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Research article Special Issues

On mean square of the error term of a multivariable divisor function

  • Let τ(n) be the Dirichlet divisor function and k2 be a fixed integer. We give an asymptotic formula of the mean square of

    Δk(x)=n1,,nkxτ(n1nk)xkPk(logx).

    Citation: Zhen Guo. On mean square of the error term of a multivariable divisor function[J]. AIMS Mathematics, 2024, 9(10): 29197-29219. doi: 10.3934/math.20241415

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  • Let τ(n) be the Dirichlet divisor function and k2 be a fixed integer. We give an asymptotic formula of the mean square of

    Δk(x)=n1,,nkxτ(n1nk)xkPk(logx).



    Let τ(n) be the Dirichlet divisor function. It is known that for a real number x2,

    nxτ(n)=xlogx+(2γ1)x+Δ(x), (1.1)

    where γ is Euler's constant. It was first proved by Dirichlet that

    Δ(x)x.

    Let θ denotes the smallest number such that

    Δ(x)xθ+ε (1.2)

    holds for any ε>0. Many authors worked on making θ smaller, such as Voronoi [1], Corput [2], Kolesnik [3], Huxley [4], etc. Until now the best result, namely θ=131/416 is due to Huxley [4].

    It is conjectured that θ=1/4, which is supported by the classical mean square result: Suppose T is a large real number, then for any ε>0, one has

    T1Δ2(x)dx=16π2C2T3/2+O(T54+ε), (1.3)

    where

    C2=n=1τ2(n)n3/2.

    This result was given by Cramér [5] in 1922. In 1956, Tong [6] showed that the error term in (1.3) can be reduced to Tlog5T. In 1988, Preissmann [7] reduced the error term to Tlog4T. In 2009, Lau and Tsang [8] proved that

    T1Δ2(x)dx=16π2C2T3/2+O(Tlog3TloglogT).

    Let k2 be a fixed integer. Tóth and Zhai [9] considered the average of divisor function of k variables. They obtain the asymptotic formula

    n1,,nkxτ(n1nk)=xkPk(logx)+O(xk1+θ+ε) (1.4)

    holds for any ε>0, where θ is the exponent in (1.2) and Pk(t) is a polynomial in t of degree k. We denote Δk(x) by

    Δk(x):=n1,,nkxτ(n1nk)xkPk(logx). (1.5)

    We have the following theorem:

    Theorem 1.1. Suppose T2 is a large real number, then we have

    T1Δk2(x)dx=k24π2T2k12L2k2(logT)+O(T2k35+ε), (1.6)

    where L2k2(u) is a polynomial in u of degree 2k2, and the implied constant about "O" depends on k and ε.

    We present some lemmas.

    Lemma 2.1. Suppose x2 is large, then for any 1Nx, we have

    Δ(x)=x1/42πnNτ(n)n3/4cos(4πnxπ4)+O(x1/2+εN1/2).

    Proof. See [10,Chapter 3.2].

    Lemma 2.2. Suppose T2, TεyT, and let

    δ1(x,y)=x1/42πnyτ(n)n3/4cos(4πnxπ4),δ2(x,y)=Δ(x)δ1(x,y),

    then we have

    2TTδ22(x,y)dxT3/2y1/2log3T+Tlog4T.

    Proof. See, for example, the following references: Lau and Tsang [8], Tsang [11], Zhai [12].

    Lemma 2.3. Suppose G0,m0 are fixed real positive numbers, let G(x) be a monotonic function defined on [a,b] such that

    |G(x)|G0

    and m(x) be a differentiable real function such that

    |m(x)|m0

    on [a,b], F()=cos() or sin() or e(), then

    baG(x)F(m(x))dxG0m01.

    Proof. See [10,Lemma 2.1].

    Lemma 2.4. Let k2 be a fixed integer and let s1,,sk be complex numbers. Then for sj>1, where j=1,2,,k, we have

    n1,,nk=1τ(n1nk)ns11nskk=ζ2(s1)ζ2(sk)Fk(s1,,sk),

    where

    Fk(s1,,sk)=n1,,nk=1f(n1,,nk)ns11nskk.

    This series is absolutely convergent provided that sj>0 and (sj+sl)>1(1j,lk), and f(n1,,nk) is multiplicative and symmetric in all variables.

    Moreover, for r1,,rk{1,2},

    n1,,nk=1f(n1,,nk)τr1(n1)τrk(nk)ns11nskk (2.1)

    is absolutely convergent provided that sj>0 and (sj+sl)>1(1j,lk).

    Proof. See Tóth and Zhai [9,Proposition 2.1], and the convergence of (2.1) is a direct corollary.

    Lemma 2.5. Suppose x,y are large real numbers, k2 is a fixed integer, s,w are given real numbers such that 0<s<1/2<w<1, f is defined in Lemma 2.4. Let M1 and M2 denote the vectors (m1,,mk) and (mk+1,,m2k), D1 and D2 denote (k1j=1mj) and (2k1j=k+1mj), respectively. Let

    Tg,k(x,y;s,w)=m1,,m2kxn1,n2ymkm2k=n1n2f(M1)f(M2)g(M1,M2)D1D2(mkm2k)sτ(n1)τ(n2)(n1n2)w,Tg,k(s,w)=M1,M2Nkn1,n2Nmkm2k=n1n2f(M1)f(M2)g(M1,M2)D1D2(mkm2k)sτ(n1)τ(n2)(n1n2)w,

    where g(M1,M2) is any function that satisfies

    g(M1,M2)(2kj=1mj)ε,

    then we have:

    (i) Tg,k(s,w) is absolutely convergent.

    (ii) We have

    Tg,k(s,w)Tg,k(x,y;s,w)x2s+ε+y12wlog3y.

    Proof. (i) Since mk/m2k=n1/n2, we can find positive integers t1,t2,g1,g2 such that

    mkm2k=n1n2=t1t2,

    where (t1,t2)=1 and t1g1=mk, t2g1=m2k, t1g2=n1, t2g2=n2. Denote by M1 and M2 the vectors (m1,,mk1) and (mk+1,,m2k1), respectively, then we have

    Tg,k(s,w)M1,M2Nk1t1,t2,g1,g2N(t1,t2)=1|f(M1,t1g1)||f(M2,t2g1)||g(M1,t1g1,M2,t2g1)|τ(t1g2)τ(t2g2)D1D2(t1g1)s(t2g1)s(t1g2)w(t2g2)wM1,M2Nk1t1,t2,g1,g2N(t1,t2)=1|f(M1,t1g1)||f(M2,t2g1)||g(M1,t1g1,M2,t2g1)|τ(t1)τ(t2)τ2(g2)D1D2(t1g1)s(t2g1)s(t1t2)wg22wM1,M2Nk1t1,t2,g1N|f(M1,t1g1)||f(M2,t2g1)||g(M1,t1g1,M2,t2g1)|τ(t1)τ(t2)D1D2g12s(t1t2)s+wg2=1τ2(g2)g22wM1,M2Nk1t1,t2,g1N|f(M1,t1g1)||f(M2,t2g1)||g(M1,t1g1,M2,t2g1)|τ(t1)τ(t2)D1D2g12s(t1t2)s+w:=U1,

    where we use the conclusion: For a given real number c, we have

    n=1τ2(n)nc<(c>1), (2.2)

    moreover, for a large real number U, using partial summation and the conclusion

    nUτ2(n)Ulog3U,

    we have

    nUτ2(n)ncU21ucd(nuτ2(n))U1clog3U. (2.3)

    Since for any real δ>0,

    τ(n)/nδ1, (2.4)

    then

    U1=M1,M2Nk1t1,t2,g1N|f(M1,t1g1)||f(M2,t2g1)||g(M1,t1g1,M2,t2g1)|D1D2(t1g1)s(t2g1)sτ(t1)τ(t2)(t1t2)wM1,M2Nk1t1,t2,g1N|f(M1,t1g1)||f(M2,t2g1)||g(M1,t1g1,M2,t2g1)|D1D2(t1g1)s(t2g1)sM1,M2Nk1t1,t2,g1N|f(M1,t1g1)||f(M2,t2g1)|(D1D2)1ε(t1g1)sε(t2g1)sεM1,M2Nk1t1,t2,g1,g1N|f(M1,t1g1)||f(M2,t2g1)|(D1D2)1ε(t1g1)sε(t2g1)sε:=U2.

    Let t1g1=mk, t2g1=m2k, by Lemma 2.4, we have

    U2=M1,M2Nk1mk,m2kN|f(M1,mk)||f(M2,m2k)|τ(mk)τ(m2k)(D1D2)1εmksεm2ksε=(M1Nk1mkN|f(M1,mk)|τ(mk)D11εmksε)21.

    Thus, we conclude that (i) holds.

    (ii) Since m1,,mk1,mk+1,,m2k1 are symmetric, we have

    Tg,k(s,w)Tg,k(x,y;s,w)T1+T2+T3, (2.5)

    where

    T1=T1(x;s,w)=M1,M2Nkn1,n2Nmkm2k=n1n2m1>x|f(M1)||f(M2)||g(M1,M2)|D1D2mksm2ksτ(n1)τ(n2)n1wn2w,T2=T2(x;s,w)=M1,M2Nkn1,n2Nmkm2k=n1n2mk>xorm2k>x|f(M1)||f(M2)||g(M1,M2)|D1D2mksm2ksτ(n1)τ(n2)n1wn2w,T3=T3(y;s,w)=M1,M2Nkn1,n2Nmkm2k=n1n2n1>yorn2>y|f(M1)||f(M2)||g(M1,M2)|D1D2mksm2ksτ(n1)τ(n2)n1wn2w.

    Similar to (i), let (t1,t2)=1 and t1g1=mk, t2g1=m2k, t1g2=n1, t2g2=n2, and denote by M1 and M2 the vectors (m1,,mk1) and (mk+1,,m2k1) respectively, then

    T1=M1,M2Nk1t1,t2,g1,g2N(t1,t2)=1m1>x|f(M1,t1g1)||f(M2,t2g1)||g(M1,t1g1,M2,t2g1)|τ(t1g2)τ(t2g2)D1D2(t1g1)s(t2g1)s(t1g2)w(t2g2)wM1,M2Nk1t1,t2,g1,g2N(t1,t2)=1m1>x|f(M1,t1g1)||f(M2,t2g1)||g(M1,t1g1,M2,t2g1)|τ(t1)τ(t2)τ2(g2)D1D2(t1g1)s(t2g1)s(t1t2)wg22wM1,M2Nk1t1,t2,g1Nm1>x|f(M1,t1g1)||f(M2,t2g1)||g(M1,t1g1,M2,t2g1)|τ(t1)τ(t2)D1D2g12s(t1t2)s+wg2=1τ2(g2)g22wM1,M2Nk1t1,t2,g1Nm1>x|f(M1,t1g1)||f(M2,t2g1)||g(M1,t1g1,M2,t2g1)|τ(t1)τ(t2)D1D2g12s(t1t2)s+w:=T1, (2.6)

    where we use (2.2).

    Let

    D1=D1/m1,

    by (2.4), we have

    T1=M1,M2Nk1t1,t2,g1Nm1>x|f(M1,t1g1)||f(M2,t2g1)||g(M1,t1g1,M2,t2g1)|τ(t1)τ(t2)D1D2(t1g1)2s3ε(t2g1)3εt1ws+3εt2s+w3εM1,M2Nk1t1,t2,g1Nm1>x|f(M1,t1g1)||f(M1,t2g1)||g(M1,t1g1,M2,t2g1)|D1D2(t1g1)2s3ε(t2g1)3εM1,M2Nk1t1,t2,g1Nm1>x|f(M1,t1g1)||f(M2,t2g1)|m11εD11εD21ε(t1g1)2s4ε(t2g1)2εM1,M2Nk1t1,t2,g1,g1Nm1>x|f(M1,t1g1)||f(M2,t2g1)|m11εD11εD21ε(t1g1)2s4ε(t2g1)2ε:=T1. (2.7)

    Let t1g1=mk, t2g1=m2k, we have

    T1=M1,M2Nk1mk,m2kNm1>x|f(M1,mk)||f(M2,m2k)|τ(mk)τ(m2k)m112s+5εD11εD21εmk2s4εm2k2ε1m12s6εx2s+6εM1,M2Nk1mk,m2kN|f(M1,mk)||f(M2,m2k)|τ(mk)τ(m2k)m112s+5εD11εD21εmk2s4εm2k2εx2s+6εM1Nk1mkN|f(M1,mk)|τ(mk)m112s+5εD11εmk2s4εM1Nk1mkN|f(M1,mk)|τ(mk)D11εmk2εx2s+ε,

    the convergence of the two series in the last step can be obtained by Lemma 2.4, and we use the arbitrariness of ε.

    For T2, if mk>x, we replace the condition "m1>x" by "t1g1>x" in (2.6), thus

    T2M1,M2Nk1t1,t2,g1Nt1g1>x|f(M1,t1g1)||f(M2,t2g1)||g(M1,t1g1,M2,t2g1)|τ(t1)τ(t2)D1D2g12s(t1t2)s+w:=T2.

    Using (2.4) we have

    T2=M1,M2Nk1t1,t2,g1Nt1g1>x|f(M1,t1g1)||f(M2,t2g1)||g(M1,t1g1,M2,t2g1)|τ(t1)τ(t2)D1D2(t1g1)2s3ε(t2g1)3εt1ws+3εt2s+w3εM1,M2Nk1t1,t2,g1Nt1g1>x|f(M1,t1g1)||f(M2,t2g1)||g(M1,t1g1,M2,t2g1)|D1D2(t1g1)2s3ε(t2g1)3εM1,M2Nk1t1,t2,g1Nt1g1>x|f(M1,t1g1)||f(M2,t2g1)|(D1D2)1ε(t1g1)2s4ε(t2g1)2εM1,M2Nk1t1,t2,g1,g1Nt1g1>x|f(M1,t1g1)||f(M2,t2g1)|(D1D2)1ε(t1g1)2s4ε(t2g1)2ε:=T2. (2.8)

    Let t1g1=mk, t2g1=m2k, we have

    T2=M1,M2Nk1mk,m2kNmk>x|f(M1,mk)||f(M2,m2k)|τ(mk)τ(m2k)(D1D2)1εmk2s4εm2k2ε1m12s6εx2s+6εM1,M2Nk1mk,m2kN|f(M1,mk)||f(M2,m2k)|τ(mk)τ(m2k)(D1D2)1εmk2s4εm2k2εx2s+6εM1Nk1mkN|f(M1,mk)|τ(mk)D11εmk2s4εM1Nk1mkN|f(M1,mk)|τ(mk)D11εmk2εx2s+ε,

    the convergence of the two series in the last step can be obtained by Lemma {2.4}, and we use the arbitrariness of ε. If m2k>x, exchange t1 and t2, we can also obtain

    T2x2s+ε.

    For T3, if n1>y, similar to (i), we obtain (t1,t2)=1 and t1g1=mk, t2g1=m2k, t1g2=n1, t2g2=n2, then

    T3=M1,M2Nk1t1,t2,g1,g2N(t1,t2)=1t1g2>y|f(M1,t1g1)||f(M2,t2g1)||g(M1,t1g1,M2,t2g1)|τ(t1g2)τ(t2g2)D1D2(t1g1)s(t2g1)s(t1g2)w(t2g2)wM1,M2Nk1t1,t2,g1,g2N(t1,t2)=1t1g2>y|f(M1,t1g1)||f(M2,t2g1)||g(M1,t1g1,M2,t2g1)|τ(t1)τ(t2)τ2(g2)D1D2(t1g1)s(t2g1)s(t1t2)wg22wM1,M2Nk1t1,t2,g1N|f(M1,t1g1)||f(M2,t2g1)||g(M1,t1g1,M2,t2g1)|τ(t1)τ(t2)D1D2g12s(t1t2)s+wg2>yt1τ2(g2)g22wy12wlog3yM1,M2Nk1t1,t2,g1N|f(M1,t1g1)||f(M2,t2g1)||g(M1,t1g1,M2,t2g1)|τ(t1)τ(t2)D1D2g12st11+swt2s+w,

    where we use (2.3) by taking U=y/t1.

    We denote the latter series by Σ, using (2.4) we have

    Σ=M1,M2Nk1t1,t2,g1N|f(M1,t1g1)||f(M2,t2g1)||g(M1,t1g1,M2,t2g1)|τ(t1)τ(t2)D1D2(t1g1)s(t2g1)st11wt2wM1,M2Nk1t1,t2,g1N|f(M1,t1g1)||f(M2,t2g1)||g(M1,t1g1,M2,t2g1)|D1D2(t1g1)s(t2g1)sM1,M2Nk1t1,t2,g1N|f(M1,t1g1)||f(M2,t2g1)|(D1D2)1ε(t1g1)sε(t2g1)sεM1,M2Nk1t1,t2,g1,g1N|f(M1,t1g1)||f(M2,t2g1)|(D1D2)1ε(t1g1)sε(t2g1)sε.

    Let t1g1=mk, t2g1=m2k, we obtain

    ΣM1,M2Nk1mk,m2kN|f(M1,mk)||f(M1,m2k)|τ(mk)τ(m2k)(D1D2)1εmksεm2ksε=(M1Nk1mkN|f(M1,mk)|τ(mk)D11εmksε)21

    by Lemma 2.4, thus

    T3y12wlog3y.

    If n2>y, exchange t1 and t2, similarly we obtain

    T3y12wlog3y.

    Above all, we obtain

    T1,T2x2s+εandT3y12wlog3y,

    then (ii) holds from (2.5).

    We shall give a more explicit expression of Δk(x). According to Lemma 2.4,

    τ(n1nk)=m1d1=n1,,mkdk=nkf(m1,,mk)τ(d1)τ(dk)

    holds for any n1,,nkN, where f is multiplicative and symmetric in all variables.

    Therefore, we deduce by (1.1) that

    \begin{align} \sum\limits_{n_1, \cdots, n_k\leqslant x}\tau(n_1\cdots n_k) = &\sum\limits_{m_1, \cdots, m_k\leqslant x}f(m_1, \cdots, m_k)\prod\limits_{j = 1}^k\left(\sum\limits_{d_j\leqslant {x/m_j}}\tau(d_j)\right)\\ = &\sum\limits_{m_1, \cdots, m_k\leqslant x}f(m_1, \cdots, m_k)\prod\limits_{j = 1}^k\left(M\left(\frac{x}{m_j}\right)+\Delta\left(\frac{x}{m_j}\right)\right)\\ = &\sum\limits_{m_1, \cdots, m_k\leqslant x}f(m_1, \cdots, m_k)\left(\prod\limits_{j = 1}^kM\left(\frac{x}{m_j}\right)\right)\\ &+k\sum\limits_{m_1, \cdots, m_k\leqslant x}f(m_1, \cdots, m_k)\left(\prod\limits_{j = 1}^{k-1}M\left(\frac{x}{m_j}\right)\right)\Delta\left(\frac{x}{m_k}\right)\\ &+\binom{k}{2}\sum\limits_{m_1, \cdots, m_k\leqslant x}f(m_1, \cdots, m_k) \left(\prod\limits_{j = 1}^{k-2}M\left(\frac{x}{m_j}\right)\right)\Delta\left(\frac{x}{m_k}\right)\Delta\left(\frac{x}{m_{k-1}}\right)\\ &+\cdots\\ &+\binom{k}{k-1}\sum\limits_{m_1, \cdots, m_k\leqslant x}f(m_1, \cdots, m_k)M\left(\frac{x}{m_1}\right)\left(\prod\limits_{j = 2}^{k}\Delta\left(\frac{x}{m_j}\right)\right)\\ &+\sum\limits_{m_1, \cdots, m_k\leqslant x}f(m_1, \cdots, m_k)\left(\prod\limits_{j = 1}^{k}\Delta\left(\frac{x}{m_j}\right)\right)\\ : = &\sum\limits_{m_1, \cdots, m_k\leqslant x}f(m_1, \cdots, m_k)\left(\prod\limits_{j = 1}^kM\left(\frac{x}{m_j}\right)\right)+\mathbf{E}(x), \end{align} (3.1)

    where

    \begin{eqnarray} M(u) = u\log u+(2\gamma-1)u. \end{eqnarray} (3.2)

    Here we have

    \begin{align} \sum\limits_{m_1, \cdots, m_k\le x}f(m_1, \cdots, m_k)\prod\limits_{j = 1}^k\left(M\left(\frac{x}{m_j}\right)\right) & = \sum\limits_{m_1, \cdots, m_k = 1}^{\infty}f(m_1, \cdots, m_k)\prod\limits_{j = 1}^k\left(M\left(\frac{x}{m_j}\right)\right)+O(x^{k-1+\varepsilon})\\ & = x^kP_k(\log x)+O\left(x^{k-1+\varepsilon}\right) \end{align} (3.3)

    by the proof of Theorem 3.4 in Tóth and Zhai [9]. From (1.2) we have

    \Delta(\frac{x}{m_j})\ll(\frac{x}{m_j})^{\theta+\varepsilon}

    for any 1\leqslant j\leqslant k , then

    \begin{eqnarray} \mathbf{E}(x)&\ll&\sum\limits_{m_1, \cdots, m_k\leqslant x}|f(m_1, \cdots, m_k)|\left(\prod\limits_{j = 1}^{k-2}M\left(\frac{x}{m_j}\right)\right)\Delta\left(\frac{x}{m_k}\right)\Delta\left(\frac{x}{m_{k-1}}\right)\\ \\ &\ll& x^{k-2+2\theta+\varepsilon}\sum\limits_{m_1, \cdots, m_k\leqslant x}\frac{|f\left(m_1, \cdots, m_k\right)|}{m_1\cdots m_{k-2}(m_{k-1}m_{k})^\theta} \cdot\frac{(m_{k-1}m_{k})^{\frac{1}{2}-\theta+\varepsilon}}{(m_{k-1}m_{k})^{\frac{1}{2}-\theta+\varepsilon}}\\ \\ &\ll& x^{k-2+2\theta+\varepsilon}(x^2)^{\frac{1}{2}-\theta+\varepsilon}\sum\limits_{m_1, \cdots, m_k = 1}^{\infty}\frac{|f(m_1, \cdots, m_k)|}{m_1\cdots m_{k-2}(m_{k-1}m_{k})^{\frac{1}{2}+\varepsilon}}\\ \\ &\ll& x^{k-1+\varepsilon}, \end{eqnarray} (3.4)

    where the convergence of latter series can be obtained by Lemma 2.4. Then we conclude that

    \begin{eqnarray} \Delta_k(x) = {\Delta_k}^*(x)+O\left(x^{k-1+\varepsilon}\right) \end{eqnarray} (3.5)

    follows by (1.4), (3.1), (3.3), and (3.4), where

    \begin{eqnarray} {\Delta_k}^*(x) = k\sum\limits_{m_1, \cdots, m_k\leqslant x}f(m_1, \cdots, m_k)\left(\prod\limits_{j = 1}^{k-1}M\left(\frac{x}{m_j}\right)\right)\Delta\left(\frac{x}{m_k}\right). \end{eqnarray}

    Suppose T\geqslant2 , we shall first estimate \int_T^{2T}({\Delta_k}^{*}(x))^2dx . Since m_1, \cdots, m_{k-1} are symmetric, we can divide {\Delta_k}^{*}(x) into three parts,

    \begin{eqnarray} {\Delta_k}^*(x)& = &M_1+O(M_2+M_3), \end{eqnarray}

    where the implied constant about " O " depends on k and \varepsilon ,

    \begin{eqnarray} &&M_1 = M_1(x, y): = k\sum\limits_{m_1, \cdots, m_k\leqslant y}f(m_1, \cdots, m_k)\prod\limits_{j = 1}^{k-1}M\left(\frac{x}{m_j}\right)\Delta\left(\frac{x}{m_k}\right), \\ &&M_2 = M_2(x, y): = \sum\limits_{\substack{m_1, \cdots, m_k\leqslant x\\m_1 > y}} |f(m_1, \cdots, m_k)|\prod\limits_{j = 1}^{k-1}M\left(\frac{x}{m_j}\right)\left|\Delta\left(\frac{x}{m_k}\right)\right|, \\ &&M_3 = M_3(x, y): = \sum\limits_{\substack{m_1, \cdots, m_k\leqslant x\\m_k > y}}|f(m_1, \cdots, m_k)|\prod\limits_{j = 1}^{k-1}M\left(\frac{x}{m_j}\right)\left|\Delta\left(\frac{x}{m_k}\right)\right|, \end{eqnarray} (4.1)

    where y is a parameter that satisfies T^{\varepsilon}\ll y\ll T . So we obtain

    \begin{align} \int_{T}^{2T}({\Delta_k}^*(x))^2dx = &\int_{T}^{2T}{M_1}^2dx+O\left(\int_{T}^{2T}(M_1M_2+M_2M_3+M_1M_3)dx\right)\\ &+ \quad O\left(\int_{T}^{2T}({M_2}^2+{M_3}^2)dx\right). \end{align} (4.2)

    First, we deal with \int_{T}^{2T}{M_1}^2dx . By Lemma 2.1 we obtain

    \begin{align} M_1(x, y)& = k\sum\limits_{m_1, \cdots, m_k\leqslant y}f(m_1, \cdots, m_k)\prod\limits_{j = 1}^{k-1}M\left(\frac{x}{m_j}\right) \left(\delta_1\left(\frac{x}{m_k}, y\right)+\delta_2\left(\frac{x}{m_k}, y\right)\right)\\ \\ &: = M_{11}(x, y)+M_{12}(x, y), \end{align}

    where

    \begin{align} M_{11}(x, y)& = k\sum\limits_{m_1, \cdots, m_k\leqslant y}f(m_1, \cdots, m_k)\prod\limits_{j = 1}^{k-1}M\left(\frac{x}{m_j}\right)\delta_1\left(\frac{x}{m_k}, y\right), \\ M_{12}(x, y)& = k\sum\limits_{m_1, \cdots, m_k\leqslant y}f(m_1, \cdots, m_k)\prod\limits_{j = 1}^{k-1}M\left(\frac{x}{m_j}\right)\delta_2\left(\frac{x}{m_k}, y\right), \end{align}

    and \delta_1(\cdot, y) and \delta_2(\cdot, y) are defined in Lemma 2.2, thus we have

    \begin{eqnarray} \int_T^{2T}{M_1}^2(x, y)dx = \int_T^{2T}{M_{11}}^2(x, y)dx+O\left(\int_T^{2T}\left(M_{11}(x, y)M_{12}(x, y)+{M_{12}}^2(x, y)\right)dx\right). \end{eqnarray} (4.3)

    Let \mathbf{M_1}, \mathbf{M_2} be defined in Lemma 2.5, using

    \cos{\alpha}\cos{\beta} = \frac{1}{2}(\cos(\alpha-\beta)+\cos(\alpha+\beta)),

    we have

    \begin{align} {M_{11}}^2(x, y) = &k^2\sum\limits_{m_1, \cdots, m_{2k}\leqslant y} f(\mathbf{M}_1) f(\mathbf{M}_2)\prod\limits_{j = 1}^{k-1}M\left(\frac{x}{m_j}\right)\prod\limits_{j = k+1}^{2k-1}M\left(\frac{x}{m_j}\right) \\ &\times\frac{x^{1/2}} {2\pi^2(m_km_{2k})^{1/4}}\sum\limits_{n_1, n_2\leqslant y}\frac{\tau(n_1)\tau(n_2)}{{n_1}^{3/4}{n_2}^{3/4}} \cos\left(4\pi\sqrt{\frac{n_1x}{m_k}}-\frac{\pi}{4}\right)\cos\left(4\pi\sqrt{\frac{n_2x}{m_{2k}}}-\frac{\pi}{4}\right)\\ = &k^2\sum\limits_{m_1, \cdots, m_{2k}\leqslant y} f(\mathbf{M}_1) f(\mathbf{M}_2)\prod\limits_{j = 1}^{k-1}M\left(\frac{x}{m_j}\right)\prod\limits_{j = k+1}^{2k-1}M\left(\frac{x}{m_j}\right) \frac{x^{1/2}}{4\pi^2(m_km_{2k})^{1/4}} \\ &\times\sum\limits_{n_1, n_2\leqslant y}\frac{\tau(n_1)\tau(n_2)}{{n_1}^{3/4}{n_2}^{3/4}} \left[\cos\left(4\pi\left(\sqrt{\frac{n_1}{m_k}}-\sqrt{\frac{n_2}{m_{2k}}}\right)\sqrt x\right) +\sin\left(4\pi\left(\sqrt{\frac{n_1}{m_k}}+\sqrt{\frac{n_2}{m_{2k}}}\right)\sqrt x\right)\right] \\ : = &k^2(S_0(x, y)+S_1(x, y)+S_2(x, y)), \end{align} (4.4)

    where

    \begin{align} S_0(x, y) = &\frac{x^{1/2}}{4\pi^2}\sum\limits_{m_1, \cdots, m_{2k}\leqslant y} \frac{f(\mathbf{M}_1)f(\mathbf{M}_2)}{(m_km_{2k})^{1/4}} \prod\limits_{j = 1}^{k-1}M\left(\frac{x}{m_j}\right)\prod\limits_{j = k+1}^{2k-1}M\left(\frac{x}{m_j}\right) \sum\limits_{\substack{n_1, n_2\le y\\n_1m_{2k} = n_2m_k}}\frac{\tau(n_1)\tau(n_2)}{{n_1}^{3/4}{n_2}^{3/4}}, \\ S_1(x, y) = &\frac{x^{1/2}}{4\pi^2}\sum\limits_{m_1, \cdots, m_{2k}\leqslant y} \frac{f(\mathbf{M}_1)f(\mathbf{M}_2)}{(m_km_{2k})^{1/4}}\prod\limits_{j = 1}^{k-1}M\left(\frac{x}{m_j}\right)\prod\limits_{j = k+1}^{2k-1}M\left(\frac{x}{m_j}\right)\\ &\times\sum\limits_{\substack{n_1, n_2\leqslant y\\n_1m_{2k}\not = n_2m_k}}\frac{\tau(n_1)\tau(n_2)}{{n_1}^{3/4}{n_2}^{3/4}} \cos\left(4\pi\left(\sqrt{\frac{n_1}{m_k}}-\sqrt{\frac{n_2}{m_{2k}}}\right)\sqrt x\right), \\ S_2(x, y) = &\frac{x^{1/2}}{4\pi^2}\sum\limits_{m_1, \cdots, m_{2k}\leqslant y} \frac{f(\mathbf{M}_1)f(\mathbf{M}_2)}{(m_km_{2k})^{1/4}}\prod\limits_{j = 1}^{k-1}M\left(\frac{x}{m_j}\right)\prod\limits_{j = k+1}^{2k-1}M\left(\frac{x}{m_j}\right)\\ &\times\sum\limits_{n_1, n_2\leqslant y}\frac{\tau(n_1)\tau(n_2)}{{n_1}^{3/4}{n_2}^{3/4}} \sin\left(4\pi\left(\sqrt{\frac{n_1}{m_k}}+\sqrt{\frac{n_2}{m_{2k}}}\right)\sqrt x\right). \end{align}

    So, it turns to evaluate

    \begin{eqnarray} \int_0 = \int_T^{2T}S_0(x, y)dx\:, \ \ \:\int_1 = \int_T^{2T}S_1(x, y)dx\:, \ \ \:\int_2 = \int_T^{2T}S_2(x, y)dx. \end{eqnarray}

    We need to give more explicit expressions for (\prod_{j = 1}^{k-1}M(\frac{x}{m_j})) and (\prod_{j = k+1}^{2k-1}M(\frac{x}{m_j})) . By the proof of Theorem 3.3 in Tóth and Zhai [9] in the case f_j(n) = \tau(n), \:a_j = 1, \:\delta_j = 1\:(1\leqslant j\leqslant k-1) we obtain

    \begin{eqnarray} \prod\limits_{j = 1}^{k-1} M\left(\frac{x}{m_j}\right) = \frac{x^{k-1}}{m_1\cdots m_{k-1}}\sum\limits_{l_1 = 0}^{k-1}C_{l_1}(\log m_1, \cdots, \log m_{k-1})(\log x)^{l_1}, \end{eqnarray} (4.5)

    where

    C_{l_1}(\log m_1, \cdots, \log m_{k-1}) = \sum\limits_{j_1, \cdots, j_{k-1} = 0, 1}c(j_1, \cdots, j_{k-1})(\log m_1)^{j_1}\cdots(\log m_{k-1})^{j_{k-1}},

    and c(j_1, \cdots, j_{k-1}) are constants (j_1, \cdots, j_{k-1} = 0, 1) . Similarly, when k+1\leqslant j\leqslant 2k-1 , we have

    \begin{eqnarray} \prod\limits_{j = k+1}^{2k-1} M\left(\frac{x}{m_j}\right) = \frac{x^{k-1}}{m_{k+1}\cdots m_{2k-1}}\sum\limits_{l_2 = k}^{2k-1}C_{l_2}(\log m_{k+1}, \cdots, \log m_{2k-1})(\log x)^{l_2-k}, \end{eqnarray} (4.6)

    where

    C_{l_2}(\log m_{k+1}, \cdots, \log m_{2k-1}) = \sum\limits_{j_{k+1}, \cdots, j_{2k-1} = 0, 1}c(j_{k+1}, \cdots, j_{2k-1})(\log m_{k+1})^{j_{k+1}}\cdots(\log m_{2k-1})^{j_{2k-1}},

    and c(j_{k+1}, \cdots, j_{2k-1}) are constants (j_{k+1}, \cdots, j_{2k-1} = 0, 1) .

    We denote C_{l_1}(\log m_1, \cdots, \log m_{k-1})C_{l_2}(\log m_{k+1}, \cdots, \log m_{2k-1}) by C_{l_1, l_2}(\mathbf{M}_1, \mathbf{M}_2) , using (4.5) and (4.6) we have

    \begin{eqnarray} S_0(x, y) = \frac{x^{2k-\frac{3}{2}}}{4\pi^2}\sum\limits_{l_1, l_2 = 0}^{k-1}(\log x)^{l_1+l_2}\sum\limits_{\substack{{m_1, \cdots, m_{2k}\leqslant y}\\n_1, n_2\leqslant y\\ \frac{m_k}{m_{2k}} = \frac{n_1}{n_2}}} \frac{f(\mathbf{M}_1)f(\mathbf{M}_2)C_{l_1, l_2}(\mathbf{M}_1, \mathbf{M}_2)}{D_1D_2(m_km_{2k})^{\frac{1}{4}}} \cdot\frac{\tau(n_1)\tau(n_2)}{(n_1n_2)^{\frac{3}{4}}}, \end{eqnarray} (4.7)

    where D_1, D_2 are defined in Lemma 2.5. Using (2.4) we obtain

    \begin{eqnarray} \:\:\:C_{l_1, l_2}(\mathbf{M}_1, \mathbf{M}_2) \ll\left(\prod\limits_{j = 1}^{k-1}m_j\prod\limits_{j = k+1}^{2k-1}m_j\right)^\varepsilon\ll\left(\prod\limits_{j = 1}^{2k}m_j\right)^{\varepsilon}. \end{eqnarray}

    Choosing

    g(\mathbf{M}_1, \mathbf{M}_2) = g_{l_1, l_2}(\mathbf{M}_1, \mathbf{M}_2) = C_{l_1, l_2}(\mathbf{M}_1, \mathbf{M}_2),

    s = 1/4 , w = 3/4 in Lemma 2.5, we obtain that S_0(x, y) can be written as

    \begin{align} S_0(x, y)& = \frac{x^{2k-\frac{3}{2}}}{4\pi^2}\sum\limits_{l_1, l_2 = 0}^{k-1}(\log x)^{l_1+l_2}\times T_{g, k}\left(y, y;\frac{1}{4}, \frac{3}{4}\right)\\ & = \frac{x^{2k-\frac{3}{2}}}{4\pi^2}\sum\limits_{l_1, l_2 = 0}^{k-1}(\log x)^{l_1+l_2}\left(T_{g, k}\left(\frac{1}{4}, \frac{3}{4}\right)+O\left({y}^{-\frac{1}{2}+\varepsilon}\right)\right). \end{align} (4.8)

    Since g(\mathbf{M}_1, \mathbf{M}_2) is related to l_1, l_2 , we denote T_{g, k}\left(\frac{1}{4}, \frac{3}{4}\right) in (4.8) by D_{k, l_1, l_2} , we conclude that

    \begin{eqnarray} S_0(x, y) = \frac{x^{2k-\frac{3}{2}}}{4\pi^2}Q_{2k-2}(\log x) +O\left(x^{2k-\frac{3}{2}+\varepsilon}{y}^{-\frac{1}{2}+\varepsilon}\right), \end{eqnarray} (4.9)

    where

    \begin{eqnarray} Q_{2k-2}(t) = \sum\limits_{l_1, l_2 = 0}^{k-1}D_{k, l_1, l_2}\:t^{\:l_1+l_2}. \end{eqnarray}

    Then it follows from (4.9) that

    \begin{eqnarray} \int_T^{2T}S_0(x, y)dx = \frac{1}{4\pi^2}\sum\limits_{l_1, l_2 = 0}^{k-1}D_{k, l_1, l_2}\int_T^{2T}x^{2k-\frac{3}{2}}(\log x)^{l_1+l_2}dx +O\left(T^{2k-\frac{1}{2}+\varepsilon}{y}^{-\frac{1}{2}}\right). \end{eqnarray} (4.10)

    Let \mathbf{M_1}, \mathbf{M_2}, D_1, D_2 be defined in Lemma 2.5, then by (4.5) and (4.6) we have

    \begin{align} \int_2 = &\int_T^{2T}\frac{x^{2k-\frac{3}{2}}}{4\pi^2}\sum\limits_{l_1, l_2 = 0}^{k-1}(\log x)^{l_1+l_2}\sum\limits_{m_1, \cdots, m_{2k}\leqslant y} \frac{C_{l_1, l_2}(\mathbf{M}_1, \mathbf{M}_2)}{D_1D_2}\\ &\times\sum\limits_{n_1, n_2\leqslant y} \frac{f(\mathbf{M}_1)f(\mathbf{M}_2)}{{m_k}^{1/4}{m_{2k}}^{1/4}}\cdot\frac{\tau(n_1)\tau(n_2)}{{n_1}^{3/4}{n_2}^{3/4}} \sin\left(4\pi\left(\sqrt{\frac{n_1}{m_k}}+\sqrt{\frac{n_2}{m_{2k}}}\right)\sqrt x\right)dx. \end{align}

    Let

    G(x) = x^{2k-\frac{3}{2}}(\log x)^{l_1+l_2}, \ \ \ F(\cdot) = \sin(\cdot), \ \ \ m(x) = 4\pi\left(\sqrt{\frac{n_1}{m_k}}+\sqrt{\frac{n_2}{m_{2k}}}\right)\sqrt x, \ \ \ a = T, \ \ b = 2T

    in Lemma 2.3, change the order of integration and summation, then we have

    \begin{align} \int_2 = &\frac{1}{4\pi^2}\sum\limits_{l_1, l_2 = 0}^{k-1}\sum\limits_{\substack{m_1, \cdots, m_{2k}\leqslant{y}\\n_1, n_2\leqslant y}} \frac{C_{l_1, l_2}(\mathbf{M}_1, \mathbf{M}_2)}{D_1D_2}\cdot\frac{f(\mathbf{M}_1)f(\mathbf{M}_2)} {{m_k}^{1/4}{m_{2k}}^{1/4}}\cdot\frac{\tau(n_1)\tau(n_2)}{{n_1}^{3/4}{n_2}^{3/4}}\\ &\times\:\int_T^{2T}x^{2k-\frac{3}{2}}(\log x)^{l_1+l_2}\sin\left(4\pi\left(\sqrt{\frac{n_1}{m_k}}+\sqrt{\frac{n_2}{m_{2k}}}\right)\sqrt x\right)dx\\ \ll&\sum\limits_{\substack{m_1, \cdots, m_{2k}\leqslant{y}\\n_1, n_2\leqslant y}} \frac{|C_{l_1, l_2}(\mathbf{M}_1, \mathbf{M}_2)|}{D_1D_2}\cdot\frac{|f(\mathbf{M}_1)||f(\mathbf{M}_2)|} {{m_k}^{1/4}{m_{2k}}^{1/4}}\cdot\frac{\tau(n_1)\tau(n_2)}{{n_1}^{3/4}{n_2}^{3/4}}\\ &\times\:T^{2k-\frac{3}{2}+\varepsilon}\cdot\frac{T^{\frac{1}{2}}}{\sqrt{\frac{n_1}{m_k}}+\sqrt{\frac{n_2}{m_{2k}}}}, \end{align}

    then we use

    a^2+b^2\geqslant2ab, \ \ \ C_{l_1, l_2}(\mathbf{M}_1, \mathbf{M}_2)\ll\left(\prod\limits_{j = 1}^{2k}m_j\right)^{\varepsilon}

    to obtain

    \begin{align} \int_2&\ll T^{2k-1+\varepsilon}\sum\limits_{\substack{m_1, \cdots, m_{2k}\leqslant{y}\\n_1, n_2\leqslant y}} \frac{\left(\prod\limits_{j = 1}^{2k}m_j\right)^{\varepsilon}}{D_1D_2}\cdot\frac{|f(\mathbf{M}_1)||f(\mathbf{M}_2)|} {{m_k}^{1/4}{m_{2k}}^{1/4}}\cdot\frac{\tau(n_1)\tau(n_2)}{{n_1}^{3/4}{n_2}^{3/4}} \cdot\left(\frac{m_km_{2k}}{n_1n_2}\right)^{\frac{1}{4}}\\ &\ll T^{2k-1+(2k+1)\varepsilon}\sum\limits_{\substack{m_1, \cdots, m_{2k}\leqslant{y}\\n_1, n_2\leqslant y}} \frac{|f(\mathbf{M}_1)||f(\mathbf{M}_2)|}{D_1D_2} \cdot\frac{\tau(n_1)\tau(n_2)}{{n_1}{n_2}}\cdot\frac{T^{2\varepsilon}}{{m_k}^{\varepsilon}{m_{2k}}^{\varepsilon}}\\ &\ll T^{2k-1+(2k+3)\varepsilon}\sum\limits_{m_1, \cdots, m_{2k} = 1}^{\infty} \frac{|f(\mathbf{M}_1)||f(\mathbf{M}_2)|}{D_1D_2{m_k}^{\varepsilon}{m_{2k}}^{\varepsilon}} \sum\limits_{n_1, n_2\leqslant y}\frac{\tau(n_1)\tau(n_2)}{{n_1}{n_2}}\\ &\ll T^{2k-1+(2k+4)\varepsilon}\left(\sum\limits_{m_1, \cdots, m_{2k} = 1}^{\infty} \frac{|f(\mathbf{M}_1)|}{D_1{m_k}^{\varepsilon}}\right)^2, \end{align}

    where by partial summation we obtain

    \sum\limits_{n_1, n_2\leqslant y}\frac{\tau(n_1)\tau(n_2)}{{n_1}{n_2}}\ll\log^4{y}\ll\log^4T\ll T^{\varepsilon}.

    Since \varepsilon > 0 is arbitrary, using Lemma 2.4 we conclude that

    \begin{eqnarray} \int_2\ll T^{2k-1+\varepsilon}. \end{eqnarray} (4.11)

    Then we turn to estimate \int_1 , similar to \int_2 , we have

    \begin{align} \int_1 = &\int_T^{2T}\frac{x^{2k-\frac{3}{2}}}{4\pi^2}\sum\limits_{l_1, l_2 = 0}^{k-1}(\log x)^{l_1+l_2}\sum\limits_{m_1, \cdots, m_{2k}\leqslant y} \frac{C_{l_1, l_2}(\mathbf{M}_1, \mathbf{M}_2)}{D_1D_2}\\ &\times\sum\limits_{\substack{n_1, n_2\leqslant y\\n_1m_{2k}\not = n_2m_k}} \frac{f(\mathbf{M}_1)f(\mathbf{M}_2)}{{m_k}^{1/4}{m_{2k}}^{1/4}}\cdot\frac{\tau(n_1)\tau(n_2)}{{n_1}^{3/4}{n_2}^{3/4}} \cos\left(4\pi\left(\sqrt{\frac{n_1}{m_k}}-\sqrt{\frac{n_2}{m_{2k}}}\right)\sqrt x\right)dx. \end{align}

    Let

    G(x) = x^{2k-\frac{3}{2}}(\log x)^{l_1+l_2}, \ \ \ F(\cdot) = \cos(\cdot), \ \ \ m(x) = 4\pi(\sqrt{\frac{n_1}{m_k}}-\sqrt{\frac{n_2}{m_{2k}}})\sqrt x, \ \ \ a = T, \ \ b = 2T

    in Lemma 2.3, change the order of integration and summation, then we have

    \begin{align} \int_1 = &\frac{1}{4\pi^2}\sum\limits_{l_1, l_2 = 0}^{k-1}\sum\limits_{m_1, \cdots, m_{2k}\leqslant{y}} \frac{C_{l_1, l_2}(\mathbf{M}_1, \mathbf{M}_2)}{D_1D_2}\cdot\frac{f(\mathbf{M}_1)f(\mathbf{M}_2)} {{m_k}^{1/4}{m_{2k}}^{1/4}}\sum\limits_{\substack{n_1, n_2\leqslant y\\n_1m_{2k}\not = n_2m_k}}\frac{\tau(n_1)\tau(n_2)}{{n_1}^{3/4}{n_2}^{3/4}}\\ &\times\:\int_T^{2T}x^{2k-\frac{3}{2}}(\log x)^{l_1+l_2}\cos\left(4\pi\left(\sqrt{\frac{n_1}{m_k}}-\sqrt{\frac{n_2}{m_{2k}}}\right)\sqrt x\right)dx\\ \ll&\sum\limits_{l_1, l_2 = 0}^{k-1}\sum\limits_{m_1, \cdots, m_{2k}\leqslant{y}} \frac{|C_{l_1, l_2}(\mathbf{M}_1, \mathbf{M}_2)|}{D_1D_2}\cdot\frac{|f(\mathbf{M}_1)||f(\mathbf{M}_2)|} {{m_k}^{1/4}{m_{2k}}^{1/4}}\sum\limits_{\substack{n_1, n_2\leqslant y_2\\n_1m_{2k}\not = n_2m_k}}\frac{\tau(n_1)\tau(n_2)}{{n_1}^{3/4}{n_2}^{3/4}}\\ &\times\:T^{2k-\frac{3}{2}+\varepsilon}\cdot\frac{T^{\frac{1}{2}}}{|\sqrt{\frac{n_1}{m_k}}-\sqrt{\frac{n_2}{m_{2k}}}|}. \end{align}

    Since

    C_{l_1, l_2}(\mathbf{M}_1, \mathbf{M}_2)\ll\left(\prod\limits_{j = 1}^{2k}m_j\right)^{\varepsilon},

    we have

    \begin{align} \int_1 &\ll T^{2k-1+\varepsilon}\sum\limits_{m_1, \cdots, m_{2k}\leqslant{y}} \frac{\left(\prod\limits_{j = 1}^{2k}m_j\right)^{\varepsilon}}{D_1D_2}\cdot\frac{|f(\mathbf{M}_1)||f(\mathbf{M}_2)|}{{m_k}^{1/4}{m_{2k}}^{1/4}} \sum\limits_{\substack{n_1, n_2\leqslant y\\n_1m_{2k}\not = n_2m_k}} \frac{\tau(n_1)\tau(n_2)}{{n_1}^{3/4}{n_2}^{3/4}}\cdot\frac{1}{|\sqrt{\frac{n_1}{m_k}}-\sqrt{\frac{n_2}{m_{2k}}}|}\\ &\ll T^{2k-1+(2k+1)\varepsilon}\sum\limits_{m_1, \cdots, m_{2k}\leqslant{y}}\frac{|f(\mathbf{M}_1)||f(\mathbf{M}_2)|} {D_1D_2{m_k}^{1/4}{m_{2k}}^{1/4}}\sum\limits_{\substack{n_1, n_2\leqslant y\\n_1m_{2k}\not = n_2m_k}} \frac{\tau(n_1)\tau(n_2)}{{n_1}^{3/4}{n_2}^{3/4}}\cdot\frac{1}{|\sqrt{\frac{n_1}{m_k}}-\sqrt{\frac{n_2}{m_{2k}}}|}\\ &: = T^{2k-1+(2k+1)\varepsilon}\sum\limits_{m_1, \cdots, m_{2k}\leqslant{y}}\frac{|f(\mathbf{M}_1)||f(\mathbf{M}_2)|} {D_1D_2{m_k}^{1/4}{m_{2k}}^{1/4}}\left(R_1+R_2\right), \end{align} (4.12)

    where

    \begin{align} R_1& = \sum\limits_{\substack{{n_1, n_2\leqslant y, \quad n_1m_{2k}\not = n_2m_k}\\ \left|\sqrt{\frac{n_1}{m_k}}-\sqrt{\frac{n_2}{m_{2k}}}\right|\le\frac{(n_1n_2)^{1/4}}{10(m_km_{2k})^{1/4}}}} \frac{\tau(n_1)\tau(n_2)}{{n_1}^{3/4}{n_2}^{3/4}}\cdot\frac{1}{|\sqrt{\frac{n_1}{m_k}}-\sqrt{\frac{n_2}{m_{2k}}}|}, \\ R_2& = \sum\limits_{\substack{{n_1, n_2\leqslant y, \quad n_1m_{2k}\not = n_2m_k}\\ \left|\sqrt{\frac{n_1}{m_k}}-\sqrt{\frac{n_2}{m_{2k}}}\right| > \frac{(n_1n_2)^{1/4}}{10(m_km_{2k})^{1/4}}}} \frac{\tau(n_1)\tau(n_2)}{{n_1}^{3/4}{n_2}^{3/4}}\cdot\frac{1}{|\sqrt{\frac{n_1}{m_k}}-\sqrt{\frac{n_2}{m_{2k}}}|}. \end{align}

    Then we have

    \begin{eqnarray} R_2\ll\sum\limits_{n_1, n_2\leqslant y}\frac{\tau(n_1)\tau(n_2)}{{n_1}^{3/4}{n_2}^{3/4}}\cdot\frac{(m_km_{2k})^{1/4}}{(n_1n_2)^{1/4}}\ll(m_km_{2k})^{\frac{1}{4}}\log^4T, \end{eqnarray} (4.13)

    where we use partial summation to obtain

    \sum\limits_{n_1, n_2\leqslant y}\frac{\tau(n_1)\tau(n_2)}{n_1n_2}\ll\log^4y\ll \log^4T.

    And for R_1 , by Lagrange's mean value theorem we have

    \sqrt{\beta_1}-\sqrt{\beta_2}\asymp(\sqrt{\beta_1\beta_2})^{-\frac{1}{2}}|\beta_1-\beta_2|

    for any

    \beta_1\asymp\beta_2\in\mathbb{R},

    thus let

    \beta_1 = n_1/m_k, \ \ \ \beta_2 = n_2/m_{2k}

    in R_1 , we obtain

    \begin{align} R_1&\ll\sum\limits_{\substack{n_1, n_2\leqslant y\\n_1m_{2k}\not = n_2m_k}} \frac{\tau(n_1)\tau(n_2)}{{n_1}^{3/4}{n_2}^{3/4}}\cdot\frac{1}{|\frac{n_1}{m_k}-\frac{n_2}{m_{2k}}|(\frac{m_km_{2k}}{n_1n_2})^{1/4}}\\ & = (m_km_{2k})^{\frac{3}{4}}\sum\limits_{\substack{n_1, n_2\leqslant y\\n_1m_{2k}\not = n_2m_k}}\frac{\tau(n_1)\tau(n_2)}{n_1^{1/2}n_2^{1/2}|n_1m_{2k}-n_2m_k|}. \end{align}

    For some real numbers N_1 , N_2 satisfying 1\leqslant N_1, N_2\leqslant y , one has

    \begin{align} R_1&\ll(m_km_{2k})^{\frac{3}{4}}\log^2y\sum\limits_{\substack{N_1 < n_1\leqslant2N_1\\N_2 < n_2\leqslant2N_2\\n_1m_{2k}\not = n_2m_k}} \frac{\tau(n_1)\tau(n_2)}{(n_1n_2)^{1/2}|n_1m_{2k}-n_2m_k|}\\ &\ll\frac{(m_km_{2k})^{\frac{3}{4}}y^{\varepsilon}}{(N_1N_2)^{1/2}}\sum\limits_{\substack{N_1 < n_1\leqslant2N_1\\N_2 < n_2\leqslant2N_2\\n_1m_{2k}\not = n_2m_k}}\frac{1}{|n_1m_{2k}-n_2m_k|}, \end{align} (4.14)

    we denote the latter sum by T(m_k, m_{2k}) . Let

    |n_1m_{2k}-n_2m_k| = r,

    we have

    r\equiv-n_1m_{2k}(\text{mod}\ {m_k}),

    so we can find a constant c_0 such that

    1\leqslant c_0 < m_k, \ \ \ r = m_kt+c_0,

    where t is an integer such that 0 < t < 2y^2 , thus

    \begin{align} T(m_k, m_{2k})& = \sum\limits_{N_1 < n_1\leqslant2N_1}\sum\limits_{\substack{N_2 < n_2\leqslant2N_2\\n_1m_{2k}\not = n_2m_k}}\frac{1}{|n_1m_{2k}-n_2m_k|}\\ &\leqslant\sum\limits_{N_1 < n_1\leqslant2N_1}\left(1+\sum\limits_{1\leqslant t\leqslant2y^2}\frac{1}{m_kt+c_0}\right)\\ &\ll N_1\log y, \end{align}

    similarly, we have

    T(m_k, m_{2k})\ll N_2\log y,

    thus

    T(m_k, m_{2k})\ll (N_1N_2)^{\frac{1}{2}}\log y.

    By (4.14) we have

    \begin{eqnarray} R_1\ll\frac{(m_km_{2k})^{\frac{3}{4}}y^{\varepsilon}}{(N_1N_2)^{1/2}}(N_1N_2)^{\frac{1}{2}}\log y\ll(m_km_{2k})^{\frac{3}{4}}T^{\varepsilon}. \end{eqnarray} (4.15)

    Finally by (4.12), (4.13) and (4.15) we obtain

    R_1+R_2\ll T^{\varepsilon}(m_km_{2k})^{3/4},

    thus

    \begin{align} \int_1&\ll T^{2k-1+(2k+1)\varepsilon}\sum\limits_{m_1, \cdots, m_{2k}\leqslant{y}}\frac{|f(\mathbf{M}_1)||f(\mathbf{M}_2)|} {D_1D_2{m_k}^{1/4}{m_{2k}}^{1/4}}\cdot T^{{\varepsilon}}{m_k}^{3/4}{m_{2k}}^{3/4}\\ &\ll T^{2k-1+(2k+2)\varepsilon}\sum\limits_{m_1, \cdots, m_{2k}\leqslant{y}}\frac{|f(\mathbf{M}_1)||f(\mathbf{M}_2)|{m_k}^{1/2}{m_{2k}}^{1/2}} {D_1D_2}\\ &\ll T^{2k-1+(2k+2)\varepsilon}y\sum\limits_{m_1, \cdots, m_{2k}\leqslant{y}}\frac{|f(\mathbf{M}_1)||f(\mathbf{M}_2)|} {D_1D_2}\cdot\frac{T^{2\varepsilon}}{{m_k}^{\varepsilon}{m_{2k}}^{\varepsilon}}\\ &\ll T^{2k-1+(2k+4)\varepsilon}y\left(\sum\limits_{m_1, \cdots, m_{k} = 1}^{\infty}\frac{|f(\mathbf{M}_1)|} {D_1{m_k}^{\varepsilon}}\right)^2\\ &\ll T^{2k-1+\varepsilon}y, \end{align} (4.16)

    since \varepsilon > 0 is arbitrary, T^\varepsilon\ll y\ll T , and the convergence of the latter series can be obtained by Lemma 2.4. Above all, by (4.4), (4.10), (4.11), and (4.16) we obtain

    \begin{align} \int_T^{2T}{M_{11}}^2(x, y)dx = &\frac{k^2}{4\pi^2}\sum\limits_{l_1, l_2 = 0}^{k-1}D_{k, l_1, l_2}\int_T^{2T}x^{2k-\frac{3}{2}}(\log x)^{l_1+l_2}dx\\ &+O\left(T^{2k-\frac{1}{2}+\varepsilon}{y}^{-\frac{1}{2}}\right)+O\left(T^{2k-1+\varepsilon}y\right). \end{align} (4.17)

    We are going to estimate \int_T^{2T}M_{12}(x, y)dx ,

    \begin{align} \begin{aligned} \int_T^{2T}{M_{12}}^2(x, y)dx & = k^2\int_T^{2T}\left(\sum\limits_{m_1, \cdots, m_{k}\leqslant y}f(\mathbf{M}_1)\prod\limits_{j = 1}^{k-1}M\left(\frac{x}{m_j}\right)\delta_2\left(\frac{x}{m_k}, y\right)\right)^2dx\nonumber\\ &\ll \int_T^{2T}\sum\limits_{m_1, \cdots, m_{2k}\leqslant y}|f(\mathbf{M}_1)||f(\mathbf{M}_2)|\left(\prod\limits_{j = 1}^{k-1}\frac{x\log x}{m_j}\prod\limits_{j = k+1}^{2k-1}\frac{x\log x}{m_j}\right) \left|\delta_2\left(\frac{x}{m_k}, y\right)\right|\left|\delta_2\left(\frac{x}{m_{2k}}, y\right)\right|dx\nonumber\\ &\ll \sum\limits_{m_1, \cdots, m_{2k}\leqslant y}\frac{|f(\mathbf{M}_1)||f(\mathbf{M}_2)|}{D_1D_2} \int_{T}^{2T}x^{2k-2+\varepsilon}\left|\delta_2\left(\frac{x}{m_k}, y\right)\right|\left|\delta_2\left(\frac{x}{m_{2k}}, y\right)\right|dx\nonumber\\ &\ll T^{2k-2+\varepsilon}\sum\limits_{m_1, \cdots, m_{2k}\leqslant y}\frac{|f(\mathbf{M}_1)||f(\mathbf{M}_2)|}{D_1D_2} \left(\int_{T}^{2T}{\delta_2}^2\left(\frac{x}{m_{k}}, y\right)dx\right)^{\frac{1}{2}}\left(\int_{T}^{2T}{\delta_2}^2\left(\frac{x}{m_{2k}}, y\right)dx\right)^{\frac{1}{2}}.\nonumber\end{aligned} \end{align} (4.18)

    By Lemma 2.2 we have

    \begin{align} \left(\int_{T}^{2T}{\delta_2}^2\left(\frac{x}{m_k}, y\right)dx\right)^{\frac{1}{2}} & = \left(m_k\int_{\frac{T}{m_k}}^{\frac{2T}{m_k}}{\delta_2}^2\left(u, y\right)du\right)^{\frac{1}{2}}\\ &\ll\left(m_k\left(\frac{T^{\frac{3}{2}}}{{m_k}^{\frac{3}{2}}{y}^{\frac{1}{2}}}\log^3T+\frac{T}{m_k}\log^4T\right)\right)^{\frac{1}{2}}\\ &\ll\frac{T^{\frac{3}{4}+\varepsilon}}{{m_k}^{\frac{1}{4}}{y}^{\frac{1}{4}}}+T^{\frac{1}{2}+\varepsilon}, \end{align}

    similarly, we have

    \begin{eqnarray} \left(\int_{T}^{2T}{\delta_2}^2\left(\frac{x}{m_{2k}}, y\right)dx\right)^{\frac{1}{2}} \ll\frac{T^{\frac{3}{4}+\varepsilon}}{{m_{2k}}^{\frac{1}{4}}{y}^{\frac{1}{4}}}+T^{\frac{1}{2}+\varepsilon}. \end{eqnarray}

    Thus,

    \begin{align} \int_T^{2T}{M_{12}}^2(x, y)dx &\ll T^{2k-2+\varepsilon}\sum\limits_{m1, \cdots, m_{2k}\leqslant y} \frac{|f(\mathbf{M}_1)||f(\mathbf{M}_2)|}{D_1D_2}\left(\frac{T^{\frac{3}{4}+\varepsilon}}{{m_k}^{\frac{1}{4}}{y}^{\frac{1}{4}}}+T^{\frac{1}{2}+\varepsilon}\right) \left(\frac{T^{\frac{3}{4}+\varepsilon}}{{m_{2k}}^{\frac{1}{4}}{y}^{\frac{1}{4}}}+T^{\frac{1}{2}+\varepsilon}\right)\\ &\ll T^{2k-\frac{1}{2}+3\varepsilon}{y}^{-\frac{1}{2}}\sum\limits_{m1, \cdots, m_{2k} = 1}^{\infty} \frac{|f(\mathbf{M}_1)||f(\mathbf{M}_2)|}{D_1D_2{m_k}^{\frac{1}{4}}{m_{2k}}^{\frac{1}{4}}}\\ &\ll T^{2k-\frac{1}{2}+3\varepsilon}{y}^{-\frac{1}{2}}, \end{align} (4.19)

    since y\ll T , and the convergence of the latter series is given by Lemma 2.4.

    Then by (4.17), (4.19), and Cauchy-Schwarz's inequality, we are able to obtain

    \begin{eqnarray} \int_T^{2T}M_{11}M_{12}dx\ll\left(\int_T^{2T}{M_{11}}^2dx\right)^{\frac{1}{2}}\left(\int_T^{2T}{M_{12}}^2dx\right)^{\frac{1}{2}}\ll\frac{T^{2k-\frac{1}{2}+\varepsilon}}{{y}^{\frac{1}{4}}}. \end{eqnarray} (4.20)

    So from (4.3) and (4.17)–(4.20), we obtain

    \begin{align} \int_T^{2T}{M_{1}}^2dx = &\frac{k^2}{4\pi^2}\sum\limits_{l_1, l_2 = 0}^{k-1}D_{k, l_1, l_2}\int_T^{2T}x^{2k-\frac{3}{2}}(\log x)^{l_1+l_2}dx\\ &+O\left(T^{2k-\frac{1}{2}+\varepsilon}{y}^{-\frac{1}{4}}\right)+O\left(T^{2k-1+\varepsilon}y\right). \end{align} (4.21)

    Let \mathbf{M_1}, \mathbf{M_2}, D_1, D_2 be defined in Lemma 2.5, then for \int_T^{2T}{M_3}^2dx we deduce that

    \begin{eqnarray} \int_T^{2T}{M_3}^2(x, y)dx = \int_{T}^{2T}\left(\sum\limits_{\substack{m_1, \cdots, m_k\leqslant x\\m_k > y}}|f(m_1, \cdots, m_k)| \prod\limits_{j = 1}^{k-1}M\left(\frac{x}{m_j}\right)\left|\Delta\left(\frac{x}{m_k}\right)\right|\right)^2dx \end{eqnarray}

    change order of integration and summation we obtain

    \begin{align} \int_T^{2T}{M_3}^2(x, y)dx &\leqslant\int_{T}^{2T}\sum\limits_{\substack{m_1, \cdots, m_{2k}\leqslant 2T\\m_k, m_{2k} > y}}|f(\mathbf{M}_1)f(\mathbf{M}_2)| \prod\limits_{j = 1}^{k-1}M\left(\frac{x}{m_j}\right) \prod\limits_{j = k+1}^{2k-1}M\left(\frac{x}{m_j}\right)\left|\Delta\left(\frac{x}{m_k}\right)\right|\left|\Delta\left(\frac{x}{m_{2k}}\right)\right|dx\\ & = \sum\limits_{\substack{m_1, \cdots, m_{2k}\leqslant{2T}\\m_k, m_{2k} > y}}|f(\mathbf{M}_1)f(\mathbf{M}_2)|\int_{T}^{2T} \prod\limits_{j = 1}^{k-1}M\left(\frac{x}{m_j}\right) \prod\limits_{j = k+1}^{2k-1}M\left(\frac{x}{m_j}\right)\left|\Delta\left(\frac{x}{m_k}\right)\right|\left|\Delta\left(\frac{x}{m_{2k}}\right)\right|dx\\ &\ll\sum\limits_{\substack{m_1, \cdots, m_{2k}\leqslant{2T}\\m_k, m_{2k} > y}}\frac{|f(\mathbf{M}_1)||f(\mathbf{M}_2)|}{D_1D_2} \int_{T}^{2T}x^{2k-2+\varepsilon}\left|\Delta\left(\frac{x}{m_k}\right)\right|\left|\Delta\left(\frac{x}{m_{2k}}\right)\right|dx\\ &\ll T^{2k-2+\varepsilon}\sum\limits_{\substack{m_1, \cdots, m_{2k}\leqslant{2T}\\m_k, m_{2k} > y}}\frac{|f(\mathbf{M}_1)||f(\mathbf{M}_2)|}{D_1D_2} \int_{T}^{2T}\left|\Delta\left(\frac{x}{m_k}\right)\right|\left|\Delta\left(\frac{x}{m_{2k}}\right)\right|dx. \end{align}

    Using Cauchy-Schwarz's inequality and (1.3), we deduce that

    \begin{align} \int_{T}^{2T}\left|\Delta\left(\frac{x}{m_k}\right)\right|\left|\Delta\left(\frac{x}{m_{2k}}\right)\right|dx &\ll\left(\int_{T}^{2T}\Delta^2\left(\frac{x}{m_k}\right)dx\right)^{\frac{1}{2}}\left(\int_{T}^{2T}\Delta^2\left(\frac{x}{m_{2k}}\right)dx\right)^{\frac{1}{2}}\\ &\ll\left(m_k\int_{\frac{T}{m_k}}^{\frac{2T}{m_k}}\Delta^2(u)du\right)^{\frac{1}{2}}\left(m_{2k}\int_{\frac{T}{m_{2k}}}^{\frac{2T}{m_{2k}}}\Delta^2(u)du\right)^{\frac{1}{2}}\\ &\ll\frac{T^{\frac{3}{2}}}{{m_k}^{\frac{1}{4}}{m_{2k}}^{\frac{1}{4}}}, \end{align}

    thus, by Lemma 2.4, we have

    \begin{align} \int_T^{2T}{M_3}^2(x, y)dx&\ll T^{2k-\frac{1}{2}+\varepsilon}\sum\limits_{\substack{m_1, \cdots, m_{2k}\leqslant2T\\m_k, m_{2k} > y}} \frac{|f(\mathbf{M}_1)||f(\mathbf{M}_2)|}{D_1D_2{m_k}^{\frac{1}{4}}{m_{2k}}^{\frac{1}{4}}}\\ &\ll T^{2k-\frac{1}{2}+\varepsilon}\sum\limits_{\substack{m_1, \cdots, m_{2k}\leqslant2T\\m_k, m_{2k} > y}}\frac{|f(\mathbf{M}_1)||f(\mathbf{M}_2)|} {D_1D_2{m_k}^{\varepsilon}{m_{2k}}^{\varepsilon}}{m_k}^{-\frac{1}{4}+\varepsilon}{m_{2k}}^{-\frac{1}{4}+\varepsilon}\\ &\ll T^{2k-\frac{1}{2}+\varepsilon}{y}^{-\frac{1}{2}+2\varepsilon}\sum\limits_{\substack{m_1, \cdots, m_{2k}\leqslant2T\\m_k, m_{2k} > y}}\frac{|f(\mathbf{M}_1)||f(\mathbf{M}_2)|} {D_1D_2{m_k}^{\varepsilon}{m_{2k}}^{\varepsilon}}\\ &\ll T^{2k-\frac{1}{2}+\varepsilon}{y}^{-\frac{1}{2}}\sum\limits_{m_1, \cdots, m_{2k} = 1}^{\infty}\frac{|f(\mathbf{M}_1)||f(\mathbf{M}_2)|} {D_1D_2{m_k}^{\varepsilon}{m_{2k}}^{\varepsilon}}\\ &\ll \frac{T^{2k-\frac{1}{2}+\varepsilon}}{{y}^{\frac{1}{2}}}. \end{align} (4.22)

    For \int_T^{2T}{M_2}^2dx , let {D_1}^{'} = D_1/m_1 , {D_2}^{'} = D_2/m_{k+1} , similar to \int_T^{2T}{M_3}^2dx , we obtain by Lemma 2.4 that

    \begin{align} \int_T^{2T}{M_2}^2(x, y)dx&\ll T^{2k-\frac{1}{2}+\varepsilon}\sum\limits_{\substack{m_1, \cdots, m_{2k}\leqslant2T\\m_1, m_{k+1} > y}} \frac{|f(\mathbf{M}_1)||f(\mathbf{M}_2)|}{D_1D_2{m_k}^{\frac{1}{4}}{m_{2k}}^{\frac{1}{4}}}\\ &\ll T^{2k-\frac{1}{2}+\varepsilon}\sum\limits_{\substack{m_1, \cdots, m_{2k}\leqslant2T\\m_1, m_{k+1} > y}}\frac{|f(\mathbf{M}_1)||f(\mathbf{M}_2)|{m_1}^{-\frac{1}{4}+\varepsilon}{m_{k+1}}^{-\frac{1}{4}+\varepsilon}} {{D_1}^{'}{D_2}^{'}{m_1}^{\frac{3}{4}+\varepsilon}{m_k}^{\frac{1}{4}}{m_{k+1}}^{\frac{3}{4}+\varepsilon}{m_{2k}}^{\frac{1}{4}}}\\ &\ll T^{2k-\frac{1}{2}+\varepsilon}{y}^{-\frac{1}{2}+2\varepsilon}\sum\limits_{\substack{m_1, \cdots, m_{2k}\leqslant2T\\m_1, m_{k+1} > y}}\frac{|f(\mathbf{M}_1)||f(\mathbf{M}_2)|} {{D_1}^{'}{D_2}^{'}{m_1}^{\frac{3}{4}+\varepsilon}{m_k}^{\frac{1}{4}}{m_{k+1}}^{\frac{3}{4}+\varepsilon}{m_{2k}}^{\frac{1}{4}}}\\ &\ll T^{2k-\frac{1}{2}+\varepsilon}{y}^{-\frac{1}{2}}\sum\limits_{m_1, \cdots, m_{2k} = 1}^{\infty}\frac{|f(\mathbf{M}_1)||f(\mathbf{M}_2)|} {{D_1}^{'}{D_2}^{'}{m_1}^{\frac{3}{4}+\varepsilon}{m_k}^{\frac{1}{4}}{m_{k+1}}^{\frac{3}{4}+\varepsilon}{m_{2k}}^{\frac{1}{4}}}\\ &\ll \frac{T^{2k-\frac{1}{2}+\varepsilon}}{{y}^{\frac{1}{2}}}. \end{align} (4.23)

    By (4.21)–(4.23) and Cauchy-Schwarz's inequality, we are able to estimate the following terms in (4.2)

    \begin{eqnarray} &&\int_T^{2T}M_1M_2dx\ll\left(\int_T^{2T}{M_1}^2dx\right)^{\frac{1}{2}}\left(\int_T^{2T}{M_2}^2dx\right)^{\frac{1}{2}}\ll\frac{T^{2k-\frac{1}{2}+\varepsilon}}{{y}^{\frac{1}{4}}}, \\ &&\int_T^{2T}M_1M_3dx\ll\left(\int_T^{2T}{M_1}^2dx\right)^{\frac{1}{2}}\left(\int_T^{2T}{M_3}^2dx\right)^{\frac{1}{2}}\ll\frac{T^{2k-\frac{1}{2}+\varepsilon}}{{y}^{\frac{1}{4}}}, \\ &&\int_T^{2T}M_2M_3dx\ll\left(\int_T^{2T}{M_2}^2dx\right)^{\frac{1}{2}}\left(\int_T^{2T}{M_3}^2dx\right)^{\frac{1}{2}}\ll\frac{T^{2k-\frac{1}{2}+\varepsilon}}{{y}^{\frac{1}{2}}}. \end{eqnarray} (4.24)

    Above all, taking y = T^{\frac{2}{5}} , then

    \begin{eqnarray} \int_T^{2T}\left({\Delta_k}^{*}(x)\right)^2dx = \frac{k^2}{4\pi^2}\sum\limits_{l_1, l_2 = 0}^{k-1}D_{k, l_1, l_2}\int_T^{2T}x^{2k-\frac{3}{2}}(\log x)^{l_1+l_2}dx +O\left(T^{2k-\frac{3}{5}+\varepsilon}\right) \end{eqnarray} (4.25)

    follows by (4.2) and (4.21)–(4.24).

    Using (3.5) and the Cauchy-Schwarz's inequality we obtain

    \begin{align} \int_T^{2T}{\Delta_k}^2(x)dx = &\int_T^{2T}\left({\Delta_k}^{*}(x)\right)^2dx+O\left(\int_T^{2T}{\Delta_k}^{*}(x)x^{k-1+\varepsilon}dx\right)+O\left(T^{2k-1+\varepsilon}\right)\\ = &\frac{k^2}{4\pi^2}\sum\limits_{l_1, l_2 = 0}^{k-1}D_{k, l_1, l_2}\int_T^{2T}x^{2k-\frac{3}{2}}(\log x)^{l_1+l_2}dx+O\left(T^{2k-\frac{3}{5}+\varepsilon}\right)\\ &+ \quad O\left(\left(\int_T^{2T}\left({\Delta_k}^{*}(x)\right)^2dx\right)^{\frac{1}{2}}\left(\int_T^{2T}x^{2k-2+\varepsilon}dx\right)^{\frac{1}{2}}\right)\\ = &\frac{k^2}{4\pi^2}\sum\limits_{l_1, l_2 = 0}^{k-1}D_{k, l_1, l_2}\int_T^{2T}x^{2k-\frac{3}{2}}(\log x)^{l_1+l_2}dx+O\left(T^{2k-\frac{3}{5}+\varepsilon}\right). \end{align}

    Then replacing T by T/2 , T/2^2 , and so on, and adding up all the results, we obtain

    \begin{align} \int_1^{T}{\Delta_k}^2(x)dx& = \frac{k^2}{4\pi^2}\sum\limits_{l_1, l_2 = 0}^{k-1}D_{k, l_1, l_2}\int_1^{T}x^{2k-\frac{3}{2}}(\log x)^{l_1+l_2}dx +O\left(T^{2k-\frac{3}{5}+\varepsilon}\right)\\ & = \frac{k^2}{4\pi^2}T^{2k-\frac{1}{2}}L_{2k-2}(\log T)+O\left(T^{2k-\frac{3}{5}+\varepsilon}\right), \end{align}

    where we use integration by part several times to obtain L_{2k-2}(u) is a polynomial in u of degree 2k-2 denoted by

    L_{2k-2}(u) = \sum\limits_{l_1, l_2 = 0}^{k-1}D_{k, l_1, l_2}\sum\limits_{r = 0}^{l_1+l_2}\frac{(-1)^r(l_1+l_2)!}{(2k-\frac{1}{2})^{r+1}(l_1+l_2-r)!}u^{l_1+l_2}.

    To sum up, this finishes the proof of the Theorem.

    In this paper, we give an asymptotic formula of the mean square of \Delta_{k}(x) , which can be viewed as an analogue of (1.3). We use the convergence of the multivariable Dirichlet series, and it can be used to show the properties of other multivariable arithmetic functions. In 2023, Tóth[13], Heyman and Tóth[14] gave some useful applications of the Dirichlet series.

    The author would like to appreciate the referee for his/her patience in refereeing this paper. This work is supported by Natural Science Foundation of Beijing Municipal (Grant No.1242003), and the National Natural Science Foundation of China (Grant Nos.12471009 and 12301006).

    No potential conflicts of interest were reported by the author.



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