Mr Littlewood, On the zeros of the Riemann zeta-function 295
On the zeros of the Riemann zeta-function*. By Mr J. E. LITTLETrinity College, Cayley Lecturer.
WOOD,
[Received 16 June 1924.]
1. Let N (T) denote, as usual, the number of zeros of £ (s)
whose imaginary part y satisfies 0 < y < T, and N (a, T) the number of these for which, in addition, the real part is greater than a.
In this definition we suppose, in the first place, that no zero actually
lies on the line t = T:ii the line contains zeros we define
N (T) = lim $ {N (T + e) + N (T - e)},
<•—o
N (a, T) = lim \ {N (a, T + e) + N (a, T - «)}f.
0
Zeros on a boundary t — constant are, in fact, only to count half the
full valuef.
We define a function / (s) = log £ (s) as follows. In the first
place when a > 1,
the familiar Dirichlet series. Next, if no zero of £ (s) has ordinate t,
we define f(a + it) to be the value obtained from / (2 + it) by
continuous variation of the argument along the straight line from
2 + it to a + it. Finally, for a t which is the ordinate of a zero we
define
f(a + it) = lim £ {/ (<r + it + ie) + f {a + it - ie)}.
e-»-0
It is known § that, as T -*• oo ,
(1.1)
N (T) = M (T) + S (T) + | + 0 (T-1)
' = M(T) + S(T) + O(l),
where
T
(1.2)
(1.3)
M(T)
=
T
T
llog£-l,
S(T)=-lf(i + iT).
77
It is known further that
(1.4)
I/(*) = 0(logO
(a>i),
* Most of the results of the paper were announced, without proof, in the Records
of Proceedings of the London Mathematical Society, February 10, 1921.
t These equations are, of course, then valid for all T.
% We adhere strictly, however, to the condition /3>(r for the real part of a zero
counted in N (a, T). A halving convention here has something against it and nothing
in its favour.
§ R. Backlund, Acta Mathematica, 41 (1918), 345-375. With our conventions
the results are evidently true for all T if true for "wurzelfrei" T.
296 Mr Littlewood, On the zeros of the Riemann zeta-function
and in particular
(1.5)
S (T) = 0 (log t).
The present paper is devoted to the study of the functions
N (a, T) and S (T): it falls naturally into two parts. In Theorem 2
we prove that
f N (a, T) da and ~ i
log | £ (a + it) I dt
AV J 0
J a
on the one hand, and in Theorem 3 that
f
I f (a + it) dt and ( log | £ (a + iT) I da
JO
J <r
on the other, are nearly equal when a > | . From the first result
we deduce upper hounds for N (a, T) when a is near the value | ;
and from the second it follows eventually that
S(t)dt=O
(log T),
an upper bound which may be improved to o (log T) if we assume
the hypothesis of Lindelof (a fortiori if we assume Riemann's hypothesis). These are the principal results of the first part. In the
second we assume the truth of the Riemann hypothesis, and prove,
among other things, that
Throughout the paper ^4's and a's denote positive absolute
constants, not always the same from one occurrence to another*.
On the other hand, an A or a affected with a suffix is the same
wherever it occurs. O's are absolute unless the contrary is statedf.
When we are concerned with approximations valid when T -> oo
we need only consider values of T greater than any convenient
constant. We shall suppose in such cases, without always making
an explicit statement, that T satisfies any inequality T > A which
is required by the run of the argument.
2. We begin by collecting some results whose proof would
interrupt the development of later arguments.
Lemma 1.
| i(a + it) \2dt<AVT
where
(T>2,
V = Min
>
3
I
a
/ •
* Thus we write, e.g.
(sin 6 +2) <(A +A) <A.
t f(x) = O {0 (x)} means \f(x)/(f> (x) \ <K (x>x0): an 0 involves two constants,
K and x0. Our convention is that both are A's.
Mr Littlewood, On the zeros of the Riemann zeta-function 297
The lemma is closely allied to known results*, but it does not
actually follow from any published theorem, and a proof must be
given de novo. We havef, for \ <j a < 2, t > 2,
+ 0(1)
2
<
n<(
2
2
2 n-3
(m\il
/•r
S w-s
2A.
(J
n<t
J2
Now
rT
rT
dt< •2
2
»<<
[T
/m\it
dt= I 2 (mn)-" (— I dt = 2
(»i»)-' (— I dt,
J 2 m,n<t
^nJ
m,n<TJTl
\nJ
J 2
where Tx = Max (m, n, 2),
+ 2
< 2
(2.2) < 2 fm- 2 ' + 2 2 ' (wiw)-ff
m
m<T
m,n<T
say, where the dash denotes that terms for which m = n are to be
omitted in 2'. Now
(2.3)
x
< T Min ( 2 m-\ 2 m-2
\r
i
< AT Min (log T,
= AVT.
-i
Also if n = w + r we have
log —
5
m
[An-"
(wz
^i i< — + A
(\r\<hn)
m
(\r\>\m).
Hence
<c
w^
^J
m<T
\
r<imr
2J
n~~
n<T
< A 2 w1"2- log (m + 1) + 4 (' 2
m<T
\m<l
<A
If a > f,
I it1-2" log (M + 1) du + A log T + A ( 2 m ~
S.,<A
j u~i Auidu + AT + AT < AT < AVT,
* See G. H. Hardy and J. E. Littlewood, Proc. London Math. Soc. (2), 21 (1921),
39-74.
t Loc. dt., 53.
VOL. XXII. PART III.
298 Mr Littlewood, On the zeros of the Biemann zeta-function
since V
= A when T is large. If a <
rT u2-2*
du
,
\x
i
T
2
4T log T . T- ('-« + AT.
The first term does not exceed AT log T on the one hand, and is
equal to
T
a
£ \
T
1<xe~'<-
~ ?)
x
(x = (2a - 1) log T)
A \P — i)
on the other. Thus
S2 < AVT + AT < AVT
also in this case, and so for J < a < 2. From this and (2.3) the
lemma follows.
3. Lemma 2. Suppose that % «wc? z2 Ke within | z | < 4, and that
z2 — zx is reaZ, positive, and not greater than TJ. T&
HI p log zcb <
log (- + 2\,
the integral being taken along a straight line.
This is, of course, nearly trivial. Suppose first that | zx \ >
Then | z | > r\ for any point of the path of integration, and so
1
1ft log 2 | <
+log4,
V
3ft Plog zcfe < r ? f l o g i + log 4) < Ar) log (- + 2).
Suppose now | z1 \ < 2rj. Then | z2 | < 3-q. Also, log z being interpreted to have its principal value, we have, writing
=
(3.1)
log zdz
I"
\z.2\,
< I zx log z11 + I z11 + I z2 log z21 + I z21
• 0
< rxJoI log rx I + (IT + 1) rx + r2 | log r2 | + (TT + 1) r2.
Since r | log r | increases to a maximum at r = e-1, then decreases
to a minimum at r = 1, and then increases, we have
rx I log rx I <
(317 >
Mr Littlewood, On the zeros of the Biemann zeta-function 299
and in either case
r
i I l°g ri I + Ari < Ar) l°g (" + 2 ) •
Since a similar result holds for r2 (3.1) leads to the desired conclusion also when | zx | < 2rj.
Lemma 3. Let C1,Ci, C3 be three simple closed curves, C3 containing C2, and C2 containing 01. Suppose that f (z) is regular and
satisfies | / 1 < M3 in the interior of C3, and satisfies | / 1 < Mx on
Gx. Then within and upon C2
\f\<Ml'MJ>,
where a, fi are positive numbers satisfying a -f- /? = 1, and depending
only on the geometrical configuration of the set of curves (not on its
absolute position or orientation).
This generalization by conformal representation of Hadaruard's
"three circles" theorem is a known result*.
4. Suppose (f> (s) is meromorphic in and upon the boundary of
a rectangle bounded by the lines t = 0, t = T; a = a, a = /3 (/? > a);
and regular and never zero on the side, I, of the rectangle for which
o- = /?. The function log <f> (s) is regular in a neighbourhood containing I, and we define
*•(«) = log ^(«)
for s of I. For other s of the rectangle we define F (s) to be the
value obtained from log <f> (ft + it) by continuous variation along
t = constant from ft + it to a + it. This provided the path does not
cross a zero or pole of <j> (s); if it does we define
F(s) = \ lira {F (a + it + ie) + F (a + it - ie)}.
Let now v (a, T) denote the excess of the number of zeros over the
number of poles in the part of the rectangle for which a > a, zeros
or poles on t = 0 or t = T counting one-half only. Then we have
Theorem 1.
j F (s) ds = - 2m ( v (a, T) da,
the integral on the left being taken round the rectangle in the positive
direction.
This formula has obvious affinities with Jensen's well-known
theorem. Various proofs are possible; the one given here depends
on the inversion of the order of a double integration. We may
suppose t = 0 and t — T to be free of zeros and poles of <f> (s); it is
* H. Bohr, E. Landau, and J. E. Littlewood, Bull. Acad. Bdgique, 15 (1913),
1144-1175, 1166.
300 Mr Littlewood, On the zeros of the Riemann zeta-function
easily verified that our halving conventions then ensure the truth
of the theorem in the general case. We have
JF(s)ds=
jfiF(a)da-
I* F (o + iT)do + j * {F tf + it)- F (a + it)}idt
•>
Ja
Ja
JO
= I F (a) da Ja
F (a + iT) da + \ idt \ %• (a + it) da*
Ja
= \ F(a)daJa
JO
F (a + iT) da +
Ja
Jo 9
do I
Ja
J (T
% (s) ds.
<P
Now by the theorem of residues the (inner) integral last written
is equal to
•
f" ^ (a) da + fT ^' (j8 + it) idt - \' % {a + iT) da - 2-rriv {a, T).
J a- <f>
J0 9
J <r <P
Hence
JF (s) ds = j\F
(a) -F(a
+ iT) + F(P)- F (a) + F (j3 + iT)
- F (ft) - F (]8 + iT) + F (a + iT) - %nv {a, T)} da
= - %ri
\\[a,T)da.
J a.
5. Theorem 2.
(5.1)
2TT f N(o,T)da=\
log\C(a + it)\dt + c(a) + lf
f(a + iT)da
(5.2)
.'o
where c (a) is a constant depending only on a.
Theorem 3.
(5.3)
fT$f(a
+ it) dt = f log \Z(o + iT)\da + c' (a) (- l < a < 2),
JO
i<T
where c' (a) is a constant depending only on a. In particular
(5.4)
\T S (t) dt = - f log |C tp + ^) | do + c^
JO
(5.5)
"• •' i
= - f log I £ (a + iT) | do + 0 (1).
* It is easily verified that J da J | <£'/0 I <^ exists: in fact near a singularity
<r0 + it0 of <t>'/<p w e h a v e
and the integral of this with respect to a converges. We may invert the order of
integration and ignore lines on which the inner integrals do not exist.
Mr Littlewood, On the zeros of the Riemann zeta-function 301
Theorems 2 and 3 are simple consequences of Theorem 1 and
known results. We take in Theorem 1 a = a, j3 large and positive,
<f> (s) = £ (s). Then (on account of the pole at s = 1)
0
Theorem 1 gives
- 2ni ( p(a,T)da=-2ni(
Jo
v {a, T) da
Jo-
= ( f{a)daJa
P'f(a + iT)da- ( f (a + it) idt + (
Jo
Jo
f(P+it)idt.
Jo
If in this we make /3 -*• oo the last term disappears, and the first
two integrals tend to integrals
. Taking real and imaginary parts
J <T
we obtain (5.1) and (5.3), and observe further that ca, cj are of
the form 0 (1) for a > — 1. (5.2) now follows from (5.1) in virtue
of (1.4)*, (5.4) is the special case a = \ of (5.3), and (5.5) follows
at once.
6. We prove next the following results.
Theorem 4.
(6.1)
N (a, T)dt = O (T) Min flog log T, log ^—]
<r
(i < a < 1).
a— •§/
V
In particular
j 1 N (a, T)da=0(T
(6.2)
\oglogT).
Theorem 5.
(6.3) N (a, T) = 0(T --j-j log —^\
From (5.1)
2TT
(£ < o- < 1).
[ N {a, T) da = [ log | £ (CT + i<) | dt + 0 (log T)
J (T
J 2
T
= i f log | I (a + it) \*dt + O (log T)
•I 2
<b(T-2)
log | | T | £ (o- + it) \2dt/(T -2)1 + 0 (log T)
< AT log JMin lo (log T), 0 (log ^ - ) l j + 0 (log T),
* Siuce/=O(2-°-)(<r^2), so that I fdff = O(l).
302 Mr Littlewood, On the zeros of the Riemann zeta-function
by Lemma 1; and this proves (6.1). We now observe that if
<*i = i + i (* - 4).
N (a, T)<—"— fN(a, T) cfo < - ^ P N (a, T) da
< AT {a - i)-i log K - J)-i < 4 T (cr - i ) - 1 log (cr - i)-i,
which is (6.3).
Theorem 6. If j> (T) is positive, and increases to <x> with T, then
all but an infinitesimal proportion of the zeros of £ (s) in the upper
half plane lie in the region
\a-i\<(f>(t)l^l,
(6.4)
t>e>.
That is to say: the number of zeros outside the region and with
imaginary part between 0 and T is of the form o (T log T) (the
total number between these limits being approximately „ - T log T).
It is clearly enough to prove that for large T the number of zeros
in the region
(6.5)
a>^ + <f>{t)
l
^Kl>
VT<t<T,
is o (T log T). The curved boundary of (6.5) lies to the right of
a = alf where
l
l
T
But evidently N (o-1; T) = o (T log T) in virtue of Theorem 5*, and
the required result is true a fortiori.
It should be observed that Theorems 4, 5 and 6 are novel only
so far as they concern a small neighbourhood of the line a = \:
ioi fixed a > \ Carlson's well-known theorem f gives results much
more precise than ours.
7. We turn now to the study of the function S (T), or rather
rT
of its integral
S (t) dt. We prove two theorems.
'o
Theorem 1%.
\ S (t) dt = 0 (log T).
o
* If Oj > 1 the result is trivial. If not, then uj — J $ i, and
J
log —,=0 (log log T).
t F. Carlson, Arkiv for Mat., Ast. och Fys., 15 (1920), No. 20. The details of the
proof can be lightened by Theorem 2, which enables us to reduce the double integration to a single one. Carlson's theorem lies much deeper than ours forfixed<r,
but it appears to throw no light on the present problem.
J H. Cramer, Math. Zeitschr., 4 (1919), 104-130, 122-130, proves, by an entirely
[T
different method, that / S(t)dt = O (T').
Mr Littlewood, On the zeros of the Riemann zeta-function 303
Theorem 8. It follows from Lindelof's hypothesis that
f S(t)dt = o (log T).
Jo
We begin with the proof of the more difficult Theorem 8.
Lindelof's hypothesis is that
| £ (i + it) | < «»a),
and this is known* to involve the proposition that the number of
zeros of £ (s) in any region, D,
<7>i + S,
\t-T\<l,
is of the form o (log T)f. We suppose 8 < -fa. Let C3 be the (boundary of the) rectangle with corners at | + 28 + i {T ± (1 — 8)},
2 ± i {T ± (1 - 8)}; C2 that with corners at
i + 38 + i {T ± (1 - 28)}, 2 - 8 + t {T ± (1 - 28)};
Cx that with corners at f + i {T ± (1 - 38)}, « + i{T ± (1 - 38)}.
We denote by S o a summation over the zeros p of £ (s) for which
| y — T | < 1, Si a summation over those zeros lying in D (the
terms of Sj are therefore contained in So). It is known that S o has
0 (log T) terms J. Let now
T VI — Y V> -*• > "I — ¥
ig\
so that ifj (s) is regular in C3.
8. Lemma 3. bis
xft (s) = 0
(uniformly) for s in G-s.
We have, for s in C3,
(8.1) + (s) = 0 (log 0 + (S - SJ ( ^
+ i) +
= 0 (log T) + (S - Sx) (-• X - + I
If S' denotes a summation over p for which | y — t \ > 1 we have||
(8.2)
S' ( —
\t>
+ - ) = 0 (log t) = O (log T).
p
p/
* R. Backlund, Ofversigt Finska Vetensk. Soc, 61 (1918-19), No. 9.
f o's are not uniform in S. O's (by our current convention) are absolute, and so
uniform in 5. We must therefore bear in mind
that, e.g., a o(logT) is not necessarily of the form 0 (log T); for example 5 - 1 is of the first form but not of the
second.
% E. Landau, Handbuch, p. 337. We refer to this treatise as H.B.
§ H.B., pp. 316-318.
|| H.B., p. 339.
304 Mr Littlewood, On the zeros of the Riemann zeta-function
Since the number of p within a vertical distance 0(1) of T is
0 (log T) we may write
where 2 2 and S 3 are exclusive and contain 0 (log T) terms, and
where further | I (s — p) | > 1 in £ 3 , and in S 2
| s — p | > distance between C2 and boundary of D > S.
Hence
by (8.2). The lemma now follows from (8.1).
9. Lemma 4.
(9.1) tp(s) = o(logT) ( i + 3 S < a < | ,
Consider 0 (s) on C1. Here a > f and so
\t-T\<l).
Since l/(s — p) = 0 (1) in Sj and S x has o (log T) terms, we have
$ (s) = o (log T). Let now
Mx = Max | 0 | , M 3 = Max \ifi\.
(72 is interior to C2 and (72 to C3 (since 8 < j^), and Eemma 3 is
applicable to ifi (s), the indices a, j8 depending only on S. Thus in
and upon C2
| 0 (s) | < M ^ ^ 8 < {o (log T)}« JO ( ^ | ^ ) } " = o (log a7).
But C2 includes the region with which the lemma is concerned.
10. We suppose now that neither t nor T is the ordinate of a p.
Then we may integrate (9.1) between s and | + it, and obtain (all
logarithms having their principal values)
/(*) - / ( # + *0 = 2 , {log(s- p) - log ($ + it-
P)}
(10.1) / (s) = 0 (1) + Sx log (s - p) - Sx 0 (1) + o (log T)
= o (log TJ + S, log (*-/>)(*+38 < < r < $ , \t-T\
<|),
since Sx has o (log T) terms. Integrating again between the same
limits (s always satisfying the conditions of (10.1))
f */ (a + it) do=o (log T) + Sx f * log (s - p) cfo.
^ <r
J tr
Mr Littlewood, On the zeros of the Riemann zeta-function 305
It is clear (and follows strictly from Lemma 2 with s — p = z, rj = 2)
that the integral last written is 0 (1). Therefore
(10.2)
f2
I
l"4
-C /
*.i\ J
f\ /"f\l
./ <7
I
J* /
'&\ J
t
f¥^\
/I
I ^^
/^\ / ~l \
/I
y» ^/T\
J <T
From this follows
Lemma 5. There exists a a0 satisfying ^ + 38 < cr0 < J + 48 <m(Z
(10.3)
log | I (s0) | = o (log T) (so = ao + iT),
(10.4)
J 2 log|£(c7 + i r ) | * 7 = . o ( l o g T ) ,
(10.5)
Sx log | s0 - P | = o (log T).
In fact (10.4) is true, by (10.2), for any <r0 > \ + 38. Also, by
the same equation,
1 ri+45
8 J i+ss
log | £ (a + iT) | da
i + 36
f{s)da
O |Ji+3S"
< o (log T) + o (log T) = o (log T).
Since the average of log | £ | over a = J + 38 to J + 48 has a modulus o (log T) (and log | £ | is continuous), the value of log | £ | itself
must be o (log T) for some a= aQ. Finally (10.5) follows from (10.1)
and (10.3).
11. For s = a + iT we have (since E = S o + S' when t = T)
= O (log T) +
(Q
\s — p
+ ) +
pj
0
p
Still supposing only a > J, integrating between s and s0, and taking
real parts, we obtain
log | £(,) | - log | £<«„) | = i a S 0 J
S
^ + O(logT)
= S o log (s-p)S o log («„ -P) + O (log T),
and so
(11.2) log | C (s) | = 0 (log T) + o (log T)
+ S o log | s ~ p | - (So - Sj) log | s0 - p |,
since
log | £ (s0) | and Sx log | s0 — p | are o (log T).
306 Mr Littlewood, On the zeros of the Riemann zeta-function
Now for a term of S o — Sx we have 3H/o < \ + 8, and so
M(so-P)>28,
and
log | s0 - p \ = 0 Nog gj .
Hence, integrating (11.2) from a = \ to a = a0 (a distance not
exceeding 48), we have
(11.3) T° log | £ («) | efo = 0 (§ log T) + o (log T)
+ £„ m | " log (s - p) cfo + (So - SJ 0 (S log | ) .
So
1R.
log (s — p) da = 3H
logzdz,
J s,
Ji
where z1; z2 satisfy the conditions of Lemma 2. with TJ = 48. Hence
(11.4)
Since S o and 2 0 — Sx have 0 (log T) terms we have finally, from
(11.3) and (11.4),
(11.5) f"° log | I (s) | da = 0 (§ log | log T) + o (log T).
Combining (11.5) and (10.4), we obtain
| 2 l o g | £ (*) I da = o (log T) + 0 (s log i log T ) ,
and so, since the left side is independent of S, and 8 log 1/8 -»- 0
with 8,
(11.6)
J log | £ (*) | cfe = o (log T).
It follows from Theorem 3 that
S (t) dt = o (log T), when, as
•?
we have supposed, T is not the ordinate of a p. But the result, if
true for such T, is true for all, and the proof of Theorem 8 is
completed.
12. The proof of Theorem 7 follows similar, but much simpler
lines. Let
We suppose t = T, or s = a + iT, throughout. Then (by (8.2))
X(*) = O (log T).
Integration gives
/ (s) - S o log (s - p) = / (2 + iT) - S o log (2 + iT - p) + 0 (log T)
= 0 (1) + S o 0 (1) + 0 (log T) = O (log T).
Mr Littlewood, On the zeros of the Riemann zeta-function 307
Integrating again and taking real parts we obtain
= 0 (log T) + S o HI 0 (1) = 0 (log T).
Theorem 7 follows from this and (5.5).
13. Theorem 9. It follows from Lindelqf's hypothesis that
S (T) = o (log T).
This is a known theorem, due to Cramer *. It is, however, interesting
to observe that it is an almost immediate consequence of Theorem 8."
Suppose the result false. Then there exists a positive h such that
either S (T) > h log T for arbitrarily large T, or else S(T)<-h
log T
for arbitrarily large T (both events, of course, can occur). We
observe that
(13.1) S(T + x)-S{T)
= {N(T + x)- N (T)} ~{M(T + x)- M (T)} + 0 (1)
> - M (T + x) + M (T) - A
> - Ax x log T - A2 (T > 2, 0 < x < T),
and similarly
(13.2) S(T-x)S (T)<A3x log T + At (T > 2, 0 < z < T).
If now S (T) > h log T for arbitrarily large T we have, writing
i
hlAlt
fT+l
rT
S(t)dt-
rl
S(t)dt=
.' o
Jo
8{T + x) dx
Jo
= £S(T)+ f {S (T + x) - S (T)} dx
.'o
[ (- AtxlogT-
A2)dx
o
7,2
= ( # - | ^ a log T + 0 (1) = ~ log T + 0 (1),
which is false, since the left side is o (log T). A similar proof dis-
poses of the other case: we consider
r?
S(T — x)dx and use (13.2).
.'o
14. We now enter upon the second part of the paper. We
always assume the Riemann hypothesis, and do not always state
it explicitly in enunciating our theorems and lemmas.
Let
So (T) = S (T),
S2n(T)=(~->
$
...f(a + iT)(da)*» (n > 1),
S,n-AT)
=
"•
J i J a J <T
,_ l\n-l
f « r » /•«>
K )
-
7T
m
...f{a
+ iT){daf^
J J j a- J v
* Math. Zeitschr., 2 (1918), 237-241.
(n > 1).
308 Mr Littleivood, On the zeros of the Riemann zeta-function
Theorem 10.
Sn (T) = I' Sn.x (t) dt + cn (n > 0),
(14.1)
Jo
where cn is a constant defending only on n.
This is deducible from Theorems 2 and 3 by repeated integration,
if we observe that N (a, T) = 0 (a > \). It is, however, unnecessary
to use these theorems. In fact, since log £ is regular in I T > {
(except at s = 1) we have, for a > | , and by continuity for a~>\,
» f Tf(° + *9 * + f"/(" + iT)da= ["/(*) ds,
the path of the last integral lying in t > 0. Deforming this path
into the real axis, and taking account of the discontinuity of / (s)
along the whole stretch J < C T < 1 , t = 0, we have, supposing
(14-2)
f (<r + iT) da = I f(s)ds + iCJ = Ca
f(a + it)dt+(
0
J <r
Jo-
where Ga, GJ depend only on a, GJ = 0 for a > 1, and Ca = 0 (2-")
as a -»- + oo . Taking real and imaginary parts in (14.2) we obtain
(14.3) I"T If (a + it) dt = f°° log | £ (or + %T) | da - GJ,
JO
(14.4)
Jo
f log | £ (a + it) \ dt = - P If {a + \T) da + Ga.
JO
J <r
If we write a = J in (14.3) we obtain (14.1) with n — 1. A second
integration gives, by means of (14.4),
rT
r<*>
S1 (t) dt =
Jo
rT
da\ log | £ {a + it) | <fe
'i
Jo
= - P° da f°{I/(a + iT) <fo + O,} AT
= -(
da I
Ji
Ja
The process may be repeated indefinitely, (14.3) and (14.4) being
used alternately.
15. Theorem 11. On the Riemann hypothesis
and, more generally,
* Results intermediate between this and the S (T) — 0 (log T) which is true
without the Riemann hypothesis have been given by Bohr, Cramer, and Landau.
Mr Littlewood, On the zeros of the Riemann zeta-function 309
The proof of this theorem is long and difficult, and depends on
a singularly varied set of ideas. I begin with some preliminary
lemmas.
Throughout the sequel T > 4. We shall adopt the abbreviations
l{t) = \ogt, A («) = log log*, /*(0 = l/A(0;
Z = Z (T) = log T, A = A (T) = log log T, M = M (T) = I/A (T).
Thus we drop the argument T in Z, A, /x, but never the argument t.
Lemma 6 (i). Letf(T) be an increasing function, f (T) > 1, and
f(T) = O (I2). Then for large T there exists a t between T and T + 1—2
for which f(t + 2) < f/(«).
(ii) Letfufz be increasing functions of T, greater than unity, and
of the form 0 (>). Let Tx (T) = 0 (I), T2 (T) = 0 (I). Then for large
T there exists a t between T and T + I* for which (simultaneously)
(i5.i)
fi{t + Tx{t)}<yi(t),
(15.2)
ft{t + Tt(t)}<yt(t).
If (i) is false we have, for some large T,
Hence for some large T and v = [|Z] — 1
which is false.
Consider now (ii).2 Let Tn = TB_1 + I2 (Tn_x) (n > 0), To = T.
Then Tn - Tn_^ < I (Tn), and if n < I
Tn-T<
nl2 (Tn) < P (Tn),
from which it follows (for large T) that Tn < 2T, and so, for
n<N=[l],
(15.3)
Tn-T < I*.
Now the pair of inequalities
(15.4)
f1(Tn)<%f1(Tn-i)
,
m
}
(15.5)
/,(?,)<!/,(?,,) (
must hold simultaneously for some n < 2V. For if not, one of them,
say the first, must be false for at least \N values of n, from which
it follows that
/i (TN) > (f )*-v/i (T) > e^
and so
ft (T + I*) > / x (TN) > V,
which is false. But for an n giving (15.4) and (15.5) we have,
taking t = Tn_t,
/i {t + T, (<)} < / a {t + I2 (t)} =A (Tn) < f/x (t)
for large T. This is (15.1), and (15.2) follows similarly.
310 Mr Littlewood, On the zeros of the Riemann zeta-function
16. Lemma 7. Suppose R (T) is continuous in T > 0 except at
isolated points, and
Rj. (T) = f1 R (t) dt + bu R2 (T) = \T R1 (t) dt + b2,
Jo
Jo
where b1} b2 are constants. Suppose further that as T -> oo
R(T) =
O{g(T)h(T)},
where g (T) is increasing and greater than unity, and T2h (T) is
increasing and greater than unity, for T > To. Then
This is practically included in a known result f.
We require also another lemma, belonging to the same order
of ideas, and embodying an argument of which we have had an
example already in § 13.
Lemma 8. Suppose ip (T) = 0 (I) is an increasing function greater
than unity (T > 4), and
R± (T) = f R (t) dt + O (1).
Suppose further that
(16.1) R(T + x)-
Jo
R{T)>-
Kxl-
K'
where K and K' are constants, and that
(16.2)
Then
(16.3)
We write, for brevity, u=^(2T).
If (16.3) is false, then,
for any positive h (however large), there are arbitrarily large T for
which either
(16.4)
R (T) >
or else
(16.5)
R(T)<-
We suppose h > K + 1 (as we may), and choose $ = x = fiu, or
* The O's here are naturally not absolute.
t See G. H. Hardy and J. E. Littlewood, Proc. London Math. Soc. (2), 11 (1913).
If R (T) is continuous the theorem is a case of Theorem 6 (a), with
r = 2. x = T, f(x)=R,JT), <t>(x)=g(T), f (x) = T*g (T)h(T).
In the actual case we can modify R (T) in a small neighbourhood of each discontinuity so as to make it continuous, altering Rt (T), R, (T) by arbitrarily little.
Mr Littlewood, On the zeros of the Riemann zeta-function 311
| = — x, according as (16.4) or (16.5) holds. Then x < \T for large
T, and we have respectively in the two cases
R(t)dt=\
R(T + y)dy = xR(T)+
-' o
IT
{R(T +
y)-R(T)}dy
.'o
> xR (T) - P (Ky log T + K') dy,
Jo
and
rT+l
rx
rx
JO
JO
R(t)dt = -\ R(T-y)dy=-xR(T)+\
JT
{R(T)-R(T-y)}dy
>-xR (T) - fX {Ky log (T-y) + K'} dy.
Jo
In either case
I rT+t
(16.6)
rx
R (t)dt > x | R (T) | - \ (Ky log T + K') dy
\J T
•' 0
> xhliiu - \KW - K'x = x (h - IK) IJXU - K'x
> x (\MyLU - K')> x (±hlfj,u) = Ihl/j.2^ (2T).
But the left side does not exceed
\R1(T + i)\ + \R1(T)\ + 0 (1)
= 0 (fjf^
>P (§T)) + 0 { W (T)} = 0 (If.) # (2T),
and it is impossible for (16.6) to hold for arbitrarily large h.
17. Let
P(T) = Max (77 | S (t) \) + l^+ 1,
P(T)= Max
A word or two on the relations of these four functions may make
them more easily distinguishable. Since S (t) is_substantially 1 / (s)
with a = £ we have, substantially, P (t) < P (t); and similarly
Q (t) < Q (t). The Q's differ from the corresponding P's (1) by the
presence of /instead of $/; and (2) by the addition of /x to a or the
lower bound of a (f, but not 5/, has infinities when o- = +, and <r
must be kept from too near an approach to | ) . A/x + 1 is added
to all the maxima to secure that the four functions exceed unity
and are of order at least Ifi. Finally we take maxima over 4 < { < T
to make our functions increasing. It will appear eventually that
all four functions are of the same order, viz. 0 (1/JL).
312 Mr Littlewood, On the zeros of the Riemann zeta-function
Lemma 9. We have
P{T) = O(l), Q(T) = O(IX).
The first of these follows from (1.4). To prove the second we
observe that on the circle with centre at s0 = •§ + it0 (t0 > 4) and
radius r = 1,
\$f(s)\<Alogt0.
Since | / (s0) | < A we have by Caratheodory's theorem
\f(8)\<{A+tAl(Q}~in and upon the concentric circle of radius p < r. Taking
p <r — fj,= 1 — /x,
we have, in particular,
\f(a + ilo)\<Al(to)X
(J + A i < « y < f ) .
We may clearly omit the restriction a < §, and the second result of
the lemma follows by taking the maximum in the range 4 < t0 < T.
18. Lemma 10. We have
P(T) = O {P (T + I)},
Q(T) = O {Q (T + I)}.
In proving this we use the method of "rotation of domains"
introduced by Lindelof*. Consider the functions
g (s) = e±^W
in the strip bounded by the vertical lines a = \, a = | . and the
horizontal line t = 4. Let s0 be a point of the strip for which
5 < t0 < T. Rotate the strip about the point s0 through an angle
— \-n, and let g1 (s) be the function that takes the value g (s) at
the point sx of the new strip corresponding to the point s of the
original. Then s0 is enclosed in a square, the length of whose side
is unity, formed by the original vertical lines and their new positions. On the vertical boundaries of the original strip
\g(s)\ < e x p | I / ( s ) |
does not exceed exp {P (t)} or exp A according as the boundary is
the left or right hand one. In either case | g (s) \ < exp {P_(t) + A}.
For internal points of the strip we have | g (s) | < exp {P (<)}. I t
follows that for any point s of a vertical boundary of the square
| g (s) | < exp {P (T + l) + A}, | 9l (s) | < exp {P (T + 1)},
and that for any point of a horizontal boundary of the square
\g(s)\< exp {P (T + 1)}, | 9l (s) \ < exp {P (T + 1) + A},
* E. Lindelof, Ada Soc. Sclent. Fenn., 46 (1915).
Mr Littlewood, On the zeros of the Riemann zeta-function 313
and so
(18.1) \ h(s) \ = \g(s)9l(s)\<ex1?{P(T
+ l) + P(T + 1) + A}
on the whole boundary of the square. Since h (s) is regular and
bounded in the interior of the square, and continuous at the boundary except at a finite number of points, we conclude, by a wellknown principle, that (18.1) is valid in the interior of the square,
and in particular at s = s0, where g (s0) = g1 (sQ). Thus
| g (s0) | 2 < exp {P (T + 1) + P (T + 1) + A}.
Since one or other of the ± signs in the definition of g gives
| <7 | = exp | 1/1 we conclude, dropping the suffix 0, that
2 |I/(s)| <P(T+ 1) + P(T+ 1) + A (i<a<-|-, 5<t<T).
In this we may further write 4 < t for 5 < t, and suppress the condition cr.< f. Next we may, by continuity, infer the inequality at
a = \ except when \ + it is a zero of £ (s). Finally, since
5 / (s) = i {/ (s + 0*) + / (s ~ 0*)}> we may infer the inequality also
in this remaining case if we replace T + 1 by T + 2. Tnus
2 | I / ( s ) | <P(T + 2) + P(T + 2) + A (a>l
4<t<T).
Taking the maximum of this for the range of t we have now
2P (T)< P (T + 2) + P (T + 2) + A + 2 ^ + 2.
By Lemma 6 (i) there exists a t = t (T) between T and T + I — 2
for which
_
_
P (t + 2) < f P (T).
Then
2P (t)<P{t + 2)+ P(t + 2) + Alfi < P (t + 2) + f P (t).+ Al/i,
P(T)<P(t)<2P(t + 2) + Alp <2P(T + l) + Alp<AP(T + I),
which proves the first part.
The second part is similar and a little more direct. Arguing
with the strip bounded by o = ^ + p, a = §, £ = 4, and with/ in place
of g, we obtain
\f(so)\^<{Q(T+l) + A}Q(T+l)<AQ(T+l)Q(T+l).
This time there is continuity at the boundary; therefore, taking
the square root and the maximum,
By Lemma 6 (i), for some t between T and T + I — 2,
Q (t + 1) < fQ (t),
Q (0 < A {Q (t) Q (t + 1)}*,
and so
Q (0 < AQ (t + 1),
Q (T) <Q (t) < AQ (T + I - 2 + 1) < AQ (T + I)
the desired result.
VOL. XXII. PART III.
2
314 Mr Littlewood, On the zeros of the Riemann zeta-function
19. Lemma 11.
(19.1) /(«) = 0{Q (T + 1) e -«i('-«*} = 0 {Q (T + 21) e-*<—*>*}
(uniformly)for
| + /* < o- < 2, t < T.
The second part follows from Lemma 10 and the first part,
which we therefore consider. We write x = a — J. It is sufficient
to prove (19.1) for \x < x < | — p. Consider three circles with a
common centre at s0 = f + /n + it and radii r3 > r2 > rx, going
through the points \ + fj. + it, | + x + if, § + it respectively. Let
Mx, M2, M3 be the upper bounds of | / ( s ) j in the circles. Then by
Hadamard's "three circles" theorem
where ^ = log - / l o g - (and 0 < ^ < 1). Now for /J, < x < 1, and
large T,
% = log {l-(x-
IJ-)}-1/log ^-i
>A{x- rf/A = A(x- /*),
M, <f(i) = A,
Ms < Max |/(«) | < Q (T + 1) + A < AQ (T + 1).
Hence
M, < AQ (T + 1) {Q (T + 1)}-* <AQ(T+
l)
the desired result.
20. Lemma 12. Q{T) = O {P (T + I5)}.
We have, from Lemmas 10 and 11,
(20.1) If (s) = 0 {P (T + 21)} (x>0,t<T),
(20.2) f (s) = 0{Q (T + 21) e-"^*}
(p<x<lt<T).
Let,
f = 2
^
v l o
so that f = o (1) (by Lemma 9), and Q (T + 21) e-*">** < /> (T + 21).
Then, for ^ < x < f, t < T, we have from (20.2)
(20.3) | / ( s ) | = 0 { Q ( r + 2J)e- a '^} = 0 { e ( T + 2Z)e-i«.Ma:+5)}
= 0 {P (T + 2Z) e - * ^ .
Consider now the range 0 < a; < £, and the circles with common
centre at s0 = -J + | 4- A, and radii £ and a;. On the outer circle
(20.4) 5 / (s) = 0{P[T + l + 2l{T + 1)]} = O {P (T + 3Z)}
by (20.1). Also, by (20.3) with x = g,
(20.5)
f(so) = O{P(T + 2l)}.
* This part of the argument is one of the main keys to the proof of Theorem 11.
Mr Littlewood, On the zeros of the Riemann zeta-function 315
By (20.4), (20.5) and Caratheodory's theorem we have, on the inner
circle and in particular at s = \ + x + it,
(20.6)
f(s) = o\f-P(T+
[X
Combining this with (20.3) we obtain
(20.7) /(«) = 0 {P (T + 21) e'a^} + 0 | | P (T + 31)1 (x > 0).
In this we take x— p, vary t and take the maximum. This gives
Q{T) = O {P (T + 21)} + 0{£\P(T + 31)} + O
since £A > 2, P (T + 2,1) > lp. Hence
(20.8)
^
< A log IV (X) + e] + A log ° V + W + AhgP
g' ( + 3 i ) .
Now by Lemma 6 (ii) there exists a t between T and T + li
for which
P [t + 3? (Q]
Q[t + 2l (Q]
QW
Then from (20.8)
<4
'
PW
<4>
U(t)<Alog[U(t)
+ e] + A,
from which it follows that U (t) < A, or
Q (t) <AP[t + Zl (*)].
Hence
Q(T)<Q (t) = 0{P(T + Z5)},
the desired result.
Lemma 13.
1 / (S) = O {P (T + i6) *-«•**} (0 < a; < | , f < T).
The result is true for a; </x by (20.1), and for /u. < x < f by
(20.2) and Lemma 12.
21. Lemma 14.
We have
P (T) = 0 (lp).
S2((T) = - -- /""(fa f"l/(i + z + tT) fc = - - Pcfe [iIfthf + 0(1)
"•Jo
Jo
"•Jo
JK
= O (^2) P (T + P) + O (1) = 0 (p2) P(T + Is)
by Lemma 13. Also S (T) = O {P (T + I6)}.
316 Mr Littlewood, On the zeros of the Riemann zeta-function
The conditions of Lemma 7 are fulfilled if -R (T) = S (T),
Rx (T) = St (T), R2 (T) = S2 (T), and we conclude that
S, (T) = 0 0*) P (T + l«).
(21.1)
Further, if R and Rt have their meanings above, and if
./- (T) = Max {r§:P\t + P (t)]) ,
if) (T) is greater than unity, increasing, and of the form 0 (A) = 0 (I).
Also, f o r O < x < | r ,
R (T + x) - R (T)
= {N (T + x)-N (T)} ~{M(T + x)- M (T)} + 0(1) .
> - {M (T + x) - M (T)} - A > - Ax log T - A.
The conditions of Lemma 8 are therefore satisfied, and we have
Since the function on the right is increasing we may write t for T
on the left and take the maximum for t < T, obtaining
P(T) = O (l^VP (2T) + 0 (I?) = 0
Hence again, for t < T,
y-^ P \f + l* (*)] = 0 {\/<p (3«)} = 0
and taking the maximum on the left,
ifi(T)<Ay/tlt($T).
(21.2)
Now (21.2) and the evident inequality
</< (T) < Al
(21.3)
together imply
</r (T) < A .
(21.4)
For we must have
(21.5)
</< (30 < 3<J> (0
for some arbitrarily large values of t; otherwise
and tp (T) > AT for some arbitrarily large T, contrary to (21.3).
But now (21.5) and (21.2) involve
for arbitrarily large t, and so, since i/> is increasing, for all t. Hence,,
since
we have
finally
P (T) = 0 (Z/x).
Mr Littlewood, On the zeros of the Riemann zeta-function 317
22. Theorem 11 now follows without difficulty. We have at
once
8{T) = 0 {P (T)} = 0 {lrf
the first of the desired results. Next we have
_
dx\
JO
\
JO
T
dx...
Jx
Jx
%
dx f dx ... fhfdx + 0 (1)
JX
<X
= 0 {/x2" P (T + I6)} = O(lp. ix2n)
by Lemma 13, the desired result for Sn when n is even. Finally the
results for odd n follow at once by Lemma 7 if we take R = S2n-2,
Rx — S2n^1> R2 =
S2n.
23. Our material enables us to prove also
Theorem 12.
In fact
log £ (i + /* + »T) = 0 {Q (T)} = 0{P(T + ¥)} = 0
Also*, for s = a + %T,
Since every term in S is positive
log j (j + »T) = m log £ (i + /* + »r) h
< 0 (iM) + /* (i log T + A) = 0
24. We prove, in conclusion, another theorem about the orders
of £ (T), Sx (T), ... as functions of log T. Let
a
m i?
i
" " - y ^ o log log T '
and a = a^. That is, on the one hand
S (T) = 0 {(log T)**},
and, on the other, this result becomes false if a is replaced by any
smaller number; and similar relations hold for Sn and an. We may
call a the " order" of S (T). It is known that a > 0; and it is in fact
true that
* H.B., p. 316.
318 Mr Littlewood, On the zeros of the Riemann zeta-function
(24.1)
a>\,
an>\;
but the proof of this would take us too far afield. We can prove,
however, within our present circle of ideas:
Theorem 13. The, number an is a convex function of n, and
1> a >o
It follows at once from Theorem 11 that an < 1. Further, supposing an > 0*, and taking R (T) = Sn (T), Rx = Sn+1, R2 = Sn+i,
we see that Lemma 7 is valid with
g (T) = (log T) W , g (T) h (T) = (log T)°»+<.
It therefore gives (0's depending on n and e)
8n+1 (T) = 0{(l0g T)i(an +
so that an is a convex function of n. Since the a's are bounded above
it follows from the convexity that an never increases, and this completes the proof.
The upper bounds for the Sn (T) given in Theorem 11 decrease
by a factor I/log log T for each addition of unity to n. Theorem 13
and the result an > 0 together show that this factor cannot be
replaced (throughout) by any constant power of I/log T, however
small.
* The modifications necessary if a n <0 or a n + a = -co can be supplied by the
reader. Since actually an > 0 it is not worth while to set out the details.