text:
\section{Connections between the results via Eisenstein and Multiple Dirichlet Series}

There are many parallels between our work and that of Conrey, Farmer and Soundararajan, and moreover both results can be interpreted as averages of Fourier coefficients of Eisenstein series.  We first mention some interesting similarities.

In both our work and that of Conrey, Farmer and Soundararajan, the result when one parameter grows more rapidly than the other can be deduced by evaluating the long sum first and then making trivial estimates on the short sum.  Indeed, if $Y^2$ is very small compared to $X$, Sarnak's solution to the shifted convolution problem \eqref{SCP} yields \[\sum_{h \leq Y}\sum_{n \leq X} \lambda_f(n) \lambda_f(n+h) \ll_{k,\varepsilon} X^{\frac{1}{2}+\varepsilon}Y.\] In the case of the Jacobi symbol, one can use the Polya-Vinogradov inequality to show that \[\sum_{\substack{n \leq X \\ n \text{ odd}}} \left(\frac{n}{m}\right) = \begin{cases} \frac{X}{2} \frac{\varphi(m)}{m} + O_\varepsilon\left(X^\varepsilon\right) & \text{if } m = \square \\ O\left(m^\frac{1}{2}\log m \right) & \text{if } m \neq \square \end{cases} \] from which it follows that \begin{eqnarray*} S(X,Y) & = & \sum_{\substack{m \leq Y\\ m = \text{odd } \square}} \left( \frac{X}{2} \frac{\varphi(m)}{m}  + O_\varepsilon\left(X^\varepsilon\right)  \right) + O\left(Y^{\frac{3}{2}}\log Y \right) \\ & = & \frac{2}{\pi^2} XY^\frac{1}{2} + O_\varepsilon \left(Y^{\frac{3}{2}}\log Y + Y^\frac{1}{2}X^\varepsilon + X \log Y\right),\end{eqnarray*} and similarly if the roles of $n,m$ are reversed, see \cite{nondifferentiable}.  The case of shifted convolution sums when $Y^2$ is much larger than $X$ follows even more simply.  Indeed, we have that \[\sum_{n \leq X} \lambda_f(n) \ll_f X^{\frac{1}{3}}.\] See for example, \cite{HafnerIvic}.  Then we find
\[\sum_{h \leq X} \sum_{n\leq X-h} \lambda_f(n) \lambda_f(n+h) = -\frac{L(1,\mathrm{sym}^2 f)}{2\zeta(2)}X +  O_f\left(X^{\frac{2}{3}}\right)\] by squaring and considering the off-diagonal terms.  Observe that these easily-derived estimates correspond to the limiting behavior of the transition regions in either of the more general theorems stated above.

At the transition phase (i.e. on the blowup) the asymptotic behavior in both theorems depends crucially on the automorphic nature of the modular form $f$ or the Jacobi symbol.  Conrey, Farmer and Soundararajan's proof is based on the Poisson summation formula, see section 3 of their paper \cite{nondifferentiable}, which is the essential ingredient in proving the functional equations of the Dirichlet $L$-functions $L(s,\chi)$, i.e. $GL_1$ automorphy.  Likewise, one might consider Lemma 1 of section 3 of this paper to be the essential step which brings out the behavior of shifted convolution sums on the blowup.  The crucial step in the proof of Lemma 1 is the application of the functional equation of the $L$-function $L(s,f \times f)$, which follows from the automorphy of $f$.

The Theorem proved by Conrey, Farmer and Soundararajan has one very surprising feature which is not apparent in our work.  The function $C(\alpha)$ appearing in their result is once continuously differentiable everywhere, but it is twice differentiable at $\alpha \mathbb{Q} $ if and only if $\alpha = 2 p/q$ with $p$ and $q$ both odd, see section 6 of their paper.  It is necessary that the sum $S(X,Y)$ has a sharp cut-off for $C(\alpha)$ to have such strange differentiability properties.  In our work, we do not prove an asymptotic result for a sharp cut-off, but we do have some flexibility in our choice of cut off functions.   In section 3, we see that the asymptotic we obtain on the blowup has a main term which depends on the particular shape of the cut-off function we choose.  This is unusual in analytic number theory.  It remains possible that averages of shifted convolution sums in the shift aspect when considered with a sharp cut-off also have strange differentiability properties at the transition region.

The most essential connection, however, is that both theorems arise from averages of Fourier coefficients of Eisenstein series, and the transition regions can be understood by studying the resulting multiple Dirichlet series.  We first discuss the case of characters.  We have that \[S(X,Y) =  \frac{1}{(2 \pi i)^2} \int_{(c)}\int_{(c)} Z(s,w) \frac{X^s}{s}\frac{Y^w}{w} \,ds\,dw\] where $c>1$ and \[Z(s,w) := \sum_{\substack{m,n \geq 1 \\ m,n \text{ odd}}} \frac{\left(\frac{m}{n}\right)}{n^s m^w}.\] This is perhaps the first example of a multiple Dirichlet series, and information about the analytic properties of $Z(s,w)$ would determine the asymptotic behavior of $S(X,Y)$.  Goldfeld and Hoffstein in \cite{GHMDS} derive the analytic properties of $Z(s,w)$, which crucially follow from studying weight $\frac{1}{2}$ Eisenstein series for the congruence subgroup $\Gamma_0(4)$.  More precisely, they study the modified multiple Dirichlet series \[Z_+(s,w) = \sum_{\substack{ m \geq 1 \\ m \text{ squarefree}}} \frac{L(s,\chi_m)}{ m^{w}},\] where \[\chi_m(n) := \begin{cases}  \left(\frac{m}{n}\right) & m \equiv 1 \pmod 4 \\ \left(\frac{4m}{n}\right) & m \equiv 2,3 \pmod 4 \end{cases} \] and find that it has poles along $w=1$ and $w=\frac{3}{2}-s$.  In fact, if $E_{\frac{1}{2}}(z,s)$ is the weight $\frac{1}{2}$ Eisenstein series at the cusp 0 (see Goldfeld and Hoffstein for details), it has the Fourier expansion \[E_{\frac{1}{2}}(z,s) =  \sum_{m \geq 1} a_m(s,y)e(nx)\] where \[a_m(s,y) = \frac{L(2s,\chi_m)(1-\chi_m(2)2^{-2s})}{\zeta(4s)(1-2^{-4s})} \frac{y^s}{4^s}  K_m(s,y)\] is essentially $L(2s,\chi_m)$ times a $K$-Bessel function.  One finds that \[ \sum_{\substack{ m \leq Y \\ m \text{ odd}}} \frac{1}{2 \pi i} \int_{(\frac{1}{2})} a_m\left(s, 1/X\right)\,ds\] evaluates to a double mellin inverse of $Z(s,w)$ with some additional factors.  Ignoring covergence issues, or after sufficient smoothing, one can see how the transition region arises directly from $Z(s,w)$.  For simplicity, we work with $Z_+(s,w)$.  We have \begin{eqnarray*} \sum_{\substack{m \leq Y \\ m \text{ squarefree} }} \sum_{n \leq X} \chi_m(n) & = &  \frac{1}{(2\pi i )^2} \int_{(c)}\int_{(c)} Z_+(s,w) \frac{X^s}{s}\frac{Y^w}{w}\,ds\,dw \\ & = & \frac{1}{2\pi i} \int_{(c)} c(w) X\frac{Y^w}{w} + c_+^*(w)  \frac{X^\frac{3}{2} }{(\frac{3}{2}-w)w} \left(\frac{Y}{X}\right)^w \,dw \\
& & + \frac{1}{(2\pi i)^2}\int_{(c)} \int_{(\frac{3}{2} - \mathop{\rm Re}(w) - \varepsilon)} Z_+(s,w)\frac{X^s}{s} \frac{Y^w}{w}\,ds \,dw \end{eqnarray*} where $c(w)$ and $c_+^*(w)$ are the meromorphic functions arising in Theorem 1 of \cite{GHMDS}.  They have poles at $w=\frac{1}{2}$, with $\mathop{\rm res}_{w=\frac{1}{2}} c(w) = - \mathop{\rm res}_{w=\frac{1}{2}} c_+^*(w)$ because of the intersection of the two singular divisors at $(s,w) = (\frac{1}{2},1)$.  Hence, if $Y/X\rightarrow \infty$, the terms involving $c(w)$ and $c_+^*(w)$ cancel, and if $Y/X \rightarrow 0$, the term involving $c(w)$ becomes a main term, and the one involving $c_+^*(w)$ becomes an error term of size $O(Y^\frac{3}{2})$.  If $Y/X$ remains constant, the term involving $c_+^*(w)$ gives the behavior on the blow-up.

Our result on the following pages can be seen to arise from an identical situation.  Sums of the form \eqref{wavy} are essentially an average of Fourier coefficients of $|f|^2$.  Our method to treat these sums is to take a spectral expansion into Eisenstein series and Maass cusp forms, so one is led to compute a sum similar to \[\sum_{m \leq Y} \frac{1}{2 \pi i} \int_{(\frac{1}{2})} b_m\left(s,1/X \right)\,ds,\] where \[E(z,s) := \sum_{\gamma \in \Gamma_\infty \backslash \Gamma} \mathop{\rm Im}(\gamma z)^s = \sum_n b_n(s,y) e(nx) \] is the standard weight 0 real analytic Eisenstein series.  The integral along $\mathop{\rm Re}(s) = \frac{1}{2}$ here arises from averaging over the continuous spectrum in our decomposition.  In this context, the analouge of $Z(s,w)$ is \[ \sum_h \sum_{ab=h} \left( \frac{a}{b}\right)^{s-\frac{1}{2}} \frac{1}{h^w} = \zeta\left(w-s+\frac{1}{2}\right)\zeta\left(w+s-\frac{1}{2}\right).\]  While not a multiple Dirichlet series, it is still a zeta function in two variables and has singular divisors at $w=s+\frac{1}{2}$ and $w= \frac{3}{2}-s$, the intersection of which gives rise to a transition behavior exactly as discussed above.  This will be carried out in explicit detail in the remainder of the paper.

Finally, one should note the connection between Riemann's nondifferentiable function and theta functions on the real axis, see the interesting paper of Duistermaat \cite{RiemannNondiffereniable}.

\section{Preliminaries}

Our main approach to the Main Theorem is to take the Petersson inner product of $y^k |f|^2$ against an incomplete Poincar\'{e} series $P_h(\cdot|\psi)$.  This approach was first introduced by Selberg \cite{SelbergSCS}, and has been successfully used by many other authors to study shifted convolution sums in the past (for an overview, see \cite{MichelLfcnsinfamilies}).  Throughout this paper we set $\Gamma = SL_2(\mathbb{Z})$, and let $\mathcal{H}$ denote the upper half plane with its hyperbolic metric.  We work in the Hilbert space $L^2(\Gamma \backslash \mathcal{H})$ of square integrable measurable functions with the Petersson inner product \[\langle u,v\rangle = \int_{\Gamma \backslash  \mathcal{H}} u(z) \overline{v(z)}\,d_\mu z.\]  The symmetric operator \[\Delta = -y^2 \left( \frac{\partial^2}{\partial x^2}+ \frac{\partial^2}{\partial y^2} \right) \] acts on the subspace of smooth functions and moreover has a unique self-adjoint extension to all $L^2(\Gamma \backslash \mathcal{H})$, see Iwaniec \cite{SpecIw} chapter 4.  Given a classical holomorphic normalized cuspidal eigenform $f$ of weight $k$, set $F(z) := y^{\frac{k}{2}}f(z).$  We have that $|F| \in L^2(\Gamma \backslash \mathcal{H}),$ but on the other hand it is no longer holomorphic.  Let $\psi(y)$ be an infinitely differentiable compactly supported function on $\mathbb{R}_{>0}$, and let $\Gamma_\infty$ denote the stabilizer in $\Gamma$ of the cusp at infinity.  Then we define the incomplete Poincar\'{e} series \[P_h(z|\psi) := \sum_{\gamma \in \Gamma_\infty \backslash \Gamma} e(h\gamma z) \psi \left(\text{Im}(\gamma z)\right),\] which is a smooth and bounded function on $\Gamma \backslash \mathcal{H}$.  By unfolding the inner product \[\langle F  P_h(\cdot | \psi),F\rangle = \int_{\Gamma \backslash \mathcal{H}} y^k|f(z)|^2 P_h(z|\psi) \,d_\mu z\] on the Poincar\'{e} series, we find that \[ \langle F P_h(\cdot | \psi),F\rangle  =  \sum_{n =1}^\infty \lambda_f(n)\lambda_f(n+h) (n(n+h))^{\frac{k-1}{2}}\int_0^\infty \psi(y) e^{-4 \pi (n+h)y} y^{k-2} \, dy,\] i.e. this inner product is a smoothed shifted convolution sum with cut-off function given in terms of an integral transform (similar to the Laplace transform) of $\psi$.  The behavior of $\psi(y)$ as $y$ tends to 0 is crucial to control the length of the shifted convolution sum.  In connection with the previous section, it should be noted that if $\psi$ were a delta function, then taking the inner product against $P_h(z|\psi)$ is equivalent to taking the $h$-th Fourier coefficient.

Let $u_j$ be a complete orthonormal system of cusp forms which are eigenfunctions of the Laplace operator and all Hecke operators.  Because we are only working in level 1, we need not worry about old forms or the Hecke operators whose index divides the level.  Define the real analytic Eisenstein series by \[E(z,s) := \sum_{\gamma \in \Gamma_\infty \backslash \Gamma } \text{Im}(\gamma z)^s\] for $\mathop{\rm Re}(s) > 1$, and in general by analytic continuation.  For each $s \neq 0,1$, the Eisenstein series are also eigenfunctions for the Laplace operator and all the Hecke operators.  We have then that \[\sum_{j = 1}^\infty \langle P_h(\cdot |\psi), u_j\rangle u_j(z) + \frac{1}{4\pi} \int_{-\infty}^\infty \langle P_h(\cdot|\psi),E(\cdot, 1/2 + it)\rangle E(z, 1/2 + it) \,dt\] converges to $P_h(z|\psi)$ in the norm topology on $L^2(\Gamma \backslash \mathcal{H})$, see Theorems 4.7 and 7.3 in \cite{SpecIw}.  Then we have that \begin{equation}\label{expansion}\langle F P_h(\cdot | \psi),F\rangle =  \sum_{j=1}^\infty \langle P_h(\cdot | \psi), u_j\rangle \langle Fu_j,F\rangle + \frac{1}{4\pi} \int_{-\infty}^\infty \langle P_h(\cdot | \psi) , E(\cdot, 1/2+ it) \rangle \langle F  E(\cdot, 1/2+ it), F \rangle\,dt,\end{equation} and we will see later that the convergence is absolute and uniform in $h$.

If $u \in L^2\left(\Gamma \backslash \mathcal{H}\right)$ and $\Delta u = \lambda u$ with $\lambda = s(1-s),$ then $u(z)$ has a Fourier expansion given in terms of the $K$-Bessel function $K_\nu(z)$,  which is an exponentially decaying solution to the differential equation \[z^2 f'' +z f' - \left(z^2 + \nu^2\right) f = 0.\]
We will primarily be interested in the $K$-Bessel function for purely imaginary $\nu$, and note some useful properties of these functions: first, that $K_{\nu}(z)$ is real for real $z$, second, that as $\mathop{\rm Im}(\nu) \rightarrow \infty$ in a fixed vertical strip, $K_\nu(z)$ is decaying exponentially, and last, that for $\nu = i t$, with $t \in \mathbb{R}$, $K_{it}(z)$ has a branch cut, which we take to be along the negative real axis in the $z$-plane.  As $z\rightarrow 0$, we have that \[|K_{it}(z)| \sim \pi  \left| \frac{\sin(t \log z/2)}{\Gamma(1+it) \sinh(\pi t)}\right|\] so long as one avoids the branch cut.

\section{Proof of Theorem}

We now proceed to the proof of the Main Theorem of the paper by summing the right side of  \eqref{expansion} over $h$.  Pointwise, the largest term comes from the discrete spectrum (see \cite{SarnakSCS}), however, on average, the continuous spectrum dominates.  We start with the continuous spectrum contribution to \eqref{expansion}.

\subsection{Eisenstein Series}

The Eisenstein series $E(z,s)$ has a Fourier expansion given by

\[E(z,s) = \varphi(0,s)+\sum_{n \neq 0} \varphi(n,s) W_{s}(nz), \]
where
\[W_{s}(z) = 2y^{\frac{1}{2}} K_{s-\frac{1}{2}}(2 \pi y) e(x).\] If $n= 0$ and $s \neq \frac{1}{2}$  then \[\varphi(0,s) = y^{s} + \frac{\xi(2s-1)}{\xi(2s)}y^{1-s},\] and if $n \neq 0$, the $n$-th Fourier coefficient of $E(z,s)$ is given by
\[\varphi(n,s) = \xi(2s)^{-1} |n|^{-1/2} \sum_{ab = |n|} \left(\frac{a}{b}\right)^{s-1/2},\] where $\xi(s) = \pi^{-s/2}\Gamma(s/2)\zeta(s)$ denotes the completed Riemann zeta function which has the functional equation $\xi(s) = \xi(1-s)$. If $s=\frac{1}{2}+it$, we find by unfolding the Poincar\'{e} series that the inner product \[\langle P_h(\cdot | \psi), E(\cdot, 1/2+it)\rangle =  \frac{2}{\xi(2s-1)}  \sum_{ab = h }\left(\frac{a}{b}\right)^{s-\frac{1}{2}} \int_0^\infty \frac{\psi(y)}{y^{3/2}}  e^{-2 \pi h y} K_{s-\frac{1}{2}}(2 \pi h y) \,dy \] where we have used that $\overline{\xi(2s)} = \xi(2 \overline{s}) = \xi(2-2s) = \xi(2s-1)$. We can also unfold the second inner product on the Eisenstein series.  Following Iwaniec \cite{ClassIw} chapter 13, set
\[L(s,f\times f) := \zeta(s) L(s,\mathrm{sym}^2 f) = \zeta(2s)L(s,f \otimes f) = \zeta(2s) \sum_{n\geq 1} \frac{\lambda_f(n)^2}{n^s},\] where the last equality is valid only for $\mathop{\rm Re}(s) >1$.  This $L$-function admits the functional equation \[\Lambda(s, f \times f) = \Lambda(1-s, f \times f),\] where \[\Lambda(s, f \times f) = L_\infty(s, f \times f)L(s, f \times f)\] with \[L_\infty(s,f \times f) := (2 \pi)^{-2s} \Gamma(s)\Gamma(s+k-1) .\] By unfolding when $\mathop{\rm Re}(s) >1$, and in general after analytic continuation,
\[ \langle F E(\cdot, s),F\rangle = \frac{\Lambda(s, f \times f)}{(4 \pi)^{k-1} \xi(2s)}.\]

Going back to the spectral expansion \eqref{expansion} and pulling these two inner products together, we have that the Eisenstein series contribution to $\langle F P_h(\cdot | \psi),F\rangle$ is
\[E_{f,h}(\psi) :=  \frac{1}{(4 \pi)^{k-1}} \frac{1}{2 \pi i} \int_{(\frac{1}{2})} \frac{\Lambda(s, f \times f)}{\xi(2s)\xi(2s-1)}\sum_{ab = h }\left(\frac{a}{b}\right)^{s-\frac{1}{2}}  \int_0^\infty \frac{\psi(y)}{y^{3/2}}  e^{-2 \pi h y} K_{s-\frac{1}{2}}(2 \pi h y) \,dy \, ds.\]
The completed $L$ and zeta functions have the same number of gamma factors in the numerator and denominator, and Bessel function decays rapidly as $|\mathop{\rm Im}(s)| \rightarrow \infty$, so the contour integral converges rapidly.  It is interesting to note that the $s$ integral only makes sense because $\xi(s)^{-1}$ has no poles and at most polynomial growth on the $\mathop{\rm Re}(s) =1$ line, i.e. due to the prime number theorem.  Indeed, one sees that the contour is constrained between the poles in the critical strips of the two $\xi(s)^{-1}$ functions appearing here.   Due to the mysterious nature of the residues at these poles, shifting contours seems to be a futile approach to understanding the asymptotic size of $E_{f,h}(\psi)$.

Introducing a sum over $h$ clears this obstruction.  We have to compute
\begin{equation}\label{eisbeginning}\sum_{h \leq Y} E_{f,h}(\psi) = \frac{1}{(4 \pi)^{k-1}} \int_0^\infty \frac{\psi(y)}{y^{\frac{3}{2}}} \frac{1}{2 \pi i} \int_{(\frac{1}{2})} \frac{\Lambda(s, f \times f)}{\xi(2s) \xi(2s-1)} \sum_{h \leq Y} \sum_{ab = h} \left(\frac{a}{b} \right)^{s-\frac{1}{2}} e^{-2 \pi y h} K_{s-\frac{1}{2}}(2 \pi y h)\,ds\,dy,\end{equation} and can evaluate the sum over $h$ by standard techniques.  We have by adapting Theorem 12.4 from \cite{TOTRZF} (due to Van Der Corput) that
\[\sum_{h \leq x} \sum_{ab = h} \left( \frac{a}{b}\right)^{s-\frac{1}{2}} = \zeta(2s) \frac{x^{s+\frac{1}{2}}}{s+\frac{1}{2}} + \zeta(2-2s) \frac{x^{-s+\frac{3}{2}}}{-s+ \frac{3}{2}} + O_{s,\varepsilon}\left(x^{\frac{27}{82}+|\mathop{\rm Re}(s) -\frac{1}{2}| +\varepsilon}\right),\] where the implied constants depend at most polynomially on $|s|$.   It should be noted that the error term here is not the best currently known, however, it is sufficiently small as to not contribute to the final result of this paper.  By partial summation, \begin{eqnarray*} & & \sum_{h \leq Y} \sum_{ab = h} \left(\frac{a}{b} \right)^{s-\frac{1}{2}} e^{-2 \pi y h} K_{s-\frac{1}{2}}(2 \pi y h) \\ & = & \zeta(2s) \int_0^Y u^{s-\frac{1}{2}} e^{-2\pi y u } K_{s-\frac{1}{2}} (2 \pi y u )\, du  + \zeta(2-2s) \int_0^Y u^{\frac{1}{2}-s} e^{-2\pi y u } K_{s-\frac{1}{2}} (2 \pi y u )\, du\\
& & + \,\,\, O_{s,\varepsilon}\left(e^{-2 \pi y Y}K_{s-\frac{1}{2}}(2 \pi y Y) Y^{\frac{27}{82}+|\mathop{\rm Re}(s) -\frac{1}{2}|  +\varepsilon}\right). \end{eqnarray*} Note that the two integrals appearing in the displayed equation are interchanged under the transformation $s \longleftrightarrow 1-s$, by symmetry of the the Bessel function.  The contour integral over $s$ in $\sum E_{f,h}(\psi)$ is also symmetric under $s \longleftrightarrow 1-s$ so that these two integrals are identical in the overall sum, and we need only work one of them out.  The first integral can be evaluated explicitly, and the answer will be in terms of hypergeometric functions.  The confluent hypergeometric function that will appear below is defined by the power series \[{}_1F_1(a,b,z) = \sum_{n=0}^\infty \frac {a^{(n)} z^n} {b^{(n)} n!} \] where \[a^{(n)}=a(a+1)(a+2)\cdots(a+n-1) = \frac{\Gamma(a+n)}{\Gamma(a)}\] is called either the `rising factorial' or `Pochhammer symbol'.
The theory of hypergeometric functions is developed in detail in \cite{HTFvol1}.  It is entire on $\mathbb{C}$ separately in each variable except for simple poles at $b=0,-1,-2,\ldots$ by the absolute and uniform convergence of the defining series, hence it is meromorphic on $\mathbb{C}^3$.  We have that the residues at these poles are given by  \[\mathop{\rm res}_{b=-n} {}_1F_1(a,b,z) = \frac{\Gamma(a+n+1)(-1)^n}{\Gamma(a)\Gamma(n+2)\Gamma(n+1)} z^{n+1} {}_1F_1(a+n+1,n+2,z),\] see Gradshteyn and Ryzhik \cite{GR}, 9.214.

We have the formulae \[K_\nu(z) = \frac{\pi}{2} \frac{i^{\nu}J_{-\nu}(iz) - i^{-\nu} J_{\nu}(iz)}{\sin \pi \nu},\] \[e^{-z} \frac{\Gamma\left( s+ \frac{1}{2}\right)}{\left(\frac{iz}{2}\right)^{s-\frac{1}{2}}}J_{s-\frac{1}{2}}(iz) = {}_1F_1(s,2s,-2z),\] \[ \frac{d}{dz} {}_1F_1(a,b,z) = \frac{a}{b}{}_1F_1(a+1,b+1,z),\] and \[ \frac{d}{dz} \left(z^{b-1}{}_1F_1(a,b,z)\right) = (b-1)z^{b-2}{}_1F_1(a,b-1,z),\] where $J_\nu(x)$ is the $J$-Bessel function, the second formula can be found in \cite{GR}, 9.215 \#3 and the last two formulae can be found in \cite{HTFvol1} section 2.1.2.  From these it follows that \[ \int_0^Y u^{s-\frac{1}{2}} e^{-2\pi y u } K_{s-\frac{1}{2}} (2 \pi y u )\, du = \frac{1}{4 (\pi y)^\frac{1}{2} s} \left(A_f(y,s)+ B_f(y,Y,s)+C_f(y,Y,s)\right),\] where \[A_f = A_f(y,s) = (\pi y)^{-s} \Gamma(1/2 +s )\] \[B_f = B_f(y,Y,s) = (\pi y)^{-s} \Gamma(1/2+s) {}_1F_1(-s,1-2s, -4 \pi y Y) \] and \[C_f = C_f(y,Y,s) =  (\pi y Y^2)^{s} \Gamma(1/2 -s) {}_1F_1(s, 1+2s, -4 \pi yY). \]

The $\zeta(2s)$ from the evaluation of the sum over $h$ cancels against the $\zeta(2s)$ in the denominator of \eqref{eisbeginning}, eliminating its poles.  Given that $A_f+B_f+C_f$ is holomorphic in $s$ past the $\mathop{\rm Re}(s) = 0$ line, we are free to pass the contour to the left.  Picking up a residue at $s=0$ we get \begin{eqnarray*}
 \sum_{h \leq Y} E_{f,h}(\psi) & = &- \frac{\Gamma(k)}{(4 \pi)^k} \frac{L(1,\mathrm{sym}^2 f)}{2 \zeta(2)} \int_0^\infty \frac{\psi(y)}{y^2}\,dy \\
& & +\,\,\, \frac{1}{(4\pi)^{k-\frac{1}{2}}} \int_0^\infty \frac{\psi(y)}{y^2}\frac{1}{2 \pi i} \int_{(-a)}\frac{\Lambda(s, f\times f)}{\xi(2s-1)}\frac{\left(A_f(y,s) + B_f(y,Y,s) + C_f(y,Y,s)\right)}{\pi^{-s}\Gamma(s+1)}\,ds\,dy \\ & &  +\,\,\, O_{k,\varepsilon}\left(Y^{\frac{27}{82}+\varepsilon} \int_0^\infty \frac{|\psi(y)|}{y^{\frac{3}{2}}}\,dy\right),
\end{eqnarray*}
where $0<a<\frac{1}{2}$, and we have again made use of the prime number theorem in estimating the error term.

The integrals over $y$ converge at $\infty$, so the asymptotic size of $\sum E_{f,h}(\psi)$ depends only on the behavior of $\psi(y)$ for small $y$.  We now make some estimates assuming that $y$ is small. The remaining contour integral \[ \frac{1}{2 \pi i} \int_{(-a)}\frac{\Lambda(s, f\times f)}{\xi(2s-1)}\frac{\left(A_f(y,s) + B_f(y,Y,s) + C_f(y,Y,s)\right)}{\pi^{-s}\Gamma(s+1)}\,ds \] is a sum of three terms coming from $A_f$, $B_f$ and $C_f$.  The term coming from $A_f$ is $\ll_k y^{\frac{1}{2}}$ as $y \rightarrow 0$.  The hypergeometric function appearing in $B_f$ is bounded above and below by universal constants in the half-plane $\mathop{\rm Re}(s)<0$ and when $0 \leq yY \leq 4 \pi$, thus term coming from $B_f$ is also $\ll y^{\frac{1}{2}}$, uniformly in $Y$.  Thus it remains to inspect the term coming from $C_f$.  Explicitly, let
\begin{eqnarray*} C_{f,1}(y,Y) & := &  \frac{1}{2 \pi i} \int_{(-a)}\frac{\Lambda(s, f\times f)}{\xi(2s-1)}\frac{C_f(y,Y,s)}{\pi^{-s}\Gamma(s+1)}\,ds \\ & = & \frac{1}{2 \pi i} \int_{(-a)}\frac{\Lambda(s, f\times f)}{\xi(2s-1)}\frac{\Gamma\left(\frac{1}{2} -s\right)}{\Gamma(s+1)} {}_1F_1(s, 1+2s, -4 \pi yY) \left(\pi^2 y Y^2\right)^{s} \,ds. \end{eqnarray*}
Thus \begin{eqnarray*} \sum_{h \leq Y} E_{f,h}(\psi)& = & - \frac{\Gamma(k)}{(4 \pi)^k} \frac{L(1,\mathrm{sym}^2 f)}{2 \zeta(2)} \int_0^\infty \frac{\psi(y)}{y^2}\,dy +  \frac{1}{(4\pi)^{k-\frac{1}{2}}} \int_0^\infty \frac{\psi(y)}{y^2}C_{f,1}(y,Y)\,dy \\ & & + \,\,\,O_{k,\varepsilon}\left(Y^{\frac{27}{82}+\varepsilon} \int_0^\infty \frac{|\psi(y)|}{y^{\frac{3}{2}}}\,dy\right).\end{eqnarray*} In the transition region, $C_{f,1}(y,Y)$ is the crucial term.

\begin{lemma}
Suppose that $y,Y \in \mathbb{R}_{>0}$ with $Y$ becoming large and $y$ becoming small.  Then
\[ C_{f,1}(y,Y) = \frac{1}{2 \sqrt{\pi}}c_f\left(4 \pi yY^2\right) + O_k\left(y^{\frac{1}{2}}\right), \]  where \[ c_f(\alpha) = =\frac{\pi^\frac{3}{2}}{2}\alpha \sum_{n \geq 1} \lambda_f(n)^2 W_k(\pi^2 n \alpha), \] and \[W_k(x) =  \frac{1}{2 \pi i}\int_{(1+a)} \frac{\Gamma(s+k-1) \Gamma(s-\frac{1}{2})}{\Gamma(2-s)}x^{-s}  \,ds\]for any fixed $a>0$.
\end{lemma}

\begin{proof}
We apply the functional equation for the $L$-function to find that \[C_{f,1}(y,Y) =  \pi^2 y Y^2 \frac{1}{2 \pi i}\int_{(1+a)}\frac{L(s,f\times f)}{\zeta(2s)} \frac{\Gamma(s+k-1) \Gamma(s-\frac{1}{2})}{\Gamma(2-s)} {}_1F_1(1-s,3-2s,-4 \pi y Y) \left(4 \pi^3 yY^2\right)^{-s}\,ds.\]  From the definition one finds that ${}_1F_1(1-s,3-2s,u) = 1 + O_s(u),$ which we use to eliminate the hypergeometric function from the above expression.  We proceed in two slightly different ways depending on whether $yY^2$ becomes large or becomes small.  If $yY^2$ remains bounded, either approach is acceptable.  First, assume that $yY^2$ is becoming small.  In this case, choose $a=\frac{1}{4}$, and observe that the $s$-dependence in the hypergeometric function is uniformly bounded along the line $\mathop{\rm Re}(s) = \frac{5}{4}$.  Together with the rapid decay of the integrand of $C_{f,1}(y,Y)$, this gives us that \[ \left|C_{f,1}(y,Y) - \frac{1}{2 \sqrt{\pi}}c_f\left(4 \pi yY^2\right)\right|  \ll_k  y^{\frac{1}{2}}(yY^2)^{\frac{1}{4}}.\]

If $yY^2$ becomes large shift the line of integration to the right, past the pole of the hypergeometric function at $s=\frac{3}{2}$ to $a = \frac{3}{4}$.  The contribution to $C_{f,1}(y,Y)$ coming from this residue is $\ll_k y^{\frac{1}{2}}$, uniformly in $Y$, and the $s$-dependence in the hypergeometric function is uniformly bounded along the line $\mathop{\rm Re}(s) = \frac{7}{4}$.  We find in this case that \[\left|C_{f,1}(y,Y) - \frac{1}{2 \sqrt{\pi}}c_f\left(4 \pi yY^2\right)\right| \ll_k y^{\frac{1}{2}}\left( 1+ (yY^2)^{-\frac{1}{4}}\right). \]
In either case, we obtain the error term stated in the Lemma.

\end{proof}
\begin{lemma}
The function $c_f(\alpha)$ defined above is $C^\infty(\mathbb{R}_{>0})$.  As $\alpha \rightarrow \infty$ it decays faster than any polynomial, and as $\alpha \rightarrow 0$
\[c_f(\alpha) = \frac{\Gamma(k)L(1,\mathrm{sym}^2 f)}{2 \zeta(2)} + E_k(\alpha)\] where $E_k(\alpha) = o_k\left(\alpha^{\frac{1}{2}}\right)$ unconditionally, and $E_k(\alpha) = O_{k,\varepsilon}\left(\alpha^{\frac{3}{4}-\varepsilon}\right)$ for any $\varepsilon>0$ assuming the Riemann hypothesis for the classical Riemann zeta function.
\end{lemma}
\begin{proof}
The $W_k(x)$ defined above is $C^\infty(\mathbb{R})$ and its integrand has no poles to the right, thus $W_k(x)$ is rapidly decaying as $x \rightarrow + \infty$.  After differentiating the series for $c_f(\alpha)$ in the second line of Lemma 1 arbitrarily many times, the resulting series for $c^{(n)}_f(\alpha)$ converges absolutely for any $\alpha >0$, so $c_f \in C^\infty(\mathbb{R}_{>0})$.  The rapid decay of $c_f(\alpha)$ follows from that of $W_k(x)$.  By shifting the line of integration in the definition of $c_f(\alpha)$ to the left, we investigate the behavior of $c_f(\alpha)$ near $\alpha = 0$.  The main term comes from the residue of the pole of $L(s, f \times f)$ at $s=1$, and the error term is estimated by pushing the contour just past the $\mathop{\rm Re}(s) = \frac{1}{2}$ line, or to the $\mathop{\rm Re}(s) = \frac{1}{4}+\varepsilon$ line if one assumes the Riemann hypothesis.
\end{proof}

Thus we have \begin{prop} For $E_{f,h}(\psi)$ and $c_f(\alpha)$ as defined above, we have that \[ \sum_{h \leq Y} E_{f,h}(\psi) = \frac{1}{(4 \pi)^k} \int_0^\infty \left( c_f\left(4 \pi yY^2\right) - \frac{\Gamma(k) L(1,\mathrm{sym}^2 f)}{2\zeta(2)} \right) \frac{\psi(y)}{y^2}\,dy + O_{k,\varepsilon}\left(Y^{\frac{27}{82}+\varepsilon} \int_0^\infty \frac{\psi(y)}{y^{\frac{3}{2}}}\,dy\right).\] \end{prop}

The term involving $\frac{27}{82}$ is smaller than the remainder terms coming from Maass forms, as we will see in the next section.

\subsection{Maass Forms}

The discrete spectrum of $\Delta$ is spanned by Maass cusp forms.  The Hecke algebra acting on $L^2( \Gamma \backslash \mathcal{H})$ is defined to be the algebra generated by the commuting self-adjoint bounded operators $T_n$, where for $u \in L^2(\Gamma \backslash \mathcal{H})$, define $T_n$ by \[(T_nu)(z) := \frac{1}{\sqrt{n}} \sum_{ad=n} \sum_{b \!\!\!\!\! \pmod d} u\left( \frac{az+b}{d}\right).\]  These operators commute with $\Delta$ as well, so in fact our basis of Maass forms can be taken to be eigenfunctions of the Hecke algebra as well, and we denote the Hecke eigenvalues of the Maass form $u_j$ by $\lambda_{u_j}(n)$.  A Maass form of Laplace eigenvalue $\lambda_j = s_j(1-s_j) = \frac{1}{4} + t_j^2$ is cuspidal, so it has a Fourier expansion of the form
\[u_j(z) = \sum_{n \neq 0} a_{u_j}(n) W_{s_j}(nz), \]
where
\[W_{s_j}(z) = 2y^{\frac{1}{2}} K_{it_j}(2 \pi y) e(x).\]

For $\Gamma = SL_2(\mathbb{Z})$, it was known to Selberg in the early 50s that the smallest Laplace eigenvalue $\lambda_1$ is $>\frac{1}{4}$, hence $t_j \in \mathbb{R}$.  For a proof of this fact, see \cite{HejhalSTFvol2}, chapter 11.  Computationally, it has been verified that $t_1 = 9.53369526\ldots$, see for example \cite{Hejhalnumerical}.  To apply the spectral theorem we must assume the normalization $||u_j||_{L^2}^2 = 1$, in which case the Fourier coefficient and Hecke eigenvalue are related by  \[a_{u_j}(n) = \left(\frac{\cosh\pi t_j}{2 |n| L(1,\mathrm{sym}^2 u_j)}\right)^{\frac{1}{2}}\lambda_{u_j}(n),\] where the symmetric square $L$-function appearing here is defined \[ L(s, \mathrm{sym}^2 u_j) = \prod_p \left(1-\frac{\alpha_{u_j}(p)^2}{p^s}\right)^{-1}  \left(1-\frac{\alpha_{u_j}(p)\beta_{u_j}(p)}{p^s}\right)^{-1}  \left(1-\frac{\beta_{u_j}(p)^2}{p^s}\right)^{-1}.\]  The Hecke eigenvalues conform to the bound $|\lambda_{u_j}(n)| \leq d(n)n^{\theta}$, where the generalized Ramanujan conjecture implies that $\theta = 0$ is admissible, and the best known unconditional bound is due to Kim and Sarnak \cite{KimSarnak}, which gives $\theta = \frac{7}{64}$.

We call the Maass form contribution to \eqref{expansion} \[M_{f,h}(\psi) := \sum_{j=1}^\infty \langle P_h(\cdot | \psi),u_j\rangle \langle F u_j,F\rangle.\] By unfolding we have \[ \langle P_h(\cdot | \psi), u_j\rangle  =  \left(\frac{\cosh \pi t_j}{L(1, \mathrm{sym}^2 u_j)}\right)^\frac{1}{2} \lambda_{u_j}(h) \int_0^\infty \frac{\psi(y)}{y^{\frac{3}{2}}} e^{-2\pi h y} K_{it_j}(2 \pi h y)\,dy \] so that \[ \sum_{h \leq Y} M_{f,h}(\psi) = \int_0^\infty  \frac{\psi(y)}{y^{\frac{3}{2}}} \sum_{j=1}^\infty \left(\frac{\cosh \pi t_j}{L(1, \mathrm{sym}^2 u_j)}\right)^\frac{1}{2} \langle F u_j,F\rangle \sum_{h \leq Y} \lambda_{u_j}(h)  e^{-2\pi h y} K_{it_j}(2 \pi h y)\,dy \]

\begin{lemma}
The spectral sum \[\sum_{j=1}^\infty \left(\frac{\cosh \pi t_j}{L(1, \mathrm{sym}^2 u_j)}\right)^\frac{1}{2} \langle F u_j,F\rangle \sum_{h \leq Y} \lambda_{u_j}(h)  e^{-2\pi h y} K_{it_j}(2 \pi h y) \]
appearing in $\sum M_{f,h}(\psi)$ converges absolutely.
\end{lemma}
\begin{proof}
There are three factors in the summand: that which involves $\cosh \pi t_j$ , the inner product, and the sum over $h$.  We show that the first of these two balance each other, and then show that the sum over $h$ decays rapidly in $|t_j|$, uniformly in the other variables.  To study $|\langle F u_j, F\rangle|$ we will use a beautiful formula of Watson \cite{Watson}, but follow a classical work-out of it from Soundararajan's paper \cite{WeakSC}.  In that paper both $f$ and $u_j$ are normalized to have mass 1, but in this paper we take $f$ to be Hecke normalized so that Watson's formula is

\[|\langle F u_j, F \rangle |^2 = \frac{1}{8} \left( \frac{\Gamma(k)}{(4 \pi)^{k}} \frac{\text{Vol}\left(\Gamma \backslash \mathcal{H}\right)}{\zeta(2)}\right)^{2} \frac{\Lambda(\frac{1}{2}, f \times f \times u_j)}{L_\infty(1,\mathrm{sym}^2 f)^2 \Lambda(1, \mathrm{sym}^2 u_j)},\] where
\[\Lambda(s,f \times f \times u_j) = L_\infty(s,f \times f \times u_j) L(s, f \times f \times u_j),\]
\[L_\infty(s, \mathrm{sym}^2 f ) =  \pi^{-\frac{3}{2}s}\Gamma\left(\frac{s+1}{2}\right)\Gamma\left(\frac{s+k-1}{2}\right)\Gamma\left(\frac{s+k}{2}\right),\]
\[ \Lambda(s,\mathrm{sym}^2 u_j) = L_\infty(s,\mathrm{sym}^2 u_j) L(s,\mathrm{sym}^2 u_j),\]
\[L_\infty(s, f \times f \times u_j) = \pi^{-4s}\prod_{\pm} \Gamma\left(\frac{s+k-1\pm i t_j}{2}\right)\Gamma\left(\frac{s+k \pm it_j}{2}\right)\Gamma\left(\frac{s+1\pm i t_j}{2}\right)\Gamma \left(\frac{s \pm it_j}{2}\right) ,\]
\[L_\infty(s,\mathrm{sym}^2 u_j) = \pi^{-3s/2}\Gamma\left(\frac{s-2it}{2}\right)\Gamma\left(\frac{s}{2}\right) \Gamma\left(\frac{s+2it}{2}\right)\]
and
\[ L(s,f \times f \times u_j) = \prod_p \left(1 - \frac{\alpha_f(p)^2\alpha_{u_j}(p)}{p^s}\right)^{-1}\left(1 - \frac{\alpha_{u_j}(p)}{p^s}\right)^{-2}\left(1 - \frac{\beta_f(p)^2\alpha_{u_j}(p)}{p^s}\right)^{-1}\]\[\times \left(1 - \frac{\alpha_f(p)^2\beta_{u_j}(p)}{p^s}\right)^{-1}\left(1 - \frac{\beta_{u_j}(p)}{p^s}\right)^{-2}\left(1 - \frac{\beta_f(p)^2\beta_{u_j}(p)}{p^s}\right)^{-1}.\] The archimedian parts evaluate to
\[ \frac{L_\infty(\frac{1}{2}, f \times f \times u_j)}{L_\infty(1,\mathrm{sym}^2 f)^2L_\infty(1,\mathrm{sym}^2 u_j)}
= 4 \pi^2\frac{ |\Gamma(k-\frac{1}{2}+ i t_j)|^2}{ \Gamma(k)^2} \]
after repeated application of the duplication formula.  Using this, Watson's formula simplifies to \begin{equation}\label{Watson2}| \langle F u_j,F\rangle | = \sqrt{2} \frac{|\Gamma(k-\frac{1}{2} + it_j)|}{(4 \pi)^k} \left(\frac{L(\frac{1}{2}, f \times f \times u_j)}{L(1,\mathrm{sym}^2 u_j)}\right)^{\frac{1}{2}}\end{equation} for $f$ Hecke normalized, and $u_j$ mass 1 normalized.  By applying Stirling's formula, we find that $|\Gamma(k-\frac{1}{2} + it_j)|(\cosh \pi t_j)^\frac{1}{2}$ is polynomially bounded as $|t_j| \rightarrow \infty$.  Together with standard convexity bounds in the $|t_j|$ aspect for $L(\frac{1}{2},f \times f \times u_j)$, we find that \[ \left|\frac{\cosh \pi t_j}{L(1, \mathrm{sym}^2 u_j)}\right|^\frac{1}{2} |\langle F u_j,F\rangle| \] is polynomially bounded as $|t_j|$ gets large.

Now we turn to the sum over $h$.  We have by a ``folklore'' result written down by Hafner and Ivi\'{c} \cite{HafnerIvic} that \[ \sum_{h \leq Y} \lambda_{u_j}(h) \ll_{u_j} Y^{\frac{1}{3}(1+ \theta)}\] where $\theta = 0$ or $=\frac{7}{64}$ as above, and where the implied constants depend at most polynomially on $|t_j|$.  As on the Eisenstein series side, the conjectural truth is $O_{u,\varepsilon}(Y^{\frac{1}{4}+\varepsilon})$, but this seems very difficult.  By partial summation \[ \sum_{h \leq Y} \lambda_u(h)  e^{-2\pi h y} K_{it_j}(2 \pi h y) \ll_{u_j}  e^{-2\pi y Y} K_{it_j}(2 \pi y Y) Y^{\frac{1}{3}(1+ \theta)} +  \int_{\frac{1}{2}}^Y u^{\frac{1}{3}(1+ \theta)} \left| \frac{\partial}{\partial u} e^{-2 \pi y u} K_{it_j}(2 \pi y u)\right| \,du, \] where the implied constants again depend at most polynomially on $|t_j|$.   We have that $|K_{it}(u)| \sim \pi \frac{|\sin(t \log u/2)|}{|\Gamma(1+it) \sinh(\pi t)|}$ for small $u$, thus after taking derivatives, changing variables, and using the power series expansion for the lower incomplete gamma function, we have that  \begin{equation}\label{maasssum} \sum_{h \leq Y} \lambda_u(h)  e^{-2\pi h y} K_{it_j}(2 \pi h y) \leq P(t_j) e^{-\frac{\pi}{2} |t_j|} Y^{\frac{1}{3}(1+\theta)},\end{equation} uniformly in $y$, where $P(t)$ is a real-valued function on $\mathbb{R}$ that grows at most polynomially as $|t|$ becomes large.  From these estimates together with Weyl's law (see, e.g. Iwaniec \cite{SpecIw}) \[ \sum_{|t_j|\leq T} 1 = \frac{\text{Vol}\left(\Gamma \backslash \mathcal{H}\right)}{4 \pi } T^2 + O_\varepsilon\left((1+T)^{1+\varepsilon}\right)\] the Lemma follows.
\end{proof}
We apply trivial estimates along with \eqref{maasssum} to obtain \begin{prop} For $M_{f,h}(\psi)$ defined above we have  \[\sum_{h \leq Y} M_{f,h}(\psi) \ll_k Y^{\frac{1}{3}(1+\theta)}\int_0^\infty \frac{|\psi(y)|}{y^{\frac{3}{2}}}\,dy. \] \end{prop} The application of trivial bounds is justified by the absolute convergence given by Lemma 3.

\section{The Main Theorem and its Corollaries}

Drawing together the propositions from the two preceding sections, we obtain
\begin{mt}
Let $\psi $ be any measurable function on $\mathbb{R}_{>0}$ such that the incomplete Poincar\'{e} series \[P_h(z|\psi) := \sum_{\gamma \in \Gamma_\infty \backslash \Gamma} e(h\gamma z) \psi(\mathop{\rm Im}(\gamma z))\] is a smooth and bounded $L^2$ function on $\Gamma \backslash \mathcal{H}$.  Denote the shifted convolution sum \[ S_{f}(\psi,Y) = \sum_{h \leq Y} \sum_{n \geq 1} \lambda_f(n)\lambda_f(n+h) (n(n+h))^{\frac{k-1}{2}}\int_0^\infty \psi(y) e^{-4 \pi (n+h)y} y^{k-2} \, dy.\]  Then we have that \[ S_f(\psi,Y) =  \frac{1}{(4 \pi)^k} \int_0^\infty \left( c_f\left(4 \pi yY^2\right) - \frac{\Gamma(k) L(1,\mathrm{sym}^2 f)}{2\zeta(2)} \right) \frac{\psi(y)}{y^2}\,dy+ O_k\left(Y^{\frac{1}{3}(1+\theta)}\int_0^\infty \frac{|\psi(y)|}{y^{\frac{3}{2}}}\,dy\right),\] where $c_f(\alpha)$ is the function defined in Lemma 1 and whose properties are given in Lemma 2, and $\theta =0$ or $\frac{7}{64}$ depending on whether we assume the truth of the generalized Ramanujan conjecture or not.
\end{mt}

It should be noted that the size of the exponent of $Y$ in the error term here depends on the sharp cut-off in $h$, and could be made smaller if we were to smooth that sum.  Now we make some choices for $\psi$, and record the results as corollaries.  First, we give the result stated in the introduction.
\begin{proof}[Proof of Corollary 1.]
Let $\psi$ be a smooth approximation to a point mass.  Specifically, let $\psi$ be smooth, non-negative, supported on a set of radius $X^{-4}$ about the point $y = \frac{1}{4 \pi X}$ and have mass $1$.  Then for any continuously differentiable function $\phi$ on $\mathbb{R}_{>0}$, we have \[\left|\phi\left(\frac{1}{4\pi X}\right) - \int_0^\infty \psi(y) \phi(y)\,dy \right| \ll \left|\phi'\left(\frac{1}{4\pi X}\right)\right| X^{-4},\] where the implied constants are absolute.  First, let $\phi(u) = u^{k-2}e^{-4\pi(n+h)u}$, so that $\left| \phi' \left(\frac{1}{4\pi X}\right)\right| \ll_k (n+h) X^{3-k} e^{-\frac{n+h}{X}}$, and thus we find that\[\left|S_f(\psi,Y) - \sum_{h \leq Y} \sum_{n \geq 1} \lambda_f(n) \lambda_f(n+h) \frac{(n(n+h))^{\frac{k-1}{2}}}{X^{k-2}} e^{-\frac{n+h}{X}}\right| \ll_{k,\varepsilon} X^{1+\varepsilon},\] hence the difference between the left hand sides of the Main Theorem and Corollary 1 is $\ll_\varepsilon X^\varepsilon$. Secondly, let \[\phi(u) = \left(c_f(4 \pi u Y^2) - \frac{\Gamma(k) L(1,\mathrm{sym}^2 f)}{2 \zeta(2)} \right) \frac{1}{u^2}.\]  From the definition of $c_f(\alpha)$ as a sum one sees that $|c'_f(\alpha)| \ll_f \alpha^{-1}$ as $\alpha \rightarrow 0$, so that $\left| \phi' \left(\frac{1}{4\pi X}\right)\right| \ll_f X^3$.  The error term can be treated similarly.   Hence, the difference between the right hand side of the Main Theorem and the right hand side of Corollary 1 is $\ll_f X^{-2}$ with this choice of $\psi$.
\end{proof}
Let \[\Gamma(s,x) = \int_x^\infty e^{-t}t^s\,\frac{dt}{t}\] denote the incomplete gamma function.
\begin{cor} Let
\[\Sigma_f(X,Y) := \sum_{h \leq Y} \sum_{n \geq 1} \lambda_f(n) \lambda_f(n+h) \left(1-\frac{h}{n+h}\right)^{\frac{k-1}{2}} \frac{\Gamma(k-1,(k-1)\frac{n+h}{X})}{\Gamma(k-1)} \]
Then \[ \Sigma_f(X,Y) = -\frac{L(1,\mathrm{sym}^2 f)}{2 \zeta(2)}X + \frac{Y^2}{\Gamma(k-1)} \int_{(k-1)\frac{Y^2}{ X}}^\infty \frac{c_f(u)}{u^2}\,du + O_k\left(X^\frac{1}{2} Y^{\frac{1}{3}(1+\theta)}\right),\] where the integral appearing here either cancels the main term if $\frac{Y^2}{X} $ approaches $0$, or decays rapidly as a function of $\frac{Y^2}{X}$ if this parameter grows without bound.
\end{cor}
\begin{proof}
Let $\mathds{1}_{>x}$ denote the indicator function of the open set $(x,\infty) \subset \mathbb{R}$.  In similar fashion to the previous proof, let $\psi(y)$ be a smooth approximation to \[\frac{1}{4\pi \Gamma(k-1)} \mathds{1}_{>\frac{k-1}{4 \pi X}}(y),\] and compute the answer.
\end{proof}

The spectral theorem as used in this paper holds for $L^2$ functions which are also $C^\infty$ and bounded, however, it should also hold for a much wider class of functions, and thus our Main Theorem should in fact hold for a much wider class of functions.  In practice however, one may always obtain a corollary for a specific choice of $\psi$ by elementary arguments similar to the proof of Corollary 1, so we do not pursue the problem of expanding the class of functions for which the Main Theorem holds.

\bibliographystyle{amsplain}

redpajama_set_name: RedPajamaArXiv


text:
NERDS AROUND TOWN
Nerds Around Town: Lady Thor, Madden and farming
Cory De La Guardia
ART BY JESUS RODRIGUEZ
Born with a comic book in one hand and a remote control in the other, Cory DLG is the talent of Conroe's very own Nerd Thug Radio and Sports. Check out the podcast replay of the FM radio show at www.nerdthugradio.com!
This week continues on and we're already at Wednesday, let's get them!
GOOD DEED OF THE DAY
Make sure to get over to The West Bar and Grill on August 4th. They are supporting 13yr old Kobee Cohen, who having previously defeated Rhabdomayosarcoma recently suffered a return of the disease and now it is in his brain. Looking towards hospice care, the family needs financial aid and support during this difficult time. There will be a raffle and donations for the raffle are still being accepted, everyone is welcome to attend and I'm encouraging them to do so!
Farming Simulator 2019 Platinum edition has been announced for the fall. You can now control over 35 different vehicles and there are now even more ways for you to tend to your crops. This gaming franchise has been a slow rolling hit for a really long time and over the last four years has really picked up steam. I know right now there are some of you reading this and thinking there's a punchline coming but there isn't. The whole idea of a farming simulator seems like the antithesis of everything that is gaming, but in actuality it's proof that gaming is whatever you want it to be in the comfort of your home in front of your PC, X Box, PS, Switch or cell phone. It's just weird to think that this many people get excited about the idea of being a farmer, but then again, for some of these people this may be their only way out of large crowded cities full of things and people they don't like. This is their escape, farming. I guess.
THORS TALKS
So one of the big announcements coming out of Comic Con was the announcement of Thor: Love and Thunder which has Natalie Portman returning and becoming Thor. This is all great news and very exciting stuff, Lady Thor is one of the best new ideas Marvel had possessed and executed in a while with the writer Jason Aaron nailing that whole storyline. Another interesting bit about the movie is it sounds like Valkyrie will be in search of a Queen to rule along side her in her newly appointed role of King of Asgard, and Jaime Alexander, known in the Marvel universe as Lady Sif has thrown her hat in the ring. That would be a great way to bring all of this full circle if now Sif and Valkyrie were ruling Asgard when in the beginning they both served other rulers of Asgardian lore. This next phase of Marvel is so exciting for all the unknowns circling around everything.
MADDEN TALK
Madden is regularly a discussion among the NFL during this time of year as the game is slated to be released soon and lots of people have questions about ratings and things of that nature and the players react accordingly. It brings up an interesting issue, so teams like the Cardinals are coming into this year with very low ratings both offensively and defensively. So overall none of these players can feel all that great about their chances this season, I mean we all know Madden has a level of accuracy to it. Athletes are driven in part by an internal ego that says they are better than everyone else and everyone else doubts them and hates them. It's their job to turn that into fuel for the fire, but how do you do that if you can't get the engine revved up at all? How do you convince yourself you're better than everyone else when they literally have every component of your game broken down and rated and they say, "yeah you're not that good." That's an impressive level of delusion.
NOT THAT YOU ASKED
So one of the things I do is host a radio show. I've had several business ventures over the years that involve radio airtime, for example I once owned an internet radio station and was selling airtime to shows. Several people accused me of "payola" because they have no idea what that word means. For those who don't know, many radio stations sell their air time, just like television stations sell their airtime, just like advertisers buy billboards at sporting venues, there is money in entertainment and sports, ok? The point of this rant though is that people have no idea what is illegal or legal and it always blows my mind. People getting up in arms about a bathroom bill, don't realize we don't have a bathroom law now. If a man uses a women's restroom right now it isn't a crime it's just inappropriate. People who are calling the immigrants law breakers don't know how Asylum works and that for many of these people it's legal for them to come here and request Asylum, it's an internationally recognized right. What blows my mind is people are so quick to be Monday Morning Prosecutor but have no basic understanding of what is or isn't even legal or illegal.
Feel free to check out my brand new comic book Another Day at the Office or buy a shirt from Side Hustle Ts where some proceeds help people struggling with cancer or listen to Nerd Thug Radio. Thoughts, complaints, events and comments can be sent to corydlg@gmail.com.
madden thor nerds around town

redpajama_set_name: RedPajamaCommonCrawl


The Perth Wildcats and New Zealand Breakers continue to be the pace setters in the 2017-18 NBL season. The Wildcats recorded two wins last weekend, while the Breakers had one win and a narrow loss to Adelaide on Sunday night.
Melbourne United continue to produce erratic basketball, getting blown out my Cairns on Saturday night, before a dominant win over the Sydney Kings on Monday night.
The Kings are clearly the league's poorest performing team, losing both of their matches last weekend, including to the lowly Brisbane Bullets.
The Adelaide 36ers and Cairns Taipans continue to put pressure on the top two teams, while the Illawarra Hawks are ravaged by injury and looking to grind out a win any way they can.
The Cairns Taipans will be seeking revenge on their intra-state rival after the Bullets have beaten the Taipans twice already this season in Brisbane.
Playing at the Snake Pit in Cairns will be a completely different proposition for the Bullets, as the Taipans have shown they can beat anyone on their home court this season.
They already boast wins against Melbourne United and the Perth Wildcats at home, as well as a narrow, buzzer-beating loss to the New Zealand Breakers.
The Bullets are in the midst of a long stretch of road games, where they will play eight of their next 10 fixtures on the road, including last weekend's two road games in New Zealand and Sydney.
With a trip to Perth also on the fixture list for this weekend, the Bullets will do well to hang in this contest. Cairns' defence and three-point shooting should be too much for the Bullets to overcome.
The current handicap on offer for Cairns is very generous and we will be taking it in spades.
The Illawarra Hawks have been beset by injuries, particularly in their forwards and centres. Demitrius Conger and A.J. Ogilvy have both missed time over the past three weeks, shifting the bulk of the scoring load onto guard Rotnei Clarke.
The Hawks were gallant in their most recent performance against the Perth Wildcats, but import Bryce Cotton put on a show, scoring 38 points to dominate the contest.
It has been the same story for Melbourne United all season. A win has been followed by an awful loss, and then another win.
Consistency is what Melbourne United fans are seeking and the team has the perfect opportunity to string together consecutive wins against an undermanned Hawks team.
Chris Goulding was coming off a dreadful 0/6 performance against Cairns before turning it on against the Sydney Kings. The challenge is for United to perform against a higher-calibre team.
We believe that they will, given the Hawks' lack of depth in this match.
The Sydney Kings have another chance to salvage their season on Sunday afternoon when they host the Cairns Taipans.
The Kings are priced at a ridiculously low $1.37 for this clash against a Cairns team that is more than capable of knocking off the bottom placed Kings.
Sydney has shown an inability to defend with any sort of intensity this season, allowing opposition teams to dictate the pace of play at the offensive end.
New imports Jerome Randle and Jeremy Tyler have added offense to this team, but have done little defensively.
This is a value play based on the fact that Cairns did beat Sydney just a fortnight ago and narrowly lost to a red-hot shooting Kings at this same arena three weeks ago.
Cairns will keep it close so a play on the handicap is there for conservative bettors, but we believe Cairns have what it takes to make it two from two this weekend.


text:
Anthony trial 'Summer of the shark' was better
Editorials Letters to the Editor Editorial Cartoons Opinion Columns
Tuesday, July 5, 2011 6:01 AM
Monday morning offered a clear reminder of the truth about cable news. MSNBC
and Fox News — ostensibly polar opposites — were airing identical coverage of the Florida trial of Casey Anthony.
The two networks are best known for their evening shows, in which talking heads yell at their viewers — MSNBC from the left, Fox from the right. Neither, however, lets ideology get in the way of a spectacle.
The Anthony trial is this year's version of "missing white woman syndrome."
That is the name given to the tendency of broadcast media to give more coverage to anything involving a white, middle-class woman — particularly an attractive one — than to a man or a non-white woman in a similar situation. Think Natalie Holloway, the young woman who disappeared in Aruba in 2005, or Chandra Levy, the woman whose 2001 murder speculation tied to a congressman.
These cases do not really matter to anyone not personally involved. They are simply sensational. The Anthony trial in particular has zero educational value, but seems to be one thing about which Fox and MSNBC can agree.
Ten years ago the networks got into a similar frenzy over shark attacks. There was no surge in shark attacks that year, just a couple of dramatic episodes during a slow news period. In one, an 8-year-old boy had his arm bitten off. His uncle then dragged the shark out of the water and retrieved the severed limb.
Other years had more shark attacks but also had other events that drew attention, such as presidential elections. So the hot months of 2001 became, as a Time magazine cover story called it, the "summer of the shark."
That ended abruptly when real news intervened on Sept. 11.
Even by cable standards, the Anthony trial is tawdry. Charged with murdering her 2-year-old daughter, Anthony is at the center of a bizarre media circus and an even stranger trial.
That the child is dead and Anthony's family is weird are about the only certainties. Her lawyers say the death was an accident, but offer an unconvincing theory — along with the assertion the defendant was sexually abused by her father and brother.
The brother broke into tears while being questioned about why he had not visited his sister in the hospital when she gave birth. Anthony's mother also testified, which all the commentator's agreed was perjury.
The prosecution seemingly cannot prove the child was murdered. The most damning evidence against Anthony apparently consists of photos of her partying taken when her daughter was missing, before the body was found. That and a tattoo.
It is a freak show and little about it is fair, balanced or forward leaning. But absent sharks, perhaps it has to suffice.
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