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A note on the fourth power moment of the Riemann zeta-function


A NOTE ON THE FOURTH POWER MOMENT OF THE RIEMANN ZETA-FUNCTION
J.B. Conrey

Abstract. We give explicit formulae for all of the terms in the asymptotic expansion of the mean fourth power of the Riemann zeta-function on the critical line.

Heath-Brown has evaluated the 4th power moment of the Riemann zeta-function on the critical line via the formula ZT T I2(T ) = j (1=2 + it)j4 dt = TP4 log 2 + O(T 7=8+ ) 0 where P4 is a 4th degree polynomial. Here we give an explicit formula for P4 . In particular, we show that the coe cients of P4 are in the eld

Q( 0; 1; 2; 3; (2); 0(2); 00 (2))
where the 's are the coe cients in the Laurent expansion (s) = s 1 1 + 0 + 1(s 1) + : : : of (s) at s = 1.

Theorem.

We have

P4(x) = g0(x) + g1(x)
2 s ( + 4 g0 (x) = Res s(s x 1)s (2s1) 2) s=0 + +

where

and

d g1(x) = ds

2

1 (xe2 0 )s ( 2 (s + 1)2 1 (2s + 1) s (s + 1) (s + 2)

(2s + 2))
s=0

:

Ivic has also written down formulae for the coe cients of P4 . Proof. We refer to Heath-Brown's paper. There he shows that P4 is naturally expressed as a sum of g0 + g1. The rst term g0 arises from the diagonal terms
Research supported in part by a grant from the NSF. Typeset by AMS-TEX

2

J.B. CONREY

and is exactly as above. The other piece arises from the o diagonal terms and it is the explicit calculation of this part that is the purpose of this paper. Following Heath-Brown, we write

X

n x

d(n)d(n + r) = m(x; r) + E (x; r)

where m(x; r)=x is a polynomial of degree 2 in x with coe cients depending on r. In fact, s m(x; r) = Res D(s; r) x s=1 s with

D(r; s) = g1 T 2
=2

1 X d(n)d(n + r)
n=1
(2 )

ns

:

Then

1 X Z T=
r=1
0

m0 (x; r) sin(Tr=x) dx:

Lemma.

With the above notation,

m(x; r) =
and

1 X (s) X log d xs + 2 s djr s
=1 2 2 2

0

1 +1
2 0

2

1 0 (x; r) = X (s) X log x + 2 m 2 d2 s2 s=1 s djr

According to Heath-Brown, m(x; r) may be calculated by considering the main terms in X X 2 R(x; q; r) R(qx1=2; q; r)
Proof.

q x1=2

q x1=2

where

R(x; q; r) = qx2

2 d (q=d ) log xq2 + 2 dj(q;r) jq=d

X X

0

1 :

Writing q = sd we have

X

q x1=2

R(x; q; r) = x

X1 X X1
y

(s) log x + 2 2 d2s2 djr d s x1=2 =d s = log y +
0

0

1

x1=2 =sd

X 1 :

Using

+ O(1=y)

A NOTE ON THE FOURTH POWER MOMENT OF THE RIEMANN ZETA-FUNCTION3

(and we note

X log
y

= 1 log2 y 2

1

+ O(log y=y); 1 log d2xs2 + 2

for future reference) we nd that the above sum is =x 2

X 1 X (s) log d xs + 2 djr d s x = =d s 0 X X + O @x =
1 2
2 2 2 1 2

0

djr s x1=2 s

1 1A :

0

Extending the sum over s to 1 the above is
1 X (s) X log d xs + 2 = s djr s
=1 2 2 2 0

1 log d2xs2 + 2 1 = y + O(1)

0

+ O(x1=2 log xd(r)):

Similarly, using and

X X
y y

log = y log y y + O(log y)

we nd that the main part of

X
q x1=2

R(x1=2q; q; r)
2 :

is given by

Putting these two calculations together, we obtain the formula for m(x; r). Di erentiating with respect to x gives the rest of the lemma. Now we compute

1 X (s) X log d xs + 2 s djr s
=1 2 2 2

0

g1 = 2

1 X Z T=
r=1
0

(2 )

m0 (x; r) sin(Tr=x) dx:

To begin with, we change variables, letting y = Tr=x. Then, substituting the expression for m0 (x; r) from the lemma gives
1 1 X Z 1 X (s) X 1 log s Tr y + 2 g =2 s djr d d rs r
1 =1 2 =1 2 2 2 0

1

2

sin y dy: y2

4

J.B. CONREY

We bring the sum over s to the front, and interchange the summation over r and the integration, and replace d by n and write r = mn. Then we have
1 X (s) Z 1 X 1 log sTm + 2 g =2 s ny s mn y= n
1 =1 2 2 2 2 2 0

sin y dy: y2
2

Next, we interchange the sum over m and the integration giving
1 1 X (s) X Z 1 X 1 g =2 log sTm + 2 s m ny m n y= m n s
1 =1 2 =1 2 2 2 2 0 0

sin y dy: y2

A change of variables in the integration, u = y=2 m gives
1 1 X (s) X Z 1 X 1 2 log T=un + 2 g =2 s m s s n un
1 =1 2 =1 1 2

sin 2 um du : 2 um u

Now,

Thus, bringing the sum over m to the inside, we get
1 =1 2 1 2

m=1

1 X sin 2 um m = u u] 1=2 = ((u)):
2 0

1 X (s) Z 1 X 1 2 log T=un + 2 g = s s s n un

((u)) du : u2 ((u)) du : u2

Interchanging the sum over n and the integration, and changing s into m leads to
1 1 X (m) X Z 1 1 T= g = log m 2 + 2 un m m n n n
1 =1 2 =1 2 2 0

Now we express the logs via di erentiation:

g1 =

d ds

2

m=1 m

1 X (m) Te
2+

s

2 0

2

1 sX 1

((u)) udus : 2+ n1+s n n=1 s=0 1 : 2ns+1

Z1

It is well known that

Z 1 ((u))
n

us+2

1 du = s + 1
2

n 1 X 1 (s + 1) sns s+1 m=1 m

!

Thus,

g1 = Nlim !1

d ds

1 Te2 0 s 2 (s + 1) (s + 2) n N X 1 1 X 1 (s + 1) sns 1+s s+1 m=1 m n=1 n

1 2ns+1 s=0 :

!

A NOTE ON THE FOURTH POWER MOMENT OF THE RIEMANN ZETA-FUNCTION5

Now so that

N n N N X 1 ! X 1 X 1 X 1 + ns : 2 ns ms = ns
2

n=1

n N N N X 1 X 1 X 1 X 1 = ns ms ms ns
+1

m=1

+1

m=1

+1

n=m

+1

Thus, the sum over n in the above expression is
n X 1 = (s + 1) ns N 1X 1 s n=1 n2s+1

n=1

+1

m=1

+1

n=1

+1

n=1

2 +2

N 1 X 1 +1 2 n=1 ns+1 n=1 If we assume that s > 0, we can let N ! 1 here, getting

! X N 1 ns :
2

n=1

2 +2

1 (s + 1)2 1 (2s + 1) (2s + 2): 2 s Inserting this into our prior expresion for g1, we obtain the Theorem.
References

H-B] D.R. Heath-Brown, The fourth power moment of the Riemann zeta function, J. London Math. Soc (3) 38 (1979), 385-422.
Department of Mathematics, Oklahoma State University, Stillwater, OK 74078


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