Some mean value results related to Hardy’s function

Abstract

Let $\zeta (s)$ and Z(t) be the Riemann zeta function and Hardy’s function respectively. We show asymptotic formulas for $\int _0^T Z(t)\zeta (1/2+it)dt$ and $\int _0^T Z^2(t) \zeta (1/2+it)dt$. Furthermore we derive an upper bound for $\int _0^T Z^3(t) \chi ^{\alpha }(1/2+it)dt$ for $-1/2<\alpha <1/2$, where $\chi (s)$ is the function which appears in the functional equation of the Riemann zeta function: $\zeta (s)=\chi (s)\zeta (1-s)$.

1 Introduction

Let Z(t) be Hardy’s function defined by

$$\begin{aligned} Z(t)=\zeta (1/2+it) \chi ^{-1/2}(1/2+it), \end{aligned}$$

where as usual $\zeta (s)$ is the Riemann zeta-function and $\chi (s)$ is the gamma factor appearing in the functional equation of $\zeta (s)$:

$$\begin{aligned} \zeta (s)=\chi (s)\zeta (1-s). \end{aligned}$$

(1)

The explicit form of $\chi (s)$ is

$$\begin{aligned} \chi (s)=2^s\pi ^{s-1}\sin \left( \frac{\pi s}{2}\right) \varGamma (1-s) \end{aligned}$$

(2)

and its asymptotic behavior is given by

$$\begin{aligned} \chi (\sigma +it)=\left( \frac{|t|}{2\pi }\right) ^{1/2-\sigma -it}e^{i(t \pm \frac{\pi }{4})} \left( 1+O\left( \frac{1}{|t|}\right) \right) \end{aligned}$$

(3)

for $|t| \ge t_0>0$, where $t \pm \frac{\pi }{4}=t+\mathrm{sgn}(t)\frac{\pi }{4}$. See Ivić [10].

From (1), it follows that Z(t) is a real-valued even function for real t and $|Z(t)|=|\zeta (1/2+it)|$. Therefore the zeros of $\zeta (s)$ on the critical line $\mathrm{Re}\,s=1/2$ coincide with the real zeros of Z(t). Historically, Hardy proved the infinity of the number of zeros of $\zeta (s)$ on the critical line in 1914. A little later Hardy and Littlewood gave another proof by showing that $\int _0^T Z(t)dt \ll T^{7/8}$ and $\int _0^T|Z(t)|dt \gg T$. See Chandrasekharan [3, Chapter II, §4 and Notes on Chapter II] or Titchmarsh [23, 10.5].

Since $Z^2(t)= |\zeta (1/2+it)|^2$, 2k-th power moment of Z(t) is equivalent to 2k-th power moment of $|\zeta (1/2+it)|$. Hardy and Littlewood first showed the asymptotic formula in the case $k=1$. In fact they showed that

$$\begin{aligned} \int _0^T |\zeta (1/2+it)|^2dt \sim T \log T \end{aligned}$$

([5, 6]). In 1926, Ingham [9] derived

$$\begin{aligned} \int _0^T |\zeta (1/2+it)|^2dt=T \log \frac{T}{2\pi }+(2\gamma _0-1)T+E(T) \end{aligned}$$

(4)

with $E(T) \ll T^{1/2}\log T$, where $\gamma _0$ is Euler’s constant. There are a lot of literatures on E(T) since then. For instance, Atkinson [1] gave an explicit formula for E(T), which becomes the fundamental tool of further researches on E(T). See Ivić [10] for more details. For $k=2$, among other things, Ingham [9] showed that

$$\begin{aligned} \int _0^T|\zeta (1/2+it)|^4dt=\frac{1}{2\pi ^4}T\log ^4T+E_2(T) \end{aligned}$$

(5)

with $E_2(T)=O(T\log ^3 T)$ by applying the famous approximate functional equation of $\zeta ^2(s)$ of Hardy and Littlewood [7]. Ingham’s result was improved by Heath-Brown [8] to $E_2(T)=T \sum _{n=0}^4 c_n \log ^n T +O(T^{7/8+\varepsilon })$. Motohashi [21] studied $E_2(T)$ by the use of spectral theory of automorphic forms. See also Ivić [11] or Titchmarsh [23, 7.20]. Other mean value theorems (of even power) were studied by Hall [4] in connection with the distribution of consecutive zeros of Z(t).

As for odd power moments of Z(t), Ivić [12] proved in 2004 that

$$\begin{aligned} \int _0^T Z(t) dt \ll T^{1/4+\varepsilon }. \end{aligned}$$

It shows that Z(t) changes sign quite often. Ivić’s result was sharpened to $\int _0^T Z(t)dt \ll T^{1/4}$ by Jutila [17, 18] and Korolev [20] independently. Moreover they showed the Omega result $\int _0^T Z(t)dt = \varOmega _{\pm }(T^{1/4})$ which was conjectured by Ivić [12]. It means that $T^{1/4}$ is the true order of $\int _0^TZ(t)dt.$ Since there are a large amount of cancellations, it is expected that the cubic power moment has an exponent less than 1. In fact, Ivić showed that

$$\begin{aligned} \int _T^{2T}Z^3(t)dt&=2\pi \sqrt{\frac{2}{3}} \sum _{(\frac{T}{2\pi })^{3/2} \le n \le (\frac{T}{\pi })^{3/2}} \frac{d_3(n)}{n^{1/6}} \cos \left( 3\pi n^{2/3}+\frac{1}{8}\pi \right) \\&\quad +O(T^{3/4+\varepsilon }) \end{aligned}$$

and conjectured that

$$\begin{aligned} \int _0^{T} Z^3(t)dt \ll T^{3/4+\varepsilon } \end{aligned}$$

(6)

([14, Chapter 11]). Here $d_3(n)$ denotes the number of triples $(k_1,k_2, k_3)$ such that $n=k_1k_2k_3, k_j \in \mathbb {Z}, k_j>0$. If we use (4), (5) and the Cauchy-Schwarz inequality we have $\int _0^T Z^3(t) dt \ll T(\log T)^{5/2}$. The best upper bound at present is due to Bettin, Chandee and Radziwiłł [2] who showed the second ineqality of the following:

$$\begin{aligned} \left| \int _0^T Z^3(t) dt \right| \le \int _0^T|Z(t)|^3 dt \ll T (\log T)^{9/4}. \end{aligned}$$

(7)

It should be noted that $T (\log T)^{9/4}$ is the correct order of $\int _0^T|Z(t)|^3dt$.

In this paper we shall prove several mean values of the functions combined with Z(t) and $\zeta (1/2+it)$.

Theorem 1

For large $T>0$, we have

$$\begin{aligned} \int _0^T Z(t)\zeta \left( \frac{1}{2}+it\right) dt&=\frac{2\sqrt{2}\pi }{3} e^{\frac{\pi i}{8}}\left( \frac{T}{2\pi }\right) ^{3/4} \left( \frac{1}{2} \log \frac{T}{2\pi }+2\gamma _0-2\log 2 -\frac{2}{3} \right) \\&\quad +O(T^{1/2}\log T). \end{aligned}$$

We recall that $\gamma _0$ is Euler’s constant which coincides with the 0-th coefficient of the Laurent expansion of $\zeta (s)$ at $s=1$.

Ivić’s conjecture (6) would follow from the bound of exponential sum

$$\begin{aligned} \sum _{N \le n \le 2\sqrt{2}N} \frac{d_3(n)}{n^{1/6}} e^{3\pi i n^{2/3}} \ll N^{1/2+\varepsilon }, \end{aligned}$$

(8)

or, as Ivić noted [15, (1.6)], from

$$\begin{aligned} \sum _{N \le n \le 2N} d_3(n) e^{3\pi i n^{2/3}} \ll N^{2/3+\varepsilon }. \end{aligned}$$

(9)

It seems that (9) (or (8)) is out of reach of the present method of exponential sums. However if we replace $d_3(n)$ by d(n) (the divisor function $d(n)=\sum _{n=d_1d_2}1$), we can prove the following theorem in the frame of Theorem 1.

Theorem 2

Let A be a parameter such that $A \gg N^{-1/4}$. Then we have

$$\begin{aligned}&\sum _{N \le k \le 2\sqrt{2}N}\frac{d(k)}{k^{1/6}}e^{3\pi i(Ak)^{2/3}} \\&\quad =\sqrt{3}A^{-4/3} \sum _{A^{4/3}N^{1/3} \le k \le \sqrt{2}A^{4/3}N^{1/3}}d(k)k^{1/2}e^{-\pi i (k/A)^2} \nonumber \\&\qquad +O(A^{-1/3}N^{1/2+\varepsilon }) + O(A^{1/3}N^{1/6} \log N)+O(A^{-1/9}N^{2/9+\varepsilon }) \nonumber \\[1ex]&\quad \ll A^{2/3}N^{1/2} \log N. \nonumber \end{aligned}$$

For another kind of mean value of Z(t) and $\zeta (1/2+it)$ we have

Theorem 3

For large $T>0$ we have

$$\begin{aligned} \int _0^T Z^2(t) \zeta (1/2+it) dt&= T \left\{ \frac{1}{2}\left( \log \frac{T}{2\pi }\right) ^2 + a_1\log \frac{T}{2\pi }+a_2 \right\} \\&\quad {}+O(T^{3/4}\log ^2 T), \end{aligned}$$

where $a_1=3\gamma _0-1, a_2=3\gamma _1+3\gamma _0^2-3\gamma _0+1$, $\gamma _j$ being the coefficients of the Laurent expansion of $\zeta (s)$ at $s=1$.

We note that the integral of the left-hand side has an asymptotic form. It may be interesting to compare with Ivić’s conjecture (6).

As for another mean value, we shall prove the following

Theorem 4

Let $\alpha$ be a real fixed constant such that $-1/2< \alpha <1/2$. Then we have

$$\begin{aligned} \int _T^{2T}Z^3(t) \chi ^{\alpha }(1/2+it) dt \ll {\left\{ \begin{array}{ll} T^{1-\frac{\alpha }{6}+\varepsilon } &{} \text {if }\, 0 \le \alpha< 1/2 , \\ T^{1+\frac{\alpha }{6}+\varepsilon } &{} \text {if }\, -1/2< \alpha \le 0. \end{array}\right. } \end{aligned}$$

The cubic moment of Hardy’s function corresponds to $\alpha =0$, but unfortunately this gives only $O(T^{1+\varepsilon })$.

2 Some Lemmas

Lemma 1

Suppose that f(x) and $\varphi (x)$ are real-valued functions on the interval [a, b] which satisfy the conditions

1) $f^{(4)}(x)$ and $\varphi ''(x)$ are continuous,

2) there exist numbers $H,A,U,0<H,A<U,0<b-a \le U$, such that

$$\begin{aligned}&A^{-1} \ll f''(x) \ll A^{-1}, \quad f^{(3)} \ll A^{-1}U^{-1}, \quad f^{(4)}(x) \ll A^{-1}U^{-2}\\&\quad \varphi (x) \ll H, \quad \varphi '(x) \ll HU^{-1}, \quad \varphi ''(x) \ll HU^{-2}, \end{aligned}$$

3) $f'(c)=0$ for some c, $a \le c \le b$.

Then

$$\begin{aligned}&\int _a^b \varphi (x)\exp (2\pi i f(x)) dx = \frac{1+i}{\sqrt{2}}\frac{\varphi (c)\exp (2\pi if(c)}{\sqrt{f''(c)}}+O(HAU^{-1})\\&\quad +O\left( H\min (|f'(a)|^{-1}, \sqrt{A})\right) +O\left( H\min (|f'(b)|^{-1}, \sqrt{A})\right) . \end{aligned}$$

This is Lemma 2 of Karatsuba and Voronin [19, p.71].

Remark 1

Here we give an important remark. As is noted in Ivić and Zhai [16], the proof actually shows that if there is no c which satisfies the condition 3, the term containing c does not appear in the right-hand side. Moreover if $c=a$ or $c=b$, then the main term is to be halved.

Lemma 2

For $\frac{1}{2} \le \sigma < 1$ fixed, $1 \ll x, y \ll t^k, s=\sigma +it, xy=(\frac{t}{2\pi })^k, t \ge t_0$ and $k \ge 1$ a fixed integer, we have

$$\begin{aligned} \zeta ^k(s)&=\sum _{m=1}^{\infty }\rho \left( \frac{m}{x}\right) d_k(m)m^{-s}+\chi ^k(s)\sum _{m=1}^{\infty }\rho \left( \frac{m}{y}\right) d_k(m)m^{s-1} \\&\quad +O(t^{k(1-\sigma )/3-1})+O(t^{k(1/2-\sigma )-2}y^{\sigma }\log ^{k-1}t). \end{aligned}$$

Here $\chi (s)$ is the function defined by (2) and $\rho (u) (\ge 0)$ is a smooth function such that $\rho (u)+\rho (1/u)=1$ for $u>0$ and $\rho (u)=0$ for $u \ge 2$.

This is Lemma 4 of [16]. See also [14, Theorem 4.16].

For the proof of Theorem 4 we need the following lemma.

Lemma 3

Let $\alpha , \beta , \gamma$ be fixed real numbers such that $\alpha (\alpha -1)\beta \gamma \ne 0$ and write $e(x)=e^{2\pi i x}$. Let

$$\begin{aligned} S=\sum _{h=H+1}^{2H}\sum _{n=N+1}^{2N} \left| \sum _{M<m \le 2M}e\left( X\frac{m^{\alpha }h^{\beta }n^{\gamma }}{M^{\alpha }H^{\beta }N^{\gamma }} \right) \right| ^{*}, \end{aligned}$$

where $*$ means that

$$\begin{aligned} \left| \sum _{N \le n \le N'}z_n\right| ^{*}=\max _{N \le N_1 \le N_2 \le N'}\left| \sum _{n=N_1}^{N_2}z_n\right| . \end{aligned}$$

Then we have

$$\begin{aligned} S \ll (HNM)^{1+\varepsilon } \left\{ \left( \frac{X}{HNM^2}\right) ^{1/4}+\frac{1}{M^{1/2}}+\frac{1}{X}\right\} . \end{aligned}$$

This is Theorem 3 of Robert and Sargos [22].

3 Proofs of Theorem 1 and 2

Proof of Theorem 1

We consider the integral

$$\begin{aligned} J=\int _T^{2T}Z(t)\zeta \left( \frac{1}{2}+it\right) dt. \end{aligned}$$

(10)

By the definition of Z(t) and applying Lemma 2 we have

$$\begin{aligned} Z(t)\zeta \left( \frac{1}{2}+it\right)&=\zeta ^2\left( \frac{1}{2}+it\right) \chi ^{-1/2}\left( \frac{1}{2}+it\right) \\&=\left( \sum _{k=1}^{\infty }\rho \left( \frac{k}{x}\right) \frac{d(k)}{k^{1/2+it}}+\chi ^2\left( \frac{1}{2}+it\right) \sum _{k=1}^{\infty }\rho \left( \frac{k}{y}\right) \frac{d(k)}{k^{1/2-it}} \right. \\&\qquad + O\left( t^{-2/3}\right) +O\left( t^{-2}y^{1/2} \log t \right) \biggl ) \chi ^{-1/2}\left( \frac{1}{2}+it\right) , \end{aligned}$$

where $xy=(t/2\pi )^2$. Substituting this expression to (10), we have

$$\begin{aligned} J=J_1+J_2+O(T^{1/3}), \end{aligned}$$

(11)

where

$$\begin{aligned} J_1&=\sum _{k=1}^{\infty }\frac{d(k)}{k^{1/2}}\int _T^{2T}\rho \left( \frac{k}{x}\right) k^{-it}\chi ^{-1/2}\left( \frac{1}{2}+it\right) dt \end{aligned}$$

(12)

and

$$\begin{aligned} J_2&=\sum _{k=1}^{\infty }\frac{d(k)}{k^{1/2}}\int _T^{2T}\rho \left( \frac{k}{y}\right) k^{it}\chi ^{3/2}\left( \frac{1}{2}+it\right) dt. \end{aligned}$$

(13)

We take

$$\begin{aligned} x=2\left( \frac{t}{2\pi }\right) , \quad y=\frac{1}{2} \left( \frac{t}{2\pi }\right) , \end{aligned}$$

and put $K=\frac{T}{\pi }.$ Then the ranges of k in the sums in (12) and (13) are, in fact, $k \le 4K$ and $k \le K$, respectively.

We first consider $J_1$. Using (3) we find that

$$\begin{aligned} k^{-it}\chi ^{-1/2}\left( \frac{1}{2}+it\right) =e^{-\frac{\pi i}{8}}e^{\frac{i}{2}(t \log \frac{t}{2\pi }-t-t\log k^2)}+O(1/t), \end{aligned}$$

hence we have

$$\begin{aligned} J_1=e^{-\frac{\pi i}{8}} \sum _{k \le 4K}\frac{d(k)}{k^{1/2}}\int _T^{2T} \rho \left( \frac{k}{x}\right) e^{\frac{i}{2}(t \log \frac{t}{2\pi }-t-t\log k^2)}dt +O(T^{1/2}\log T). \end{aligned}$$

We evaluate the above integral by applying Lemma 1 with $\varphi (t)=\rho \left( k\left( \frac{\pi }{t}\right) \right)$ and $f(t)=\frac{1}{4\pi }(t \log \frac{t}{2\pi }-t-t\log k^2)$. Note that $\varphi (t)$ satisfies the conditions of Lemma 1 with $H=1, U=T$. Since $f'(t_0)=0$ if and only if $t_0=2\pi k^2$, the main term of the integral appears for k such that

$$\begin{aligned} \left( \frac{T}{2\pi }\right) ^{1/2} \le k \le \left( \frac{T}{\pi }\right) ^{1/2}. \end{aligned}$$

(14)

Thus we get

$$\begin{aligned}&\int _T^{2T}\rho \left( \frac{k}{x}\right) e^{\frac{i}{2}(t \log \frac{t}{2\pi }-t-t\log k^2)}dt \\&\quad =M(k)+O\left( 1+\min \Bigl (\sqrt{T}, \frac{1}{|\log (\frac{T}{2\pi k^2})|}\Bigr )+ \min \Bigl (\sqrt{T}, \frac{1}{|\log (\frac{T}{\pi k^2})|}\Bigr ) \right) , \nonumber \end{aligned}$$

where

$$\begin{aligned} M(k)= e^{\frac{\pi i}{4}}\rho \left( \frac{1}{2k}\right) 2\sqrt{2} \pi k e^{-\pi i k^2}=2\sqrt{2} \pi e^{\frac{\pi i }{4}}k (-1)^k \end{aligned}$$

for k satisfying the condition (14) and 0 otherwise. This yields that

$$\begin{aligned} J_1&=2\sqrt{2}\pi e^{\frac{\pi i}{8}}\sum _{(\frac{T}{2\pi })^{1/2} \le k \le (\frac{T}{\pi })^{1/2}} ' \qquad (-1)^kd(k)k^{1/2} \\&\quad +\sum _{k \le 4K}\frac{d(k)}{k^{1/2}}\ O\left( 1+ \min \Bigl (\sqrt{T}, \frac{1}{|\log (\frac{T}{2\pi k^2})|}\Bigr )+ \min \Bigl (\sqrt{T}, \frac{1}{|\log (\frac{T}{\pi k^2})|}\Bigr )\right) \\&\quad +O(T^{1/2}\log T) \\&=:R_0+R_1+R_2+R_3 +O(T^{1/2}\log T), \end{aligned}$$

where $\sum '$ means that the terms for $k=(T/2\pi )^{1/2}$ and $k=(T/\pi )^{1/2}$ are to be halved if they are integers. It is clear that $R_1 \ll T^{1/2} \log T.$ To estimate $R_2$, we divide the sum into four parts:

$$\begin{aligned} \sum _{k \le 4K}&=\sum _{1 \le k< \frac{1}{2}\left( \frac{T}{2\pi }\right) ^{1/2}} +\sum _{\frac{1}{2}\left( \frac{T}{2\pi }\right) ^{1/2} \le k< \left( \frac{T}{2\pi }\right) ^{1/2}} +\sum _{\left( \frac{T}{2\pi }\right) ^{1/2} \le k \le 2\left( \frac{T}{2\pi }\right) ^{1/2}} + \sum _{2\left( \frac{T}{2\pi }\right) ^{1/2} < k \le 4K} \\&=:S_1+S_2+S_3+S_4. \end{aligned}$$

For $S_1$ and $S_4$ we have $\min (\sqrt{T}, \frac{1}{|\log (\frac{T}{2\pi k^2})|}) \ll \frac{1}{\log 4}$, hence we get $S_1 \ll T^{1/4}\log T$ and $S_4 \ll T^{1/2} \log T$. For $S_2$, we write $k=[(\frac{T}{2\pi })^{1/2}]-j$ for k in this range and divide the sum over j as $S_2=S_{2,1}+S_{2,2}$, where $S_{2,1}$ is the sum for $j=0,1,2$ and $S_{2,2}$ is the sum for $3 \le j \le [(\frac{T}{2\pi })^{1/2}]-\frac{1}{2}(\frac{T}{2\pi })^{1/2}$. For $S_{2,1}$ we use $\min (\sqrt{T}, \frac{1}{|\log (\frac{T}{2\pi k^2})|}) \le \sqrt{T}$ and hence $S_{2,1} \ll T^{1/4+\varepsilon }$ since the sum is finite. As for $S_{2,2}$, from

$$\begin{aligned} \log \frac{\left( \frac{T}{2\pi }\right) ^{1/2}}{k} =\left| \log \frac{\left[ \left( \frac{T}{2\pi }\right) ^{1/2}\right] -j}{\left( \frac{T}{2\pi }\right) ^{1/2}}\right| \asymp \frac{j}{\left( \frac{T}{2\pi }\right) ^{1/2}}, \end{aligned}$$

we have

$$\begin{aligned} S_{2,2} \ll T^{-1/4+\varepsilon }\sum _{j} \frac{\left( \frac{T}{2\pi }\right) ^{1/2}}{j} \ll T^{1/4+\varepsilon }. \end{aligned}$$

Thus we get $S_2 \ll T^{1/4+\varepsilon }$. It is the same for $S_3$. Combining these estimates we find that $R_2 \ll T^{1/2}\log T.$ Similarly we have $R_3 \ll T^{1/2}\log T.$ As a result, we get

$$\begin{aligned} J_1 = 2\sqrt{2}\pi e^{\frac{\pi i}{8}}\sum _{\left( \frac{T}{2\pi }\right) ^{1/2} \le k \le \left( \frac{T}{\pi }\right) ^{1/2}} ' \quad (-1)^kd(k)k^{1/2}+O(T^{1/2}\log T). \end{aligned}$$

(15)

Next we consider $J_2$. Similarly to the case of $J_1$, we have by (3) that

$$\begin{aligned} J_2&=e^{\frac{3\pi i}{8}}\sum _{k \le K} \frac{d(k)}{k^{1/2}} \int _T^{2T}\rho \left( \frac{k}{y}\right) e^{-\frac{3}{2} i (t \log \frac{t}{2\pi }-t-t\log k^{2/3})}dt \nonumber \\&\quad + O(T^{1/2}\log T). \end{aligned}$$

(16)

We apply Lemma 1 to the above integral with $\varphi (t)=\rho (2k(2\pi /t))$ and $f(t)=-\frac{3}{4\pi }(t \log \frac{t}{2\pi }-t-t\log k^{2/3})$. In this case $f'(t_0)=0$ if and only if $t_0=2\pi k^{2/3}$ and $t_0$ is contained in the interval [T, 2T] if and only if

$$\begin{aligned} \left( \frac{T}{2\pi }\right) ^{3/2} \le k \le \left( \frac{T}{\pi }\right) ^{3/2}. \end{aligned}$$

Since the range of the sum over k is $1 \le k \le K$, there are no such k, that is, the integral in (16) does not have a main term. Considering the error term by Lemma 1 we find that

$$\begin{aligned} J_2&\ll \sum _{k \le K}\frac{d(k)}{k^{1/2}}\left( 1+\min \Bigl ((\sqrt{T},\frac{1}{|\log \frac{T}{2\pi k^{2/3}}|}\Bigr ) +\min \Bigl (\sqrt{T}, \frac{1}{|\log \frac{T}{\pi k^{2/3}}|}\Bigr )\right) \\&=: R_1'+R_2'+R_3'. \end{aligned}$$

We have clearly $R_1' \ll T^{1/2}\log T$. For $R_2'$ and $R_3'$ we note that $|\log \frac{T}{k^{2/3}}| \gg 1$ since $k \le K$, which implies that $R_2', R_3' \ll T^{1/2}\log T$. Hence

$$\begin{aligned} J_2 \ll T^{1/2} \log T. \end{aligned}$$

(17)

From (11), (15) and (17), we get

$$\begin{aligned} J = 2\sqrt{2}\pi e^{\frac{\pi i}{8}}\sum _{(\frac{T}{2\pi })^{1/2} \le k \le (\frac{T}{\pi })^{1/2}} ' \quad (-1)^kd(k)k^{1/2}+O(T^{1/2}\log T). \end{aligned}$$

Now dividing the interval [0, T] as $\cup _j [T/2^j, T/2^{j-1}]$ and summing the above evaluations we have

$$\begin{aligned} \int _0^T Z(t)\zeta \left( \frac{1}{2}+it\right) dt&=2\sqrt{2}\pi e^{\frac{\pi i}{8}}\sum _{k \le (\frac{T}{2\pi })^{1/2}}(-1)^kd(k)k^{1/2} \nonumber \\&\quad +O(T^{1/2}\log T). \end{aligned}$$

(18)

To evaluate the sum on the right-hand side of (18) we recall that

$$\begin{aligned} \sum _{k \le x}(-1)^k d(k)=\frac{x}{2}(\log x+2\gamma _0-1-2\log 2)+O(x^{1/3+\varepsilon }) \end{aligned}$$

for $x \gg 1$ (see, e.g., Ivić [13]), so by partial summation we have

$$\begin{aligned} \sum _{k \le x}(-1)^k d(k)k^{1/2}=\frac{1}{3} x^{3/2} \left( \log x+2\gamma _0-2\log 2-\frac{2}{3}\right) +O(x^{5/6+\varepsilon }). \end{aligned}$$

Substituting this form to (18) we finally get

$$\begin{aligned} \int _0^T Z(t)\zeta \left( \frac{1}{2}+it\right) dt&=\frac{2\sqrt{2}\pi }{3} e^{\frac{\pi i}{8}}\left( \frac{T}{2\pi }\right) ^{3/4} \left( \frac{1}{2} \log \frac{T}{2\pi }+2\gamma _0-2\log 2 -\frac{2}{3} \right) \\&\quad +O(T^{1/2}\log T). \end{aligned}$$

This proves the assertion of Theorem 1. $\square$

Proof of Theorem 2

Let A be a parameter such that $T^{-1/2} \ll A \ll T^{3/2}$. We shall consider the integral

$$\begin{aligned} J_A=\int _T^{2T}Z(t)\zeta \left( \frac{1}{2}+it\right) A^{it}dt \end{aligned}$$

by the same way as in the proof of Theorem 1. Applying Lemma 2 we get

$$\begin{aligned} J_A= J_{A,1}+J_{A,2}+O(T^{1/3}), \end{aligned}$$

(19)

where

$$\begin{aligned} J_{A,1}=\int _T^{2T} \chi ^{-1/2}\left( \frac{1}{2}+it\right) \sum _{k=1}^{\infty } \rho \left( \frac{k}{x}\right) \frac{d(k)}{k^{1/2+it}}A^{it}dt \end{aligned}$$

(20)

and

$$\begin{aligned} J_{A,2}=\int _T^{2T} \chi ^{3/2}\left( \frac{1}{2}+it\right) \sum _{k=1}^{\infty } \rho \left( \frac{k}{y}\right) \frac{d(k)}{k^{1/2-it}}A^{it}dt, \end{aligned}$$

(21)

where $xy=(\frac{t}{2\pi })^2$. Hereafter we put $K_0=\left( \frac{T}{\pi }\right) ^{1/2}$. $\square$

Now we shall evaluate $J_{A,1}$ and $J_{A,2}$ by taking two different choices of x and y, that is,

Case 1 : we take $x=8A(\frac{t}{2\pi })^{1/2}$ and $y=\frac{1}{8A}(\frac{t}{2\pi })^{3/2}$,

Case 2 : we take $x=\frac{A}{4}(\frac{t}{2\pi })^{1/2}$ and $y=\frac{4}{A}(\frac{t}{2\pi })^{3/2}$.

3.1 Case 1

The ranges of the sums in $J_{A,1}$ and $J_{A,2}$ are at most $k \le 16AK_0$ and $k \le \frac{1}{4A}K_0^3$, respectively. By (3) and the trivial estimate for the error term we get

$$\begin{aligned} J_{A,1}&=e^{-\frac{\pi i}{8}}\sum _{k \le 16AK_0} \frac{d(k)}{k^{1/2}}\int _T^{2T} \rho \left( \frac{k}{x}\right) e^{\frac{i}{2}(t \log \frac{t}{2\pi }-t-t\log (\frac{k}{A})^2)} dt \nonumber \\&\quad + O\left( A^{1/2}T^{1/4+\varepsilon }\right) . \end{aligned}$$

(22)

We shall evaluate the integral by Lemma 1. Let $f(t)=\frac{1}{4\pi }(t \log \frac{t}{2\pi }-t-t\log (\frac{k}{A})^2)$. Then $f'(t_0)=0$ if and only if $t_0=2\pi (\frac{k}{A})^2$ and $T \le t_0 \le 2T$ if and only if

$$\begin{aligned} A\left( \frac{T}{2\pi }\right) ^{1/2} \le k \le A\left( \frac{T}{\pi }\right) ^{1/2}. \end{aligned}$$

(23)

We find that all k satisfing (23) are contained in the range $k \le 16AK_0$. Therefore the integral in (22) has a main term which is given by

$$\begin{aligned} M_A(k)= e^{\frac{\pi i}{4}}\rho \left( \frac{1}{8}\right) 2\sqrt{2}\pi \frac{k}{A} e^{-\pi i (k/A)^2} \end{aligned}$$

for $A(\frac{T}{2\pi })^{1/2} \le k \le A(\frac{T}{\pi })^{1/2}$ and $M_A(k)=0$ otherwise. We note that $\rho (1/8)=1$ in the above formula. It follows from Lemma 1 and (22) that

$$\begin{aligned} J_{A,1}&=e^{-\frac{\pi i}{8}}\sum _{A\left( \frac{T}{2\pi }\right) ^{1/2} \le k \le A\left( \frac{T}{\pi }\right) ^{1/2}}\frac{d(k)}{k^{1/2}}M_A(k) \\&+ \sum _{k \le 4AK_0} \frac{d(k)}{k^{1/2}} \ O\left( 1+ \min \biggl (\sqrt{T}, \frac{1}{\bigl |\log \bigl (\frac{\big (\frac{T}{2\pi }\big )^{\frac{1}{2}}}{k/A}\bigr )\big |}\biggr ) + \min \biggl (\sqrt{T}, \frac{1}{\bigl |\log \bigl (\frac{\big (\frac{T}{\pi }\big )^{\frac{1}{2}}}{k/A}\bigr )\big |}\biggr )\right) \\&+O(A^{1/2}T^{1/4+\varepsilon }). \end{aligned}$$

Similarly to the proof of Theorem 1, we see that the contributions from the O-terms are bounded by $O(A^{1/2}T^{1/4+\varepsilon }+A^{-1/2}T^{1/4+\varepsilon })$. Hence we get

$$\begin{aligned} J_{A,1}&=e^{\frac{\pi i}{8}}\frac{2\sqrt{2}\pi }{A}\sum _{A\left( \frac{T}{2\pi }\right) ^{\frac{1}{2}}\le k \le A\left( \frac{T}{\pi }\right) ^{\frac{1}{2}}} d(k)k^{1/2}e^{-\pi i (k/A)^2} \nonumber \\&\quad {} +O(A^{1/2}T^{1/4+\varepsilon })+O(A^{-1/2}T^{1/4+\varepsilon }). \end{aligned}$$

(24)

Next we consider $J_{A,2}$. Similarly to $J_{A,1}$ we have

$$\begin{aligned} J_{A,2}&=e^{\frac{3\pi i}{8}}\sum _{k \le \frac{1}{4A}K_0^3}\frac{d(k)}{k^{1/2}}\int _T^{2T} \rho \left( \frac{k}{y}\right) e^{-\frac{3}{2}i(t \log \frac{t}{2\pi }-t-t\log (Ak)^{2/3})}dt \\&\quad +O(A^{-1/2}T^{3/4+\varepsilon }). \end{aligned}$$

If we put $f(t)=-\frac{3}{4\pi }(t \log \frac{t}{2\pi }-t-t\log (Ak)^{2/3})$ this time, $f'(t_0)=0$ if and only if $t_0=2\pi (Ak)^{2/3}$ and so $T \le t_0 \le 2T$ if and only if

$$\begin{aligned} \frac{1}{A}\left( \frac{T}{2\pi }\right) ^{3/2} \le k \le \frac{1}{A}\left( \frac{T}{\pi }\right) ^{3/2}. \end{aligned}$$

(25)

Since k runs in the range $1 \le k \le \frac{1}{4A}K_0^3$, there is no main term in the integral of $J_{A,2}$. Hence by Lemma 1, we get similarly that

$$\begin{aligned}&J_{A,2} \ll \sum _{k \le \frac{1}{4A}K_0^3}\frac{d(k)}{k^{1/2}} \left( 1+\min \biggl (\sqrt{T}, \frac{1}{|\log \left( \frac{(T/2\pi )^{3/2}}{Ak}\right) |}\biggr ) \right. \nonumber \\&\quad \left. +\min \big (\sqrt{T}, \frac{1}{|\log \big (\frac{(T/\pi )^{3/2}}{Ak}\big )|}\biggr )\right) \nonumber \\&\quad \ll A^{-1/2}T^{3/4+\varepsilon }+A^{1/2}T^{-1/4+\varepsilon }. \end{aligned}$$

(26)

From (19), (24) and (26), we obtain

$$\begin{aligned} J_{A}&=e^{\frac{\pi i}{8}}\frac{2\sqrt{2}\pi }{A}\sum _{A(\frac{T}{2\pi })^{\frac{1}{2}}\le k \le A(\frac{T}{\pi })^{\frac{1}{2}}} d(k)k^{1/2}e^{-\pi i (k/A)^2} \nonumber \\&\quad {}+O(A^{1/2}T^{1/4+\varepsilon })+O(A^{-1/2}T^{3/4+\varepsilon })+O(T^{1/3}). \end{aligned}$$

(27)

3.2 Case 2

In this choice of x and y, the sums in (20) and (21) are actually over $k \le \frac{1}{2} A K_0$ and $k \le \frac{8}{A}{K_0}^3$, respectively. Thus

$$\begin{aligned} J_{A,1}&=e^{-\frac{\pi i}{8}}\sum _{k \le \frac{A}{2} K_0} \frac{d(k)}{k^{1/2}}\int _T^{2T} \rho \left( \frac{k}{x}\right) e^{\frac{i}{2}(t \log \frac{t}{2\pi }-t-t\log (\frac{k}{A})^2)}dt \\&\quad +O\left( A^{1/2}T^{1/4+\varepsilon }\right) \end{aligned}$$

and

$$\begin{aligned} J_{A,2}&=e^{\frac{3\pi i}{8}}\sum _{k \le \frac{8}{A} K_0^3} \frac{d(k)}{k^{1/2}}\int _T^{2T} \rho \left( \frac{k}{y}\right) e^{-\frac{3}{2}i(t \log \frac{t}{2\pi }-t-t\log (Ak)^{2/3})}dt \\&\quad + O(A^{-1/2}T^{3/4+\varepsilon }). \end{aligned}$$

As for $J_{A,1}$, the integral has a main term if and only if k satisfies (23). Since k runs over $1 \le k \le \frac{A}{2} K_0$, there are no such k. The contribution from the error term of the integral is the same as in the previous case since the range of the sum has the same order, hence we get

$$\begin{aligned} J_{A,1} \ll A^{1/2}T^{1/4+\varepsilon }+A^{-1/2}T^{1/4+\varepsilon }. \end{aligned}$$

(28)

On the other hand, the integral of $J_{A,2}$ has a main term if and only if k satisfies (25), and in fact all k are in the range $k \le \frac{8}{A}K_0^3$. Hence by Lemma 1, $J_{A,2}$ has the following form:

$$\begin{aligned} J_{A,2}&=e^{\frac{3\pi i}{8}}\sum _{\frac{1}{A}\big (\frac{T}{2\pi }\big )^{3/2} \le k \le \frac{1}{A}\big (\frac{T}{\pi }\big )^{3/2}} \frac{d(k)}{k^{1/2}}\widetilde{M}_A(k) \\&\quad + \sum _{k \le \frac{8}{A} K_0^3} \frac{d(k)}{k^{1/2}} \, O\bigg (1+\min \biggl (\sqrt{T}, \frac{1}{|\log \big (\frac{(T/2\pi )^{3/2}}{Ak}\big )|}\biggr ) \\&\quad +\min \biggl (\sqrt{T}, \frac{1}{|\log \big (\frac{(T/\pi )^{3/2}}{Ak}\big )|}\biggr ) \bigg )\\&\quad +O(A^{-1/2}T^{3/4+\varepsilon }), \end{aligned}$$

where

$$\begin{aligned} \widetilde{M}_A(k)=e^{-\frac{\pi i}{4}}\rho \left( \frac{1}{4}\right) \frac{2\sqrt{2}\pi }{\sqrt{3}}(Ak)^{1/3}e^{3\pi i (Ak)^{2/3}} \end{aligned}$$

for $\frac{1}{A}(\frac{T}{2\pi })^{3/2} \le k \le \frac{1}{A}(\frac{T}{\pi })^{3/2}$ and 0 otherwise. We see that the contribution from the O-term is the same as the previous case, therefore

$$\begin{aligned} J_{A,2}&=e^{\frac{\pi i}{8}}\frac{2\sqrt{2}\pi }{\sqrt{3}}A^{1/3} \sum _{\frac{1}{A}\left( \frac{T}{2\pi }\right) ^{3/2} \le k \le \frac{1}{A}\left( \frac{T}{\pi }\right) ^{3/2}}\frac{d(k)}{k^{1/6}}e^{3\pi i(Ak)^{2/3}} \nonumber \\&\quad + O(A^{-1/2}T^{3/4+\varepsilon })+O(A^{1/2}T^{-1/4+\varepsilon }). \end{aligned}$$

(29)

From (28) and (29) we obtain that

$$\begin{aligned} J_A&=e^{\frac{\pi i}{8}}\frac{2\sqrt{2}\pi }{\sqrt{3}}A^{1/3} \sum _{\frac{1}{A}\left( \frac{T}{2\pi }\right) ^{3/2} \le k \le \frac{1}{A}\left( \frac{T}{\pi }\right) ^{3/2}}\frac{d(k)}{k^{1/6}}e^{3\pi i(Ak)^{2/3}} \nonumber \\&\quad + O(A^{-1/2}T^{3/4+\varepsilon })+O(A^{1/2}T^{1/4+\varepsilon })+O(T^{1/3}). \end{aligned}$$

(30)

Now we have two expressions of $J_A$: (27) and (30). Comparing these expressions we obtain

$$\begin{aligned}&\sum _{\frac{1}{A}\left( \frac{T}{2\pi }\right) ^{3/2} \le k \le \frac{1}{A}\left( \frac{T}{\pi }\right) ^{3/2}}\frac{d(k)}{k^{1/6}}e^{3\pi i(Ak)^{2/3}} \nonumber \\&\quad =\sqrt{3}A^{-4/3} \sum _{A\left( \frac{T}{2\pi }\right) ^{\frac{1}{2}}\le k \le A\left( \frac{T}{\pi }\right) ^{\frac{1}{2}}}d(k)k^{1/2}e^{-\pi i (k/A)^2} \nonumber \\&\quad +O(A^{-5/6}T^{3/4+\varepsilon }) +O(A^{1/6}T^{1/4+\varepsilon })+O(A^{-1/3}T^{1/3+\varepsilon }) \nonumber \\&\quad \ll A^{1/6}T^{3/4} \log T, \end{aligned}$$

(31)

where the last inequality is obtained by the trivial estimate. In (31), we take $T=2\pi (AN)^{2/3}$. Then (31) is transformed to

$$\begin{aligned}&\sum _{N \le k \le 2\sqrt{2}N}\frac{d(k)}{k^{1/6}}e^{3\pi i(Ak)^{2/3}} \\&\quad =\sqrt{3}A^{-4/3} \sum _{A^{4/3}N^{1/3} \le k \le \sqrt{2}A^{4/3}N^{1/3}}d(k)k^{1/2}e^{-\pi i (k/A)^2} \nonumber \\&\quad +O(A^{-1/3}N^{1/2+\varepsilon }) + O(A^{1/3}N^{1/6+\varepsilon })+O(A^{-1/9}N^{2/9+\varepsilon }) \nonumber \\&\quad \ll A^{2/3}N^{1/2} \log N \nonumber \end{aligned}$$

for $A \gg N^{-1/4}$. This proves the assertion of Theorem 2.

4 Proof of Theorem 3

Since the method is similar to Theorem 1, we shall only give an outline of proof. Let

$$\begin{aligned} I=\int _T^{2T} Z^2(t)\zeta \left( \frac{1}{2}+it\right) dt. \end{aligned}$$

This time we have

$$\begin{aligned}&Z^2(t)\zeta \left( \frac{1}{2}+it\right) =\zeta ^3\left( \frac{1}{2}+it\right) \chi ^{-1}\left( \frac{1}{2}+it\right) \nonumber \\&\quad =\chi ^{-1}\left( \frac{1}{2}+it\right) \sum _{k=1}^{\infty }\rho \left( \frac{k}{x}\right) \frac{d_3(k)}{k^{1/2+it}} + \chi ^2\left( \frac{1}{2}+it\right) \sum _{k=1}^{\infty }\rho \left( \frac{k}{y}\right) \frac{d_3(k)}{k^{1/2-it}} \nonumber \\&\quad + O\left( t^{-1/2}\right) +O\left( t^{-2}y^{1/2}\log ^ 2 t\right) , \end{aligned}$$

(32)

where $xy=(\frac{t}{2\pi })^3$.

We take $x=2(\frac{t}{2\pi })^{3/2}$ and $y=\frac{1}{2}(\frac{t}{2\pi })^{3/2}$ in (32) and put $K_3=(T/\pi )^{3/2}$. Then the ranges of k in the above two sums are at most $k \le 4K_3$ and $k \le K_3$, respectively. Hence

$$\begin{aligned} I&=\sum _{k \le 4K_3}\frac{d_3(k)}{k^{1/2}}\int _T^{2T}\rho \left( \frac{k}{x}\right) k^{-it}\chi ^{-1}\left( \frac{1}{2}+it\right) dt \nonumber \\&\quad +\sum _{k \le K_3}\frac{d_3(k)}{k^{1/2}}\int _T^{2T}\rho \left( \frac{k}{y}\right) k^{it}\chi ^{2}\left( \frac{1}{2}+it\right) dt + O(T^{1/2}) \nonumber \\&=: I_1+I_2+O(T^{1/2}). \end{aligned}$$

(33)

As for $I_2$, using (3), we get

$$\begin{aligned} I_2&=e^{\frac{\pi i}{2}}\sum _{k \le K_3}\frac{d_3(k)}{k^{1/2}} \int _T^{2T} \rho \left( \frac{k}{y}\right) e^{-2i(t \log \frac{t}{2\pi }-t-t\log \sqrt{k})} dt+ O(T^{3/4}\log ^2 T). \end{aligned}$$

As in the pevious case, we apply Lemma 1 to the above integral with $\varphi (t)=\rho \left( 2k \left( \frac{2\pi }{t}\right) ^{3/2}\right)$ and $f(t)=-\frac{1}{\pi }(t \log \frac{t}{2\pi }-t-t\log \sqrt{k})$. Then we find that $f'(t_0)=0$ if and only if $t_0=2\pi \sqrt{k}$, and this $t_0$ is contained in the interval [T, 2T] if and only if

$$\begin{aligned} \left( \frac{T}{2\pi }\right) ^2 \le k \le \left( \frac{T}{\pi }\right) ^2. \end{aligned}$$

(34)

Since k runs over the range $1 \le k \le K_3$, there is no k which satisfies (34), hence the main term does not appear in this integral. On the other hand, the error term of this integral is given by

$$\begin{aligned} 1+ \min \left( \sqrt{T}, \frac{1}{|\log \frac{T}{2\pi \sqrt{k}}|}\right) +\min \left( \sqrt{T}, \frac{1}{|\log \frac{T}{\pi \sqrt{k}}|}\right) \ll 1, \end{aligned}$$

hence

$$\begin{aligned} I_2 \ll \sum _{k \le K_3} \frac{d_3(k)}{k^{1/2}}+T^{3/4} \log ^2 T \ll T^{3/4} \log ^2 T. \end{aligned}$$

(35)

Next we treat $I_1$. By (3) again, we have

$$\begin{aligned} I_1&=e^{-\frac{\pi i}{4}}\sum _{k \le 4K_3}\frac{d_3(k)}{k^{1/2}} \int _T^{2T} \rho \left( \frac{k}{x}\right) e^{i(t \log \frac{t}{2\pi }-t-t\log k)} dt+O(T^{3/4}\log ^2 T). \end{aligned}$$

In this case $\varphi (t)=\rho (k(\frac{2\pi }{t})^{3/2}/2)$ and $f(t)=\frac{1}{2\pi }(t \log \frac{t}{2\pi }-t-t\log k)$. We see that $f'(t_0)=0$ if and only if $t_0=2\pi k$ and this $t_0$ is contained in [T, 2T] if and only if

$$\begin{aligned} \frac{T}{2\pi } \le k \le \frac{T}{\pi }. \end{aligned}$$

Therefore we have

$$\begin{aligned}&\int _T^{2T}\rho \left( \frac{k}{x}\right) e^{i(t \log \frac{t}{2\pi }-t-t\log k)}dt \\&\quad =M(k)+O\left( 1+\min \left( \sqrt{T}, \frac{1}{|\log \frac{T}{2\pi k}|}\right) +\min \left( \sqrt{T}, \frac{1}{|\log \frac{T}{\pi k}|}\right) \right) , \end{aligned}$$

where M(k) is the main term given by

$$\begin{aligned} M(k)&=e^{\frac{\pi i}{4}}\rho \left( \frac{1}{2\sqrt{k}}\right) (2\pi t_0)^{1/2} e^{-2\pi i k} =2\pi e^{\frac{\pi i }{4}}k^{1/2} \end{aligned}$$

for k such that $\frac{T}{2\pi } \le k \le \frac{T}{\pi }$ and 0 otherwise. Therefore we get

$$\begin{aligned} I_1&=2\pi \sum _{\frac{T}{2\pi } \le k \le \frac{T}{\pi }}{ '} \quad d_3(k) \nonumber \\&\quad + \sum _{k \le 4K_3}\frac{d_3(k)}{k^{1/2}}\left( 1+\min \left( \sqrt{T}, \frac{1}{|\log \frac{T}{2\pi k}|}\right) +\min \left( \sqrt{T}, \frac{1}{|\log \frac{T}{\pi k}|}\right) \right) \nonumber \\&= 2\pi \sum _{\frac{T}{2\pi } \le k \le \frac{T}{\pi }}{ '} \quad d_3(k)+O(T^{3/4} \log ^2T). \end{aligned}$$

(36)

Here we can get the last O-term by the same way as in the previous case. Combining (33), (35) and (36), we obtain

$$\begin{aligned} I=2\pi \sum _{\frac{T}{2\pi } \le k \le \frac{T}{\pi }} { '} \quad d_3(k)+O(T^{3/4}\log ^2T). \end{aligned}$$

Now dividing the interval [0, T] as $\cup _j [T/2^{j}, T/2^{j-1}]$ and summing the above estimate we obtain that

$$\begin{aligned} \int _0^T Z^2(t) \zeta \left( \frac{1}{2}+it\right) dt= 2\pi \sum _{k \le \frac{T}{2\pi }}d_3(k)+O(T^{3/4}\log ^2 T). \end{aligned}$$

Theorem 3 follows from the well-known formula:

$$\begin{aligned} \sum _{n \le x}d_3(n)=x\left( \frac{1}{2} \log ^2 x+(3\gamma _0-1)\log x+3\gamma _1+3\gamma _0^2-3\gamma _0+1\right) +O(x^{1/2}), \end{aligned}$$

where $\gamma _j$ is the coefficients of the Laurent expansion of $\zeta (s)$ at $s=1$.

5 Proof of Theorem

To prove Theorem , we put

$$\begin{aligned} I(\alpha )=\int _T^{2T}Z^3\left( \frac{1}{2}+it\right) \chi ^{\alpha }\left( \frac{1}{2}+it\right) dt, \end{aligned}$$

where $\alpha$ is a fixed constant such that $-1/2<\alpha <1/2$. This time we have

$$\begin{aligned} I=I_1+I_2+O(T^{1/2}), \end{aligned}$$

where

$$\begin{aligned} I_1(\alpha )&=\sum _{k=1}^{\infty }\frac{d_3(k)}{k^{1/2}}\int _T^{2T}\rho \left( \frac{k}{x}\right) k^{-it} \chi ^{\alpha -\frac{3}{2}}\left( \frac{1}{2}+it\right) dt \end{aligned}$$

(37)

and

$$\begin{aligned} I_2(\alpha )&=\sum _{k=1}^{\infty }\frac{d_3(k)}{k^{1/2}}\int _T^{2T}\rho \left( \frac{k}{y}\right) k^{it} \chi ^{\alpha +\frac{3}{2}}\left( \frac{1}{2}+it\right) dt, \end{aligned}$$

(38)

where $xy=(\frac{t}{2\pi })^3$. We only sketch an outline of evaluations of $I_j(\alpha )$.

Assume that $0 \le \alpha < \frac{1}{2}$. We take $x=2(\frac{t}{2\pi })^{1/2}$ and $y=\frac{1}{2}(\frac{t}{2\pi })^{1/2}$ and put $K_4=(\frac{T}{\pi })^{3/2}$. Then the range of k in the sum of (37) and (38) are at most $1 \le k \le 4K_4$ and $1 \le k \le K_4$, respectively.

The integral in (37) becomes

$$\begin{aligned} e^{\frac{\pi i}{4}\left( \alpha -\frac{3}{2}\right) }\int _T^{2T}\rho \left( \frac{k}{x}\right) e^{\left( \frac{3}{2}-\alpha \right) i\left( t \log \frac{t}{2\pi }-t-t\log k^{\frac{1}{3/2-\alpha }}\right) }dt +O(1). \end{aligned}$$

The main term of this integral appears only when

$$\begin{aligned} \left( \frac{T}{2\pi }\right) ^{\frac{3}{2}-\alpha } \le k \le \left( \frac{T}{\pi }\right) ^{\frac{3}{2}-\alpha }, \end{aligned}$$

in which case it is given by

$$\begin{aligned} M_{\alpha }(k)=e^{\frac{\pi i}{4}}\rho \left( k^{\frac{-2\alpha }{3-2\alpha }}/2\right) \frac{2\pi }{\sqrt{3/2-\alpha }}\ k^{\frac{1}{3-2\alpha }} \ e^{-\left( \frac{3}{2}-\alpha \right) i k^{\frac{1}{3/2-\alpha }}}. \end{aligned}$$

By Lemma 1 again, we get

$$\begin{aligned} I_1(\alpha )&=e^{\frac{\pi i}{4}(\alpha -\frac{1}{2})}\frac{2\pi }{\sqrt{3/2-\alpha }} \sum _{\left( \frac{T}{2\pi }\right) ^{3/2-\alpha } \le k \le \left( \frac{T}{\pi }\right) ^{3/2-\alpha }} \frac{d_3(k)}{k^{1/2}}k^{\frac{1}{3-2\alpha }}e^{-\left( \frac{3}{2}-\alpha \right) ik^{\frac{1}{3/2-\alpha }}} \nonumber \\&\quad + \sum _{k \le K_4}\frac{d_3(k)}{k^{1/2}} \ O \left( 1+\min \biggl (\sqrt{T}, \frac{1}{|\log \frac{(T/2\pi )^{3/2-\alpha }}{k}|}\biggr ) \right. \nonumber \\&\quad \left. +\min \biggl (\sqrt{T}, \frac{1}{|\log \frac{(T/\pi )^{3/2-\alpha }}{k}|}\biggr ) \right) . \end{aligned}$$

(39)

Just in the same way as the previous cases, we can see easily that the above O-term is estimated as $O(T^{3/4}\log ^2 T)$.

On the other hand, for $I_2(\alpha )$, the main term does not appear from the integral by the assumption $0 \le \alpha <1/2$ and the sum over k is estimated as $O(T^{3/4} \log ^2 T)$.

Now it remains to evaluate the sum over k in (39). Let

$$\begin{aligned} S= \sum _{\left( \frac{T}{2\pi }\right) ^{3/2-\alpha } \le k \le \left( \frac{T}{\pi }\right) ^{3/2-\alpha }} \frac{d_3(k)}{k^{1/2}} k^{\frac{1}{3-2\alpha }}e^{-\left( \frac{3}{2}-\alpha \right) ik^{\frac{1}{3/2-\alpha }}}. \end{aligned}$$

By partial summation we may have

$$\begin{aligned} S \ll T^{\frac{\alpha }{2}-\frac{1}{4}} \max _{\left( \frac{T}{2\pi }\right) ^{3/2-\alpha } \le T' \le \left( \frac{T}{\pi }\right) ^{3/2-\alpha }} \left| \sum _{\left( \frac{T}{2\pi }\right) ^{3/2-\alpha } \le k \le T'} d_3(k)e^{-(3/2-\alpha )ik^{\frac{1}{3/2-\alpha }}}\right| . \end{aligned}$$

(40)

Considering the definition of $d_3(k)$, it is reduced to the estimate of the sum of the form

$$\begin{aligned} S_1:=\sum _{T_1 \le k_1k_2k_3 \le 2T_1} e^{2\pi ic(k_1k_2k_3)^{\delta }}, \end{aligned}$$

where $\delta =\frac{1}{3/2-\alpha }$, c is a real constant and $(\frac{T}{2\pi })^{3/2-\alpha } \le T_1 \le \frac{1}{2}(\frac{T}{\pi })^{3/2-\alpha }$. Since $\delta \ne 0, 1$ we can apply Lemma 3. Divide the summation condition in $S_1$ into $O(\log ^3 T)$ subintervals of the form $(k_1,k_2,k_3) \in [H, 2H]\times [N,2N]\times [M,2M]$. By symmetry of $k_j$, we can assume that M is the largest, hence $M \gg T_1^{1/3}$. Now applying Lemma 3 to the sum $S_1$ by taking $X=(HNM)^{\delta } \asymp T_1^{\delta }$, we find that

$$\begin{aligned} S_1 \ll T_1^{1+\varepsilon }(T_1^{(\delta -\frac{4}{3})/4}+T_1^{-1/6}+T_1^{-\delta }) \ll T_1^{2/3+\delta /4+\varepsilon }. \end{aligned}$$

(41)

Here the last inequality follows from the assumption $0 \le \alpha <1/2.$ By (40), (41) and $T_1 \asymp T^{3/2-\alpha }, \ \delta =\frac{1}{3/2-\alpha }$ we find that

$$\begin{aligned} S \ll T^{1-\frac{\alpha }{6}+\varepsilon }. \end{aligned}$$

This proves the assertion in the case $0 \le \alpha <1/2$.

In the case $-1/2< \alpha \le 0$, we take $x=\frac{1}{2}(\frac{t}{2\pi })^{3/2}$ and $y=2(\frac{t}{2\pi })^{3/2}$. Then the main term arises from the integral corresponding $I_2(\alpha )$ and the assertion is proved similarly. We omit the details in this case.