sum_{n=1}^{\infty}\sigma_1(n)e^{-2\pi n y},$$
we have that
$$F(iy)=\frac{i^2}{y^2}F(i/y)+\frac{12}{2\pi i (iy)},$$
and hence by analytic continuation that
$$z^2F(z)=F(-1/z)+\frac{12 z}{2\pi i}$$
for all $z\in \H.$
This is the transformation formula for $E_2$.  Integrating, exponentiating, and taking $24^{\text{th}}$ powers shows that $\Delta$
(as given by the product $q\prod (1-q^n)^{24}$) is a cusp form of weight 12.

\end{exer}


\end{document}
