THE FOURTH POWER MOMENT OF ζ( 12 + it)
Notes by Tim Jameson
Introduction
The following result was obtained by Ingham in 1926 [Ing]:
THEOREM 1. We have
Z T
1
|ζ( 21 + it)|4 dt = 2 T log4 T + O(T log3 T ).
2π
0
(1)
Another proof is given in [Iv, chap. 5]. At the expense of a lot more effort, a more
specific estimation was obtained by Heath-Brown in 1979 [HB].
Theorem 1 obviously incorporates the following weaker statement (cf. [Ti, p. 146–7]):
THEOREM 2. We have
Z
T
0
|ζ( 21 + it)|4 dt T log4 T.
(2)
A discrete analogue (eg. [Iv, sect. 8.3]) is:
THEOREM 3. Suppose that
0 ≤ t1 < t2 < . . . < tR ≤ T
Then
R
X
and
tr − tr−1 ≥ 1 for each r.
|ζ( 12 + itr )|4 T log5 T.
(3)
(4)
r=1
Here I present another proof of these results. In common with Ingham, it uses the
approximate functional equation for ζ(s)2 . Where it differs is in the use of an averaging
technique which opens the way to an application of the Montgomery-Vaughan mean-value
theorem for Dirichlet polynomials, and circumvents an estimation for the divisor function.
The method is particularly efficient for Theorems 2. A partial version of it is enough for
Theorem 3; this is outlined in [Iv, p. 227], and we repeat it here.
For Theorems 1 and 2, we will actually show that
Z 2T
I =:
|ζ( 21 + it)|4 dt
T
1
(5)
satisfies an estimate similar to the one stated. The result for [0, T ] then follows by application
to the intervals [2−j T, 2−j+1 T ]. Similarly, in Theorem 3, we will assume that the points tr
lie in [T, 2T ].
The approximate functional equation for ζ(s)2
As usual, write
χ(s) =
ζ(s)
πs
= 2s π s−1 sin Γ(1 − s).
ζ(1 − s)
2
The approximate functional equation for ζ(s)2 , specialised to the case s =
1
2
+ it, states:
Let τ (n) be the divisor function. Let x t, and define y by 4π 2 xy = t2 . Then
ζ( 12 + it)2 = S(x, t) + χ( 12 + it)2 S(y, −t) + O(log t),
(6)
where
S(x, t) =
X
1
τ (n)n− 2 −it .
(7)
n≤x
A pleasantly simple proof, restricted to the case y = x, was given by Motohashi [Moto].
This method actually adapts readily to the general case: details are given in my notes [Jam].
Note that by the elementary estimation
P
n≤x
τ (n)n−1/2 x1/2 log x (see below), we
have |S(x, t)| x1/2 log x t1/2 log t.
For Theorem 1 (but not Theorems 2 and 3), we will need following modified version of
(6). We use the known fact that for t > 0,
χ( 21 − it) = eiφ(t)−iπ/4 + O(1/t),
where
φ(t) = t log
t
,
2πe
so that
χ( 21 − it)2 = ψ(t) + O(1/t),
where ψ(t) = −ie2iφ(t) . Clearly, we have
ζ( 21 + it)2 = S(x, t) + ψ(t)S(y, −t) + O(log t),
(8)
Summation of τ (n)2
The following estimation of
P
n≤x
τ (n)2 is originally due to Ramanujan. It can be
seen in many books, e.g. [Nath, Theorem 7.8] or [Hux, p. 9]; we include a proof here for
2
completeness. We assume familiarity with convolutions and Abel summation. We start with
P
the trivial estimate n≤x τ (n) = x log x + O(x). By Abel summation, it follows that
X τ (n)
n≤x
and
= 12 log2 x + O(log x),
n
X τ (n)
n≤x
n1/2
x1/2 log x
(as already assumed for the estimation of |S(x, t)|). Of course, more accurate estimations
are possible, but they are not needed for our purposes.
Denote the convolution of arithmetic functions a, b by a ∗ b. Define u by u(n) = 1 for
all n. As usual, denote the Möbius function by µ, and let τ3 = τ ∗ u, so that τ3 (n) is the
number of triples (i, j, k) with ijk = n.
LEMMA 1. We have τ 2 = τ3 ∗ µ2 .
Proof. Both sides are multiplicative, so it is sufficient to consider the values at a prime
power pk . We have
τ3 (pk ) = (τ ∗ u)(pk ) =
k
X
τ (pr ) =
r=0
k
X
(r + 1) = 21 (k + 1)(k + 2),
r=0
hence
(τ3 ∗ µ2 )(pk ) = τ3 (pk ) + τ3 (pk−1 ) = 12 (k + 1)(k + 2) + 21 k(k + 1) = (k + 1)2 = τ (pk )2 .
LEMMA 2. We have
X
τ (n)2 =
n≤x
X τ (n)2
n≤x
n
1
x log3 x + O(x log2 x),
π2
(9)
1
log4 x + O(log3 x).
2
4π
(10)
=
Proof. By the formula for partial sums of convolutions,
X
X
X τ (j)
x
τ3 (n) =
τ (j)
=x
+ O(x log x) = 21 x log2 x + O(x log x).
j
j
n≤x
j≤x
j≤x
By Abel summation, it follows that
X τ3 (n)
n≤x
n
= 16 log3 x + O(log2 x),
3
X τ3 (n)
n≤x
P
Let M2 (x) =
n≤x
n1/2
= O(x1/2 log2 x).
µ(n)2 . Assuming the well-known estimate M2 (x) = (6/π 2 )x +
O(x1/2 ), we deduce from Lemma 1 that
x
X
X
2
τ (n) =
τ3 (n)M2
n
n≤x
n≤x
1/2 X
x
6 x
+O
=
τ3 (n)
2
π n
n1/2
n≤x
1
x log3 x + O(x log2 x).
π2
This establishes (9), and (10) follows by Abel summation.
=
Integrals and sums of |S(x, t)|2
We use the following mean-value theorem for Dirichlet polynomials, an easy consequence of the Montgomery-Vaughan generalisation of Hilbert’s inequality (see [Mont, p.
140] or [Iv, Theorem 5.2]):
Let f (t) =
PN
n=1
Z
an n−it . Then
T2
2
|f (t)| dt = (T2 − T1 )
T1
N
X
2
|an | + O
N
X
!
2
n|an |
.
(11)
n=1
n=1
A weaker version, proved rather more easily, has error term N
PN
n=1
|an |2 .
LEMMA 3. Suppose that x T2 T . Write 1/(4π 2 ) = K and log T = L. Then
Z T2
|S(x, t)|2 dt = K(T2 − T1 )L4 + O(T L3 ).
(12)
T1
Proof. By (9), (10) and (11), with an = τ (n)/n1/2 and N = [x], we have
Z T2
|S(x, t)|2 dt = K(T2 − T1 ) log4 x + (T2 − T1 )O(log3 x) + O(x log3 x).
T1
Since x T , we have log x = L + O(1), hence log4 x = L4 + O(L3 ).
In particular, the integral is O(T L4 ); for this conclusion, the weaker version of (11)
would have been sufficient.
LEMMA 4. If x T2 T , then
Z T2
Z
2
2
4
t |S(x, t)| dt = KL
T1
T2
T1
4
t2 dt + O(T 3 L3 ).
Proof. Write |S(x, t)|2 − KL4 = g(t) and G(t) =
Rt
T1
g(u) du. By (12), G(t) T L3 for
T1 ≤ t ≤ T2 . Integrating by parts, we have
Z T2
Z T2
h
iT2 Z
2
2
2
2
4
t g(t) dt = t G(t) −
t (|S(x, t)| − KL ) dt =
T1
T1
T1
T2
2tG(t) dt T 3 L3 .
T1
For Theorem 3, we will apply instead the following “large values estimate”, which can
be deduced easily from (11) by Gallagher’s method [Iv, Theorem 5.3]:
Let f (t) =
PN
n=1
a−it
n , and let tr (1 ≤ r ≤ R) satisfy (3). Then
R
X
2
|f (tr )| (T + N ) log N
r=1
N
X
|an |2 .
n=1
For our case, we deduce:
LEMMA 5. Suppose that x T and tr (1 ≤ r ≤ R) satisfy (3). Then
R
X
|S(x, tr )|2 T L5 .
r=1
Proofs of Theorems 2 and 3
We consider both x and t in the interval [T, 2T ]. Write K = 1/4π 2 and y = Kt2 /x, so
that 21 KT ≤ y ≤ 4KT and x y T . Also, write L = log T .
Since (a + b + c)2 ≤ 3(a2 + b2 + c2 ) and |χ( 21 + it)| = 1, (6) gives
|ζ( 21 + it)|4 |S(x, t)|2 + |S(y, t)|2 + L2 .
(13)
Proof of Theorem 2. Let I0 (x) = I1 (x) + I2 (x), where
Z 2T
I1 (x) =
|S(x, t)|2 dt,
T
Z
2T
Z
2
|S(y, t)| dt =
I2 (x) =
T
T
2T
2 2
S Kt , t dt.
x
Then for any x ∈ [T, 2T ],
I I0 (x) + T L2 .
By Lemma 3, I1 (x) T L4 for each x. However, since y varies with t, I2 (x) is not of the
same form. This is where we introduce our averaging technique. Instead of estimating I0 (x)
for one fixed x, we consider its average over x in [T, 2T ], as follows. For j = 0, 1, 2, let
Z
1 2T
Aj =
Ij (x) dx,
(14)
T T
5
Clearly, A0 = A1 + A2 and I A0 + T L2 .
Clearly, A1 T L4 ; of course, the averaging step was not needed for this term. Now
consider A2 . Reversing the order of integration, substituting x = Kt2 /y and reversing again,
we have
A2
1
=
T
Z
1
=
T
Z
1
=
T
Z
2T
2T
Z
T
|S(y, t)|2 dx dt
T
2T
Z
Kt2
T
|S(y, t)|2
Kt2
2T
T
4KT
Z
K
y2
1
KT
2
Kt2
dy dt
y2
T2
t2 |S(y, t)|2 dt dy,
(15)
T1
where for present purposes we only need to know that [T1 , T2 ] is contained in [T, 2T ], so that
by Lemma 3 again,
Z
T2
|S(y, t)|2 dt T L4 ,
T1
Since t2 /y 2 = O(1), it follows that
A2 1
T.T L4 = T L4 ,
T
so I T L4 .
Proof of Theorem 3. This time, we average the expression for |ζ( 21 + it)|4 itself, not its
integral. Let
2T
1
A1 (t) =
T
Z
1
A2 (t) =
T
Z
|S(x, t)|2 dx,
T
2T
|S(y, t)|2 dx.
T
By (13), for each t,
|ζ( 21 + it)|4 A1 (t) + A2 (t) + L2 .
Since R ≤ T + 1,
R
X
|ζ( 21
4
+ itr )| r=1
By Lemma 5,
PR
r=1
R
X
[A1 (tr ) + A2 (tr )] + T L2 .
r=1
A1 (tr ) T L5 . As in the proof of Theorem 2 (without the integration
with respect to t), we have
1
A2 (t) T
for each t in [T, 2T ], hence also
PR
r=1
Z
4KT
|S(y, t)|2 dy,
1
KT
2
A2 (tr ) T L5 .
6
Note. For Theorems 2 and 3 (but not Theorem 1), we can work with the approximate
functional equation for ζ(s) itself instead of ζ(s)2 :
ζ( 21 + it) = Z(x, t) + χ( 12 + it)Z(y, −t) + O(1).
where Z(x, t) =
1
P
n≤x
n− 2 −it and 2πxy = t, so that
|ζ( 12 + it)|4 |Z(x, t)|4 + |Z(y, t)|4 + 1.
Now
X
|Z(x, t)|2 =
1
τ (n, x)n− 2 −it ,
n≤x2
where τ (n, x) counts factorisations jk of n with j, k ≤ x, so that τ (n, x) ≤ τ (n), and hence
Z
2T
|Z(x, t)|4 dt (T + x2 )
T
X τ (n)2
T L4 .
n
2
n≤x
The steps are now as before, but with the averaging applied over T 1/2 ≤ x ≤ 2T 1/2 .
Proof of Theorem 1
We must replace the upper estimate (13) used in Theorem 2 by an asymptotic estimate.
Recall from (8) that ζ( 12 + it)|2 = W (x, t) + O(L), where
W (x, t) = S(x, t) + ψ(t)S(y, −t)).
Just writing W and ζ, we have |W | = |ζ|2 + ρ, where ρ = O(L). Hence
|W |2 = |ζ|4 + 2ρ|ζ|2 + ρ2 .
By Theorem 2,
R 2T
T
|ζ|4 T L4 . So by the Cauchy-Schwarz inequality,
2T
Z
|ζ|2 T 1/2 (T 1/2 L2 ) = T L2 .
T
(In fact, it is well known that
R 2T
T
|ζ|2 T L: see [Ti, p. 141].) So if
Z
I(x) =
2T
|W (x, t)|2 dt,
T
then
I = I(x) + O(T L3 ),
for each x in [T, 2T ].
7
(16)
Now apply the averaging process: let
1
A=
T
2T
Z
I(x) dx.
T
Then I = A + O(T L3 ), so Theorem 1 will follow if we can show that
A = 2KT L4 + O(T L3 ).
(17)
Now
|W (x, t)|2 = W (x, t)W (x, t) = |S(x, t)|2 + |S(y, t)|2 + 2Re (ψ(t)S(x, t)S(y, t)) ,
so
I(x) = I1 (x) + I2 (x) + 2I3 (x),
where I1 (x) and I2 (x) are as before, and
2T
Z
I3 (x) =
Re (ψ(t)S(x, t)S(y, t)) dt.
T
With Aj defined by (14) for j = 1, 2, 3, we have
A = A1 + A2 + 2A3 .
By Lemma 3, I1 (x) = KT L4 + O(T L3 ) for each x, hence
A1 = KT L4 + O(T L3 ).
(18)
Now consider A2 . In the expression (15) for A2 , we now need to specify
!
r
r !
2T y
Ty
, T2 = min 2T,
.
T1 = max T,
K
K
By Lemma 4,
Z
T2
2
2
t |S(y, t)| dt = KL
4
Z
T1
T2
t2 dt + O(T 3 L3 ).
T1
Now reversing the steps that led to (15), with |S(y, t)|2 replaced by 1, we have
1
T
also
Z
4KT
1
KT
2
K
y2
Z
T2
T1
1
t dt dy =
T
2
1
T
Z
4KT
1
KT
2
Z
2T
Z
1 dx dt = T,
T
T
K
2
dy < 2 ,
2
y
T
8
2T
hence
A2 = KT L4 + O(T L3 ).
(19)
The required estimation (17) will follow if we can establish that A3 T L3 . Since
ψ(t) = −ie2iφ(t) ,
Re (ψ(t)S(x, t)S(y, t)) = Im e2iφ(t) S(x, t)S(y, t) .
Now
X X τ (m)τ (n)
X X τ (m)τ (n)
=
eifm,n (t) ,
1/2+it
1/2
(mn)
(mn)
m≤x
m≤x
Kt2
Kt2
e2iφ(t) S(x, t)S(y, t) = e2iφ(t)
n≤
n≤
x
x
where
fm,n (t) = 2φ(t) − t log mn = t(2 log t − 2 − 2 log 2π − log mn).
For a given n, integration is on the interval [T0 , 2T ], where
r nx
.
T0 = max T,
K
So
I3 (x) =
X X τ (m)τ (n)
Im Jm,n ,
(mn)1/2
m≤x
4KT 2
n≤
where
(20)
x
Z
2T
eifm,n (t) dt.
Jm,n =
T0
Denote by I3,1 (x) the contribution to I3 (x) of the terms with m ≤ x − 1, and by I3,2 (x)
the contribution of the single term with x − 1 < m ≤ x, with corresponding averages A3,1
and A3,2 .
Since
P
n≤x
τ (n)n−1/2 x1/2 log x, we have
X τ (n)
T 1/2 L.
1/2
n
4KT 2
n≤
(21)
x
Consider A3,2 first: for this, we have m = [x]. By (21) and the trivial bound |Jm,n | ≤ T ,
I3,2 (x) Hence
Z
A3,2 L
τ (m) 3/2
T L τ (m)T L.
m1/2
2T
τ ([x]) dx ≤ L
T
X
m≤2T
9
τ (m) T L2 .
(Note that we cannot deduce such an estimate for I3,2 (x) for a fixed x, since τ (m) is not
bounded by a power of L.)
Now consider A3,1 . We use the following Lemma [Ti, Lemma 4.3] to estimate Jm,n :
LEMMA 6. If f is a real-valued function with f 0 (t) increasing on [a, b] and f 0 (a) =
µ > 0, then
Z b
2
if (t)
e
dt ≤ .
µ
a
Proof. Let h(t) = 1/f 0 (t). Integrating by parts, we have
Z b
Z b
if (t)
e
dt =
h(t)f 0 (t)eif (t) dt = J1 + J2 ,
a
where
a
h1
ib
Z
b
h0 (t)eif (t) dt.
i
a
a
Clearly, |J1 | ≤ h(a) + h(b), and since |h0 (t)| = −h0 (t),
Z b
|J2 | ≤ −
h0 (t) dt = h(a) − h(b).
J1 =
−f (t)
h(t)e
,
J2 =
a
We have
Kt2
,
mn
0
which is clearly increasing. Also, KT02 ≥ nx, so for m < x, fm,n
(T0 ) ≥ log x/m. Now
Z x
1
x−m
x
=
dt >
,
log
m
x
m t
0
fm,n
(t) = 2 log t − 2 log 2π − log mn = log
so
1
x
m
<
=1+
.
log x/m
x−m
x−m
Hence for each x, we have
I3,1 (x) T
1/2
T
1/2
X τ (m) m
L
1+
m1/2
x−m
m≤x−1
L.T
1/2
T L2 + T L
L+T
1/2
X m1/2 τ (m)
L
x−m
m≤x−1
τ (m)
.
x−m
m≤x−1
X
Now integrate with respect to x on [T, 2T ]. For a fixed m, we integrate 1/(x − m) on an
interval contained in [m + 1, 2T ]: the contribution is less than log 2T . So
X
A3,1 T L2 + L2
τ (m) T L3 .
m<2T
10
So A3 T L3 , as required.
Comparison with Ingham’s method
Ingham’s method takes x = y = t/(2π), requiring (in our notation) the estimation of
RT
0
|S(t/2π, t)|2 dt. This in turn requires the estimation
N X
X
n=1 m<n
τ (m)τ (n)
N log3 N,
− log m)
(mn)1/2 (log n
(22)
which is achieved by calculations specific to the divisor function. For Theorem 2, this
estimation would still be needed, with bound N log4 N ; this is essentially the method implied
in [Ti, p. 146–7]. We remark that there is no short cut to (22) by the Montgomery-Vaughan
theorems, since it really involves | log n − log m|, not log n − log m. Indeed, it is not hard to
show that the best constant C in the bilinear form estimate
N
X
xm xn
≤C
|xn |2
log n − log m
n=1
n=1 m<n
N X
X
satisfies C ∼ N log N . This would only lead to the bound N log5 N in (22).
References
[HB]
D. R. Heath-Brown, The fourth power moment of the Riemann zeta function,
Proc. London Math. Soc. (3) 38 (1979), 385–422.
[Hux]
M. N. Huxley, The Distribution of Prime Numbers, Oxford Univ. Press (1972).
[Ing]
[Iv]
[Jam]
A. E. Ingham, Mean-value theorems in the theory of the Riemann zeta function,
Proc. London Math. Soc. 27 (1928), 273–300.
A. Ivić, The Riemann Zeta Function, Wiley (1985).
Tim Jameson, The approximate functional equation for ζ(s)2 , at
www.maths.lancs.ac.uk/~jameson
[Mont]
H. L. Montgomery, Ten Lectures on the Interface Between Analytic Number
Theory and Harmonic Analysis, CBMS no. 84, Amer. Math. Soc. (1990).
[Moto]
Y. Motohashi, A note on the approximate functional equation for ζ 2 (s), Proc.
Japan Acad. 59A (1983), 392–396.
[Nath]
Melvyn B. Nathanson, Elementary Methods in Number Theory, Springer (2000).
[Ti]
E. C. Titchmarsh, The Theory of the Riemann Zeta Function, 2nd ed., Oxford
Univ. Press (1986).
11