A note on the pair correlation of Farey fractions

The Farey fractions sequence $\mathfrak{F}_{Q}:=\{\frac{a}{q}:0<a\leq q\leq Q,(a,q)=1\}$ arises in several problems in mathematics. The elements of $\mathfrak{F}_{Q}$ are well known to be uniformly distributed in $[0,1]$ as $Q\rightarrow\infty$ [19], with discrepancy exactly $\frac{1}{Q}$ [11]. The distribution of Farey fractions is of major interest, due in part to the connection with the distribution of zeros of the Riemann zeta function [13, 17] or of Dirichlet $L$-functions [16].

Although the major problems in the area remain widely open, the spacing statistics of Farey fractions are more accessible. The gap distribution of $h$-tuples of consecutive gaps between elements of $\mathfrak{F}_{Q}$ was computed in [14] for $h=1$ and in [3] for $h\geq 2$. More recently, the correlations of $\mathfrak{F}_{Q}$, shown to exist and explicitly computed by Zaharescu and the first author [9], turned out to play a key role in the study of the moments of eigenvalues of large sieve matrices [8].

Motivated by Huxley’s work [16], a number of papers investigated various features (such as discrepancy or gap distribution) of the distribution of Farey fractions with denominators subjected to various constraints [1, 2, 4, 7, 15, 18].

For every finite set $F\subseteq\mathbb{R}$ of cardinality $N(F)$ and every interval $I$, define

$${\mathcal{G}}_{F}(I):=\frac{1}{N(F)}\,\#\bigg{\{}(x,y)\in F^{2}:y\neq x,\,y-x\in\frac{1}{N(F)}\,I+\mathbb{Z}\bigg{\}}.$$

(1.1)

The pair correlation measure of an increasing sequence $(F_{n})$ of finite subsets of $\mathbb{R}$ is defined (when it exists) by

$${\mathcal{G}}(I):=\lim\limits_{n}{\mathcal{G}}_{F_{n}}(I)\qquad(\text{$I$ interval}).$$

If, in addition,

$$G(\lambda):={\mathcal{G}}([0,\lambda])=\int_{0}^{\lambda}g(x)\,dx,$$

then $g$ is called the pair correlation function of $(F_{n})$.

The pair correlation function of $\mathfrak{F}_{Q}$ was shown in [9] to be given by

$$g(\lambda)=\frac{1}{\zeta(2)\lambda^{2}}\sum\limits_{1\leq\Delta\leq 2\zeta(2)\lambda}\varphi(\Delta)\log\frac{2\zeta(2)\lambda}{\Delta},\qquad\forall\lambda>0.$$

(1.2)

This formula was useful in [8] to recognize the connection between the pair correlation of $\mathfrak{F}_{Q}$ and the expression of the main term of the second moment of the large sieve matrix, provided in [20].

The proof of formula (1.2) given in [9] relies essentially on the Poisson summation formula. The original motivation of this note was to re-prove (1.2) using some different counting arguments that also provide effective estimates. Our direct approach turns out to also work well in the case of two important subsets of $\mathfrak{F}_{Q}$, obtained by imposing congruence conditions on the denominators:

	$\displaystyle\mathfrak{F}_{Q}^{(m)}$	$\displaystyle:=\bigg{\{}\gamma=\frac{a}{q}\in\mathfrak{F}_{Q}:(q,m)=1\bigg{\}},$
	$\displaystyle\mathfrak{F}_{Q}^{(m,b)}$	$\displaystyle:=\bigg{\{}\frac{a}{q}\in\mathfrak{F}_{Q}:q\equiv b\pmod{m}\bigg{\}},$

where $m\in\mathbb{N}$, $b\in\mathbb{Z}$ and $(b,m)=1$.

Set

$$N_{Q,m;\alpha,\beta}:=\#(\mathfrak{F}_{Q}^{(m)}\cap(\alpha,\beta]),\quad\text{and }\quad N_{Q,m}:=N_{Q,m;0,1}.$$

The following constant will appear several times in this paper:

$$C_{m}:=\frac{\varphi(m)}{\zeta(2)m}\prod_{\begin{subarray}{c}p|m\\ p\text{ prime}\end{subarray}}\bigg{(}1-\frac{1}{p^{2}}\bigg{)}^{-1}=\frac{\varphi(m)}{m}\prod_{\begin{subarray}{c}p\nmid m\\ p\text{ prime}\end{subarray}}\bigg{(}1-\frac{1}{p^{2}}\bigg{)}.$$

As noticed at the beginning of Section 2, for every $0\leq\alpha<\beta\leq 1$ we have

	$\displaystyle N_{Q,m;\alpha,\beta}$	$\displaystyle=(\beta-\alpha)N_{Q,m}+O_{\delta}(Q^{1+\delta})$
		$\displaystyle=\frac{(\beta-\alpha)C_{m}}{2}\,Q^{2}+O_{\delta}(Q^{1+\delta}),\quad\forall\delta>0,$

which gives an effective estimate for the uniform distribution of $\mathfrak{F}_{Q}^{(m)}$.

In the first part of Section 2 we prove

In particular, the support of the function $g_{(m)}$ is the interval $[\frac{1}{2}C_{m},\infty)$.

Next, we extend the equality $\lim\limits_{\lambda\rightarrow\infty}g_{(1)}(\lambda)=1,$ due to R. R. Hall and presented in [9], by proving that

$$\lim\limits_{\lambda\rightarrow\infty}g_{(m)}(\lambda)=1,\quad\forall m\in\mathbb{N}.$$

(1.3)

In Section 3 we investigate the pair correlation of $\mathfrak{F}_{Q}^{(m,b)}$ under the assumption $(b,m)=1$. The cardinality of $\mathfrak{F}_{Q}^{(m,b)}$ is given by

$$N_{Q,(m,b)}=\frac{C_{m}}{2\varphi(m)}\,Q^{2}+O_{m}(Q\log Q).$$

(1.4)

Furthermore, we have

$$\#(\mathfrak{F}_{Q}^{(m,b)}\cap(\alpha,\beta])=(\beta-\alpha)N_{Q,(m,b)}+O_{\delta}(Q^{1+\delta}),\quad\forall\delta>0,$$

showing effectively that the elements of $\mathfrak{F}_{Q}^{(m,b)}$ are uniformly distributed.

We prove

Our approach also allows us to prove that the pair correlation function of $\mathfrak{F}_{Q}^{(m)}\cap I$, respectively $\mathfrak{F}_{Q}^{(m,b)}\cap I$, coincides with $g_{(m)}$, respectively $\widetilde{g}_{(m)}$, for every interval $I\subseteq[0,1]$. It also gives effective asymptotic formulas in $Q$ for the quantities ${\mathcal{G}}_{\mathfrak{F}_{Q}}([0,\lambda])$, ${\mathcal{G}}_{\mathfrak{F}_{Q}^{(m)}}([0,\lambda])$ and ${\mathcal{G}}_{\mathfrak{F}_{Q}^{(m,b)}}([0,\lambda])$.

A related result is contained in [21, case $n=1$ of Theorem 3.2]. However, our result involves the extra coprimality condition $(a,q)=1$ in the definition of $\mathfrak{F}_{Q}^{(m,b)}$, which is not included in formulas (3.3.1) and (3.3.2) of [21].

Lemma 2.1 of [6] gives

$$N_{Q,m}=1+\sum\limits_{\begin{subarray}{c}k=1\\ (k,m)=1\end{subarray}}^{Q}\varphi(k)=C_{m}\frac{Q^{2}}{2}\ +\ O(Q\log Q).$$

When restricting to $[0,\beta]$, the number of new fractions in the $k$th step is not $\varphi(k)$ anymore, but rather

$$\sum\limits_{\begin{subarray}{c}n=1\\ (n,k)=1\end{subarray}}^{\lfloor k\beta\rfloor}1=\frac{\varphi(k)}{k}\lfloor k\beta\rfloor+O_{\delta}(k^{\delta}),$$

where one can use, for example, [10, Lemma A.1]. Thus

	$\displaystyle N_{Q,m;0,\beta}$	$\displaystyle=\beta N_{Q,m}+O_{\delta}(Q^{1+\delta}),$
	$\displaystyle N_{Q,m;\alpha,\beta}$	$\displaystyle=N_{Q,m;0,\beta}-N_{Q,m;0,\alpha}=(\beta-\alpha)N_{Q,m}+O_{\delta}(Q^{1+\delta}).$

Set

	$\displaystyle H_{Q,m;\beta}(\lambda)$	$\displaystyle:=\#\bigg{\{}(\gamma,\gamma^{\prime}):\gamma,\gamma^{\prime}\in\mathfrak{F}_{Q}^{(m)},0<\gamma^{\prime}-\gamma\leq\frac{\lambda}{Q^{2}},\ \gamma^{\prime}\leq\beta\bigg{\}},$
	$\displaystyle\overline{H}_{Q,m;\alpha}(\lambda)$	$\displaystyle:=\#\bigg{\{}(\gamma,\gamma^{\prime}):\gamma,\gamma^{\prime}\in\mathfrak{F}_{Q}^{(m)},0<\gamma^{\prime}-\gamma\leq\frac{\lambda}{Q^{2}},\ \gamma\leq\alpha\bigg{\}}.$

If the limit exists, set

$$G_{(m;\alpha,\beta)}(\lambda):=\lim_{Q\rightarrow\infty}\frac{1}{N_{Q,m;\alpha,\beta}(Q)}\,\#\left\{(\gamma,\gamma^{\prime}):\begin{matrix}\gamma,\gamma^{\prime}\in\mathfrak{F}_{Q}^{(m)}\\ 0<\gamma^{\prime}-\gamma\leq\frac{\lambda}{N_{Q,m;\alpha,\beta}}\end{matrix}\right\}.$$

Then

	$\displaystyle H_{Q,m;\beta}(\lambda)$	$\displaystyle=\sum\limits_{1\leq\Delta\leq\lambda}\#S_{Q,m;\beta}(\Delta,\lambda)=\sum\limits_{1\leq\Delta\leq\lambda}\#\widetilde{S}_{Q,m;\beta}(\Delta,\lambda),$
	$\displaystyle\overline{H}_{Q,m;\alpha}(\lambda)$	$\displaystyle=\sum\limits_{1\leq\Delta\leq\lambda}\#\overline{S}_{Q,m;\alpha}(\Delta,\lambda),$

where we used the variables $x=a^{\prime},v=q^{\prime},u=q,y=a$ to get

	$\displaystyle S_{Q,m;\beta}(\Delta,\lambda)=\left\{(u,v,x,y)\in\mathbb{N}^{4}:\begin{matrix}xu-yv=\Delta,\ \begin{subarray}{c}x\leq\beta v,\ v\leq Q\\ y\leq u\leq Q\end{subarray},\ \begin{subarray}{c}(x,v)=1\\ (y,u)=1\end{subarray}\\ \frac{\Delta}{uv}\leq\frac{\lambda}{Q^{2}},\ \begin{subarray}{c}(u,m)=1\\ (v,m)=1\end{subarray}\end{matrix}\right\},$
	$\displaystyle\widetilde{S}_{Q,m;\beta}(\Delta,\lambda)=\left\{(u,v,x)\in\mathbb{N}^{3}:\begin{array}[]{c}\frac{\Delta Q}{\lambda}\leq v\leq Q,\ \frac{\Delta Q^{2}}{\lambda v}\leq u\leq Q,\ \begin{subarray}{c}x\leq\beta v\\ (x,v)=1\end{subarray}\\ xu\equiv\Delta\mod v,\ (\frac{xu-\Delta}{v},u)=1,\ \begin{subarray}{c}(u,m)=1\\ (v,m)=1\end{subarray}\end{array}\right\},$
	$\displaystyle\overline{S}_{Q,m;\alpha}(\Delta,\lambda)=\left\{(u,v,x,y)\in\mathbb{N}^{4}:\begin{matrix}xu-yv=\Delta,\ \begin{subarray}{c}x\leq v\leq Q\\ y\leq\alpha u,\ u\leq Q\end{subarray},\ \begin{subarray}{c}(x,v)=1\\ (y,u)=1\end{subarray}\\ \frac{\Delta}{uv}\leq\frac{\lambda}{Q^{2}},\ \begin{subarray}{c}(u,m)=1\\ (v,m)=1\end{subarray}\end{matrix}\right\}.$

Observe that $xu-yv=\Delta\Leftrightarrow\frac{x}{v}=\frac{y}{u}+\frac{\Delta}{uv}$ implies

$$\#S_{Q,m;\beta}(\Delta,\lambda)\leq\#\overline{S}_{Q,m;\beta}(\Delta,\lambda)\leq\#S_{Q,m;\beta+\frac{\lambda}{Q^{2}}}(\Delta,\lambda),$$

so that $\overline{H}_{Q,m;\beta}(\lambda)$ is asymptotically the same as $H_{Q,m;\beta}(\lambda)$ as $Q\rightarrow\infty$. Thus it suffices to estimate $\#\widetilde{S}_{Q,m;\beta}(\Delta,\lambda)$ as follows:

	$\displaystyle\#\widetilde{S}_{Q,m;\beta}(\Delta,\lambda)$	$\displaystyle=\sum\limits_{\begin{subarray}{c}\frac{\Delta Q}{\lambda}\leq v\leq Q\\ (v,m)=1\end{subarray}}\sum\limits_{\begin{subarray}{c}\frac{\Delta Q^{2}}{\lambda v}\leq u\leq Q,\ (u,m)=1\\ x\leq\beta v,\ (x,v)=1\\ xu\equiv\Delta\mod v\\ (\frac{xu-\Delta}{v},u)=1\end{subarray}}1$
		$\displaystyle=\sum\limits_{\begin{subarray}{c}\frac{\Delta Q}{\lambda}\leq v\leq Q\\ (v,m)=1\end{subarray}}\sum\limits_{\begin{subarray}{c}\frac{\Delta Q^{2}}{\lambda v}\leq u\leq Q,\ (u,m)=1\\ x\leq\beta v,\ (x,v)=1\\ xu\equiv\Delta\mod v\end{subarray}}\sum\limits_{\begin{subarray}{c}d\mid\frac{xu-\Delta}{v}\\ d\mid u\end{subarray}}\mu(d)$
		$\displaystyle\stackrel{{\scriptstyle u=dw}}{{=}}\sum\limits_{\begin{subarray}{c}d\mid\Delta\\ (d,m)=1\end{subarray}}\mu(d)\sum\limits_{\begin{subarray}{c}\frac{\Delta Q}{\lambda}\leq v\leq Q\\ (v,m)=1\end{subarray}}\ \ \sum\limits_{\begin{subarray}{c}\frac{\Delta Q^{2}}{\lambda dv}\leq w\leq\frac{Q}{d},\ (w,m)=1\\ x\leq\beta v,\ (x,v)=1\\ xw\equiv\frac{\Delta}{d}\mod v\end{subarray}}1.$		(2.1)

To estimate the innermost sum on the right hand side in (2), we need to check that [10, Proposition A.3] carries over with the additional condition $(q,m)=1$. Set

$$\mathcal{N}_{q,h,m}(I_{1},I_{2})=\{(x,y)\in I_{1}\times I_{2}:(x,q)=1,\ (y,m)=1,\ xy\equiv h\mod q\}.$$

Thus, under the correspondence $v\leftrightarrow q,x\leftrightarrow x,w\leftrightarrow y,\frac{\Delta}{d}\leftrightarrow h$, relation (2) becomes

	$\displaystyle\#\widetilde{S}_{Q,m;\beta}(\Delta,\lambda)=\sum\limits_{\begin{subarray}{c}d\mid\Delta\\ (d,m)=1\end{subarray}}\mu(d)\sum\limits_{\begin{subarray}{c}\frac{\Delta Q}{\lambda}\leq v\leq Q\\ (v,m)=1\end{subarray}}\bigg{(}\mathcal{N}_{v,\frac{\Delta}{d},m}\Big{(}[0,\beta v],\big{[}0,\tfrac{Q}{d}\big{]}\Big{)}-\mathcal{N}_{v,\frac{\Delta}{d},m}\Big{(}[0,\beta v],\big{[}0,\tfrac{\Delta Q^{2}}{\lambda dv}\big{]}\Big{)}\bigg{)}$
	$\displaystyle\qquad=\sum\limits_{\begin{subarray}{c}d\mid\Delta\\ (d,m)=1\end{subarray}}\mu(d)\sum\limits_{\begin{subarray}{c}\frac{\Delta Q}{\lambda}\leq v\leq Q\\ (v,m)=1\end{subarray}}\Bigg{(}\frac{\varphi(v)}{v^{2}}\cdot\frac{\varphi(m)}{m}\,\beta v\bigg{(}\frac{Q}{d}-\frac{\Delta Q^{2}}{\lambda vd}\bigg{)}+O_{\delta,m,\Delta}\bigg{(}v^{1/2+\delta}\Big{(}\frac{Q}{v}+\frac{Q^{2}}{\lambda v^{2}}\Big{)}\bigg{)}\Bigg{)}$
	$\displaystyle\qquad=\beta Q\,\frac{\varphi(m)}{m}\sum\limits_{\begin{subarray}{c}d\mid\Delta\\ (d,m)=1\end{subarray}}\frac{\mu(d)}{d}\sum\limits_{\begin{subarray}{c}\frac{\Delta Q}{\lambda}\leq v\leq Q\\ (v,m)=1\end{subarray}}\frac{\varphi(v)}{v}\bigg{(}1-\frac{\Delta Q}{\lambda v}\bigg{)}\ +\ O_{\delta,m,\Delta}(Q^{3/2+\delta}).$

Note that there is no dependence of $\lambda$ in the error term because the inequality $\gamma^{\prime}-\gamma\geq\frac{1}{Q^{2}}$, $\forall\gamma,\gamma^{\prime}\in\mathfrak{F}_{Q}$, $\gamma<\gamma^{\prime}$ allows for $\lambda\geq 1$.

A simple calculation shows that the function

$$K_{m}(n)=\sum\limits_{\begin{subarray}{c}d\mid n\\ (d,m)=1\end{subarray}}\frac{\mu(d)}{d}$$

is multiplicative for every $m$, and that at prime powers we have

$$K_{m}(p^{\ell})=\sum\limits_{\begin{subarray}{c}d\mid p^{\ell}\\ (d,m)=1\end{subarray}}\frac{\mu(d)}{d}=\begin{cases}1,&\text{ if }p\mid m\\ 1-\frac{1}{p}=\frac{\varphi(p^{\ell})}{p^{\ell}},&\text{ if }p\nmid m\end{cases}.$$

Therefore

$$K_{m}(\Delta)=\prod_{\begin{subarray}{c}p\mid\Delta\\ p\nmid m\end{subarray}}\bigg{(}1-\frac{1}{p}\bigg{)}=\frac{\prod\limits_{\begin{subarray}{c}p\mid\Delta\end{subarray}}\big{(}1-\frac{1}{p}\big{)}}{\prod\limits_{\begin{subarray}{c}p\mid\Delta\\ p|m\end{subarray}}\big{(}1-\frac{1}{p}\big{)}}=\frac{\prod\limits_{\begin{subarray}{c}p\mid\Delta\end{subarray}}\big{(}1-\frac{1}{p}\big{)}}{\prod\limits_{\begin{subarray}{c}p\mid(\Delta,m)\end{subarray}}\big{(}1-\frac{1}{p}\big{)}}=\frac{\frac{\varphi(\Delta)}{\Delta}}{\frac{\varphi\big{(}(\Delta,m)\big{)}}{(\Delta,m)}},$$

and consequently

$$\#\widetilde{S}_{Q,m;\beta}(\Delta,\lambda)=\beta Q\,\frac{\varphi(m)}{m}\cdot\frac{\varphi(\Delta)}{\Delta}\cdot\frac{(\Delta,m)}{\varphi\big{(}(\Delta,m)\big{)}}\sum\limits_{\begin{subarray}{c}\frac{\Delta Q}{\lambda}\leq v\leq Q\\ (v,m)=1\end{subarray}}\frac{\varphi(v)}{v}\bigg{(}1-\frac{\Delta Q}{\lambda v}\bigg{)}+\ O_{\delta,m,\Delta}(Q^{3/2+\delta}).$$

Using [6, Lemma 2.1] twice, we get

$$\#\widetilde{S}_{Q,m;\beta}(\Delta,\lambda)=\beta C_{m}Q^{2}\,\frac{\varphi(m)}{m}\cdot\frac{\varphi(\Delta)}{\Delta}\cdot\frac{(\Delta,m)}{\varphi\big{(}(\Delta,m)\big{)}}\bigg{(}1-\frac{\Delta}{\lambda}-\frac{\Delta}{\lambda}\log\frac{\lambda}{\Delta}\bigg{)}+\ O_{\delta,m,\Delta}(Q^{3/2+\delta}).$$

Thus, for every $K\geq 1$ we have, uniformly in $\lambda\in[1,K]$,

$$H_{Q,m;\beta}(\lambda)=\beta C_{m}Q^{2}\ \frac{\varphi(m)}{m}\sum\limits_{1\leq\Delta\leq\lambda}\frac{\varphi(\Delta)}{\Delta}\cdot\frac{(\Delta,m)}{\varphi\big{(}(\Delta,m)\big{)}}\bigg{(}1-\frac{\Delta}{\lambda}-\frac{\Delta}{\lambda}\log\frac{\lambda}{\Delta}\bigg{)}+\ O_{\delta,m,K}(Q^{3/2+\delta}).$$

This leads in turn to

$$G_{(m;\alpha,\beta)}(\lambda)=2\,\frac{\varphi(m)}{m}\sum\limits_{1\leq\Delta\leq\frac{2\lambda}{C_{m}}}\frac{\varphi(\Delta)}{\Delta}\bigg{(}1-\frac{\Delta C_{m}}{2\lambda}-\frac{\Delta C_{m}}{2\lambda}\log\frac{2\lambda}{\Delta C_{m}}\bigg{)}.$$

(2.3)

Finally, Theorem 1 follows by differentiating in (2.3).

For $m=1$, $\alpha=0$, $\beta=1$, we retrieve Theorem 1 in [9]. For $m=2$ we get

$$g_{(2)}(\lambda)=\frac{1}{3\zeta(2)\lambda^{2}}\sum\limits_{1\leq\Delta\leq 3\zeta(2)\lambda}\varphi(\Delta)(\Delta,2)\log\frac{3\zeta(2)\lambda}{\Delta}.$$

In the remaining part of this section we prove equality (1.3). Consider the Dirichlet series

$$D_{m}(s):=\sum\limits_{\Delta=1}^{\infty}\frac{K_{m}(\Delta)}{\Delta^{s-1}}=\prod\limits_{p}\bigg{(}1+\frac{K_{m}(p)}{p^{s-1}}+\frac{K_{m}(p^{2})}{p^{2s-2}}+\cdots\bigg{)}\quad\text{if }\operatorname{Re}s>2.$$

Take $\displaystyle\zeta_{m}(s):=\prod\limits_{p\nmid m}\bigg{(}1-\frac{1}{p^{s}}\bigg{)}^{-1}$, $\operatorname{Re}s>1$. We have

$$\frac{\zeta_{m}(s-1)}{\zeta_{m}(s)}=\prod\limits_{p\nmid m}\frac{1-\frac{1}{p^{s}}}{1-\frac{1}{p^{s-1}}}=\prod\limits_{p\nmid m}\frac{p^{s}-1}{p^{s}-p}$$

and

$$\sum\limits_{\ell=0}^{\infty}\frac{K_{m}(p^{\ell})}{p^{\ell(s-1)}}=\begin{cases}1+\big{(}1-\frac{1}{p}\big{)}\frac{1}{p^{s-1}}\cdot\frac{1}{1-\frac{1}{p^{s-1}}}=1+\frac{p-1}{p^{s}-p}=\frac{p^{s}-1}{p^{s}-p}&\text{if }p\nmid m\\ \big{(}1-\frac{1}{p^{s-1}}\big{)}^{-1}&\text{if }p\mid m\end{cases},$$

leading to

	$\displaystyle D_{m}(s)$	$\displaystyle=\frac{\zeta_{m}(s-1)}{\zeta_{m}(s)}\prod\limits_{p\mid m}\bigg{(}1-\frac{1}{p^{s-1}}\bigg{)}^{-1}=\frac{\zeta(s-1)}{\zeta(s)}\prod\limits_{p\nmid m}\frac{p^{s}-p}{p^{s}-1}\prod\limits_{p\mid m}\frac{p^{s-1}}{p^{s-1}-1}$
		$\displaystyle=\frac{\zeta(s-1)}{\zeta(s)}\,c_{m}(s),\quad\text{where }c_{m}(s):=\prod\limits_{p\mid m}\frac{1}{1-\frac{1}{p^{s}}}.$

Next we follow closely the final part of [9]. By Perron’s formula [9, (4.14)] we infer

	$\displaystyle\sum\limits_{1\leq\Delta\leq x}\Delta K_{m}(\Delta)\log\frac{x}{\Delta}$	$\displaystyle=\frac{1}{2\pi i}\int\limits_{\sigma_{0}-i\infty}^{\sigma_{0}+i\infty}D_{m}(s)\,\frac{x^{s}}{s^{2}}\,ds$
		$\displaystyle=\frac{1}{2\pi i}\int\limits_{\sigma_{0}-i\infty}^{\sigma_{0}+i\infty}\frac{\zeta(s-1)}{\zeta(s)}\,c_{m}(s)\,\frac{x^{s}}{s^{2}}\,ds\qquad(\sigma_{0}>2).$

Moving the contour at $\operatorname{Re}s=1$ and employing the notation from [9] we get

	$\displaystyle\frac{\zeta(it)}{\zeta(1+it)}\prod\limits_{p\mid m}\bigg{(}1-\frac{1}{p^{it}}\bigg{)}$	$\displaystyle=\chi(it)\,\frac{\zeta(1-it)}{\zeta(1+it)}\prod\limits_{p\mid m}\bigg{(}1-\frac{1}{p^{it}}\bigg{)},$
	$\displaystyle\zeta(1-it)$	$\displaystyle=\overline{\zeta(1+it)},$
	$\displaystyle\underset{s=2}{\operatorname{Res}}\,\frac{\zeta(s-1)}{\zeta(s)}\,c_{m}(s)\,\frac{x^{s}}{s^{2}}$	$\displaystyle=\lim\limits_{s\rightarrow 2}\frac{(s-2)\zeta(s-1)}{\zeta(s)}\,c_{m}(s)\,\frac{x^{s}}{4}$
		$\displaystyle=\frac{c_{m}(2)}{\zeta(2)}\cdot\frac{x^{2}}{4}\quad\text{($s=2$ is a simple pole)}.$

Estimating the error as in [9] we find

	$\displaystyle\sum\limits_{1\leq\Delta\leq x}\Delta K_{m}(\Delta)\log\frac{x}{\Delta}$	$\displaystyle=\underset{s=2}{\operatorname{Res}}\,\frac{\zeta(s-1)}{\zeta(s)}\,c_{m}(s)\,\frac{x^{s}}{s^{2}}+O_{m}(x)$
		$\displaystyle=\frac{c_{m}(2)}{\zeta(2)}\cdot\frac{x^{2}}{4}+O_{m}(x).$

Setting $\mu:=\frac{2\lambda}{C_{m}}$ we get $\lambda=\frac{C_{m}\mu}{2}$ and (as $\lambda\rightarrow\infty$)

	$\displaystyle g_{(m)}(\lambda)$	$\displaystyle=g_{(m)}(\tfrac{1}{2}C_{m}\mu)=\frac{\varphi(m)}{m}\,C_{m}\,\frac{4}{C_{m}^{2}\mu^{2}}\sum\limits_{1\leq\Delta\leq\mu}\Delta K_{m}(\Delta)\log\frac{\mu}{\Delta}$
		$\displaystyle=\frac{4\varphi(m)}{mC_{m}\mu^{2}}\left(\prod\limits_{p\mid m}\bigg{(}1-\frac{1}{p^{2}}\bigg{)}^{-1}\frac{\mu^{2}}{4\zeta(2)}+O_{m}(\mu)\right)$
		$\displaystyle=\frac{1}{\zeta(2)}\prod\limits_{p\nmid m}\bigg{(}1-\frac{1}{p^{2}}\bigg{)}^{-1}\prod\limits_{p\mid m}\bigg{(}1-\frac{1}{p^{2}}\bigg{)}^{-1}+O_{m}(\mu^{-1})$
		$\displaystyle=1+O_{m}(\lambda^{-1}).$