Effective Estimates for a Class of Farey Fraction Sums
and Bounds for Mundici-Type Constants
June 11, 2026
Let \(D_{2}(Q)\) denote the sum of squared distances between consecutive Farey fractions in the full interval \((0, 1]\). Daniele Mundici conjectured that \(C(Q):=D_{2}(Q)\cdot Q^2/\log Q\) is less than 3 for all \(Q\geq 2\), which is confirmed true in [1]. In this paper, we generalize this result to subintervals of \((0, 1]\) and to \(h\)-spacings. As applications, we obtain Mundici-type bounds in these two settings, extending the full-interval consecutive-spacing case of Mundici’s conjecture.
Let \(Q\geq 2\) be an integer. The \(Q\)-th Farey sequence \(F_Q\) is defined as follows: \[\begin{align} F_Q :=\left\{\frac{a}{q}: 1\leq a\leq q\leq Q,\;(a,q)=1\right\}. \end{align}\]
Write \(|F_Q|=N(Q)\), and enumerate \[\begin{align} F_Q :=\{\gamma_1,\gamma_2,\dots,\gamma_{N(Q)}\} \end{align}\] with \(\gamma_1<\gamma_2<\dots<\gamma_{N(Q)}\). For simplicity, we write \(N=N(Q)\), while keeping the dependence of \(Q\) in mind. Moreover, we set \(\gamma_0=0\). Whenever an index \(j\) exceeds \(N\), we use the periodic extension \[\gamma_j = \gamma_{j \bmod N}+\left\lfloor \frac{j}{N}\right\rfloor.\] In particular, \(\gamma_{j+N} = \gamma_j+1\). For general facts about Farey fractions, the reader may refer to Chapter 3 of Hardy and Wright’s book [2] and the survey [3] by Cobeli and one of the authors.
In [1], Li and two of the present authors study the distribution of spacings between consecutive Farey fractions and obtain an effective/explicit formula for \(D_{2,1}(Q)\), which is the sum of squared distances between consecutive Farey fractions in the full interval \((0, 1]\). This formula is then used in [1] to prove Mundici’s conjectural bound for the normalized quantity \[\begin{align} C(Q) = \frac{D_{2,1}(Q) \cdot Q^2}{\log Q}, \end{align}\] which states that \[\begin{align} C(Q)<3 \text{ for all Q\geq 2.} \end{align}\] We refer to this normalized quantity, and to its analogues below, as Mundici-type constants. At the end of [1], the authors raise some open problems concerning two further directions: \(h\)-th level consecutive spacings for \(h\geq 2\), previously considered by Augustin, Boca, Cobeli, and one of the authors [4], [5], and localized sums over subintervals \(I\) of \((0,1]\), a direction previously studied in [5].
Our main goal in this paper is to tackle these open problems. These two directions introduce additional complications beyond those appearing in the full-interval consecutive-spacing problem. In the short-interval setting, we need to provide and use effective estimates for Kloosterman sums. Moreover, as established in [5], a subinterval \(I\) comes with an associated quantity \(c_I\), called the defect; whereas for the full interval \((0,1]\), this defect vanishes. On the other hand, the \(h\)-spacing problem involves longer chains of consecutive Farey fractions as opposed to the original problem’s consecutive pairs. These chains are intrinsically related to the behavior of the so-called BCZ-map introduced by Boca, Cobeli, and one of the authors [6]. For further work on this map, see, for example, Athreya and Cheung [7].
We first consider the \(h\)-spacing problem. For \(h\geq 1\), define \[\begin{align} D_{2,h}(Q):= \sum_{j=1}^{N} (\gamma_{j+h}-\gamma_j)^2. \end{align}\] The corresponding Mundici-type constant is \[\begin{align} C_h(Q) := \frac{D_{2,h}(Q)Q^2}{\log Q}.\label{def:Ch40Q41} \end{align}\tag{1}\] Our first result gives a uniform bound for \(C_h(Q)\).
Theorem 1. For any integer \(h\geq 1\), we have \[\begin{align} C_h(Q)\leq\frac{2h^2}{\log 2} \text{\quad for all Q\geq 2}.\label{eq:32bound32of32Ch40Q41} \end{align}\qquad{(1)}\] Moreover, such an upper bound is always attained: for any \(h\geq 1\), we have \(C_h(2)=2h^2/\log 2\).
Notice that when \(h\) grows, the bound in ?? is quadratic in \(h\). However, such a bound can be significantly improved when \(Q\) is large enough. With some additional work, our next result gives a linear bound in \(h\) with an explicit threshold.
Theorem 2. For any integer \(h\geq 1\), there exists an effectively computable integer \(Q_h\) such that \[\begin{align} C_h(Q)< 3h\text{\quad for all Q\geq Q_h}.\label{eq:Ch40Q41603h} \end{align}\qquad{(2)}\] Moreover, one can take \(Q_1=2\), \(Q_2 = 19397\), and for \(h\geq 3\), \[\begin{align} Q_h = \exp\left\{\max\left\{ \frac{2D(h)}{\Delta_h},(h+1)W\left(\frac{1}{h+1}\left(\frac{h\cdot2^{3h+9}}{\Delta_h}\right)^{\frac{h+2}{h+1}}\right)\right\}\right\}, \end{align}\] where the constant \(D(h)\) is defined in ?? below, \(W(x)\) is the Lambert \(W\) function (see [8]) defined as the function satisfying \[\begin{align} W(x)e^{W(x)} = x, \end{align}\] and \[\begin{align} \Delta_h = 3h-\frac{12(2h-1)}{\pi^2}. \end{align}\]
Remark 3. We remark that the value \(Q_2 = 19397\) in Theorem 2 is the best possible in the sense that the inequality in ?? fails if one takes \(Q_2=19396\).
We next turn to the short-interval problem over a subinterval \(I\) of \((0,1]\). In this setting, the relevant quantity is the contribution of the squared-spacing sum from Farey fractions in the interval \(I\). Denote \[\begin{align} \label{def:S040Q44I41} S_0(Q, I) := \sum_{\gamma_j \in F_I(Q)} (\gamma_{j+1} - \gamma_j)^2, \end{align}\tag{2}\] where \[F_I(Q) = \left\{ \dfrac{a}{q} \in I, 1 \leq a \leq q \leq Q, (a, q) = 1 \right\}.\] Define \[\begin{align} C_0(Q,I):=\frac{S_0(Q,I)Q^2}{|I|\log Q}.\label{def:C040Q44I41} \end{align}\tag{3}\] Our next theorem gives an explicit Mundici-type bound after normalizing by \(|I|\).
Theorem 4. For any integer \(Q\geq \exp\{{e^{22}}\}\) and any subinterval \(I=(\alpha,\beta]\subseteq (0,1]\), we have \[\begin{align} C_0(Q,I)&<\frac{12}{\pi^2} + \frac{2.01}{\log Q}+\frac{7.06}{|I|\log Q}, \end{align}\] where \(C_0(Q,I)\) is defined in 3 . If we further assume that \((\beta-\alpha)\geq 4/\log Q\), then \[\begin{align} C_0(Q,I)&<3. \end{align}\]
Remark 5. An upper bound of \(C_0(Q,I)\) which holds true for any \(Q\geq 2\) is \[\begin{align} C_0(Q,I)\leq \frac{2}{|I|\log 2}. \end{align}\] Such a bound follows easily from Theorem 1 (which we will show later), but it blows up when \(|I|\) tends to \(0\). However, the bound in Theorem 4 suggests that, for any fixed interval \(I\), when \(Q\) is sufficiently large, the upper bound for \(C_0(Q,I)\) approaches \(12/\pi^2\). Moreover, in the proof of Theorem 4, we obtain an explicit asymptotic formula for \(S_0(Q,I)\), which indicates that there does not exist an absolute constant \(K\), independent of \(I\), such that \(C_0(Q,I)<K\) for all subintervals \(I\).
The paper is organized as follows. In Section 2, we prove Theorem 1 and reduce Theorem 2 to explicit asymptotic estimates for \(D_{2,h}(Q)\), \(h\geq 2\). We then show that these estimates follow from asymptotic formulas for \(S_r(Q)\), \(r\geq 1\). Assuming these auxiliary estimates, we complete the proof of Theorem 2. In Section 3, we reduce Theorem 4 to an explicit asymptotic formula for \(S_0(Q,I)\). Sections 4 and 5 are devoted to the asymptotic formulas required in Section 2 with explicit error bounds. Finally, in Section 6, we prove the explicit asymptotic formula for \(S_0(Q,I)\) needed in Section 3.
In this section, we give a complete proof of Theorems 1 and 2, which concern results in \(h\)-spacings. Theorem 1 requires no additional lemma, whereas the proof of Theorem 2 rests on two supporting theorems.
It will be shown later in 7 that \[D_{2,h}(Q) = \sum_{j=1}^N \left(\sum_{r=0}^{h-1}\ell_{j+r}\right)^2.\] By the Cauchy–Schwarz inequality, \[\left(\sum_{r=0}^{h-1}\ell_{j+r}\right)^2\leq h \sum_{r=0}^{h-1} \ell_{j+r}^2.\] Therefore, \[\begin{align} D_{2,h}(Q)\leq h \sum_{r=0}^{h-1}\sum_{j=1}^N\ell_{j+r}^2= h^2 \sum_{j=1}^N \ell_j^2 = h^2D_{2,1}(Q), \end{align}\] which gives \[C_h(Q)\leq h^2C_1(Q).\] By the numerical computation in [1], \[C_1(Q)\leq \frac{2}{\log 2},\] and therefore, for all \(Q\geq 2\), \[C_h(Q) \leq \frac{2h^2}{\log 2}.\] In fact, this maximum is attained at \(Q=2\) for all \(h\geq 1\). For \(h\geq 1\), we have \[\begin{align} \gamma_{j+h}-\gamma_j =\frac{h}{2} \end{align}\] for both \(j=1,2\). Therefore, \[C_h(2) = \frac{D_{2,h}(2)\times2^2}{\log 2} = \frac{2h^2}{\log 2}.\]
We now turn to Theorem 2. To prove Theorem 2, it suffices to have the following theorem, which is a version of [5] with concrete error bounds.
Theorem 6. Let \(Q\) be a positive integer, and let \(\gamma_1,\gamma_2,\cdots,\gamma_N\) be the \(Q\)-th Farey sequence. For \(h\geq 2\) and \(Q\geq \max\{6163, 2^{(h+2)^2}\}\), \[\begin{align} \sum_{j=1}^{N} (\gamma_{j+h}-\gamma_j)^2 = \frac{12(2h-1)\log Q}{\pi^2Q^2}+\frac{D(h)}{Q^2}+E_h(Q), \end{align}\] where \(\gamma\) is Euler’s constant, \[B= \frac{1}{2}+\log 2 +2\sum_{h=1}^\infty\frac{\zeta(2h)-1}{2h-1}=2.546277\dots,\] \[\begin{align} D(h) = \frac{12}{\pi^2} \left[ (2h-1)\left( \gamma-\frac{\zeta'(2)}{\zeta(2)} \right) +\frac{h}{2}+ (h-1)B + \sum_{k=2}^{h-1}(h-k)I_k \right],\label{def:D40h41} \end{align}\qquad{(3)}\] \[I_r := \iint\limits_{\mathscr{T}}\frac{dxdy}{xyL_r(x,y)L_{r+1}(x,y)},\] with \(\mathscr{T}\) the Farey triangle defined by \(0<x,y\leq 1\) and \(x+y>1\) in the plane, \(L_r(x,y)\) defined in 20 and 21 , and \[\begin{align} |E_h(Q)|\leq h\cdot 2^{3h+8} \frac{(\log Q)^{1/(h+2)}}{Q^{2+1/(h+2)}}. \end{align}\] When \(h=2\), the bound in the error term can be improved to \[\begin{align} |E_2(Q)| \leq \frac{138\log Q}{Q^{5/2}}+\frac{216(\log Q)^{3/2}}{Q^{11/4}}+\frac{146(\log Q)^2+212\log Q+538}{Q^3}. \end{align}\]
Remark 7. The analogous result to Theorem 6 for \(h=1\) has been obtained in [1], and states that \[\begin{align} \sum_{j=1}^{N} (\gamma_{j+1}-\gamma_j)^2 &= \frac{12\log Q}{\pi^2 Q^2}-\frac{2}{Q^2}\frac{\zeta'(2)}{\zeta(2)^2} +(2\gamma+1)\frac{6}{Q^2\pi^2}+E_{1},\label{eq:D4024414140Q4132result} \end{align}\qquad{(4)}\] where \[|E_{1}|\leq \frac{64(\log Q)^2+106\log Q+269}{Q^3}.\]
Proof of Theorem 2.. By the definition in 1 and Theorem 6, \[\begin{align} C_h(Q) =\frac{12(2h-1)}{\pi^2}+\frac{D(h)}{\log Q}+\frac{Q^2E_h(Q)}{\log Q}. \end{align}\] When \(h=1\), by [1], \(C_1(Q)<3\) for all \(Q>1\).
For \(h=2\), \[\begin{align} |E_2| \leq \frac{138\log Q}{Q^{5/2}}+\frac{216(\log Q)^{3/2}}{Q^{11/4}}+\frac{146(\log Q)^2+212\log Q+538}{Q^3}. \end{align}\] Define \[G_2(Q):= \frac{36}{\pi^2} + \frac{D(2)}{\log Q} + \frac{138}{Q^{1/2}} + \frac{216(\log Q)^{1/2}}{Q^{3/4}} + \frac{146\log Q+212+\frac{538}{\log Q}}{Q}\] Observe that each nonconstant summand in \(G_2(Q)\) is decreasing when \(Q\) increases, and therefore, it takes the largest value at \(Q=6163\), which evaluates to be \[G_2(6163)=7.54749759\dots\] Thus, to finish the proof for \(h=2\), it suffices to find the smallest \(Q_2\geq 6163\) such that \[G_2(Q_2)<6.\] A numerical computation shows that \(Q_2=19397\).
For \(h\geq 3\), we have \[\begin{align} |E_h(Q)|\leq h\cdot 2^{3h+8} \frac{(\log Q)^{1/(h+2)}}{Q^{2+1/(h+2)}}. \end{align}\] Therefore, \[\begin{align} C_h(Q)\leq G_h(Q), \end{align}\] where \[\begin{align} G_h(Q) = \frac{12(2h-1)}{\pi^2} + \frac{D(h)}{\log Q} + h2^{3h+8} \frac{1}{Q^{1/(h+2)}(\log Q)^{(h+1)/(h+2)}}. \end{align}\] Observe that \(D(h)\) is positive, since each summand in \(D(h)\) is positive. In order to complete the proof of the theorem, we need to find a \(Q_h\geq\max\{6163,2(h-1),2^{(h+2)^2}\}\) such that \[\begin{align} \frac{D(h)}{\log Q_h} + \frac{h2^{3h+8}}{Q_h^{1/(h+2)}(\log Q_h)^{(h+1)/(h+2)}}<3h-\frac{12(2h-1)}{\pi^2}.\label{eq:32optimization32Qh} \end{align}\tag{4}\] For simplicity, denote \[\begin{align} \Delta_h:=3h-\frac{12(2h-1)}{\pi^2}. \end{align}\] A sufficient choice of \(Q_h\) can be obtained by forcing each term in 4 to be less than \(\Delta_h/2\). Then we have \[\begin{align} Q_h&>\exp\left\{ \frac{2D(h)}{\Delta_h}\right\}, \shortintertext{and} \log Q_h \exp\left\{ \frac{\log Q_h}{h+1}\right\}&>\left(\frac{h\cdot2^{3h+9}}{\Delta_h}\right)^{\frac{h+2}{h+1}}, \end{align}\] which is equivalent to \[\begin{align} Q_h>\exp\left\{(h+1)W\left(\frac{1}{h+1}\left(\frac{h\cdot2^{3h+9}}{\Delta_h}\right)^{\frac{h+2}{h+1}}\right)\right\},\label{eq:upper32bound32of32Qh} \end{align}\tag{5}\] where \(W(x)\) is the Lambert \(W\) function. Since the lower threshold for \(Q_h\) in 5 is larger than \(\max\{6163, 2^{(h+2)^2}\}\) for all \(h\geq 3\), we can safely take \[Q_h = \exp\left\{\max\left\{ \frac{2D(h)}{\Delta_h},(h+1)W\left(\frac{1}{h+1}\left(\frac{h\cdot2^{3h+9}}{\Delta_h}\right)^{\frac{h+2}{h+1}}\right)\right\}\right\}.\] This finishes the proof of Theorem 2. ◻
In this subsection, we further reduce the proof of Theorem 6, which is the auxiliary theorem used to prove Theorem 2, to the following result. Denote \[\begin{align} \label{def:Sr40Q41} S_r(Q) = \sum_{j = 1}^{N} (\gamma_{j+1} - \gamma_j)(\gamma_{j+r+1} - \gamma_{j+r}) \end{align}\tag{6}\] for each \(r \geq 0\).
Theorem 8. Let \(Q\geq 2, r\geq 1\) be integers, and \(S_r(Q)\) be defined as in 6 . For \(Q\geq 6163\), we have \[S_1(Q) = \frac{6}{\pi^2}Q^{-2} \log Q + AQ^{-2}+R_1,\] where \[A = \frac{6}{\pi^2}\left(\gamma-\frac{\zeta'(2)}{\zeta(2)}+B\right),\] the constant \(\gamma\) is Euler’s constant, \[\begin{align} B&= \frac{1}{2}+\log 2 +2\sum_{h=1}^\infty\frac{\zeta(2h)-1}{2h-1}=2.546277\dots, \shortintertext{and} |R_1|&\leq \frac{69\log Q}{Q^{5/2}}+\frac{108(\log Q)^{3/2}}{Q^{11/4}}+\frac{9(\log Q)^2}{Q^3}. \end{align}\] Moreover, for \(r\geq 2\) and \(Q/\log Q\geq 2^{(r+3)^2}\), we have \[S_r(Q) = \frac{6I_r}{\pi^2Q^2} + R_r,\] with \[\begin{align} |R_r| &\leq \left(3\cdot 2^{3r+6} +3\cdot 2^{3r+8} +\frac{9\cdot 2^{3r+8}r}{\pi^2}\right) \frac{\log^{1/(r+3)}Q}{Q^{2+1/(r+3)}} +7\cdot 2^{3r+6}\frac{(1+4\log Q)}{(\log Q)^{1/(r+3)}Q^{3-1/(r+3)}}\\ & +2^{3r+6}\frac{1}{(\log Q)^{1/(r+3)}Q^{3-1/(r+3)}} +\frac{2^{3r+6}r}{Q^3} \left(\log Q+2\right), \end{align}\] and where \[I_r := \iint\limits_{\mathscr{T}}\frac{dxdy}{xyL_r(x,y)L_{r+1}(x,y)},\] with \(\mathscr{T}\) the Farey triangle defined by \(0<x,y\leq 1\) and \(x+y>1\) in the plane, and \(L_r(x,y)\) defined in 20 and 21 .
Proof of Theorem 6.. In the definition of \(S_r(Q)\) in 6 , set \(\ell_j=\gamma_{j+1}-\gamma_j\) for \(j\geq 0\). It is clear that \(\ell_{j+N}=\ell_j = \ell_{N-j-1}\). Then, \[S_r(Q) = \sum_{j=1}^{N}\ell_j\ell_{j+r}.\] With this, we may recursively write \[\begin{align} D_{2,h}(Q)&=\sum_{j=1}^{N} (\ell_{j+h-1}+\ell_{j+h-2}+\cdots+\ell_j)^2\notag\\ &=hS_0(Q)+2\sum_{k=1}^{h-1} (h-k)S_k(Q)\label{def:D244h40Q41} \end{align}\tag{7}\] for \(h\geq 2\).
Applying ?? and Theorem 8 for \(r=1\) to 7 , we obtain for \(Q\geq 6163\), \[\begin{align} \sum_{j=1}^{N}(\gamma_{j+2}-\gamma_j)^2 &= 2(S_0(Q)+S_1(Q))\\ &=\frac{36\log Q}{\pi^2 Q^2} + \frac{12}{Q^2\pi^2}\left(3\gamma-3\frac{\zeta'(2)}{\zeta(2)}+B+1\right)+E_2, \end{align}\] where \[\begin{align} |E_2| \leq \frac{138\log Q}{Q^{5/2}}+\frac{216(\log Q)^{3/2}}{Q^{11/4}}+\frac{146(\log Q)^2+212\log Q+538}{Q^3}. \end{align}\] This finishes the proof for \(h=2\).
Similarly, for \(h\geq 3\), applying ?? and Theorem 8 for \(r=1,2,3,\cdots,h-1\), we obtain for \(Q\geq \max\{6163, 2(h-1)\}\), \[\begin{align} \sum_{j=1}^{N} (\gamma_{j+h}-\gamma_j)^2 = \frac{12(2h-1)\log Q}{\pi^2Q^2}+\frac{D(h)}{Q^2}+E_h(Q), \end{align}\] where \[\begin{align} D(h) &= \frac{12}{\pi^2} \left[ (2h-1)\left( \gamma-\frac{\zeta'(2)}{\zeta(2)} \right) +\frac{h}{2}+ (h-1)B + \sum_{k=2}^{h-1}(h-k)I_k \right] \shortintertext{and} |E_h(Q)|&\leq \frac{138(h-1)\log Q}{Q^{5/2}} + \frac{216(h-1)(\log Q)^{3/2}}{Q^{11/4}} + \frac{ (82h-18)(\log Q)^2 }{Q^3} +\frac{106h\log Q}{Q^3}\\ & \quad+\frac{269h}{Q^3}+ 2\sum_{k=2}^{h-1}(h-k)2^{3k+6} \Bigg[ \left( 15+\frac{36k}{\pi^2} \right) \frac{(\log Q)^{1/(k+3)}}{Q^{2+1/(k+3)}} \\ &\quad+ \frac{28\log Q+8}{(\log Q)^{1/(k+3)}Q^{3-1/(k+3)}} + \frac{k(\log Q+2)}{Q^3} \Bigg]\leq h\cdot 2^{3h+8} \frac{(\log Q)^{1/(h+2)}}{Q^{2+1/(h+2)}}. \end{align}\] This finishes the proof of Theorem 6. ◻
Therefore, what remains for the results in the direction of \(h\)-spacings is to prove Theorem 8, which we present in Sections 4 and 5. We split the proof into two parts. In Section 4, we provide an asymptotic formula for \(S_1(Q)\), and leave the formula of \(S_r(Q)\) for \(r\geq 2\) to Section 5.
In this section, we lay out the proof of Theorem 4. We show that it suffices to prove the following result.
Theorem 9. For any integer \(Q>1024\) and any subinterval \(I = (\alpha, \beta] \subseteq (0,1]\), we have \[\begin{align} S_0(Q,I)= \frac{12|I|}{\pi^2} \frac{\log Q}{Q^2} + |I|E_{1,I} +E_{2,I},\label{eq:asymptotic32for32S040Q44I4132in32Lem321468} \end{align}\qquad{(5)}\] where \(S_0(Q,I)\) is defined in 2 , \[\begin{align} |E_{1,I}|\leq\frac{6(2\gamma+1)}{\pi^2Q^2}-\frac{2\zeta'(2)}{Q^2\zeta(2)^2}+\frac{64(\log Q)^2+106\log Q+269}{Q^3}+\frac{(4Q^{1/10}-2)\log Q}{(Q^{1/10}-1)^2Q^2}, \end{align}\] and \[\begin{align} |E_{2,I}| &\leq \frac{\pi^4}{18Q^2}+Q^{-5/2}+4Q^{-21/10+2.1322/(\log\log Q-0.1054)}(1+Q^{-1/10}) (2\log Q+\log^2Q)\notag\\ &\quad+\frac{(4Q^{1/10}-2)\log Q}{(Q^{1/10}-1)^2Q^2}+\frac{\pi^2Q^{-9/5}}{6(Q^{1/10}-1)^2}. \end{align}\]
Proof of Theorem 4.. Recall that the Mundici-type constant \(C_0(Q,I)\) is defined by \[C_0(Q,I)=\frac{S_0(Q,I)Q^2}{|I|\log Q}.\] Note that by definition, we have \[S_0(Q,I)\leq S_0(Q).\] Since the numerical computation in [1] gives \(S_0(Q)\leq 2/\log 2\), we trivially have \[C_0(Q,I)\leq \frac{2}{|I|\log 2}\] for any \(Q>1\). To obtain the sharper bound in Theorem 4, we proceed as follows.
By the asymptotic formula in ?? , we obtain that for \(Q>1024\), \[\begin{align} C_0(Q,I)\leq \frac{12}{\pi^2} + \frac{E_{1,I}(Q)Q^2}{\log Q}+\frac{E_{2,I}(Q)Q^2}{|I|\log Q}.\label{eq:C95040Q44I4132explicit} \end{align}\tag{8}\] Observe that for \(Q\geq\exp\{{e^{22}}\}\), we have \[\upsilon(Q):=\frac{2.1322}{\log\log Q-0.1054}-\frac{1}{10}<0.\] Substituting \(Q\geq\exp\{{e^{22}}\}\) in 8 , we have \[E_{1,I}(Q)Q^2<2.01 \quad\text{and}\quad E_{2,I}(Q)Q^2<7.06.\]
Therefore, we conclude that for \(Q\geq \exp\{{e^{22}}\}\), \[C_0(Q,I)< \frac{12}{\pi^2} + \frac{2.01}{\log Q}+\frac{7.06}{|I|\log Q}.\] In particular, when \(|I|\geq 4/\log Q\), we have \[C_0(Q,I)<3,\] which finishes the proof of Theorem 4. ◻
Thus, what remains in the direction of short intervals is the proof of Theorem 9, whose proof is postponed until Section 6.
In [9], Hall proved that \[S_1(Q) = \frac{6}{\pi^2}Q^{-2} \log Q + AQ^{-2}+O\left(\frac{\log Q}{Q^2\sqrt{Q}}\right),\] where \[A = \frac{6}{\pi^2}\left(\gamma-\frac{\zeta'(2)}{\zeta(2)}+B\right),\] the constant \(\gamma\) is Euler’s constant, and \[B= \frac{1}{2}+\log 2 +2\sum_{h=1}^\infty\frac{\zeta(2h)-1}{2h-1}=2.546277\dots\] Our goal in this section is to trace Hall’s proof in [9] and obtain the asymptotic formula for \(S_1(Q)\) in Theorem 8. The result is as follows.
Theorem 10. Let \(Q\geq 6163\) be an integer, and \(S_1(Q)\) be defined as in 6 . Then, \[S_1(Q) = \frac{6}{\pi^2}Q^{-2} \log Q + AQ^{-2}+R_1,\] where \[A = \frac{6}{\pi^2}\left(\gamma-\frac{\zeta'(2)}{\zeta(2)}+B\right),\] the constant \(\gamma\) is Euler’s constant, \[\begin{align} B&= \frac{1}{2}+\log 2 +2\sum_{h=1}^\infty\frac{\zeta(2h)-1}{2h-1}=2.546277\dots, \shortintertext{and} |R_1|&\leq \frac{69\log Q}{Q^{5/2}}+\frac{108(\log Q)^{3/2}}{Q^{11/4}}+\frac{9(\log Q)^2}{Q^3}. \end{align}\]
Proof. Following [10], \[\begin{align} S_1(Q) = \sum_{s=1}^Q s^{-2}\sum_{\substack{n=Q-s+1\\(n,s)=1}}^Q\frac{1}{nt},\label{eq:S140Q4132full32sum} \end{align}\tag{9}\] where \(t = t(n,s,Q) := s[(Q+n)/s]-n\). Choose an integer \(K\geq 2\). For \(2\leq k<K\), define \(s_{k}= (2Q+1)/k\), and set \(z=(2Q+1)/K\). Now split the sum \(S_1(Q)\) into two parts, \(U_Q\) and \(V_Q\), according to whether \(s\leq s_K =z\). Hall picked \(K=[Q^{1/4}\log^{-1/2}Q]\) at this point. Therefore, keep in mind that \(z\) can be computed to only depend on \(Q\) from now on.
We first consider \(U_Q\). For \(s\leq z\), put \(n=Q-n'\), \(t=Q-t'\) with \(0\leq n',t'\leq s-1\), then \[\begin{align} \frac{1}{nt} = \frac{1}{Q^2}\frac{1}{(1-n'/Q)(1-t'/Q)}. \end{align}\] Upon expansion, we have \[\begin{align} \left|\frac{1}{nt}-\frac{1}{Q^2}-\frac{n'+t'}{Q^3}\right| &\leq \frac{s^2}{Q^4}+\frac{2s^2}{Q^4}\frac{1+s/Q}{(1-s/Q)}\\ &\leq \frac{s^2}{Q^4}+\frac{2s^2}{Q^4}\times 6 = \frac{13s^2}{Q^4} \end{align}\] when \(Q\geq 6163\) (which forces \(K\geq 3\)). Therefore, substituting back in 9 , \[\begin{align} U_Q = \frac{1}{Q^2}\sum_{s\leq z} \frac{\varphi(s)}{s^2}+\frac{1}{Q^3}\sum_{s\leq z} \frac{1}{s^2}\sum_{\substack{Q-s+1\leq n\leq Q\\(n,s)=1}}(n'+t')+R_{1,1},\label{eq:UQ} \end{align}\tag{10}\] where \[\begin{align} |R_{1,1}|\leq \sum_{s\leq z}\frac{\varphi(s)}{s^2}\cdot \frac{13s^2}{Q^4}\leq \frac{6z^2}{Q^4}. \end{align}\] The first term in 10 can be written as \[\begin{align} \sum_{s\leq z}\frac{\varphi(s)}{s^2} &= \sum_{s\leq z}\frac{1}{s^2}\sum_{d|s}\mu(d)\frac{s}{d}=\frac{1}{\zeta(2)}\sum_{s\leq z}\frac{1}{s}-\sum_{s\leq z}\frac{1}{s}\sum_{d>z/s}\frac{\mu(d)}{d^2}\\ &=\frac{6}{\pi^2}(\log z+\gamma)-\sum_{2\leq d\leq z}\frac{\mu(d)}{d^2}\log d+R_{1,2}\\ &=\frac{6}{\pi^2}(\log z+\gamma)-\frac{\zeta'(2)}{\zeta(2)^2}+R_{1,3}, \end{align}\] where \[\begin{align} |R_{1,2}|&\leq \frac{1}{z}+\frac{1}{z}\sum_{2\leq d\leq z}\frac{1}{d} \shortintertext{and} |R_{1,3}|&\leq \frac{1}{z}+\frac{1}{z}\sum_{2\leq d\leq z}\frac{1}{d}+\frac{4\log z}{z}\leq \frac{6\log z}{z}, \end{align}\] when \(Q\geq 6163\). Following the discussion in [9], the second term in 10 is equivalent to \[\begin{align} &\frac{1}{Q^3}\sum_{s\leq z} \frac{\varphi(s)}{s}+R_{1,4}\notag\\ =& \frac{z}{Q^3}\sum_{d\leq z}\frac{\mu(d)}{d^2} -\frac{1}{Q^3}\sum_{d\leq z}\frac{\mu(d)}{d}\{z/d\}+R_{1,4}=\frac{6z}{\pi^2Q^3}+R_{1,5},\label{eq:sum32of32phi40s4147s} \end{align}\tag{11}\] where \[\begin{align} |R_{1,4}|&\leq \frac{1}{Q^3}\sum_{s\leq z} s^{-2}\varphi(s) + \frac{1}{Q^3}\sum_{s\leq z} s^{-1} \tau(s)\leq \frac{\log z+(\log z)^2}{Q^3}, \shortintertext{and} |R_{1,5}| &\leq \frac{(\log z)^2+2\log z+2}{Q^3}. \end{align}\]
Combining all, we obtain \[\begin{align} U_Q = \frac{6}{\pi^2Q^2}\left(\log z+\gamma-\frac{\zeta'(2)}{\zeta(2)}+\frac{z}{Q}\right)+R_{1,6},\label{eq:final32estimate32of32U40Q41} \end{align}\tag{12}\] where \[\begin{align} |R_{1,6}|\leq \frac{6z^2}{Q^4}+\frac{6\log z}{Q^2z}+ \frac{(\log z)^2+2\log z+2}{Q^3}. \end{align}\]
We now shift our attention to \(V_Q\). Following the original proof, \[\begin{align} V_Q=\sum_{z<s\leq Q}\frac{2}{s^3}\Big\{\frac{1}{k(s)-1}\sum_{\substack{n=Q-s+1\\(n,s)=1}}^Q\frac{1}{n}-\frac{1}{k(s)(k(s)-1)}\sum_{\substack{n=sk(s)-Q\\(n,s)=1}}^Q\frac{1}{n}\Big\},\label{eq:VQ} \end{align}\tag{13}\] where \(k(s)=[(2Q+1)/s]\) and the right-hand inner sum is empty when \(s \mid (2Q+1)\). For \(u\leq v\) positive integers, \[\begin{align} \sum_{\substack{n=u}^v} \frac{1}{n} = \frac{\varphi(s)}{s}\log \frac{v}{u}+R_{1,7}, \end{align}\] where \[|R_{1,7}|\leq \frac{2\tau(s)}{u}.\] In 13 , take \(u=Q-s+1,v = Q\) for the first inner sum, and take \(u=sk(s)-Q,v=Q\) for the second inner sum respectively, we see that the contributions from the total error terms, call it \(R_{1,8}\), is bounded by \[\begin{align} |R_{1,8}|&\leq \sum_{z<s\leq Q}\frac{2}{s^3}\Big\{\frac{1}{k(s)-1}\frac{2\tau(s)}{Q-s+1}-\frac{1}{k(s)(k(s)-1)}\frac{2\tau(s)}{sk(s)-Q}\Big\}\\ &\leq \frac{16}{Q^2}\sum_{s>z}\frac{\tau(s)}{s^2}+\frac{32}{Q^3}\sum_{Q/2<s\leq Q}\frac{\tau(s)}{Q-s+1}+4\sum_{z<s\leq Q}\frac{\tau(s)}{s^2Q(Q-s+1)}. \end{align}\] Using \[\begin{align} \sum_{s>z}\frac{\tau(s)}{s^2} &= \sum_{d=1}^\infty\sum_{\substack{s>z\\d|s}}\frac{1}{s^2}\leq \frac{\log z+3}{z} \shortintertext{and} \sum_{Q/2<s\leq Q}\frac{\tau(s)}{Q-s+1}&\leq 2\sqrt{Q}\sum_{1\leq s\leq Q/2}\frac{1}{s}\leq 2\sqrt{Q}(1+\log Q), \end{align}\] where the first inequality comes from trivially bounding \(\tau(s)\) by \(2\sqrt{s}\), we arrive at \[\begin{align} |R_{1,8}|\leq \frac{20(\log z+3)}{Q^2z}+\frac{64(1+\log Q)}{Q^{5/2}}. \end{align}\] Therefore, we are left to approximate the main term in 13 , which is \[\begin{align} M_{V,Q}:=\sum_{z<s\leq Q}\frac{2}{s^4}\Big\{\frac{\varphi(s)}{k(s)-1}\log\frac{Q}{Q-s+1}-\frac{\varphi(s)}{k(s)(k(s)-1)}\log\frac{Q}{sk(s)-Q}\Big\}.\label{eq:M40V44Q41} \end{align}\tag{14}\]
Split the sum into ranges \((s_{k+1},s_k]\) for \(2\leq k<K\), so \(k(s)=k\). Using 11 and partial summation, we obtain \[\begin{align} \sum_{s_{k+1}<s\leq s_k}\frac{\varphi(s)}{s^4}\log\frac{Q}{Q-s+1} = \frac{6}{\pi^2}\int_{s_{k+1}}^{s_k}s^{-3}\log\frac{Q}{Q-s+1}\;ds+R_{1,9},\label{eq:first32summand32in32M40V44Q41} \end{align}\tag{15}\] where \[\begin{align} |R_{1,9}|&\leq 2(\log s_k+2)s_{k+1}^{-3}\log\frac{Q}{Q-s_{k}+1}\\ &\quad+\int_{s_{k+1}}^{s_k}3(\log s+2)s^{-4}\log\frac{Q}{Q-s+1}+(\log s+2)\frac{s^{-3}}{Q-s+1}\;ds\\ &\leq \frac{62(\log Q)^2}{Qs_k^2} \end{align}\] by bounding \(\log Q/(Q-s+1)\) by \(3s/Q\) for \(k\geq 3\) and by \(\log (Q+1/2)\) when \(k=2\). Similarly, \[\begin{align} \sum_{s_{k+1}<s\leq s_k}\frac{\varphi(s)}{s^4}\log\frac{Q}{sk-Q} = \frac{6}{\pi^2}\int_{s_{k+1}}^{s_k}s^{-3}\log\frac{Q}{sk-Q}\;ds+R_{1,10},\label{eq:second32summand32in32M40V44Q41} \end{align}\tag{16}\] where \[\begin{align} |R_{1,10}| \leq \frac{16k(\log Q)^2}{Qs_k^2}. \end{align}\] Substituting 15 and 16 in 14 together with taking \(R_{1,8}\) into account, we arrive at \[\begin{align} V_Q = \frac{12}{\pi^2}\int_{z}^{Q+1/2}\Big\{\frac{1}{(k(s)-1)}\log\frac{Q}{Q-s+1}-\log\frac{Q}{sk(s)-Q}\Big\}s^{-3}\;ds+R_{1,11},\label{eq:VQ32main32term324332error32term} \end{align}\tag{17}\] where \[\begin{align} |R_{1,11}|&\leq \sum_{k=2}^{K-1}\frac{124(\log Q)^2}{(k-1)Qs_k^2}+\frac{32k(\log Q)^2}{k(k-1)Qs_k^2}\\ &\leq 39K^2Q^{-3}(\log Q)^2. \end{align}\]
Denote the main term in 17 by \(I_Q\). Upon substituting \(s=(2Q+1)/x\), we have \[\begin{align} (2Q+1)^2I_Q &= \frac{12}{\pi^2}\int_2^K\Big\{\frac{x}{[x]-1}\log\frac{Qx}{(Q+1)x-2Q-1}\\ &\quad+\frac{x}{[x]([x]-1)}\log\frac{(2Q+1)[x]-Qx}{Qx}\Big\} \;dx. \end{align}\]
We have \[\begin{align} \log\frac{Qx}{(Q+1)x-2Q-1} = \log\frac{x}{x-2}-\log \left(1+\frac{x-1}{Q(x-2)}\right) = \log\frac{x}{x-2} + R_{1,12} \end{align}\] for \(x\geq 3\), where \[\begin{align} |R_{1,12}|&\leq \log (1+2/Q)\leq 2/Q, \shortintertext{and} \log\frac{(2Q+1)[x]-Qx}{Qx} &= \log \left(2\frac{[x]}{x}-1\right)+R_{1,13} \end{align}\] for \(x\geq 2\), where \[|R_{1,13}|\leq 2/Q.\] Therefore, \[\begin{align} I_Q = \frac{12}{\pi^2(2Q+1)^2}\int_2^K \left(f(x)+\frac{2}{x}\right)\;dx + R_{1,14}, \end{align}\] where \[f(x):=-\frac{2}{x}+\frac{x}{[x]-1}\log\frac{x}{x-2}+\frac{x}{[x]([x]-1)}\log \left(2\frac{[x]}{x}-1\right),\] and \[\begin{align} |R_{1,14}|&\leq \frac{12}{\pi^2(2Q+1)^2} \int_3^K\left\{\frac{x}{[x]-1}\frac{2}{Q}+\frac{x}{[x]([x]-1)}\frac{2}{Q}\right\}\;dx\\ &\leq \frac{18(K-3+\log Q)}{\pi^2Q^3}+\frac{27+18\log Q}{\pi^2Q^3}, \end{align}\] where the last summand in the last line comes from approximating the integral by \(f(x)+2/x\) over \(2\leq x\leq 3\). Putting everything together, we obtain \[\begin{align} V_Q = \frac{24}{\pi^2(2Q+1)^2}\left(\log\frac{K}{2}+B-B(K) \right)+R_{1,15},\label{eq:V40Q4132final32estimate} \end{align}\tag{18}\] where \[\begin{align} B = \frac{1}{2}\int_2^\infty f(x)dx, \quad B(K) = \frac{1}{2}\int_K^\infty f(x)\;dx, \end{align}\] and \[\begin{align} |R_{1,15}|&\leq 39K^2Q^{-3}(\log Q)^2+\frac{18K-27+36\log Q}{\pi^2Q^3}+\frac{24\log K}{\pi^2(2Q+1)^2}\\ &\leq \frac{39\log Q}{Q^{5/2}}+\frac{18}{\pi^2Q^{11/4}(\log Q)^{1/2}}+\frac{36\log Q-26}{\pi^2Q^3}. \end{align}\] The constant \(B\) was computed in [9] exactly. To estimate \(B(K)\), a calculation shows that if \(x\geq 3\), then \[f(x)=\frac{4}{x^2}+\left(\frac{20}{3}+4\theta(1-\theta)\right)\frac{1}{x^3}+R_{1,16},\] where \(|R_{1,16}|\leq 1000x^{-4}\). Therefore, \[\begin{align} B(K) &= \frac{1}{2}\int_K^\infty\frac{4}{x^2}\;dx+\frac{1}{2}\int_K^\infty\left(\frac{20}{3}+4\theta(1-\theta)\right)x^{-3}\;dx+R_{1,17}\notag\\ &=\frac{2}{K} +\frac{11}{6K^2}-2\int_K^\infty B_2(\{x\})x^{-3}\;dx+R_{1,17}\notag\\ & = \frac{2}{K} +\frac{11}{6K^2}+R_{1,18},\label{eq:estimate32of32B40K41} \end{align}\tag{19}\] where \[\begin{align} |R_{1,17}|&\leq \frac{1}{2}\int_K^\infty \frac{1000}{x^4}\;dx = \frac{500}{3K^3}, \shortintertext{and} |R_{1,18}|&\leq \frac{500}{3K^3}+\frac{\sqrt{3}}{54K^3}\leq \frac{167}{K^3}. \end{align}\] Substituting 19 back into the estimate for \(V(Q)\) in 18 , and combining with the asymptotic formulas for \(U_Q\) in 12 , we finally arrive at \[\begin{align} S_1(Q) &= \frac{6}{\pi^2Q^2}\left(\log z+\gamma-\frac{\zeta'(2)}{\zeta(2)}+\frac{z}{Q}\right)+\frac{24}{\pi^2(2Q+1)^2}\left(\log\frac{K}{2}+B-\frac{2}{K}-\frac{11}{6K^2} \right)\\ &\quad+R_{1,19}, \end{align}\] where \[\begin{align} R_{1,19}&\leq \frac{6z^2}{Q^4}+\frac{6\log z}{Q^2z}+ \frac{(\log z)^2+2\log z+2}{Q^3}\\ &\quad+ \frac{39\log Q}{Q^{5/2}}+\frac{18}{\pi^2Q^{11/4}(\log Q)^{1/2}}+\frac{36\log Q-26}{\pi^2Q^3}+\frac{24}{\pi^2(2Q+1)^2}\cdot\frac{167}{K^3}. \end{align}\] Substituting the values of \(z\) and \(K\), we will obtain Theorem 10. ◻
In this section, our goal is to prove the asymptotic formula of \(S_r(Q)\) for \(r\geq 2\) presented in Theorem 8. The result is as follows.
Theorem 11. Let \(r\geq 2\). Define \[I_r := \iint\limits_{\mathscr{T}}\frac{dxdy}{xyL_r(x,y)L_{r+1}(x,y)},\] where \(\mathscr{T}\) denotes the Farey triangle defined by \(0<x,y\leq 1\) and \(x+y>1\) in the plane, and \(L_r(x,y)\) is defined in 20 and 21 . Then, for \(Q\) such that \(Q/\log Q\geq 2^{(r+3)^2}\), we have \[S_r(Q) = \frac{6I_r}{\pi^2Q^2} + R_r,\] where \[\begin{align} |R_r| &\leq \left(3\cdot 2^{3r+6} +3\cdot 2^{3r+8} +\frac{9\cdot 2^{3r+8}r}{\pi^2}\right) \frac{\log^{1/(r+3)}Q}{Q^{2+1/(r+3)}} +7\cdot 2^{3r+6}\frac{(1+4\log Q)}{(\log Q)^{1/(r+3)}Q^{3-1/(r+3)}}\\ & +2^{3r+6}\frac{1}{(\log Q)^{1/(r+3)}Q^{3-1/(r+3)}} +\frac{2^{3r+6}r}{Q^3} \left(\log Q+2\right). \end{align}\]
We follow the proof structure in Sections 3–7 of [5], while obtaining concrete bounds for the error terms in all necessary lemmas.
We follow the same notation in [5]. We introduce the functions \(L_i\) defined on \(\mathscr{T} = \{(x,y): 0 < x, y \leq 1, x+y>1\}\) by \[\begin{align} L_0(x,y) = x, \quad L_1(x,y) = y\label{def:32Li40x44y4132def321} \end{align}\tag{20}\] and \[\begin{align} \label{def:32Li40x44y4132def322} L_i(x,y) = \begin{cases} \left[ \dfrac{1+L_{i-2}(x,y)}{L_{i-1}(x,y)}\right]L_{i-1}(x,y)-L_{i-2}(x,y) & \text{ if } i\geq 2 \\ \left[ \dfrac{1+L_{i+2}(x,y)}{L_{i+1}(x,y)}\right]L_{i+1}(x,y)-L_{i+2}(x,y) & \text{ if } i\leq -1. \end{cases} \end{align}\tag{21}\] Additionally, consider the function \[\begin{align} f_r(x,y) = \dfrac{1}{xyL_r(x,y)L_{r+1}(x,y)}.\label{def:fr40x44y41} \end{align}\tag{22}\]
Now, for a fixed \(\mathbf{k}\in\mathbb{N}^r\), we define \[\begin{align} L_{\mathbf{k}, 0}(x,y) = x, L_{\mathbf{k}, 1}(x,y) = y, \end{align}\] then recursively, for \(i \in \{ 2, \dots, r+1\}\), the linear function \[\begin{align} \label{def: L[k,i](x,y)} L_{\mathbf{k},i}(x,y)= k_{i-1}L_{\mathbf{k}, i-1}(x,y) - L_{\mathbf{k}, i-2}(x,y), \quad (x,y) \in \mathbb{R}^2. \end{align}\tag{23}\] We also want to consider the set of indices \[\begin{align} \mathscr{L}_{\mathbf{k}} := \{ 1, \dots, N(Q)\} \cap \{j;q_{j+1} = L_{\mathbf{k},i}(q_j, q_{j+1}) \text{ for all } i \in \{2, \dots, r+1\}\}. \end{align}\] In other words, \[\mathscr{L}_{\mathbf{k}} =\{\text{index j}: q_{j+i+1} = k_iq_{j+i}-q_{j+i-1}\},\] where \(k_i = [\frac{Q+q_{i-1}}{q_i}]\).
For each \(r \geq 0\) and \(\mathbf{k} \in (\mathbb{N}^{*})^r\), \[\begin{align} S_{r,\mathbf{k}} &= \sum_{j \in \mathscr{L}_\mathbf{k}} (\gamma_{j+1} - \gamma_j)(\gamma_{j+r+1} - \gamma_{j+r}) = \sum_{j \in \mathscr{L}_\mathbf{k}} \dfrac{1}{q_jq_{j+1}}\cdot\dfrac{1}{q_{j+r}q_{j+r+1}}\\ &= \sum_{j \in \mathscr{L}_\mathbf{k}} \dfrac{1}{q_jq_{j+1}L_{\mathbf{k},r}(q_j,q_{j+1})L_{\mathbf{k},r+1}(q_j,q_{j+1})}. \end{align}\] Then, we have \[\begin{align} S_r(Q) = \sum_{\mathbf{k} \in (\mathbb{N}^*)^r} S_{r,\mathbf{k}}(Q). \end{align}\] These sums can be truncated to \[\begin{align} S_{r,T}(Q) = \sum_{1\leq k_1, k_2, \dots, k_r \leq T} S_{r,\mathbf{k}}(Q), \end{align}\] where \(T \geq 1\). Also, define \[\begin{align} f_{r, \mathbf{k}}(x,y) = \dfrac{1}{xyL_{\mathbf{k},r}(x,y)L_{\mathbf{k},r+1}(x,y)}. \end{align}\]
Let \(\Omega\subseteq \mathbb{R}^2\) be a convex bounded region with rectifiable boundary \(\partial\Omega\) and assume that \(f\) is a \(C^1\) function on \(\Omega\). Denote \(\|f\|_\infty = \sup_{(x,y)\in\Omega} |f(x,y)|\) and set \[S = S(f,\Omega) = \sum_{(a,b)\in\Omega\cap\mathbb{Z}^2} f(a,b).\]
Lemma 12. (Effective version of Lemma 1 in [5])Suppose that \(\Omega\) and \(f\) are as above. Then \[\begin{align} \left|S-\iint\limits_\Omega f(x,y)dxdy\right|\leq \left(\Big\|\frac{\partial f}{\partial x}\Big\|_\infty+\Big\|\frac{\partial f}{\partial y}\Big\|_\infty\right)\text{Area}(\Omega)+7\|f\|_\infty (1+\text{length}(\partial\Omega)). \end{align}\]
Proof. Denote \(R_{a,b}=[a,a+1]\times[b,b+1]\) for \(a,b\in\mathbb{Z}\). We have \[\left|S-\iint\limits_\Omega f(x,y)dxdy\right|\leq \|f\|_\infty E(\Omega),\] where \[E(\Omega) := \sum_{\substack{a,b\in\mathbb{Z}\\ R_{a,b}\cap\partial\Omega\neq \emptyset}} 1 = \text{Area}(\bigcup_{\substack{a,b\in\mathbb{Z}\\ R_{a,b}\cap\partial\Omega\neq \emptyset}} R_{a,b}).\] For any unit square \(R_{a,b}\) to overlap with \(\partial\Omega\), we will have \[\text{dist}(z,\partial\Omega)\leq \sqrt{1^2+1^2}=\sqrt{2}.\] Therefore, by the Steiner Formula (for example, see [11]), we have \[\begin{align} E(\Omega)\leq 2\times\sqrt{2}\times\text{length}(\partial\Omega)+\pi(\sqrt{2})^2. \end{align}\] Combining with the rest of the proof in [5], we obtain Lemma 12. ◻
Define \[\begin{align} S' = S'(f,\Omega) = \sum_{\substack{(a,b)\in\Omega\cap\mathbb{Z}^2\\\gcd(a,b)=1}} f(a,b).\label{def:S39} \end{align}\tag{24}\]
Lemma 13. (Effective version of Lemma 2 and Corollary 2 in [5])Suppose that \(\Omega\) and \(f\) are as above. In addition, suppose \(\Omega\subseteq [1,R]\times [1,R]\). Then \[\begin{align} \left|S'-\frac{6}{\pi^2}\iint\limits_\Omega f(x,y)dxdy\right|&\leq \left(\Big\|\frac{\partial f}{\partial x}\Big\|_\infty+\Big\|\frac{\partial f}{\partial y}\Big\|_\infty\right)\text{Area}(\Omega)\log R\\ &\quad+7\|f\|_\infty (R+4R\log R)+\|f\|_\infty R. \end{align}\]
Proof. Following the proof of [5] with Lemma 12, we end up with \[\begin{align} \left|S'-\sum_{d=1}^R\frac{\mu(d)}{d^2}\iint\limits_\Omega f(x,y)dxdy\right|&\leq \left(\Big\|\frac{\partial f}{\partial x}\Big\|_\infty+\Big\|\frac{\partial f}{\partial y}\Big\|_\infty\right)\text{Area}(\Omega)\log R\notag\\ &\quad+7\|f\|_\infty (R+\text{length}(\partial\Omega)\log R).\label{eq:estimation32of32S39} \end{align}\tag{25}\]
Now \[\begin{align} \sum_{d=1}^R\frac{\mu(d)}{d^2} = \sum_{d=1}^\infty\frac{\mu(d)}{d^2} + E_1(R), \label{eq:approximation32of32zeta40241} \end{align}\tag{26}\] where \[\begin{align} |E_1(R)|\leq \sum_{d=R+1}^\infty \frac{1}{d^2}\leq \int_R^\infty \frac{1}{t^2}dt\leq \frac{1}{R}. \end{align}\] Note that the main term 26 is \(1/\zeta(2) =6/\pi^2\). Combining 25 and 26 , we arrive at \[\begin{align} \left|S'-\frac{6}{\pi^2}\sum_{d=1}^R\iint\limits_\Omega f(x,y)dxdy\right|&\leq \left(\Big\|\frac{\partial f}{\partial x}\Big\|_\infty+\Big\|\frac{\partial f}{\partial y}\Big\|_\infty\right)\text{Area}(\Omega)\log R\notag\\ &\quad+7\|f\|_\infty (R+\text{length}(\partial\Omega)\log R)+\frac{1}{R}\iint\limits_\Omega f(x,y)dxdy. \end{align}\] Observe that \[\begin{align} \frac{1}{R}\iint\limits_\Omega f(x,y)dxdy\leq \frac{\|f\|_\infty \text{Area}(\Omega) }{R}\leq \|f\|_\infty R. \end{align}\] Moreover, since \(\Omega\) is convex, the projection of \(\partial\Omega\) to \(x\) and \(y\)-axis is of length at most \(R\). Therefore, \[\text{length}(\partial\Omega)\leq 4R.\] This finishes the proof of Lemma 13. ◻
Corollary 14. (Effective version of Lemma 5 in [5])Suppose that \(r\geq 2\), \(\mathbf{k}\in\mathbb{N}^r, M\geq 1\) and \((x,y)\in\Omega_{\mathbf{k},M}\). Then \[|f_{r,\mathbf{k}}(x,y)|\leq 2^{3r+6}\frac{M}{Q^4}.\]
Proof. By definition of \(\Omega_{\mathbf{k},M}\), we have \(x,y,L_{\mathbf{k},r}(x,y)\), and \(L_{\mathbf{k},r+1}(x,y)\) all \(\geq Q/M\). By [5], we have at least three of these four expressions \(\geq 2^{-r-2}Q\). Therefore, \[xyL_{\mathbf{k},r}(x,y)L_{\mathbf{k},r+1}(x,y)\geq \frac{Q}{M}(2^{-r-2}Q)^3 = 2^{-3r-6}\frac{Q^4}{M}.\] Therefore, \[|f_{r,\mathbf{k}}(x,y)| = \frac{1}{xyL_{\mathbf{k},r}(x,y)L_{\mathbf{k},r+1}(x,y)}\leq 2^{3r+6}\frac{M}{Q^4}.\] This finishes the proof. ◻
Corollary 15. (Effective version of Corollary 3 in [5])Suppose that \(r\geq 2\), \(\mathbf{k}\in\mathbb{N}^r, M\geq 1\) and \((x,y)\in\Omega_{\mathbf{k},M}\). Then \[\max \left(\Big|\frac{\partial f_{r,\mathbf{k}}}{\partial x}(x,y)\Big|, \Big|\frac{\partial f_{r,\mathbf{k}}}{\partial y}(x,y)\Big|\right)\leq 3\cdot 2^{3r+6}\frac{k_1k_2\cdots k_{r}M^2}{Q^5}.\]
Proof. Denote \(g = f_{r,\mathbf{k}}^{-1}\). Then \[\begin{align} \frac{\partial f_{r,\mathbf{k}}}{\partial x} &= -g^{-2}\frac{\partial g}{\partial x} = -g^{-2}y(L_{\mathbf{k},r}L_{\mathbf{k},r+1}+x(L_{\mathbf{k},r}\cdot\partial_x L_{\mathbf{k},r+1}+\partial_x L_{\mathbf{k},r}\cdot L_{\mathbf{k},r+1}))\\ &=-f_{r,\mathbf{k}}\left(\frac{1}{x}+\frac{\partial_xL_{\mathbf{k},r}}{L_{\mathbf{k},r}}+\frac{\partial_x L_{\mathbf{k},r+1}}{L_{\mathbf{k},r+1}}\right). \end{align}\] Similarly, \[\frac{\partial f_{r,\mathbf{k}}}{\partial y} = -f_{r,\mathbf{k}}\left(\frac{1}{y}+\frac{\partial_yL_{\mathbf{k},r}}{L_{\mathbf{k},r}}+\frac{\partial_y L_{\mathbf{k},r+1}}{L_{\mathbf{k},r+1}}\right).\] By Corollary 14, \[|f_{r,\mathbf{k}}|\leq 2^{3r+6}\frac{M}{Q^4}.\]
Moreover, by the recursive definition of \(L_{\mathbf{k},i}(x,y)\) in @{eq:def: Lk,i} , \[\partial_xL_{\mathbf{k},r},\partial_yL_{\mathbf{k},r}\leq k_1k_2\cdots k_{r-1}.\] Combining with the property of \(\Omega_{\mathbf{k},M}\), we obtain \[\begin{align} \left|\frac{\partial f_{r,\mathbf{k}}}{\partial x} \right|&\leq 2^{3r+6}\frac{M}{Q^4}\left(\frac{M}{Q}+\frac{k_1k_2\cdots k_{r-1}M}{Q}+\frac{k_1k_2\cdots k_{r}M}{Q}\right)\\ &\leq 3\cdot 2^{3r+6}\frac{k_1k_2\cdots k_{r}M^2}{Q^5}. \end{align}\] Same bound works for \(\partial_y f_{r,\mathbf{k}}\). This finishes the proof of the lemma. ◻
Let \[\begin{align} \Omega_k = \{ (x,y) \in \mathbb{R}^2 : &0 <L_{\mathbf{k},i}(x,y)\leq Q \text{ for all } 0 \leq i \leq r+1,\\ &Q <L_{\mathbf{k},i}(x,y) + L_{\mathbf{k},i+1}(x,y) \text{ for all } 0 \leq i \leq r\}. \end{align}\] For each \(M \geq 1\) and \(\mathbf{k} \in (\mathbb{N}^*)^r\), we consider its convex subset \[\begin{align} \Omega_{\mathbf{k},M} &= \{ (x,y) \in \Omega_\mathbf{k}: \min\left(x,y,L_{\mathbf{k},r}(x,y), L_{\mathbf{k},r+1}(x,y)\right) \geq Q/M \}\\ &= \Omega_\mathbf{k} \cap [Q/M, \infty)^2 \cap \bigcap_{i \in \{r, r+1\}} \{(x,y) \in \mathbb{R}^2: L_{\mathbf{k},i}(x,y) \geq Q/M\}. \end{align}\]
Also, let \[\begin{align} \mathscr{M}_{\mathbf{k}} = \{(a,b) \in \Omega_\mathbf{k}\cap \mathbb{Z}^2: \gcd(a,b) = 1\}. \end{align}\] We then consider its subset \[\begin{align} \mathscr{M}_{\mathbf{k}, M} = \mathscr{M}\cap \Omega_{\mathbf{k},M} \end{align}\] of \(\mathscr{M}_\mathbf{k}\) and the sum \[\begin{align} S_{r,\mathbf{k},M}(Q) = \sum_{(a,b) \in \mathscr{M}_{\mathbf{k},M}} f_{r, \mathbf{k}}(a,b).\label{def:S40r44k44M41} \end{align}\tag{27}\]
Lemma 16. (Effective version of Lemma 6 in [5])Suppose that \(r\geq 2\) and \(M\geq 1\). Then \[\sum_{\mathbf{k}\in \mathbb{N}^r} |S_{r,\mathbf{k}}(Q)-S_{r,\mathbf{k},M}(Q)| \leq \frac{2^{3r+8}}{MQ^2}.\]
Proof. By [5], among \(q_j, q_{j+1},q_{j+r}, q_{j+r+1}\), we have at least three of these four numbers \(\geq 2^{-r-2}Q\). Therefore, following the original proof, \[\begin{align} \sum_{\mathbf{k}\in \mathbb{N}^r} |S_{r,\mathbf{k}}(Q)-S_{r,\mathbf{k},M}(Q)|&\leq \sum_{\substack{1\leq j\leq N\\\min(q_j,q_{j+1}, q_{j+r}, q_{j+r+1})<Q/M}}\frac{2^{3r+6}}{Q^3\min(q_j,q_{j+1}, q_{j+r}, q_{j+r+1})}\\ &\leq \frac{2^{3r+8}}{MQ^2}. \end{align}\] ◻
Now we are ready to approximate \(S_{r}(Q)\) using Lemma 13. For \(T,M \in\mathbb{N}\) and \(k\in\mathbb{N}^r\) with \(r\geq 2\), define \[\begin{align} \mathscr{D}(T,M) = \frac{1}{Q}\bigcup_{1\leq k_1,\cdots, k_r\leq T}\Omega_{\mathbf{k},M}.\label{def:32D40T44M41} \end{align}\tag{28}\] An important remark is that \(\mathscr{D}(T,M)\) is indeed independent of \(Q\). To that end, we have the following lemma.
Lemma 17. Suppose \(r\geq 2\), \(M\geq 1\) and \(2Q\geq T\geq 2^{r+3}\). Then, \[\begin{align} |Q^2S_r(Q)-\frac{6}{\pi^2}\iint\limits_{\mathscr{D}(T,M)}f_r(x,y)\;dxdy|&\leq 3\cdot 2^{3r+6}\frac{T^r M^2 \log Q}{Q}+7\cdot 2^{3r+6}\frac{M(1+4\log Q)}{Q}\notag\\ &\quad+ 2^{3r+6}\frac{M}{Q}+\frac{2^{3r+8}}{M}+\frac{2^{3r+6} r}{Q}\left(\frac{12Q}{\pi^2 T}+\log\frac{2Q}{T}+2\right). \end{align}\]
Proof. Apply Lemma 13 with \(R=Q, \;\Omega =\Omega_{\mathbf{k},M}\) and \(f=f_{r,\mathbf{k}}\), we obtain \[\begin{align} \Big|S_{r,\mathbf{k},M}(Q)-\frac{6}{\pi^2}\iint\limits_{\Omega_{\mathbf{k},M}}f_{r,\mathbf{k}}(x,y)\;dxdy\Big|&\leq \left(\Big\|\frac{\partial f_{r,\mathbf{k}}}{\partial x}\Big\|_\infty+\Big\|\frac{\partial f}{\partial y}\Big\|_\infty\right)\text{Area}(\Omega_{\mathbf{k},M})\log Q\\ &\quad+7\|f_{r,\mathbf{k}}\|_\infty (Q+4Q\log Q)+\|f_{r,\mathbf{k}}\|_\infty Q. \end{align}\] where \(S_{r,\mathbf{k},M}\) is defined in 27 . Applying Corollaries 14 and 15, and bounding \(\text{Area}(\Omega_{\mathbf{k},M})\) by \(\text{Area}(\Omega_{\mathbf{k}})\), we obtain \[\begin{align} \Big|S_{r,\mathbf{k},M}(Q)-\frac{6}{\pi^2}\iint\limits_{\Omega_{\mathbf{k},M}}f_{r,\mathbf{k}}(x,y)\;dxdy\Big|&\leq 6\cdot 2^{3r+6}\frac{k_1k_2\cdots k_r M^2\text{Area}(\Omega_{\mathbf{k}})\log Q}{Q^5}\\ &\quad+7\cdot 2^{3r+6}\frac{M(1+4\log Q)}{Q^3}+ 2^{3r+6}\frac{M}{Q^3}. \end{align}\] Therefore, \[\begin{align} &\Big|\sum_{1\leq k_1,\cdots,k_r\leq T}S_{r,\mathbf{k},M}(Q)-\frac{6}{\pi^2}\sum_{1\leq k_1,\cdots,k_r\leq T}\iint\limits_{\Omega_{\mathbf{k},M}}f_{r,\mathbf{k}}(x,y)\;dxdy\Big|\label{eq:approximate32S40r44k44M4132with32double32integral}\\ \leq&\;3\cdot 2^{3r+6}\frac{T^r M^2 \log Q}{Q^3}+7\cdot 2^{3r+6}\frac{M(1+4\log Q)}{Q^3}+ 2^{3r+6}\frac{M}{Q^3}.\notag \end{align}\tag{29}\] Recall the definition of \(f_{r}(x,y)\) and \(\mathscr{D}(T,M)\) in 22 and 28 The second summand in 29 can be reformulated as \[\frac{6}{\pi^2}\sum_{1\leq k_1,\cdots,k_r\leq T}\iint\limits_{\Omega_{\mathbf{k},M}}f_{r,\mathbf{k}}(x,y)\;dxdy = \frac{6}{\pi^2}\iint\limits_{\mathscr{D}(T,M)}f_{r}(x,y)\;dxdy.\] Applying Lemma 16 to the first summand in 29 , we arrive at \[\begin{align} \Big|S_{r,T}(Q)-\sum_{1\leq k_1,\cdots,k_r\leq T}S_{r,\mathbf{k},M}(Q)\Big|\leq \frac{2^{3r+8}}{MQ^2}. \end{align}\]
Combining all, we get \[\begin{align} \Big|Q^2S_{r,T}(Q)-\frac{6}{\pi^2}\iint\limits_{\mathscr{D}(T,M)}f_{r}(x,y)\;dxdy\Big|&\leq 3\cdot 2^{3r+6}\frac{T^r M^2 \log Q}{Q}+7\cdot 2^{3r+6}\frac{M(1+4\log Q)}{Q}\notag\\ &\quad+ 2^{3r+6}\frac{M}{Q}+\frac{2^{3r+8}}{M}.\label{eq:approximate32SrT40Q4132by32double32integral} \end{align}\tag{30}\] Finally, we approximate \(S_r(Q)\) by \(S_{r,T}(Q)\). Similar to the proof of Lemma 16, following the proof of [5], we have \[\begin{align} \Big|S_{r,T}(Q)-S_r(Q) \Big|&\leq \frac{2^{3r+6}}{Q^3}\sum_{j_0=1}^r\sum_{\mathbf{k}, k_{j_0}>T}\sum_{j\in\mathscr{L}_\mathbf{k}}\frac{1}{\min(q_j,q_{j+1}, q_{j+r}, q_{j+r+1})}\\ &\leq \frac{2^{3r+6}\cdot r}{Q^3}\sum_{q=1}^{[2Q/T]}\frac{\varphi(q)}{q}. \end{align}\] Using Möbius inversion, we have \[\frac{\varphi(q)}{q} = \sum_{d|q}\frac{\mu(d)}{d}.\] Therefore, \[\begin{align} \sum_{q=1}^{[2Q/T]}\frac{\varphi(q)}{q} &= \sum_{q=1}^{[2Q/T]}\sum_{d|q}\frac{\mu(d)}{d} = \sum_{d\leq [2Q/T]}\frac{\mu(d)}{d}\left\lfloor \frac{[2Q/T]}{d}\right\rfloor\\ &\leq [2Q/T]\sum_{d\leq [2Q/T]}\frac{\mu(d)}{d^2}+\sum_{d\leq [2Q/T]}\frac{1}{d}\\ &\leq \frac{12Q}{\pi^2 T}+\frac{T}{Q}+\log\frac{2Q}{T}+\gamma+\frac{T}{2Q}\leq \frac{12Q}{\pi^2 T}+\log\frac{2Q}{T}+2, \end{align}\] provided that \([2Q/T]\geq 1\). Substituting back, we obtain \[\begin{align} \Big|S_{r,T}(Q)-S_r(Q) \Big|&\leq \frac{2^{3r+6}\cdot r}{Q^3}\left(\frac{12Q}{\pi^2 T}+\log(2Q/T)+2\right).\label{eq:approximate32SrT40Q4132by32Sr40Q41} \end{align}\tag{31}\]
Now we are ready to prove Theorem 11.
Proof of Theorem 11.. Observe that \[\begin{align} \bigcup_{T,M\geq 1}\mathscr{D}(T, M) = \mathscr{T} \quad\text{ and }\quad \mathscr{D}(T, M)\subset \mathscr{D}(T_1, M_1)\text{ for T_1\geq T and M_1\geq M}. \end{align}\] Therefore, \[I_r =\lim_{T_1,M_1\rightarrow\infty} \iint\limits_{\mathscr{D}(T_1,M_1)}f_r(x,y)\;dxdy,\] and \[\begin{align} \Big| I_r - \iint\limits_{ \mathscr{D}(T, M)} f_r(x,y)dx\;dy \Big| = \Big| \lim_{T_1,M_1\rightarrow\infty} \iint\limits_{ \mathscr{D}(T_1, M_1)\backslash \mathscr{D}(T, M)} f_r(x,y)dx\;dy \Big|. \end{align}\] Recall the definition of \(f_r(x,y)\) in 22 and the independence of \(\mathscr{D}(T, M)\) from \(Q\). For every \(N\) with \(2N\geq T\geq 2^{r+3}\), Lemma 17 gives \[\begin{align} \iint\limits_{\mathscr{D}(T_1, M_1)\backslash \mathscr{D}(T, M)} f_r(x,y)dx\;dy &\leq \pi^2\cdot 2^{3r+6}\frac{T_1^r M_1^2 \log N}{N}+7\pi^2\cdot 2^{3r+6}\frac{M_1(1+4\log N)}{3N}\notag\\ &\quad+ 2^{3r+6}\frac{M_1\pi^2}{3N}+\frac{2^{3r+8}\pi^2}{3M}+\frac{2^{3r+6}\pi^2 r}{3N}\left(\frac{12N}{\pi^2 T}+\log\frac{2N}{T}+2\right). \end{align}\] Since the left side is independent of \(N\), we may let \(N\rightarrow\infty\) while keeping \(T_1,T,M_1,M\) fixed. This gives \[\begin{align} \iint\limits_{\mathscr{D}(T_1, M_1)\backslash \mathscr{D}(T, M)} f_r(x,y)dx\;dy &\leq \frac{2^{3r+8}\pi^2}{3M}+\frac{2^{3r+8}r}{T}. \end{align}\] Now letting \(T_1,M_1\rightarrow\infty\), we obtain \[\Big| I_r - \iint\limits_{ \mathscr{D}(T, M)} f_r(x,y)dx\;dy \Big| \leq \frac{2^{3r+8}\pi^2}{3M}+\frac{2^{3r+8}r}{T}.\]
Choose \(M=T=\lceil (Q/\log Q)^{1/(r+3)}\rceil\). Combining with Lemma 17, for \(Q/\log Q\geq 2^{(r+3)^2}\), we arrive at \[\begin{align} |R_r| &\leq \left(3\cdot 2^{3r+6} +3\cdot 2^{3r+8} +\frac{9\cdot 2^{3r+8}r}{\pi^2}\right) \frac{\log^{1/(r+3)}Q}{Q^{2+1/(r+3)}}+7\cdot 2^{3r+6}\frac{(1+4\log Q)}{(\log Q)^{1/(r+3)}Q^{3-1/(r+3)}} \\ &\quad+2^{3r+6}\frac{1}{(\log Q)^{1/(r+3)}Q^{3-1/(r+3)}}+\frac{2^{3r+6}r}{Q^3} \left(\log Q+2\right). \end{align}\] ◻
Let \(I=(\alpha,\beta]\) be a subinterval of \((0,1])\). We devote this section to the estimation of \(S_0(Q,I)\) defined in 2 , with explicit error bounds. To this end, we give a proof of Theorem 4, which is the generalization of Mundici’s original conjecture to short intervals.
The general idea of the proof follows that in Section 9 of [5], except that we obtain concrete error bounds for the implied constants. In this subsection, we present several lemmas needed for the asymptotic formula of \(S_0(Q,I)\). To do so, similarly to Section 3, we need to estimate sums of the type \[\begin{align} S_I' = S_I'(f,\Omega) = \sum_{\substack{(a,b)\in\Omega\cap\mathbb{Z}^2\\\gcd(a,b)=1\\\overline{b}\in I_a}} f(a,b),\label{def:S39I} \end{align}\tag{32}\] where \(I_q := [q(1-\beta),\;q(1-\alpha))\).
Lemma 18. (Effective version of Lemma 1.6 in [6])For any positive integer \(q\), any integers \(m\) and \(n\), and any subinterval \(I\) of \([1,q]\), denote \[\begin{align} S_{I}(m,n,q):=\sum_{\substack{x\in I\\(x,q)=1}}e\left(\frac{mx+n\bar{x}}{q}\right).\label{def:SI40m44n44q41} \end{align}\qquad{(6)}\] Then for \(q\geq e^3\), \[|S_I(m,n,q)|\leq (n,q)^{\frac{1}{2}}q^{\frac{1}{2}+\frac{1.0661}{\log \log q}}(2+\log q).\]
Proof. Let \(\sigma_0(q)\) denote the number of divisors of \(q\). Following the proof of [6], we have \[\begin{align} |S_{I}(m,n,q)|&\leq \frac{1}{q}\sum^{q-1}_{k=1}\frac{1}{2\|\frac{k}{q}\|}|S(m-k,n,q)|+ \frac{|I|}{q}|S(m,n,q)|\\ &\leq \sigma_0(q)(n,q)^{\frac{1}{2}}q^{\frac{1}{2}}\left(\frac{1}{2q}\sum^{q-1}_{k=1}\frac{1}{\|\frac{k}{q}\|}+ \frac{|I|}{q}\right)\\ &\leq \sigma_0(q)(n,q)^{\frac{1}{2}}q^{\frac{1}{2}}(2+\log q), \end{align}\] where the second inequality follows from the explicit upper bound of \(S(m,n,q)\) in [12]. For \(q\geq e^3\), using the upper bound of \(\sigma_0(q)\) in [13], we obtain \[|S_I(m,n,q)|\leq (n,q)^{\frac{1}{2}}q^{\frac{1}{2}+\frac{1.0661}{\log \log q}}(2+\log q).\] ◻
Lemma 19. (Effective version of Lemma 9 in [5])Let \(f\) be a \(C^1\) function on \(\Omega\), where \(\Omega\subseteq \mathbb{R}^2\) is still a convex bounded region with rectifiable boundary \(\partial\Omega\). \[\begin{align} S_{f,J}(l,a):=\sum_{\substack{b\in J\\\gcd(a,b)=1}}f(a,b)e\left(\frac{l\bar{b}}{a}\right) \label{def:Sf44J40l44a41} \end{align}\qquad{(7)}\] with \(J\) a bounded interval in \(\mathbb{R}\). Then, for \(a\geq e^3\), \[|S_{f,J}(l,a)|\leq 2m\|f\|_\infty a^{\frac{1}{2}+\frac{1.0661}{\log \log a}}(2+\log a)\left(\frac{|J|}{a}+1\right)(l,a)^{1/2},\] where \(m=m_f\) is an upper bound for the number of intervals of monotonicity of the function \(J \ni y \mapsto f(a,y)\).
Proof. By Lemma 18, for any \(J_0\) subinterval of \([1,a]\) with \(a\geq e^3\), we have \[\begin{align} |S_{1,J_0}(l,a)|=|S_{J_0}(0,l,a)|\leq a^{\frac{1}{2}+\frac{1.0661}{\log \log a}}(2+\log a)(l,a)^{1/2}. \end{align}\] Applying partial summation as in [5], we obtain the lemma. ◻
Lemma 20. (Effective version of Lemma 10 in [5])Suppose that \(\Omega\) is a convex subset of the rectangle \[[A,A+R]\times[B,B+R]\] for some \(A,B\geq e^3\) and \(R\geq 1.\) Then, \[\begin{align} |S'_I-|I|S'|&\leq \|f\|_{\infty}\frac{(R+1)R}{A}\\ &\quad+4 m\|f\|_{\infty}R(A+R)^{\frac{1}{2}+\frac{2.1322}{\log \log A}} \Big[2\log (A+R) + \log^2 (A+R) \Big], \end{align}\] where \(S'\) and \(S_I'\) are defined in 32 and 24 respectively, and \(m=m_f\) is an upper bound for the number of intervals of monotonicity of the function \(J \ni y \mapsto f(x,y)\).
Proof. Observe that \[S'_I=\sum_{\substack{(a,b)\in\Omega\cap\mathbb{Z}^2\\\gcd(a,b)=1}}f(a,b)\sum_{x\in I_a}\frac{1}{a}\sum^a_{l=1}e\left(\frac{l(\bar{b}-x)}{a}\right):= S_1+S_2,\] where \(S_1\) is the sum of terms with \(l=a\) and \(S_2\) is the sum of the remaining terms.
Since \(|I_a|=|I|a\), we then have \[S_1=\sum_{\substack{(a,b)\in\Omega\cap\mathbb{Z}^2\\ \gcd(a,b)=1}}f(a,b)\sum_{x\in I_a}\frac{1}{a}\leq \sum_{\substack{(a,b)\in\Omega\cap\mathbb{Z}^2\\ \gcd(a,b)=1}}f(a,b)\frac{1}{a}(|I_a|+1)=|I|S'+ \sum_{\substack{(a,b)\in\Omega\cap\mathbb{Z}^2\\ \gcd(a,b)=1}}f(a,b)\frac{1}{a}.\] The second sum on the right side can be bounded by \[\begin{align} \sum_{\substack{(a,b)\in\Omega\cap\mathbb{Z}^2\\ \gcd(a,b)=1}}f(a,b)\frac{1}{a} &\leq \|f\|_{\infty}\sum_{\substack{A\leq a\leq A+ R_1\\B\leq b\leq B+R_2}}\frac{1}{a}\#\{b:(a,b)\in\Omega\cap\mathbb{Z}^2,\gcd(a,b)=1\}\\ &\leq \|f\|_{\infty}\sum_{A\leq a\leq A+R}\frac{1}{a}R\leq \|f\|_{\infty}\frac{(R+1)R}{A}. \end{align}\]
Now, it remains to upper bound \(S_2\). By definition, \[S_2=\sum_{\substack{(a,b)\in\Omega\cap\mathbb{Z}^2\\\gcd(a,b)=1}}f(a,b)\sum_{x\in I_a}\frac{1}{a}\sum^{a-1}_{l=1}e\left(\frac{l(\bar{b}-x)}{a}\right)=\sum_{a\in \text{pr}_1(\Omega)}\frac{1}{a}\sum^{a-1}_{l=1}\left(\sum_{x\in I_a} e\left(-\frac{lx}{a}\right)\right)S_{f,I'_a}(l,a),\] where \(\text{pr}_1(\Omega)\) is the projection of \(\Omega\) onto the first coordinate and \(I_a' =I_a\cap \{b\in\mathbb{R}:(a,b)\in\Omega\}\). Observe that the sum over \(x\) in \(S_2\) is a geometric sum, and thus we have \[\begin{align} |S_2| &\leq \sum_{A\leq a\leq A+ R}\frac{1}{a}\sum_{l=1}^{a-1}\frac{a}{2\min(l,a-l)} \cdot |S_{f,I_a'}(l,a)|\\ &\leq \sum_{A\leq a\leq A+R}\frac{1}{a}\sum_{l=1}^{a-1} \frac{a}{\min(l,a-l)}\cdot m\|f\|_{\infty}a^{\frac{1}{2}+\frac{1.0661}{\log \log a}}(2+\log a)\left(\frac{|I_a'|}{a}+1\right)(l,a)^{1/2}\\ &\leq 4m\|f\|_{\infty}\sum_{A\leq a\leq A+ R}a^{\frac{1}{2}+\frac{1.0661}{\log \log a}}(2+\log a)\sum_{l=1}^{a-1}\frac{(l,a)^{1/2}}{l}, \end{align}\] where the second inequality follows from Lemma 19 and the third follows from the fact that \((l,a) = (a-l,a)\). We can further simplify the innermost sum as \[\begin{align} \sum^{a-1}_{l=1}\frac{(l,a)^{1/2}}{l}\leq \sum_{d|a}d^{-1/2}\log a \leq \sigma_0(a)\log a\leq a^{\frac{1.0661}{\log\log a}}\log a. \end{align}\]
Plugging this back into \(|S_2|\), we have \[\begin{align} |S_2| &\leq 4m\|f\|_{\infty}\sum_{A\leq a\leq A+ R} a^{\frac{1}{2}+\frac{2.1322}{\log \log a}}(2\log a+\log^2 a)\\ &\leq 4 m\|f\|_{\infty}R(A+R)^{\frac{1}{2}+\frac{2.1322}{\log \log A}} \Big[2\log (A+R) + \log^2 (A+R) \Big]. \end{align}\] Therefore, combining both parts, we have \[\begin{align} |S'_I-|I|S'|&\leq \|f\|_{\infty}\frac{(R+1)R}{A}\\ &\quad+4 m\|f\|_{\infty}R(A+R)^{\frac{1}{2}+\frac{2.1322}{\log \log A}} \Big[2\log (A+R) + \log^2 (A+R) \Big]. \end{align}\] ◻
Recall in 2 the definition of \(S_0(Q,I)\): \[S_0(Q,I)= \sum_{\gamma_j\in F_I(Q)}(\gamma_{j+1}-\gamma_j)^2= \sum_{\gamma_j\in F_I(Q)}\frac{1}{q^2_jq^2_{j+1}}.\]
Denote \(T=Q^c\) for small \(c\in (0,1)\). In [5], the constant \(c\) is optimized to be \(1/10\). We decompose \(S_0(Q,I)\) as the sum of \(T_1(Q,I)+ T_2(Q,I) +T_3(Q,I)\), where \[\begin{align} T_1(Q,I)&=\sum_{\substack{\gamma_j\in F_I(Q)\\q_j,q_{j+1}\geq Q/T}}\frac{1}{q^2_jq^2_{j+1}},\\ T_2(Q,I)&=\sum_{\substack{\gamma_j\in F_I(Q)\\q_j<Q/T}}\frac{1}{q^2_jq^2_{j+1}},\\ T_3(Q,I)&=\sum_{\substack{\gamma_j\in F_I(Q)\\q_{j+1}<Q/T}}\frac{1}{q^2_jq^2_{j+1}}. \end{align}\]
Now we introduce the defect constant \(c_I\) that depends on the endpoints of the interval \(I\). Denote \[c_I=\sum_{q\geq 1}\frac{\#\{a\in qI;(a,q)=1\}-|I|\varphi(q)}{q^2}.\] Notice that \[\begin{align} \left|\#\{a\in qI; (a,q)=1\}-|I|\varphi(q)\right|&= \Big|\sum_{a\in qI}\sum_{\substack{d|q\\d|a}}\mu(d)-|I|\varphi(q)\Big|\\ &=\Big|\sum_{d|q}\mu(d)\cdot\#\{a\in qI: d|a\}-|I|\varphi(q)\Big|\\ &\leq |\sum_{d|q}\mu(d)|\leq \sigma_0(q), \end{align}\] where the first inequality in the previous line follows from \(\sum_{d|q}\mu(d)/d = \varphi(q)/q\). Using the upper bound of \(\sigma_0(q)\) in [13], we can bound \(c_I\) by \[\begin{align} |c_I|\leq\sum_{q\geq 1}\frac{|\#\{a\in qI;(a,q)=1\}-|I|\varphi(q)|}{q^2}\leq \sum_{q\geq 1}\frac{\sigma_0(q)}{q^2}\leq \zeta^2(2)=\frac{\pi^4}{36}.\label{eq:bound32for32cI} \end{align}\tag{33}\]
Now we evaluate the three components of \(S_0(Q,I)\). For \(T_1(Q,I)\), apply Lemma 20 with \[\begin{align} A=B=\frac{Q}{T}, \qquad &R=\left(1-\frac{1}{T}\right)Q,\qquad \Omega = \{(x,y)\in Q\mathscr{T}:\min(x,y)\geq Q/T\},\\ &f(a,b)=\frac{1}{a^2b^2}, \qquad \text{and\quad}m=1. \end{align}\] Then, we obtain \(\|f\|_{\infty}\leq T^4/Q^4\) and so \[\begin{align} T_1(Q,I)-|I|T_1(Q) = R_{I,1},\label{eq:difference32from32I42T140Q41} \end{align}\tag{34}\] where \(T_1(Q) := T_1(Q,(0,1])\) and \[\begin{align} |R_{I,1}| &\leq \|f\|_{\infty}\frac{(R+1)R}{A}+4m\|f\|_{\infty}R(A+R)^{\frac{1}{2}+\frac{2.1322}{\log\log A}} \Big[2\log(A+R)+\log^2(A+R)\Big]\notag\\ &\leq \frac{T^5}{Q^3}\left(1-\frac{1}{T}\right)^2 +\frac{T^5}{Q^4}\left(1-\frac{1}{T}\right)+4(T^4-T^3)Q^{-\frac{5}{2}+\frac{2.1322}{\log\log Q/T}} (2\log Q+\log^2 Q). \end{align}\]
We now turn to \(T_2(Q,I)\). Following [5], we obtain \[\begin{align} T_2(Q,I)&\leq \left(1-\frac{1}{T}\right)^{-2}Q^{-2}\sum_{q\leq Q/T}\frac{\#\{a\in qI;(a,q)=1\}}{q^2}\\ &= \left(\frac{1}{Q^2}+\frac{2T-1}{(T-1)^2Q^2}\right)\sum_{q\leq Q/T}\frac{\#\{a\in qI;(a,q)=1\}}{q^2}. \end{align}\] Moreover, trivially we have \[T_2(Q,I)\geq \frac{1}{Q^2}\sum_{q\leq Q/T}\frac{\#\{a\in qI;(a,q)=1\}}{q^2}.\] Hence, combining both upper and lower bounds of \(T_2(Q,I)\), we have \[\begin{align} \Big|T_2(Q,I)-\frac{1}{Q^2}\sum_{q\leq Q/T}\frac{\#\{a\in qI;(a,q)=1\}}{q^2}\Big|&\leq \frac{(2T-1)}{(T-1)^2Q^2}\sum_{q\leq Q/T}\frac{\#\{a\in qI;(a,q)=1\}}{q^2}\notag\\ &\leq \frac{(2T-1)}{(T-1)^2Q^2}\sum_{q\leq Q/T}\frac{\varphi(q)}{q^2}\notag\\ &\leq \frac{(2T-1)(\log (Q/T) + 1)}{(T-1)^2Q^2}.\label{eq:error32term32from32T240Q44I41} \end{align}\tag{35}\]
Finally, we estimate \(T_3(Q,I)\). Following the arguments in [5], we have \[\begin{align} T_3(Q,I)&\leq \left(1-\frac{1}{T}\right)^{-2}Q^{-2}\sum_{q\leq Q/T}\frac{\#\{a\in qI';(a,q)=1\}}{q^2} \shortintertext{and} T_3(Q,I)&\geq\frac{1}{Q^2}\sum_{q\leq Q/T}\frac{\#\{a\in qI'';(a,q)=1\}}{q^2}, \end{align}\] where \[I'=\left(\alpha,\beta+\dfrac{2}{Q}\right]\quad\text{ and }\quad I''=\left(\alpha+\frac{2}{Q},\beta\right] .\] Notice that the interval \(qI'\setminus qI= (q\beta, q\beta +(2q/Q)]\) contains at most one integer for \(Q>2^{1/c}\) and \(q\leq Q/T=Q^{1-c}\). Similar arguments apply to \(qI\setminus qI''\). Thus, for \(Q>2^{1/c}\), we obtain \[\begin{align} T_3(Q,I) &\leq \left(1-\frac{1}{T}\right)^{-2}Q^{-2}\sum_{q\leq Q/T}\frac{\#\{a\in qI;(a,q)=1\}}{q^2} + \left(1-\frac{1}{T}\right)^{-2}Q^{-2}\sum_{q\leq Q/T}\frac{1}{q^2}\\ &\leq \left(1-\frac{1}{T}\right)^{-2}Q^{-2}\sum_{q\leq Q/T}\frac{\#\{a\in qI;(a,q)=1\}}{q^2}+ \frac{\pi^2}{6}\left(1-\frac{1}{T}\right)^{-2}Q^{-2}, \end{align}\] and \[\begin{align} T_3(Q,I)&\geq \frac{1}{Q^2}\sum_{q\leq Q/T}\frac{\#\{a\in qI;(a,q)=1\}}{q^2}- \frac{1}{Q^2}\sum_{q\leq Q/T}\frac{1}{q^2} \\ &\geq \frac{1}{Q^2}\sum_{q\leq Q/T}\frac{\#\{a\in qI;(a,q)=1\}}{q^2}-\frac{\pi^2}{6Q^2}. \end{align}\] Hence, together with 35 , we obtain \[\Big|T_3(Q,I)- \frac{1}{Q^2}\sum_{q\leq Q/T}\frac{\#\{a\in qI;(a,q)=1\}}{q^2}\Big|\leq \frac{(2T-1)(\log (Q/T) + 1)}{(T-1)^2Q^2}+ \frac{\pi^2T^2}{6(T-1)^2Q^2}.\]
Thus, combining the estimations for \(T_2(Q,I)\) and \(T_3(Q,I)\), we have \[\begin{align} T_2(Q,I)+T_3(Q,I)= \frac{2}{Q^2}\sum_{q\leq Q/T}\frac{\#\{a\in qI;(a,q)=1\}}{q^2}+ R_{I,2},\label{eq:T243T343error32term} \end{align}\tag{36}\] where \[|R_{I,2}|\leq \frac{(4T-2)(\log (Q/T) + 1)}{(T-1)^2Q^2}+ \frac{\pi^2T^2}{6(T-1)^2Q^2}.\]
Define \(T_2(Q) := T_2(Q,(0,1])\) and \(T_3(Q) := T_3(Q,(0,1])\). Using 36 with \(I=(0,1]\), we have \[\begin{align} T_2(Q)+T_3(Q)= \frac{2}{Q^2}\sum_{q\leq Q/T}\frac{\varphi(q)}{q^2} + R_{I,3},\label{eq:T243T3} \end{align}\tag{37}\] where \[|R_{I,3}|\leq \frac{(4T-2)(\log (Q/T) + 1)}{(T-1)^2Q^2}.\] Notice that the \(\pi^2/6\)-term is not included in \(R_{I,3}\), because when \(I\) is the full interval, \(qI'\setminus qI\) and \(qI\setminus qI''\) don’t contain any integer for \(Q>2^{1/c}\). Recall that \(T=Q^c\). Substituting \(c=1/10\) and combining all restrictions for \(Q\), 33 , 34 , 36 , and 37 , we obtain that for \(Q> 1024\), \[\begin{align} S_0(Q,I)&=T_1(Q,I)+T_2(Q,I)+T_3(Q,I)\notag\\ &= |I|S_0(Q) + \frac{2}{Q^2}\sum_{q\leq Q/T}\frac{\#\{a\in qI;(a,q)=1\}-|I|\varphi(q)}{q^2} +R_{I,1}+R_{I,2}+|I|R_{I,3}\notag\\ &=|I| S_0(Q) + R_{I,4},\label{eq:S040Q44I4132final32asymptotic} \end{align}\tag{38}\] where \[\begin{align} |R_{I,4}|&\leq \frac{\pi^4}{18Q^2}+Q^{-5/2}\left(1-Q^{-1/10}\right)^2\\ &\quad+4Q^{-21/10+2.1322/(\log\log Q-0.1054)}(1+Q^{-1/10}) (2\log Q+\log^2Q)\\ &\quad+\frac{(|I|+1)(4Q^{1/10}-2)((9\log Q)/10+1)}{(Q^{1/10}-1)^2Q^2}+\frac{\pi^2Q^{-9/5}}{6(Q^{1/10}-1)^2}. \end{align}\] Note that the constant \(-0.1054\) comes from bounding \(\log\log (Q/T)\geq \log\log Q-0.1054\).
Substituting the explicit bound of \(S_0(Q)\) from [1] into 38 , we arrive at the desired result.
Remark 21. In [5], the analogue of our Theorem 9 is Theorem 2. Their Theorem 2 includes terms of order \(Q^{-2}\) in the main term rather than in the error term. This is not achievable in our explicit setting. Indeed, when estimating \(T_2(Q,I)\) and \(T_3(Q,I)\), we don’t have an explicit upper bound on \(Q\) which guarantees that the interval \[qI'\setminus qI= \left(q\beta, q\beta +\dfrac{2q}{Q}\right]\] contains no integer. This is precisely where the error term of order \(Q^{-2}\) arises. Moreover, since the contribution from terms involving the defect \(c_I\) is absorbed in an error term of order \(Q^{-2}\), the restriction that the endpoints of the subinterval \(I\) be rational can be removed.
A.D. is supported by the Shaff–Andrews Fellowship, Department of Mathematics, University of Illinois Urbana-Champaign.