1 Introduction
The asymptotic evaluation of moments of
$L$
-functions is not only an important tool for solving problems in number theory and arithmetic geometry, but also a subject of independent interest. Various conjectures predict the shape of the main terms for all moments of
$L$
-functions within a family of certain type of symmetry (see, for example, [Reference Conrey, Farmer, Keating, Rubinstein and Snaith5, Reference Diaconu, Goldfeld and Hoffstein7]). Even though exact results for moments are hitherto known only for small values, this is already sufficient for many applications (see, for example, [Reference Iwaniec and Sarnak10]). In this regard, the quality of the asymptotic error estimates plays a crucial role.
In general, there are three main techniques used for evaluating the moments: we can apply the approximate functional equation, the Rankin–Selberg method or the method of analytic continuation, each of which has certain advantages. The method of the approximate functional equation, which allows the bypassing of convergence problems, is the most common approach. The method of analytic continuation and the Rankin–Selberg method, however, reveal the structure of the mean values and yield exact formulas for the moments. Consequently, the theory of special functions can be used in order to prove sharp error estimates.
The problem we consider here requires the combination of all three methods. We study the asymptotic behaviour of the first moment of the product of the Hecke
$L$
-function
$L(f,1/2)$
and the symmetric square
$L$
-function
$L(\operatorname{sym}^{2}f,1/2)$
in the weight aspect on average, where
$f$
runs over the space
$H_{4k}$
of primitive forms of weight
$4k$
,
$k\in \mathbf{N}$
.
The most interesting phenomenon of this moment is the presence of the non-diagonal main term, which is smaller in size and depends on the special value of the double Dirichlet series
$$\begin{eqnarray}\displaystyle & \displaystyle L_{g}^{-}(s)=\frac{\unicode[STIX]{x1D6E4}(3/4)}{4\sqrt{\unicode[STIX]{x1D70B}}}\mathop{\sum }_{n<0}\frac{\mathscr{L}_{4n}(1/2)}{|n|^{s+1/2}}, & \displaystyle\end{eqnarray}$$
$$\begin{eqnarray}\displaystyle & \displaystyle \mathscr{L}_{n}(z)=\frac{\unicode[STIX]{x1D701}(2z)}{\unicode[STIX]{x1D701}(z)}\mathop{\sum }_{q=1}^{\infty }\frac{1}{q^{z}}\biggl(\mathop{\sum }_{1\leqslant t\leqslant 2q;t^{2}\equiv n~(\text{mod}~4q)}1\biggr). & \displaystyle\end{eqnarray}$$
Isolating this non-diagonal main term and applying the Liouville–Green method for error estimates, we prove an asymptotic formula with an arbitrary power-saving error term.
Before stating the main theorem we introduce some notation. Let
$\langle f,f\rangle _{1}$
be the Petersson inner product on the space of level-1 holomorphic modular forms. We denote the standard harmonic weight by
$$\begin{eqnarray}\unicode[STIX]{x1D714}(f):=\frac{\unicode[STIX]{x1D6E4}(2k-1)}{(4\unicode[STIX]{x1D70B})^{2k-1}\langle f,f\rangle _{1}}.\end{eqnarray}$$
Then we have the following asymptotic formula.
Theorem 1.1. Let
$h\in C_{0}^{\infty }(\mathbf{R}^{+})$
be a non-negative, compactly supported function on the interval
$[\unicode[STIX]{x1D703}_{1},\unicode[STIX]{x1D703}_{2}]$
,
$\unicode[STIX]{x1D703}_{2}>\unicode[STIX]{x1D703}_{1}>0$
, and
$$\begin{eqnarray}\Vert h^{(n)}\Vert _{1}\ll 1\quad \text{for all}~n\geqslant 0.\end{eqnarray}$$
Then, for any fixed
$A>0$
, we have
$$\begin{eqnarray}\displaystyle & & \displaystyle \mathop{\sum }_{k}h\biggl(\frac{4k}{K}\biggr)\mathop{\sum }_{f\in H_{4k}}\unicode[STIX]{x1D714}(f)L(f,1/2)L(\operatorname{sym}^{2}f,1/2)\nonumber\\ \displaystyle & & \displaystyle \!\quad =\frac{HK}{4}\unicode[STIX]{x1D701}(3/2)\biggl(\!2\log K-3\log 2\unicode[STIX]{x1D70B}-2\log 2+\frac{\unicode[STIX]{x1D70B}}{2}+3\unicode[STIX]{x1D6FE}+\frac{2\unicode[STIX]{x1D701}^{\prime }(3/2)}{\unicode[STIX]{x1D701}(3/2)}+\frac{2H_{1}}{H}\biggr)\nonumber\\ \displaystyle & & \displaystyle \!\qquad +\,P_{h}(1/K)+L_{g}^{-}(1/4)\sqrt{K}Q_{h}(1/K)+O(K^{-A}),\end{eqnarray}$$
where
$\unicode[STIX]{x1D701}(s)$
is the Riemann zeta function and
$\unicode[STIX]{x1D6FE}$
denotes the Euler constant. Furthermore,
$$\begin{eqnarray}H:=\int _{0}^{\infty }h(y)\,dy,\qquad H_{1}:=\int _{0}^{\infty }h(y)\log y\,dy\end{eqnarray}$$
and
$P_{h}(x)$
,
$Q_{h}(x)$
are polynomials in
$x$
of degree
$A-1$
and
$A$
, respectively, with coefficients depending on the function
$h$
.
This is an example of mixed moments that were previously investigated, for example, in [Reference Lau, Royer and Wu13, Reference Munshi and Sengupta16, Reference Ng17, Reference Royer and Wu21].
The proof of Theorem 1.1 consists of several steps. We start by combining the exact formula for the twisted first moment of symmetric square
$L$
-functions and the approximate functional equation for the Hecke
$L$
-function. Consequently, the average
$$\begin{eqnarray}\mathop{\sum }_{f\in H_{4k}}\unicode[STIX]{x1D714}(f)L(f,1/2)L(\operatorname{sym}^{2}f,1/2)\end{eqnarray}$$
splits into a diagonal main term, a non-diagonal main term and a smaller contribution expressed in terms of the special functions
$$\begin{eqnarray}\displaystyle \displaystyle \unicode[STIX]{x1D6F9}_{k}(x) & := & \displaystyle x^{k}\frac{\unicode[STIX]{x1D6E4}(k-1/4)\unicode[STIX]{x1D6E4}(k+1/4)}{\unicode[STIX]{x1D6E4}(2k)}\,_{2}F_{1}\biggl(k-\frac{1}{4},k+\frac{1}{4};2k;x\biggr),\end{eqnarray}$$
$$\begin{eqnarray}\displaystyle \displaystyle \unicode[STIX]{x1D6F7}_{k}(x) & := & \displaystyle \frac{\unicode[STIX]{x1D6E4}(k-1/4)\unicode[STIX]{x1D6E4}(3/4-k)}{\unicode[STIX]{x1D6E4}(1/2)}\,_{2}F_{1}\biggl(k-\frac{1}{4},\frac{3}{4}-k;1/2;x\biggr),\end{eqnarray}$$
where
$\unicode[STIX]{x1D6E4}(s)$
is the Gamma function and
$_{2}F_{1}(a,b;c;x)$
is the Gauss hypergeometric function.
The diagonal main term is evaluated in Corollary 4.3. The Rankin–Selberg method serves to isolate the non-diagonal main term, as shown in Corollary 5.2. Lemma 6.1 provides an estimate for the error term of the first type. Using the Liouville–Green method, we approximate the error term of the second type by the series in
$Y_{0}$
and
$J_{0}$
Bessel functions, average the result over
$k$
and, as a consequence, prove estimate (6.5). To sum up, we obtain the full asymptotic expansion with an arbitrary power-saving error term.
Note that the second main term phenomenon does not appear in Khan’s [Reference Khan11] evaluation of the moment
$$\begin{eqnarray}\mathop{\sum }_{k}h\biggl(\frac{2k}{K}\biggr)\mathop{\sum }_{f\in H_{2k}}\unicode[STIX]{x1D714}(f)L^{2}(\operatorname{sym}^{2}f,1/2).\end{eqnarray}$$
Namely, Khan computed the diagonal main term of size
$K\log ^{3}K$
and estimated the remaining terms as
$O(K^{\unicode[STIX]{x1D716}})$
for any
$\unicode[STIX]{x1D716}>0$
. This fact can be easily explained by the exact formula (see [Reference Balkanova and Frolenkov2, Theorem 2.1]) for the twisted first moment of symmetric square
$L$
-functions associated to
$H_{2k}$
. The off-diagonal term in this formula is
$$\begin{eqnarray}(-1)^{k}\frac{\sqrt{2\unicode[STIX]{x1D70B}}}{2l^{1/2}}\frac{\unicode[STIX]{x1D6E4}(k-1/4)}{\unicode[STIX]{x1D6E4}(k+1/4)}\mathscr{L}_{-4l^{2}}(1/2).\end{eqnarray}$$
To obtain asymptotics for (1.8), it is required to average (1.9) over all values of
$k$
. Consequently, the contribution of this term is rather small due to the oscillating multiple
$(-1)^{k}$
. In the case of the mixed moment
$$\begin{eqnarray}\mathop{\sum }_{k}h\biggl(\frac{4k}{K}\biggr)\mathop{\sum }_{f\in H_{4k}}\unicode[STIX]{x1D714}(f)L(f,1/2)L(\operatorname{sym}^{2}f,1/2),\end{eqnarray}$$
the summation is taken only over even values of
$k$
because
$L(f,1/2)$
is identically zero for
$f\in H_{2k}$
with odd
$k$
. As a result, the oscillating multiple
$(-1)^{k}$
disappears and the summand corresponding to (1.9) becomes the second main term.
Note that the absolute value estimate for the non-diagonal main term in (1.10) is
$K^{3/4+\unicode[STIX]{x1D703}}$
, where
$\unicode[STIX]{x1D703}$
denotes the subconvexity exponent for the Dirichlet series (1.2). Using the best subconvexity result due to Conrey and Iwaniec [Reference Conrey and Iwaniec6] gives a power-saving bound
$K^{11/12+\unicode[STIX]{x1D716}}$
. The Rankin–Selberg method allows one to remove the dependence on
$\unicode[STIX]{x1D703}$
and improve the estimate to
$K^{3/4}$
. We finally observe that there are some cancellations between the diagonal and the non-diagonal main terms. This can be shown by direct computations, but the most convenient way is to choose a suitable form of the approximate functional equation for
$L(f,s)$
. Accordingly, we obtain the second main term of size
$K^{1/2}$
.
In the level aspect similar results for the Hecke congruence subgroup of prime level
$q$
were obtained by Munshi and Sengupta [Reference Munshi and Sengupta16] with the error estimate
$O(q^{-1/8})$
. The authors of [Reference Munshi and Sengupta16] isolate only the diagonal main term. Our expectation is that there is a second main term of size
$q^{-1/2}$
.
We remark that the additional average over
$k$
in (1.5) is only required to estimate the error term of the second type, which involves the highly oscillatory special function
$\unicode[STIX]{x1D6F7}_{k}(x)$
. It might be possible to smooth out the oscillations of
$\unicode[STIX]{x1D6F7}_{k}(x)$
by using instead the average over
$l$
in (6.2), and, hence, prove an asymptotic formula for
$$\begin{eqnarray}\mathop{\sum }_{f\in H_{4k}}\unicode[STIX]{x1D714}(f)L(f,1/2)L(\operatorname{sym}^{2}f,1/2).\end{eqnarray}$$
A consequence of this result is the simultaneous non-vanishing of the corresponding
$L$
-functions.
Finally, the methods of the present work shed some light on the structure of the second moment of symmetric square
$L$
-functions. In the level aspect the upper bound
$$\begin{eqnarray}\mathop{\sum }_{f\in H_{2k}(N)}L^{2}(\operatorname{sym}^{2}f,1/2)\ll N^{1+\unicode[STIX]{x1D716}}\quad \text{for any}~\unicode[STIX]{x1D716}>0~\text{and fixed}~k,\end{eqnarray}$$
was proved by Iwaniec and Michel [Reference Iwaniec and Michel9] in 2001. In the weight aspect this is a challenging unsolved problem in the theory of
$L$
-functions. It is believed that finding an asymptotic formula or even an upper bound for
$$\begin{eqnarray}\mathop{\sum }_{f\in H_{2k}}\unicode[STIX]{x1D714}(f)L^{2}(\operatorname{sym}^{2}f,1/2),\quad \text{as}~k\rightarrow \infty ,\end{eqnarray}$$
is out of reach by standard tools. Our results suggest the following approach. Combining the exact formula (3.15) and the approximate functional equation for
$L(\operatorname{sym}^{2}f,1/2)$
, one can isolate the diagonal and the off-diagonal main terms in (1.13). The main difference with (1.11) is the presence of the off-off-diagonal main terms in sums (6.1) and (6.2). Technically, this can be explained as follows. The approximate functional equation for symmetric square
$L$
-functions is longer and, therefore, the required range includes a transition point at which the special functions
$\unicode[STIX]{x1D6F7}_{k}(x)$
and
$\unicode[STIX]{x1D6F9}_{k}(x)$
change their behaviour. This produces the third off-off-diagonal main term and creates additional technical difficulties in estimating error terms.
2 Double Dirichlet series
The goal of this subsection is to obtain functional equations for double Dirichlet series of type (1.1). To this end, we use the results of Müller [Reference Müller15].
2.1 Some transformation properties
As the first step, we prove certain transformation properties. The main references for this subsection are [Reference Kohnen12, Reference Pitale20].
For a Hecke congruence subgroup
$\unicode[STIX]{x1D6E4}_{0}(4)$
of level 4, we denote by
$\mathscr{F}(\unicode[STIX]{x1D6E4}_{0}(4),\unicode[STIX]{x1D712},1/2,\unicode[STIX]{x1D706})$
the space of all non-holomorphic automorphic forms of weight
$1/2$
, multiplier system
$\unicode[STIX]{x1D712}$
and Laplace eigenvalue
$\unicode[STIX]{x1D706}=1/4-\unicode[STIX]{x1D70C}^{2}$
,
$\Re \unicode[STIX]{x1D70C}\geqslant 0$
. Any
$f\in \mathscr{F}(\unicode[STIX]{x1D6E4}_{0}(4),\unicode[STIX]{x1D712},1/2,\unicode[STIX]{x1D706})$
has Fourier–Whittaker expansion of the form
$$\begin{eqnarray}f(z)=A_{0}(y)+\mathop{\sum }_{n\neq 0}a_{n}W_{\operatorname{sgn}(n)/4,\unicode[STIX]{x1D70C}}(4\unicode[STIX]{x1D70B}|n|y)\exp (2\unicode[STIX]{x1D70B}\text{i}nx),\end{eqnarray}$$
where
$z=x+\text{i}y\in \mathbf{H}$
and
$W_{\unicode[STIX]{x1D708},\unicode[STIX]{x1D707}}(y)$
is the classical Whittaker function (see, for example, [Reference Müller15, §3]) normalized such that
$W_{\unicode[STIX]{x1D708},\unicode[STIX]{x1D707}}(y)\sim \text{e}^{-y/2}y^{\unicode[STIX]{x1D708}}$
as
$y\rightarrow \infty$
. Assume that
$$\begin{eqnarray}a_{n}=0\quad \text{for}~n\equiv 2,3~(\text{mod}~4).\end{eqnarray}$$
Furthermore, we define the operators
$$\begin{eqnarray}\displaystyle & \displaystyle (f|U)(z)=\frac{1}{\sqrt{2}}\biggl(f\biggl(\frac{z}{4}\biggr)+f\biggl(\frac{z+2}{4}\biggr)\biggr), & \displaystyle\end{eqnarray}$$
$$\begin{eqnarray}\displaystyle & \displaystyle (f|W)(z)=\biggl(\frac{-\text{i}z}{|z|}\biggr)^{-1/2}f\biggl(\frac{-1}{4z}\biggr). & \displaystyle\end{eqnarray}$$
These operators leave the space
$\mathscr{F}(\unicode[STIX]{x1D6E4}_{0}(4),\unicode[STIX]{x1D712},1/2,\unicode[STIX]{x1D706})$
stable and preserve property (2.2). If
$f$
has the expansion (2.1), then the function
$$\begin{eqnarray}g(z):=\frac{1}{\sqrt{2}}(f|U)(z)\end{eqnarray}$$
satisfies
$$\begin{eqnarray}g(z)=A_{0}(y/4)+\mathop{\sum }_{n\neq 0}a_{4n}W_{\operatorname{sgn}(n)/4,\unicode[STIX]{x1D70C}}(4\unicode[STIX]{x1D70B}|n|y)\exp (2\unicode[STIX]{x1D70B}\text{i}nx).\end{eqnarray}$$
By [Reference Pitale20, Proposition 4.1] we have
$$\begin{eqnarray}(f|U|W)(z)=f(z).\end{eqnarray}$$
Lemma 2.1. Let
$f$
and
$g$
be as above. For
$z=x+\text{i}y\in \mathbf{H}$
we have
$$\begin{eqnarray}\exp (\unicode[STIX]{x1D70B}\text{i}/4)(f|_{1/2}J)(z)=\sqrt{2}g(z/4),\qquad J=\left(\begin{array}{@{}cc@{}}0 & -1\\ 1 & 0\end{array}\right),\end{eqnarray}$$
where the slash operator is defined by the relation
$$\begin{eqnarray}(f|_{1/2}M)(z)=\biggl(\frac{cz+d}{|cz+d|}\biggr)^{-1/2}f(Mz),\qquad M=\left(\begin{array}{@{}cc@{}}a & b\\ c & d\end{array}\right).\end{eqnarray}$$
Proof. Applying equations (2.4), (2.7) and (2.9), we find
$$\begin{eqnarray}\displaystyle \hspace{24.0pt}\sqrt{2}g(z/4) & = & \displaystyle (f|U)(z/4)=\text{i}^{1/2}\biggl(\frac{z}{|z|}\biggr)^{-1/2}(f|U|W)(-1/z)\nonumber\\ \displaystyle & = & \displaystyle \text{i}^{1/2}\biggl(\frac{z}{|z|}\biggr)^{-1/2}f(-1/z)=\exp (\unicode[STIX]{x1D70B}\text{i}/4)(f|_{1/2}J)(z).\hspace{24.0pt}\square\nonumber\end{eqnarray}$$
2.2 Eisenstein series of weight 1/2 and level 4
The connection between quadratic Dirichlet
$L$
-functions and Fourier coefficients of half-integral weight forms was first discovered by Maaß in 1937 [Reference Maaß14]. Similar results were obtained by Shimura [Reference Shimura22], Shintani [Reference Shintani23], Cohen [Reference Cohen4], Goldfeld–Hoffstein [Reference Goldfeld and Hoffstein8], and others.
We study the more general Dirichlet series (see [Reference Bykovskii3] and [Reference Zagier24] for details)
$$\begin{eqnarray}\mathscr{L}_{n}(s)=\frac{\unicode[STIX]{x1D701}(2s)}{\unicode[STIX]{x1D701}(s)}\mathop{\sum }_{q=1}^{\infty }\frac{1}{q^{s}}\biggl(\mathop{\sum }_{1\leqslant t\leqslant 2q;t^{2}\equiv n~(\text{mod}~4q)}1\biggr).\end{eqnarray}$$
Only if
$n\equiv 0,1~(\text{mod}~4)$
does the function
$\mathscr{L}_{n}(s)$
, considered as a function of
$s$
, not vanish. We can think of
$\mathscr{L}_{n}(s)$
as a certain generalization of the Riemann zeta function and quadratic Dirichlet
$L$
-functions. Indeed,
$$\begin{eqnarray}\mathscr{L}_{0}(s)=\unicode[STIX]{x1D701}(2s-1).\end{eqnarray}$$
If
$n$
is a fundamental discriminant, then
$$\begin{eqnarray}\mathscr{L}_{n}(s)=\mathop{\sum }_{q=1}^{\infty }\frac{\unicode[STIX]{x1D712}_{n}(q)}{q^{s}},\end{eqnarray}$$
where
$\unicode[STIX]{x1D712}_{n}$
is a primitive quadratic character mod
$|n|$
. According to [Reference Balkanova and Frolenkov2, Lemma 4.2], for any
$\unicode[STIX]{x1D716}>0$
,
$$\begin{eqnarray}\mathscr{L}_{n}(1/2)\ll |n|^{\unicode[STIX]{x1D703}+\unicode[STIX]{x1D716}},\end{eqnarray}$$
where
$\unicode[STIX]{x1D703}$
is a subconvexity exponent for Dirichlet
$L$
-functions. The best known result
$\unicode[STIX]{x1D703}=1/6$
is due to Conrey and Iwaniec [Reference Conrey and Iwaniec6]. The Lindelöf hypothesis asserts that
$\unicode[STIX]{x1D703}=0$
. The completed
$L$
-function
$$\begin{eqnarray}\mathscr{L}_{n}^{\ast }(s)=(\unicode[STIX]{x1D70B}/|n|)^{-s/2}\unicode[STIX]{x1D6E4}(s/2+1/4-\operatorname{sgn}n/4)\mathscr{L}_{n}(s)\end{eqnarray}$$
satisfies the functional equation
$$\begin{eqnarray}\mathscr{L}_{n}^{\ast }(s)=\mathscr{L}_{n}^{\ast }(1-s).\end{eqnarray}$$
The function
$\mathscr{L}_{n}^{\ast }(s)$
appears in the Fourier–Whittaker expansion of the combination of the Maaß–Eisenstein series of weight
$1/2$
and level 4, namely
$$\begin{eqnarray}E_{(0;\infty )}^{(1/2)}(z;s)=\unicode[STIX]{x1D701}(4s-1)\biggl(E_{\infty }(z;s;1/2)+\frac{1+\text{i}}{4^{s}}E_{0}(z;s;1/2)\biggr),\end{eqnarray}$$
where for a cusp
$\unicode[STIX]{x1D6FC}$
the series
$E_{\unicode[STIX]{x1D6FC}}(z;s;k)$
is defined in [Reference Petridis, Raulf and Risager19, §3]. Note that, according to [Reference Petridis, Raulf and Risager19, §3A],
$$\begin{eqnarray}E_{(0;\infty )}^{(1/2)}(z;s)\in \mathscr{F}(\unicode[STIX]{x1D6E4}_{0}(4),\unicode[STIX]{x1D708},1/2,1/4-s^{2}),\end{eqnarray}$$
where
$\unicode[STIX]{x1D708}$
is the weight-
$1/2$
multiplier system related to the theta series (see [Reference Petridis, Raulf and Risager19, p. 1570]) such that
$$\begin{eqnarray}\unicode[STIX]{x1D708}(\unicode[STIX]{x1D6FE})=\biggl(\frac{c}{d}\biggr)\unicode[STIX]{x1D716}_{d}^{-1}\quad \text{for}~\unicode[STIX]{x1D6FE}=\left(\begin{array}{@{}cc@{}}a & b\\ c & d\end{array}\right)\in \unicode[STIX]{x1D6E4}_{0}(4),\end{eqnarray}$$
with
$(c/d)$
being the Jacobi symbol and
$\unicode[STIX]{x1D716}_{d}=(\unicode[STIX]{x1D712}_{-4}(d))^{1/2}$
for the quadratic character
$\unicode[STIX]{x1D712}_{-4}(d)=(-1)^{(d-1)/2}$
.
Lemma 2.2. We have
$$\begin{eqnarray}\displaystyle E_{(0;\infty )}^{(1/2)}(z;s) & = & \displaystyle \unicode[STIX]{x1D701}(4s-1)y^{s}+\frac{\sqrt{\unicode[STIX]{x1D70B}}\unicode[STIX]{x1D6E4}(2s-1)\unicode[STIX]{x1D701}(4s-2)}{4^{2s-1}\unicode[STIX]{x1D6E4}(2s-1/2)}y^{1-s}+\frac{\unicode[STIX]{x1D70B}^{2s-3/4}}{4^{s}\unicode[STIX]{x1D6E4}(2s-1/2)}\nonumber\\ \displaystyle & & \displaystyle \times \,\mathop{\sum }_{n\neq 0}\frac{\mathscr{L}_{n}^{\ast }(2s-1/2)}{|n|^{3/4}}W_{\operatorname{sgn}n/4,s-1/2}(4\unicode[STIX]{x1D70B}|n|y)\exp (2\unicode[STIX]{x1D70B}\text{i}nx).\nonumber\end{eqnarray}$$
Proof. The Fourier–Whittaker expansion for
$E_{\infty }(z;s;1/2)$
is given in [Reference Petridis, Raulf and Risager19, §3A]. Computations for
$E_{0}(z;s;1/2)$
are similar.◻
Computing the limit as
$s\rightarrow 1/2$
, we find
$$\begin{eqnarray}\displaystyle E_{(0;\infty )}^{(1/2)}(z;1/2) & = & \displaystyle \frac{1}{2}y^{1/2}\log y+(\unicode[STIX]{x1D6FE}-\log 4\unicode[STIX]{x1D70B})y^{1/2}+\frac{1}{2\sqrt{\unicode[STIX]{x1D70B}}}\nonumber\\ \displaystyle & & \displaystyle \times \,\mathop{\sum }_{n\neq 0}\frac{\mathscr{L}_{n}(1/2)}{|n|^{1/2}}\unicode[STIX]{x1D6E4}\biggl(\frac{1}{2}-\frac{\operatorname{sgn}n}{4}\biggr)\nonumber\\ \displaystyle & & \displaystyle \times \,W_{\operatorname{sgn}n/4,0}(4\unicode[STIX]{x1D70B}|n|y)\exp (2\unicode[STIX]{x1D70B}\text{i}nx).\end{eqnarray}$$
Now we can apply the results of §2.1. Note that condition (2.2) is satisfied since
$\mathscr{L}_{n}(1/2)$
vanishes for
$n\equiv 2,3~(\text{mod}~4)$
. For
$$\begin{eqnarray}f(z):=E_{(0;\infty )}^{(1/2)}(z;1/2)\in \mathscr{F}(\unicode[STIX]{x1D6E4}_{0}(4),\unicode[STIX]{x1D708},1/2,0),\end{eqnarray}$$
define
$g(z)$
by equation (2.5). Then
$g$
has the expansion
$$\begin{eqnarray}\displaystyle g(z) & = & \displaystyle \frac{1}{4}y^{1/2}\log y+\frac{1}{2}(\unicode[STIX]{x1D6FE}-\log 8\unicode[STIX]{x1D70B})y^{1/2}+\frac{1}{4\sqrt{\unicode[STIX]{x1D70B}}}\nonumber\\ \displaystyle & & \displaystyle \times \,\mathop{\sum }_{n\neq 0}\frac{\mathscr{L}_{4n}(1/2)}{|n|^{1/2}}\unicode[STIX]{x1D6E4}\biggl(\frac{1}{2}-\frac{\operatorname{sgn}n}{4}\biggr)\nonumber\\ \displaystyle & & \displaystyle \times \,W_{\operatorname{sgn}n/4,0}(4\unicode[STIX]{x1D70B}|n|y)\exp (2\unicode[STIX]{x1D70B}\text{i}nx).\end{eqnarray}$$
2.3 Functional equations for the double Dirichlet series
We now consider the associated Dirichlet series:
$$\begin{eqnarray}\displaystyle & \displaystyle L_{f}^{+}(s)=\frac{\unicode[STIX]{x1D6E4}(1/4)}{2\sqrt{\unicode[STIX]{x1D70B}}}\mathop{\sum }_{n>0}\frac{\mathscr{L}_{n}(1/2)}{n^{s+1/2}},\qquad L_{f}^{-}(s)=\frac{\unicode[STIX]{x1D6E4}(3/4)}{2\sqrt{\unicode[STIX]{x1D70B}}}\mathop{\sum }_{n<0}\frac{\mathscr{L}_{n}(1/2)}{|n|^{s+1/2}},\qquad \quad & \displaystyle\end{eqnarray}$$
$$\begin{eqnarray}\displaystyle & \displaystyle L_{g}^{+}(s)=\frac{\unicode[STIX]{x1D6E4}(1/4)}{4\sqrt{\unicode[STIX]{x1D70B}}}\mathop{\sum }_{n>0}\frac{\mathscr{L}_{4n}(1/2)}{n^{s+1/2}},\qquad L_{g}^{-}(s)=\frac{\unicode[STIX]{x1D6E4}(3/4)}{4\sqrt{\unicode[STIX]{x1D70B}}}\mathop{\sum }_{n<0}\frac{\mathscr{L}_{4n}(1/2)}{|n|^{s+1/2}}.\qquad \quad & \displaystyle\end{eqnarray}$$
Theorem 2.3. The functions
$L_{f}^{\pm }(s)$
and
$L_{g}^{\pm }(s)$
have a meromorphic continuation to the whole complex plane and satisfy the functional equations
$$\begin{eqnarray}\displaystyle L_{g}^{+}(s) & = & \displaystyle \frac{-\unicode[STIX]{x1D70B}^{2s+2}}{\sqrt{2}\unicode[STIX]{x1D6E4}^{2}(1/2+s)\sin ^{2}\unicode[STIX]{x1D70B}s}\nonumber\\ \displaystyle & & \displaystyle \times \,\biggl(\frac{\sin \unicode[STIX]{x1D70B}(-s-1/4)}{\unicode[STIX]{x1D70B}}L_{f}^{+}(-s)-\frac{L_{f}^{-}(-s)}{\unicode[STIX]{x1D6E4}^{2}(3/4)}\biggr),\end{eqnarray}$$
$$\begin{eqnarray}\displaystyle \displaystyle L_{g}^{-}(s) & = & \displaystyle \frac{\unicode[STIX]{x1D70B}^{2s+2}}{\sqrt{2}\unicode[STIX]{x1D6E4}^{2}(1/2+s)\sin ^{2}\unicode[STIX]{x1D70B}s}\nonumber\\ \displaystyle & & \displaystyle \times \,\biggl(-\frac{\sin \unicode[STIX]{x1D70B}(-s+1/4)}{\unicode[STIX]{x1D70B}}L_{f}^{-}(-s)+\frac{L_{f}^{+}(-s)}{\unicode[STIX]{x1D6E4}^{2}(1/4)}\biggr).\end{eqnarray}$$
Furthermore,
$L_{f}^{\pm }(s)$
and
$L_{g}^{\pm }(s)$
are holomorphic in
$\mathbf{C}$
except for a double pole at
$s=1/2$
.
Proof. Note that, according to Lemma 2.1, the first condition of [Reference Müller15, Theorem 4.1] is satisfied by taking
$k=1/2$
,
$C=\sqrt{2}$
,
$\unicode[STIX]{x1D6FE}=1/4$
. Therefore, the statement of the theorem is a direct consequence of [Reference Müller15, Theorem 4.1] for
$\unicode[STIX]{x1D70C}=0$
with a small correction that the functions
$L_{f}^{\pm }(s)$
and
$L_{g}^{\pm }(s)$
have poles of order two at
$s=1/2$
.◻
3 Preliminary evaluation
Let
$\mathbf{H}$
be the Poincaré upper half-plane and denote by
$H_{2k}$
the normalized Hecke basis for the space of holomorphic cusp forms of even weight
$2k\geqslant 2$
with respect to the full modular group. If the function
$f\in H_{2k}$
has the Fourier expansion
$$\begin{eqnarray}f(z)=\mathop{\sum }_{n\geqslant 1}\unicode[STIX]{x1D706}_{f}(n)n^{k-1/2}\exp (2\unicode[STIX]{x1D70B}\text{i}nz),\quad \unicode[STIX]{x1D706}_{f}(1)=1,\end{eqnarray}$$
the associated Hecke
$L$
-function is defined by
$$\begin{eqnarray}L(f,s)=\mathop{\sum }_{n\geqslant 1}\frac{\unicode[STIX]{x1D706}_{f}(n)}{n^{s}},\qquad \Re s>1.\end{eqnarray}$$
Let
$\unicode[STIX]{x1D6E4}(s)$
be the gamma function. Then the completed
$L$
-function
$$\begin{eqnarray}\unicode[STIX]{x1D6EC}(f,s)=\biggl(\frac{1}{2\unicode[STIX]{x1D70B}}\biggr)^{s}\unicode[STIX]{x1D6E4}\biggl(s+\frac{2k-1}{2}\biggr)L(f,s)\end{eqnarray}$$
satisfies the functional equation
$$\begin{eqnarray}\unicode[STIX]{x1D6EC}(f,s)=\unicode[STIX]{x1D716}_{f}\unicode[STIX]{x1D6EC}(f,1-s),\quad \unicode[STIX]{x1D716}_{f}=\text{i}^{2k},\end{eqnarray}$$
and can be analytically continued to the entire complex plane. Note that, by equation (3.4), we have
$L_{f}(1/2)=0$
for odd
$k$
.
For
$\Re s>1$
the symmetric square
$L$
-function is defined by
$$\begin{eqnarray}L(\operatorname{sym}^{2}f,s)=\unicode[STIX]{x1D701}(2s)\mathop{\sum }_{n=1}^{\infty }\frac{\unicode[STIX]{x1D706}_{f}(n^{2})}{n^{s}}.\end{eqnarray}$$
Let
$$\begin{eqnarray}L_{\infty }(s):=\unicode[STIX]{x1D70B}^{-3s/2}\unicode[STIX]{x1D6E4}\biggl(\frac{s+1}{2}\biggr)\unicode[STIX]{x1D6E4}\biggl(\frac{s-1}{2}+k\biggr)\unicode[STIX]{x1D6E4}\biggl(\frac{s}{2}+k\biggr).\end{eqnarray}$$
The completed
$L$
-function
$$\begin{eqnarray}\unicode[STIX]{x1D6EC}(\operatorname{sym}^{2}f,s):=L_{\infty }(s)L(\operatorname{sym}^{2}f,s)\end{eqnarray}$$
is entire and satisfies the functional equation
$$\begin{eqnarray}\unicode[STIX]{x1D6EC}(\operatorname{sym}^{2}f,s)=\unicode[STIX]{x1D6EC}(\operatorname{sym}^{2}f,1-s).\end{eqnarray}$$
In this section we combine the techniques of analytic continuation and the approximate functional equation in order to express the average
$$\begin{eqnarray}\mathop{\sum }_{f\in H_{4k}}\unicode[STIX]{x1D714}(f)L(f,1/2)L(\operatorname{sym}^{2}f,1/2)\end{eqnarray}$$
as a sum of three parts.
First, we use the exact formula for the twisted moment of symmetric square
$L$
-functions shifted from the critical point by a small parameter
$u\neq 0$
. The role of the shift
$u$
is to simplify the evaluation of the diagonal main term in §4.
Lemma 3.1. The formula
$$\begin{eqnarray}\mathop{\sum }_{f\in H_{4k}}\unicode[STIX]{x1D714}(f)\unicode[STIX]{x1D706}_{f}(l)L(\operatorname{sym}^{2}f,1/2+u)=M^{D}(u,l)\unicode[STIX]{x1D6FF}_{l=\Box }+M^{\mathit{ND}}(u,l)+ET(u,l)\end{eqnarray}$$
holds, where
$$\begin{eqnarray}\unicode[STIX]{x1D6FF}_{l=\Box }=\left\{\begin{array}{@{}ll@{}}1\quad & \text{if}~l~\text{is a full square,}\\ 0\quad & \text{otherwise},\end{array}\right.\end{eqnarray}$$
$$\begin{eqnarray}\displaystyle M^{D}(u,l^{2}) & = & \displaystyle \frac{\unicode[STIX]{x1D701}(1+2u)}{l^{1/2+u}}+\sqrt{2}(2\unicode[STIX]{x1D70B})^{3u}\cos \unicode[STIX]{x1D70B}(1/4+u/2)\nonumber\\ \displaystyle & & \displaystyle \times \,\frac{\unicode[STIX]{x1D701}(1-2u)}{l^{1/2-u}}\frac{\unicode[STIX]{x1D6E4}(2k-1/4-u/2)\unicode[STIX]{x1D6E4}(2k+1/4-u/2)\unicode[STIX]{x1D6E4}(1-2u)}{\unicode[STIX]{x1D6E4}(2k+1/4+u/2)\unicode[STIX]{x1D6E4}(2k-1/4+u/2)\unicode[STIX]{x1D6E4}(1-u)},\nonumber\\ \displaystyle & & \displaystyle\end{eqnarray}$$
$$\begin{eqnarray}M^{\mathit{ND}}(0,l)=\frac{\sqrt{2\unicode[STIX]{x1D70B}}}{2l^{1/4}}\frac{\unicode[STIX]{x1D6E4}(2k-1/4)}{\unicode[STIX]{x1D6E4}(2k+1/4)}\mathscr{L}_{-4l}(1/2),\end{eqnarray}$$
$$\begin{eqnarray}\displaystyle ET(0,l) & = & \displaystyle \frac{1}{l^{1/4}}\mathop{\sum }_{1\leqslant n<2\sqrt{l}}\mathscr{L}_{n^{2}-4l}(1/2)\unicode[STIX]{x1D6F7}_{2k}\biggl(\frac{n^{2}}{4l}\biggr)\nonumber\\ \displaystyle & & \displaystyle +\,\frac{1}{\sqrt{2}l^{1/2}}\mathop{\sum }_{n>2\sqrt{l}}\mathscr{L}_{n^{2}-4l}(1/2)\sqrt{n}\unicode[STIX]{x1D6F9}_{2k}\biggl(\frac{4l}{n^{2}}\biggr),\end{eqnarray}$$
where the functions
$\unicode[STIX]{x1D6F7}_{2k}(x)$
and
$\unicode[STIX]{x1D6F9}_{2k}(x)$
are defined by (1.7) and (1.6), respectively.
Proof. See [Reference Balkanova and Frolenkov2, equation (2.9), (5.6)]. ◻
Second, we obtain an approximate functional equation for the Hecke
$L$
-function at the central point. Let
$$\begin{eqnarray}g(s,u):=\frac{(s^{2}-(-1/4-u/2)^{2})(s^{2}-(-1/4+u/2)^{2})}{(1/4+u/2)^{2}(1/4-u/2)^{2}}\exp (s^{2}).\end{eqnarray}$$
Lemma 3.2. For
$f\in H_{4k}$
, we have
$$\begin{eqnarray}L(f,1/2)=2\mathop{\sum }_{l=1}^{\infty }\frac{\unicode[STIX]{x1D706}_{f}(l)}{\sqrt{l}}V_{k}(l,u),\end{eqnarray}$$
where
$$\begin{eqnarray}V_{k}(l,u)=\frac{1}{2\unicode[STIX]{x1D70B}\text{i}}\int _{(\unicode[STIX]{x1D70E})}g(s,u)\frac{\unicode[STIX]{x1D6E4}(2k+s)}{\unicode[STIX]{x1D6E4}(2k)}\frac{ds}{(2\unicode[STIX]{x1D70B}l)^{s}s},\quad \unicode[STIX]{x1D70E}>0.\end{eqnarray}$$
Proof. Consider
$$\begin{eqnarray}I(s)=\int _{(\unicode[STIX]{x1D70E})}g(s,u)\unicode[STIX]{x1D6EC}(f,1/2+s)\frac{ds}{s},\quad \unicode[STIX]{x1D70E}>0.\end{eqnarray}$$
Note that
$g(s,u)$
is an even function and
$g(0,u)=1$
. Therefore, moving the contour of integration from
$\unicode[STIX]{x1D70E}$
to
$-\unicode[STIX]{x1D70E}$
, applying the functional equation (3.4) and picking up a simple pole at
$s=0$
, we conclude that
$$\begin{eqnarray}2I(s)=\unicode[STIX]{x1D6EC}(f,1/2).\end{eqnarray}$$
The assertion follows. ◻
As a consequence of Lemmas 3.1 and 3.2 we get the following decomposition:
$$\begin{eqnarray}\displaystyle & & \displaystyle \mathop{\sum }_{f\in H_{4k}}\unicode[STIX]{x1D714}(f)L(f,1/2)L(\operatorname{sym}^{2}f,1/2+u)\nonumber\\ \displaystyle & & \displaystyle \quad =2\mathop{\sum }_{l=1}^{\infty }\frac{V_{k}(l,u)}{\sqrt{l}}M^{D}(u,l)\unicode[STIX]{x1D6FF}_{l=\Box }\nonumber\\ \displaystyle & & \displaystyle \qquad +\,2\mathop{\sum }_{l=1}^{\infty }\frac{V_{k}(l,u)}{\sqrt{l}}M^{\mathit{ND}}(u,l)+2\mathop{\sum }_{l=1}^{\infty }\frac{V_{k}(l,u)}{\sqrt{l}}ET(u,l).\end{eqnarray}$$
4 Diagonal main term
In this section we evaluate asymptotically the diagonal term in (3.15), namely,
$$\begin{eqnarray}M^{D}:=2\mathop{\sum }_{l=1}^{\infty }\frac{V_{k}(l,0)}{\sqrt{l}}M^{D}(0,l)\unicode[STIX]{x1D6FF}_{l=\Box }.\end{eqnarray}$$
Lemma 4.1. For any
$\unicode[STIX]{x1D716}>0$
and any real fixed number
$0<a<k$
, we have
$$\begin{eqnarray}\displaystyle & & \displaystyle 2\mathop{\sum }_{l=1}^{\infty }\frac{V_{k}(l,u)}{\sqrt{l}}M^{D}(u,l)\unicode[STIX]{x1D6FF}_{l=\Box }=2\biggl(\unicode[STIX]{x1D701}(1+2u)\unicode[STIX]{x1D701}(3/2+u)\nonumber\\ \displaystyle & & \displaystyle \quad +\,\sqrt{2}(2\unicode[STIX]{x1D70B})^{3u}\cos \unicode[STIX]{x1D70B}(1/4+u/2)\unicode[STIX]{x1D701}(1-2u)\unicode[STIX]{x1D701}(3/2-u)\nonumber\\ \displaystyle & & \displaystyle \quad \times \,\frac{\unicode[STIX]{x1D6E4}(2k-1/4-u/2)}{\unicode[STIX]{x1D6E4}(2k+1/4+u/2)}\frac{\unicode[STIX]{x1D6E4}(2k+1/4-u/2)\unicode[STIX]{x1D6E4}(1-2u)}{\unicode[STIX]{x1D6E4}(2k-1/4+u/2)\unicode[STIX]{x1D6E4}(1-u)}\biggr)+O(k^{-a+\unicode[STIX]{x1D716}}).\qquad\end{eqnarray}$$
Proof. Consider
$$\begin{eqnarray}2\mathop{\sum }_{l=1}^{\infty }\frac{V_{k}(l,u)}{\sqrt{l}}M^{D}(u,l)\unicode[STIX]{x1D6FF}_{l=\Box }=\frac{2}{2\unicode[STIX]{x1D70B}\text{i}}\int _{(\unicode[STIX]{x1D70E})}\frac{\unicode[STIX]{x1D6E4}(2k+s)}{\unicode[STIX]{x1D6E4}(2k)}\frac{g(s,u)}{(2\unicode[STIX]{x1D70B})^{s}}\mathop{\sum }_{l=1}^{\infty }\frac{M^{D}(u,l^{2})}{l^{2s+1}}\frac{ds}{s}.\end{eqnarray}$$
According to (3.10), this is equal to
$$\begin{eqnarray}\displaystyle & & \displaystyle \frac{2}{2\unicode[STIX]{x1D70B}\text{i}}\int _{(\unicode[STIX]{x1D70E})}\frac{\unicode[STIX]{x1D6E4}(2k+s)}{\unicode[STIX]{x1D6E4}(2k)}g(s,u)\biggl[\frac{\unicode[STIX]{x1D701}(1+2u)\unicode[STIX]{x1D701}(3/2+u+2s)}{2}\nonumber\\ \displaystyle & & \displaystyle \quad +\,\sqrt{2}(2\unicode[STIX]{x1D70B})^{3u}\cos \unicode[STIX]{x1D70B}(1/4+u/2)\unicode[STIX]{x1D701}(1-2u)\unicode[STIX]{x1D701}(3/2-u+2s)\nonumber\\ \displaystyle & & \displaystyle \quad \times \,\frac{\unicode[STIX]{x1D6E4}(2k-1/4-u/2)\unicode[STIX]{x1D6E4}(2k+1/4-u/2)\unicode[STIX]{x1D6E4}(1-2u)}{\unicode[STIX]{x1D6E4}(2k+1/4+u/2)\unicode[STIX]{x1D6E4}(2k-1/4+u/2)\unicode[STIX]{x1D6E4}(1-u)}\biggr]\frac{ds}{(2\unicode[STIX]{x1D70B})^{s}s}.\nonumber\end{eqnarray}$$
The integrand has poles at
$$\begin{eqnarray}s=0,\qquad s=-k-j,~j=0,1,\ldots .\end{eqnarray}$$
Note that the poles at
$s=-1/4\pm u/2$
are compensated by the zeros of
$g(s,u)$
. Crossing the pole at
$s=0$
, we can move the line of integration to any real number
$-a$
such that
$0<a<k$
. Consequently, the resulting integral is bounded by
$O(k^{-a+\unicode[STIX]{x1D716}})$
. The assertion follows by calculating the residue at
$s=0$
.◻
Computing the limit as
$u\rightarrow 0$
, we evaluate the main term at the central point.
Corollary 4.2. For any
$\unicode[STIX]{x1D716}>0$
and any real fixed number
$0<a<k$
, we have
$$\begin{eqnarray}\displaystyle M^{D} & = & \displaystyle \unicode[STIX]{x1D701}(3/2)\biggl(-3\log 2\unicode[STIX]{x1D70B}+\frac{\unicode[STIX]{x1D70B}}{2}+3\unicode[STIX]{x1D6FE}+2\frac{\unicode[STIX]{x1D701}^{\prime }(3/2)}{\unicode[STIX]{x1D701}(3/2)}\nonumber\\ \displaystyle & & \displaystyle +\,\unicode[STIX]{x1D713}(2k-1/4)+\unicode[STIX]{x1D713}(2k+1/4)\biggr)+O(k^{-a+\unicode[STIX]{x1D716}}),\end{eqnarray}$$
where
$\unicode[STIX]{x1D713}(x)$
is the logarithmic derivative of the gamma function.
Finally, the diagonal main term can be averaged over the weight
$k$
with a suitable test function.
Corollary 4.3. For any fixed
$A>0$
, we have
$$\begin{eqnarray}\displaystyle & & \displaystyle \mathop{\sum }_{k}h\biggl(\frac{4k}{K}\biggr)M^{D}\nonumber\\ \displaystyle & & \displaystyle \!\!\quad =\frac{HK}{4}\unicode[STIX]{x1D701}(3/2)\biggl(2\log K-3\log 2\unicode[STIX]{x1D70B}-2\log 2+\frac{\unicode[STIX]{x1D70B}}{2}+3\unicode[STIX]{x1D6FE}+2\frac{\unicode[STIX]{x1D701}^{\prime }(3/2)}{\unicode[STIX]{x1D701}(3/2)}+2\frac{H_{1}}{H}\biggr)\nonumber\\ \displaystyle & & \displaystyle \!\!\qquad +\,P_{h}(1/K)+O(K^{-A}),\end{eqnarray}$$
where
$$\begin{eqnarray}H=\int _{0}^{\infty }h(y)\,dy,\qquad H_{1}=\int _{0}^{\infty }h(y)\log y\,dy\end{eqnarray}$$
and
$P_{h}(x)$
is a polynomial in
$x$
of degree
$A-1$
and with coefficients depending on the function
$h$
.
Proof. This is derived from Corollary 4.2 by using the asymptotic formula
$$\begin{eqnarray}\unicode[STIX]{x1D713}(2k-1/4)+\unicode[STIX]{x1D713}(2k+1/4)=2\log k+2\log 2+P(1/k)+O\biggl(\frac{1}{k^{A+1}}\biggr),\end{eqnarray}$$
where
$P(x)$
is a polynomial of degree
$A$
such that
$P(0)=0$
, and results of [Reference Balkanova and Frolenkov1, §7], namely
$$\begin{eqnarray}\mathop{\sum }_{k}h\biggl(\frac{4k}{K}\biggr)=\frac{HK}{4}+O\biggl(\frac{1}{K^{b}}\biggr)\end{eqnarray}$$
and
$$\begin{eqnarray}\mathop{\sum }_{k}h\biggl(\frac{4k}{K}\biggr)\log k=\frac{HK}{4}(\log K-\log 4)+\frac{H_{1}K}{4}+O\biggl(\frac{1}{K^{b}}\biggr)\end{eqnarray}$$
for any
$b>0$
.◻
5 Non-diagonal main term
Consider the non-diagonal term
$$\begin{eqnarray}\displaystyle M^{\mathit{ND}} & := & \displaystyle 2\mathop{\sum }_{l=1}^{\infty }\frac{V_{k}(l,0)}{\sqrt{l}}M^{\mathit{ND}}(0,l)\nonumber\\ \displaystyle & = & \displaystyle \sqrt{2\unicode[STIX]{x1D70B}}\frac{\unicode[STIX]{x1D6E4}(2k-1/4)}{\unicode[STIX]{x1D6E4}(2k+1/4)}\mathop{\sum }_{l=1}^{\infty }\frac{V_{k}(l,0)}{l^{3/4}}\mathscr{L}_{-4l}(1/2).\nonumber\end{eqnarray}$$
Lemma 5.1. For any fixed
$A>0$
, we have
$$\begin{eqnarray}M^{\mathit{ND}}=\frac{4\unicode[STIX]{x1D70B}\sqrt{2}}{\unicode[STIX]{x1D6E4}(3/4)}L_{g}^{-}(1/4)\frac{\unicode[STIX]{x1D6E4}(2k-1/4)}{\unicode[STIX]{x1D6E4}(2k+1/4)}+O(k^{-A}).\end{eqnarray}$$
Proof. Consider
$$\begin{eqnarray}\mathop{\sum }_{l=1}^{\infty }\frac{V_{k}(l,0)}{l^{3/4}}\mathscr{L}_{-4l}(1/2)=\frac{1}{2\unicode[STIX]{x1D70B}\text{i}}\int _{(\unicode[STIX]{x1D70E}_{1})}\frac{\unicode[STIX]{x1D6E4}(2k+s)}{\unicode[STIX]{x1D6E4}(2k)}\mathop{\sum }_{l=1}^{\infty }\frac{\mathscr{L}_{-4l}(1/2)}{l^{s+3/4}}\frac{g(s,0)\,ds}{s(2\unicode[STIX]{x1D70B})^{s}}.\end{eqnarray}$$
We assume that
$\unicode[STIX]{x1D70E}_{1}>1$
to justify the change of order of summation and integration. Equation (2.17) implies that
$$\begin{eqnarray}\frac{\unicode[STIX]{x1D6E4}(3/4)}{4\sqrt{\unicode[STIX]{x1D70B}}}\mathop{\sum }_{l=1}^{\infty }\frac{\mathscr{L}_{-4l}(1/2)}{l^{s+3/4}}=L_{g}^{-}(s+1/4).\end{eqnarray}$$
According to Theorem 2.3, the function
$L_{g}^{-}(s+1/4)$
is holomorphic in
$\mathbf{C}$
except for a double pole at
$s=1/4$
, which is compensated by the zeros of
$g(s,0)$
. Therefore, moving the contour of integration from
$\unicode[STIX]{x1D70E}_{1}$
to
$\unicode[STIX]{x1D70E}_{2}=-A$
for any
$A>0$
, we cross only a simple pole at
$s=0$
. Consequently,
$$\begin{eqnarray}\mathop{\sum }_{l=1}^{\infty }\frac{V_{k}(l,0)}{l^{3/4}}\mathscr{L}_{-4l}(1/2)=\operatorname{res}_{s=0}F(s)+\frac{1}{2\unicode[STIX]{x1D70B}\text{i}}\int _{(-A)}F(s)\,ds,\end{eqnarray}$$
where
$$\begin{eqnarray}F(s):=\frac{4\sqrt{\unicode[STIX]{x1D70B}}}{\unicode[STIX]{x1D6E4}(3/4)}g(s,0)\frac{\unicode[STIX]{x1D6E4}(2k+s)}{\unicode[STIX]{x1D6E4}(2k)}\frac{L_{g}^{-}(s+1/4)}{(2\unicode[STIX]{x1D70B})^{s}s}.\end{eqnarray}$$
The functional equation (see (2.19))
$$\begin{eqnarray}\displaystyle L_{g}^{-}(s+1/4) & = & \displaystyle \frac{\unicode[STIX]{x1D70B}^{2s+5/2}}{\sqrt{2}\unicode[STIX]{x1D6E4}^{2}(s+3/4)\sin ^{2}\unicode[STIX]{x1D70B}(s+1/4)}\nonumber\\ \displaystyle & & \displaystyle \times \,\biggl(\frac{1}{\unicode[STIX]{x1D6E4}^{2}(1/4)}L_{f}^{+}(-s-1/4)+\frac{\sin \unicode[STIX]{x1D70B}s}{\unicode[STIX]{x1D70B}}L_{f}^{-}(-s-1/4)\biggr)\nonumber\end{eqnarray}$$
and the estimate
$$\begin{eqnarray}\frac{\unicode[STIX]{x1D6E4}(2k+s)}{\unicode[STIX]{x1D6E4}(2k)}\ll \left\{\begin{array}{@{}ll@{}}k^{\Re s},\quad & |\Im s|\leqslant k^{1+\unicode[STIX]{x1D716}},\\ k^{\Re s}\exp ((-\unicode[STIX]{x1D70B}/2+\unicode[STIX]{x1D716}_{1})|\Im s|),\quad & |\Im s|>k^{1+\unicode[STIX]{x1D716}},\end{array}\right.\end{eqnarray}$$
imply that
$$\begin{eqnarray}\int _{(-A)}\frac{\unicode[STIX]{x1D6E4}(2k+s)}{\unicode[STIX]{x1D6E4}(2k)}L_{g}^{-}(s+1/4)\frac{g(s,0)\,ds}{(2\unicode[STIX]{x1D70B})^{s}s}\ll k^{-A}.\end{eqnarray}$$
The residue at the origin is equal to
$$\begin{eqnarray}\operatorname{res}_{s=0}F(s)=\frac{4\sqrt{\unicode[STIX]{x1D70B}}}{\unicode[STIX]{x1D6E4}(3/4)}L_{g}^{-}(1/4).\end{eqnarray}$$
Therefore, for any
$A>0$
, we have
$$\begin{eqnarray}\mathop{\sum }_{l=1}^{\infty }\frac{V_{k}(l,0)}{l^{3/4}}\mathscr{L}_{-4l}(1/2)=\frac{4\sqrt{\unicode[STIX]{x1D70B}}}{\unicode[STIX]{x1D6E4}(3/4)}L_{g}^{-}(1/4)+O(k^{-A}).\end{eqnarray}$$
The assertion follows. ◻
Corollary 5.2. For any fixed
$A>0$
, we have
$$\begin{eqnarray}\mathop{\sum }_{k}h\biggl(\frac{4k}{K}\biggr)M^{\mathit{ND}}=L_{g}^{-}(1/4)\sqrt{K}Q_{h}(1/K)+O(K^{-A}),\end{eqnarray}$$
where
$Q_{h}(x)$
is a polynomial in
$x$
of degree
$A$
with coefficients depending on the function
$h$
.
Proof. The statement follows from Lemma 5.1 and [Reference Olver, Lozier, Boisvert and Clarke18, equation (5.11.13)]. ◻
6 Error terms
Finally, we estimate the last term appearing in (3.15), which we split into two terms
$$\begin{eqnarray}ET:=2\mathop{\sum }_{l=1}^{\infty }\frac{V_{k}(l,0)}{\sqrt{l}}ET(0,l)=ET_{1}+ET_{2},\end{eqnarray}$$
where (see (3.12))
$$\begin{eqnarray}\displaystyle \displaystyle ET_{1} & := & \displaystyle \sqrt{2}\mathop{\sum }_{l=1}^{\infty }\frac{V_{k}(l,0)}{l}\mathop{\sum }_{n>2\sqrt{l}}\mathscr{L}_{n^{2}-4l}(1/2)\sqrt{n}\unicode[STIX]{x1D6F9}_{2k}\biggl(\frac{4l}{n^{2}}\biggr),\end{eqnarray}$$
$$\begin{eqnarray}\displaystyle \displaystyle ET_{2} & := & \displaystyle 2\mathop{\sum }_{l=1}^{\infty }\frac{V_{k}(l,0)}{l^{3/4}}\mathop{\sum }_{n<2\sqrt{l}}\mathscr{L}_{n^{2}-4l}(1/2)\unicode[STIX]{x1D6F7}_{2k}\biggl(\frac{n^{2}}{4l}\biggr).\end{eqnarray}$$
Note that if
$l>k^{1+\unicode[STIX]{x1D716}}$
, then
$$\begin{eqnarray}V_{k}(l,0)\ll \biggl(\frac{k}{l}\biggr)^{A}\quad \text{for any}~A>0.\end{eqnarray}$$
Using inequality (6.3) and estimating the sums over
$n$
in (6.1) and (6.2) by [Reference Balkanova and Frolenkov2, Lemmas 7.1 and 7.3], we can assume that
$l<k^{1+\unicode[STIX]{x1D716}}$
in (6.1) and (6.2) at the cost of a negligible error term.
6.1 Error term of the first type
Lemma 6.1. For any
$A>0$
, we have
$$\begin{eqnarray}\mathop{\sum }_{k}h\biggl(\frac{4k}{K}\biggr)ET_{1}\ll K^{-A}.\end{eqnarray}$$
Proof. It follows from [Reference Balkanova and Frolenkov2, Lemma 7.3] that
$$\begin{eqnarray}\hspace{7.20007pt}\mathop{\sum }_{k}h\biggl(\frac{4k}{K}\biggr)ET_{1}\ll \mathop{\sum }_{k}h\biggl(\frac{4k}{K}\biggr)\mathop{\sum }_{l<k^{1+\unicode[STIX]{x1D716}}}\frac{1}{\sqrt{l}}\frac{l^{-1/24}}{\sqrt{k}}\exp \biggl(-c\frac{k}{l^{1/4}}\biggr)\ll K^{-A}.\hspace{3.60004pt}\square\end{eqnarray}$$
6.2 Error term of the second type
In this subsection we prove the estimate
$$\begin{eqnarray}\mathop{\sum }_{k}h\biggl(\frac{4k}{K}\biggr)ET_{2}\ll K^{-A},\end{eqnarray}$$
for any fixed
$A>0$
. The idea is to apply the Liouville–Green method for the function
$\unicode[STIX]{x1D6F7}_{2k}$
. In order to obtain the estimate
$K^{-A}$
it is required to take
$N$
terms in the Liouville–Green approximation [Reference Balkanova and Frolenkov2, Theorems 6.5, 6.10] for sufficiently large
$N=N(A)$
. In fact, each of these terms can be treated similarly. Therefore, for the sake of simplicity we consider only the first term corresponding to
$N=1$
in the lemma below. This yields the estimate
$K^{1/4+\unicode[STIX]{x1D703}}$
. The remark following the lemma provides further explanations as to how to modify the arguments in order to sharpen this to
$K^{-A}$
.
Lemma 6.2. We have
$$\begin{eqnarray}\mathop{\sum }_{k}h\biggl(\frac{4k}{K}\biggr)ET_{2}\ll K^{1/4+\unicode[STIX]{x1D703}}.\end{eqnarray}$$
Proof. We decompose the sum over
$l$
in (6.2) into two parts:
$$\begin{eqnarray}\mathop{\sum }_{l=1}^{\infty }=\mathop{\sum }_{l\neq \Box }+\mathop{\sum }_{l=\Box }.\end{eqnarray}$$
Suppose that
$l$
is not a full square. To approximate the function
$\unicode[STIX]{x1D6F7}_{2k}$
we apply [Reference Balkanova and Frolenkov2, Theorem 6.10] with
$$\begin{eqnarray}\cos ^{2}\sqrt{\unicode[STIX]{x1D709}}:=n^{2}/(4l),\end{eqnarray}$$
namely,
$$\begin{eqnarray}\unicode[STIX]{x1D6F7}_{2k}(\cos ^{2}\sqrt{\unicode[STIX]{x1D709}})=\frac{-\unicode[STIX]{x1D70B}}{\unicode[STIX]{x1D709}^{1/4}(\sin \sqrt{\unicode[STIX]{x1D709}})^{1/2}}[(1+C_{J})Z_{J}(\unicode[STIX]{x1D709})+C_{Y}Z_{Y}(\unicode[STIX]{x1D709})].\end{eqnarray}$$
Using [Reference Balkanova and Frolenkov2, Theorem 6.5] to evaluate
$Z_{J}(\unicode[STIX]{x1D709})$
,
$Z_{Y}(\unicode[STIX]{x1D709})$
and [Reference Balkanova and Frolenkov2, Corollary 6.9] to approximate the constants
$C_{J}$
,
$C_{Y}$
, we have
$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D6F7}_{2k}(\cos ^{2}\sqrt{\unicode[STIX]{x1D709}}) & = & \displaystyle \frac{-\unicode[STIX]{x1D70B}}{\unicode[STIX]{x1D709}^{1/4}(\sin \sqrt{\unicode[STIX]{x1D709}})^{1/2}}\biggl[\sqrt{\unicode[STIX]{x1D709}}Y_{0}((4k-1)\sqrt{\unicode[STIX]{x1D709}})\nonumber\\ \displaystyle & & \displaystyle +\sqrt{\unicode[STIX]{x1D709}}J_{0}((4k-1)\sqrt{\unicode[STIX]{x1D709}})+O\biggl(\frac{1}{k}|\sqrt{\unicode[STIX]{x1D709}}Y_{0}((4k-1)\sqrt{\unicode[STIX]{x1D709}})|\biggr)\biggr].\qquad\end{eqnarray}$$
In order to apply [Reference Olver, Lozier, Boisvert and Clarke18, equation (10.7.8)] to approximate the Bessel functions above we show that the argument is large, namely,
$$\begin{eqnarray}((4k-1)\sqrt{\unicode[STIX]{x1D709}})\gg k^{1/2-\unicode[STIX]{x1D716}}.\end{eqnarray}$$
Indeed, since
$l$
is not a full square there exist
$u$
and
$m$
such that
$4l=m^{2}+u$
,
$1\leqslant u\leqslant 2m$
. Therefore,
$$\begin{eqnarray}2\sqrt{l}-m=\frac{u}{\sqrt{m^{2}+u}+m}<1\end{eqnarray}$$
and
$$\begin{eqnarray}[2\sqrt{l}]=m,\qquad \{2\sqrt{l}\}=\frac{u}{\sqrt{m^{2}+u}+m}.\end{eqnarray}$$
Consequently,
$$\begin{eqnarray}\frac{\{2\sqrt{l}\}}{4\sqrt{l}}=\frac{u}{m+\sqrt{m^{2}+u}}\frac{1}{2\sqrt{m^{2}+u}}\geqslant \frac{1}{4m^{2}}.\end{eqnarray}$$
This inequality can be applied to estimate
$$\begin{eqnarray}(4k-1)\sqrt{\unicode[STIX]{x1D709}}=(4k-1)\arccos \frac{n}{2\sqrt{l}},\quad 1\leqslant n\leqslant [2\sqrt{l}].\end{eqnarray}$$
Changing the variable
$n$
to
$[2\sqrt{l}]-n$
, we have
$$\begin{eqnarray}(4k-1)\sqrt{\unicode[STIX]{x1D709}}=(4k-1)\arccos \frac{[2\sqrt{l}]-n}{2\sqrt{l}},\quad 0\leqslant n\leqslant [2\sqrt{l}]-1.\end{eqnarray}$$
Further, trigonometric identities give
$$\begin{eqnarray}\displaystyle (4k-1)\sqrt{\unicode[STIX]{x1D709}} & = & \displaystyle 2(4k-1)\arcsin \sqrt{\frac{n+\{2\sqrt{l}\}}{4\sqrt{l}}}\nonumber\\ \displaystyle & {\geqslant} & \displaystyle 2(4k-1)\arcsin \sqrt{\frac{\{2\sqrt{l}\}}{4\sqrt{l}}}\gg \frac{k}{m}\gg \frac{k}{\sqrt{l}}\gg k^{1/2-\unicode[STIX]{x1D716}},\nonumber\end{eqnarray}$$
as required.
Next, we insert the asymptotic expansion (6.8) into
$ET_{2}$
. The contribution of the error term in (6.8) is majorized by
$$\begin{eqnarray}\displaystyle ET_{2,1} & := & \displaystyle \mathop{\sum }_{k}h\biggl(\frac{4k}{K}\biggr)\frac{1}{k}\mathop{\sum }_{\substack{ l\leqslant k^{1+\unicode[STIX]{x1D716}} \\ l\neq \Box }}\frac{1}{l^{3/4}}\mathop{\sum }_{n<2\sqrt{l}}(4l-n^{2})^{\unicode[STIX]{x1D703}}\frac{\unicode[STIX]{x1D709}^{1/4}}{(\sin \sqrt{\unicode[STIX]{x1D709}})^{1/2}}\nonumber\\ \displaystyle & & \displaystyle \times \,|Y_{0}((4k-1)\sqrt{\unicode[STIX]{x1D709}})|\nonumber\\ \displaystyle & \ll & \displaystyle \mathop{\sum }_{k}h\biggl(\frac{4k}{K}\biggr)\frac{1}{k}\mathop{\sum }_{l\leqslant k^{1+\unicode[STIX]{x1D716}}}\frac{1}{l^{3/4}}\mathop{\sum }_{n<2\sqrt{l}}\frac{(4l-n^{2})^{\unicode[STIX]{x1D703}}}{k^{1/2}(\sin \sqrt{\unicode[STIX]{x1D709}})^{1/2}}.\nonumber\end{eqnarray}$$
According to (6.7), we have
$$\begin{eqnarray}\sin \sqrt{\unicode[STIX]{x1D709}}=\frac{\sqrt{4l-n^{2}}}{2\sqrt{l}},\end{eqnarray}$$
and therefore
$$\begin{eqnarray}ET_{2,1}\ll \mathop{\sum }_{k}h\biggl(\frac{4k}{K}\biggr)\frac{1}{k^{3/2}}\mathop{\sum }_{l\leqslant k^{1+\unicode[STIX]{x1D716}}}\frac{1}{l^{3/4}}\mathop{\sum }_{n<2\sqrt{l}}(4l-n^{2})^{\unicode[STIX]{x1D703}-1/4}l^{1/4}\ll K^{\unicode[STIX]{x1D703}+1/4}.\end{eqnarray}$$
It remains to estimate the contribution of the main term in (6.8) given by
$$\begin{eqnarray}\displaystyle ET_{2,2} & := & \displaystyle \mathop{\sum }_{k}h\biggl(\frac{4k}{K}\biggr)\mathop{\sum }_{l}\frac{V_{k}(l,0)}{l^{3/4}}\mathop{\sum }_{n<2\sqrt{l}}\mathscr{L}_{n^{2}-4l}(1/2)\nonumber\\ \displaystyle & & \displaystyle \times \,\frac{\sin ((4k-1)\sqrt{\unicode[STIX]{x1D709}}-\unicode[STIX]{x1D70B}/4)}{\sqrt{4k-1}(\sin \sqrt{\unicode[STIX]{x1D709}})^{1/2}}=\mathop{\sum }_{l\ll K^{1+\unicode[STIX]{x1D716}}}\frac{1}{l^{1/2}}\mathop{\sum }_{n<2\sqrt{l}}\frac{\mathscr{L}_{n^{2}-4l}(1/2)}{(4l-n^{2})^{1/4}}\nonumber\\ \displaystyle & & \displaystyle \times \,\mathop{\sum }_{k}h\biggl(\frac{4k}{K}\biggr)V_{k}(l,0)\frac{\sin ((4k-1)\arccos (n/2\sqrt{l})-\unicode[STIX]{x1D70B}/4)}{\sqrt{4k-1}}.\nonumber\end{eqnarray}$$
The inner sum over
$k$
can be evaluated similarly to [Reference Balkanova and Frolenkov1, Lemma 7.3], as we now show. Using the Poisson summation formula,
$$\begin{eqnarray}\mathop{\sum }_{k}h\biggl(\frac{4k}{K}\biggr)V_{k}(l,0)\frac{\sin ((4k-1)\arccos (n/2\sqrt{l})-\unicode[STIX]{x1D70B}/4)}{\sqrt{4k-1}}=\mathop{\sum }_{m}I(m),\end{eqnarray}$$
where
$$\begin{eqnarray}\displaystyle I(m) & = & \displaystyle \int _{-\infty }^{\infty }h\biggl(\frac{4y}{K}\biggr)V_{y}(l,0)\frac{\sin ((4y-1)\arccos (n/2\sqrt{l})-\unicode[STIX]{x1D70B}/4)}{\sqrt{4y-1}}\nonumber\\ \displaystyle & & \displaystyle \times \exp (-my)\,dy\ll K\biggl|\int _{-\infty }^{\infty }h(y)\exp (\text{i}g(y))V_{yK/4}(l,0)\frac{dy}{\sqrt{yK-1}}\biggr|\nonumber\end{eqnarray}$$
with
$$\begin{eqnarray}g(y)=\frac{K}{4}y\biggl(-2\unicode[STIX]{x1D70B}m\pm 4\arccos \frac{n}{2\sqrt{l}}\biggr).\end{eqnarray}$$
Integrating
$a$
times by parts, we obtain
$$\begin{eqnarray}I(m)\ll \left\{\begin{array}{@{}ll@{}}\sqrt{K}/(Km)^{a},\quad & m\neq 0,\\ \displaystyle \sqrt{K}/\biggl(K\arccos {\displaystyle \frac{n}{2\sqrt{l}}}\biggr)^{a},\quad & m=0.\end{array}\right.\end{eqnarray}$$
Finally, since
$$\begin{eqnarray}\arccos \frac{n}{2\sqrt{l}}\gg \frac{1}{\sqrt{l}},\end{eqnarray}$$
we have
$$\begin{eqnarray}\mathop{\sum }_{m}I(m)\ll \frac{\sqrt{K}}{(K\arccos (n/2\sqrt{l}))^{a}}\ll \frac{\sqrt{K}}{(K/\sqrt{l})^{a}}\end{eqnarray}$$
and
$$\begin{eqnarray}ET_{2,2}\ll K^{-a}.\end{eqnarray}$$
The last step is to evaluate the sum over
$l=\Box$
. Making the change of variables from
$l$
to
$l^{2}$
, we have to estimate
$$\begin{eqnarray}E_{2,3}:=\mathop{\sum }_{k}h\biggl(\frac{4k}{K}\biggr)\mathop{\sum }_{l}\frac{V_{k}(l^{2},0)}{l^{3/2}}\mathop{\sum }_{n<2l}\mathscr{L}_{n^{2}-4l^{2}}(1/2)\unicode[STIX]{x1D6F7}_{2k}\biggl(\frac{n^{2}}{4l^{2}}\biggr).\end{eqnarray}$$
Using the subconvexity estimate (2.11), we have
$$\begin{eqnarray}ET_{2,3}\ll \mathop{\sum }_{l\ll K^{1/2+\unicode[STIX]{x1D716}}}\frac{1}{l^{3/2}}\mathop{\sum }_{n<2l}(4l^{2}-n^{2})^{\unicode[STIX]{x1D703}}\biggl|\mathop{\sum }_{k}h\biggl(\frac{4k}{K}\biggr)V_{k}(l^{2},0)\unicode[STIX]{x1D6F7}_{2k}\biggl(\frac{n^{2}}{4l^{2}}\biggr)\biggr|.\end{eqnarray}$$
Changing the variable
$n$
to
$2l-n$
gives
$$\begin{eqnarray}ET_{2,3}\ll \mathop{\sum }_{l\ll K^{1/2+\unicode[STIX]{x1D716}}}\mathop{\sum }_{n<2l}\frac{(nl)^{\unicode[STIX]{x1D703}}}{l^{3/2}}\biggl|\mathop{\sum }_{k}h\biggl(\frac{4k}{K}\biggr)V_{k}(l^{2},0)\unicode[STIX]{x1D6F7}_{2k}\biggl(\biggl(1-\frac{n^{2}}{4l^{2}}\biggr)^{2}\biggr)\biggr|.\end{eqnarray}$$
Let
$$\begin{eqnarray}\unicode[STIX]{x1D709}:=4\biggl(\arcsin \sqrt{\frac{n}{4l}}\biggr)^{2}.\end{eqnarray}$$
Consequently,
$$\begin{eqnarray}\biggl(1-\frac{n}{2l}\biggr)^{2}=\cos ^{2}\sqrt{\unicode[STIX]{x1D709}}.\end{eqnarray}$$
Applying [Reference Balkanova and Frolenkov2, Theorem 6.10], we have
$$\begin{eqnarray}\unicode[STIX]{x1D6F7}_{2k}(\cos ^{2}\sqrt{\unicode[STIX]{x1D709}})=\frac{-\unicode[STIX]{x1D70B}}{\unicode[STIX]{x1D709}^{1/4}(\sin \sqrt{\unicode[STIX]{x1D709}})^{1/2}}[(1+C_{J})Z_{J}(\unicode[STIX]{x1D709})+C_{Y}Z_{Y}(\unicode[STIX]{x1D709})].\end{eqnarray}$$
Note that now
$$\begin{eqnarray}(4k-1)\sqrt{\unicode[STIX]{x1D709}}=2(4k-1)\arcsin \sqrt{\frac{n}{4l}}\gg \frac{k}{\sqrt{l}}\gg k^{3/4}.\end{eqnarray}$$
In the same way as in the case
$l\neq \Box$
, we prove
$$\begin{eqnarray}ET_{2,3}\ll K^{-1/4+\unicode[STIX]{x1D703}}.\end{eqnarray}$$
Finally,
$$\begin{eqnarray}\hspace{72.0pt}ET_{2}\ll ET_{2,1}+ET_{2,2}+ET_{2,3}\ll K^{1/4+\unicode[STIX]{x1D703}}.\hspace{60.0pt}\square\end{eqnarray}$$
Remark.
Using [Reference Balkanova and Frolenkov2, Theorem 6.5] with sufficiently large
$N$
depending on
$A$
and taking more terms in the asymptotics for
$C_{Y},C_{J}$
, we obtain equation (6.8) with additional series of main terms plus the error term
$$\begin{eqnarray}O\biggl(\frac{1}{k^{A+2}}|\sqrt{\unicode[STIX]{x1D709}}Y_{0}((4k-1)\sqrt{\unicode[STIX]{x1D709}})|\biggr).\end{eqnarray}$$
As a result, following the proof of Lemma 6.2, we obtain the estimate (6.5).
Acknowledgements
The authors thank the anonymous referee for helpful comments. The first author thanks Viktor A. Bykovskii for providing his handwritten notes with a full proof of Lemma 2.2 and for introducing her to the Rankin–Selberg method.
