The Mellin transform of the square of Riemann's zeta-function

0.11 MB·English

by  Aleksandar Ivić

The Mellin transform of the square of Riemann’s zeta-function
Aleksandar Ivic´
Abstract. Let (s) = ∞ ζ(1 + ix) 2x−sdx (σ = es > 1). A result concerning
Z1 1 | 2 | ℜ
analytic continuation of Z (s) to C is proved, and also a result relating the order of
1
R
(σ +it) (1 σ 1, t t ) to the order of (1 +it).
Z1 2 ≤ ≤ ≥ 0 Z1 2
5
0
1. Introduction
0
2
n Let Zk(s), the (modified) Mellin transform of |ζ(21 +ix)|2k, denote the analytic con-
a tinuation of the function defined initially by
J

4
∞
1 (1.1) (s) := ζ(1 +ix) 2kx−sdx (k N, σ = es > c(k) (> 1)).
Zk | 2 | ∈ ℜ
] Z1
T
N This function, when k = 2, was introduced by Y. Motohashi [15] (see also [16]), and its
h. properties were further studied in [10] and [11]. The latter work also contains some results
t on the function (s), which is the principal object of the study in this paper. It was
a 1
Z
m shown that (s) is regular for σ > 3/4, except for a double pole at s = 1. The principal
1
Z −
[ part of the Laurent expansion of (s) at s = 1 is
Z1
2
v 1 2γ log(2π)
0 (1.2) + − ,
4 (s 1)2 s 1
− −
0
1
where γ = Γ′(1) = 0.577215... is Euler’s constant. M. Jutila [13] continued the study of
1
−
4 (s) and proved that (s) continues meromorphically to C, having only a double pole
1 1
0 Z Z
at s = 1 and at most double poles for s = 1, 2,....
/ − −
h
t The object of this note is to prove some new results on (s). First we make more
a 1
Z
m precise Jutila’s result on the analytic continuation of (s). We have the following
1
Z
:
v
i THEOREM 1. The function 1(s) continues meromorphically to C, having only a
X Z
double pole at s = 1, and at most simple poles at s = 1, 3,.... The principal part of its
r − −
a Laurent expansion at s = 1 is given by (1.2).
Our second aim is to prove an order result for (s). This is a result M. Jutila
1
Z
mentioned in [13], but the term that I initially claimed, namely t1−2σ+ε is too optimistic.
It appears that what can be proved is contained in
THEOREM 2. We have, for 1 σ 1, t t > 0,
2 ≤ ≤ ≥ 0
(1.3) (σ +it) t21−σ+ε max (1 +iv) +(t9−716σ +t−1)logt.
Z1 ≪ε t−tε≤v≤t+tε|Z1 2 |
2001 Mathematics Subject Classification. 11M06
1
2 Aleksandar Ivi´c
Corollary (see M. Jutila [13]). For 1 σ 1, t t we have
2 ≤ ≤ ≥ 0
(1.4) (σ +it) t56−σ+ε.
1 ε
Z ≪
The bound (1 +it) t1/3+ε was mentioned in [11], and its proof was elaborated
Z1 2 ≪ε
by M. Jutila in [13]. If one inserts this bound in (1.3), then (1.4) follows immediately. We
note that here and later ε denotes arbitrarily small constants, not necessarily the same
ones at each occurrence. It can be also proved that, for any given ε > 0,
T
(1.5) (σ +it) 2dt T2−2σ−ε (k = 1,2; 1 < σ < 1).
|Zk | ≫ε 2
Z1
The bound (1.5) for k = 2 appeared in my paper [10], and the proof of the bound when
k = 1 is on similar lines, so that the details will not be given here. It is plausible that
(1.6) max (1 +iv) 1
t−tε≤v≤t+tε|Z1 2 | ≫
holds, but at present I am unable to prove (1.6).
2. Proof of Theorem 1
Let
∞
(2.1) L (s) := ζ(1 +ix) 2ke−sxdx (σ = es > 0, k N)
k | 2 | ℜ ∈
Z0
be the Laplace transform of ζ(1 + ix) 2k. We are interested in using the expression for
| 2 |
L (s) when s = 1/T, T . Such a formula has been known for a long time and is due
1
→ ∞
to H. Kober [14] (see also a proof in [17, Chapter 9]). The functions L (s) when k = 1,2
k
were studied by F.V. Atkinson [1], [2]. The function L (s) was considered by the author
2
[8], [9], and the approach to mean value of Dirichlet series by Laplace transforms by M.
Jutila [12]. Kober’s result on L (s) is that, for any integer N 0,
1
≥
N
γ log(2πσ)
(2.2) L (σ) = − + d σn +O (σN+1) (σ 0+),
1 2sin σ n N →
2 n=0
X
where d are suitable constants. Note that for σ = 1/T, T we have
n
→ ∞
γ log 2π log T +γ
− T = 2π
2sin 1 1 1 + 1
2(cid:0)T (cid:1) T − 24T3(cid:0) 1(cid:1)6·5!T5 −···
∞
(cid:0) (cid:1)T
= log +γ c T1−2n
n
2π
(cid:18) (cid:18) (cid:19) (cid:19)n=0
X
2
The Mellintransform of the square of Riemann’s zeta-function 3
with the coefficients c that may be explicitly evaluated. Therefore for any integer N 0
n
≥
N N
1 T
(2.3) L = log +γ a T1−2n + b T−2n +O (T−1−2N logT)
1 n n N
T 2π
(cid:18) (cid:19) (cid:18) (cid:18) (cid:19) (cid:19)n=0 n=0
X X
with the coefficients a ,b that may be explicitly evaluated. In particular, we have
n n
(2.4) a = 1, b = π.
0 0
It is clear from Kober’s formula (2.2) that a = 1. To see that b = π one can use the work
0 0
of Hafner–Ivi´c [4] (this was mentioned by Conrey et al. [3]), which will be stated now in
detail, as it will be needed also for the proof of Theorem 2. Let, as usual, for T 0,
≥
T T
E(T) = ζ(1 +it) 2dt T log +2γ 1 (E(0) = 0)
| 2 | − 2π −
Z0 (cid:18) (cid:18) (cid:19) (cid:19)
denote the error term in the mean square formula for ζ(1 +it) . Let further
| 2 |
T T
(2.5) G(T) := E(t)dt πT, G (T) := G(t)dt.
1
−
Z0 Z0
Then Hafner–Ivi´c [4] proved (see also [6])
(2.6) G(T) = S (T;N) S (T;N)+O(T1/4)
1 2
−
with
−2 −1/4
πn T 1
S (T;N) = 2−3/2 ( 1)nd(n)n−1/2 arsinh + sin(f(T,n)),
1
− 2T 2πn 4
n≤N (cid:18) r (cid:19) (cid:18) (cid:19)
X
−2
T
S (T;N) = d(n)n−1/2 log sin(g(T,n)),
2
2πn
n≤N′ (cid:18) (cid:19)
X
πn π
f(T,n) = 2T arsinh + 2πnT +π2n2 ,
2T − 4
r
p
T π
g(T,n) = T log T + , arsinhx = log(x+ x2 +1),
2πn − 4
(cid:18) (cid:19)
p
T N N2 NT
AT < N < A′T (0 < A < A′constants), N′ = + + .
2π 2 − 4 2π
r
We use Taylor’s formula (see [7, Lemma 3.2]) to simplify (2.6). Then we obtain
∞
π
G(T) = 2−1/4π−3/4T3/4 ( 1)nd(n)n−5/4sin(√8πnT )+O(T2/3logT),
− − 4
n=1
X
3
4 Aleksandar Ivi´c
so that
(2.7) G(T) = O(T3/4), G(T) = Ω (T3/4).
±
We use (2.6) with T = t,N = T,T t 2T and apply the first derivative test ([5, Lemma
≤ ≤
2.1]) to deduce that
2T
G(t)dt T5/4,
≪
ZT
since the term O(1) in [5, eq. (15.29)] in Atkinson’s formula is in fact O(1/T). Hence
T
(2.8) G (T) = G(t)dt T5/4.
1
≪
Z0
To prove that b = π (cf. (2.4)) we note that, using the definition of E(T) and
0
applying integration by parts,
1 ∞ t
L = e−t/T log +2γ +E′(t) dt
1
T 2π
(cid:18) (cid:19) Z0 (cid:18) (cid:18) (cid:19) (cid:19)
∞ ∞
T 1
= T e−x(logx+log +2γ)dx+ E(t)e−t/T dt
2π T
Z0 Z0
T 1 ∞ t
(2.9) = T log +γ + E(u)du e−t/T dt
2π T2 ·
(cid:18) (cid:18) (cid:19) (cid:19) Z0 Z0
T 1 ∞
= T log +γ + (πt+O(t3/4))e−t/T dt
2π T2
(cid:18) (cid:18) (cid:19) (cid:19) Z0
T
= T log +γ +π +O(T−1/4),
2π
(cid:18) (cid:18) (cid:19) (cid:19)
where we also used the O-bound of (2.7) and
∞
e−xlogxdx = Γ′(1) = γ.
−
Z0
A comparison of (2.3) and (2.9) proves that b = π, as asserted by (2.4).
0
We return now to the proof of Theorem 1. Let
∞
(2.10) L¯ (s) := ζ(1 +iy) 2ke−ysdy (k N, σ = es > 0).
k | 2 | ∈ ℜ
Z1
Then we have by absolute convergence, taking σ sufficiently large and making the change
of variable xy = t,
∞ ∞ ∞
L¯ (x)xs−1dx = ζ(1 +iy) 2ke−yxdy xs−1dx
k | 2 |
Z0 Z0 (cid:18)Z1 (cid:19)
∞ ∞
(2.11) = ζ(1 +iy) 2k xs−1e−xydx dy
| 2 |
Z1 (cid:18)Z0 (cid:19)
∞ ∞
= ζ(1 +iy) 2ky−sdy e−tts−1dt = (s)Γ(s).
| 2 | Zk
Z1 Z0
4
The Mellintransform of the square of Riemann’s zeta-function 5
Further we have
∞ 1 ∞
L¯ (x)xs−1dx = L¯ (x)xs−1dx+ L¯ (x)xs−1dx
1 1 1
Z0 Z0 Z1
∞
= L¯ (1/x)x−1−sdx+A(s) (σ > 1),
1
Z1
where
∞
A(s) := L¯ (x)xs−1dx
1
Z1
is an entire function. Since
1
L¯ (1/x) = L (1/x) ζ(1 +iy) 2e−y/xdy (x 1),
1 1 − | 2 | ≥
Z0
it follows from (2.11) with k = 1 by analytic continuation that, for σ > 1,
(2.12)
∞ ∞ 1
(s)Γ(s) = L (1/x)x−1−sdx ζ(1 +iy) 2e−y/xdy x−1−sdx+A(s)
Z1 1 − | 2 |
Z1 Z1 Z0
(cid:0) (cid:1)
= I (s) I (s)+A(s),
1 2
−
say. Clearly for any integer M 1
≥
∞ 1 M ( 1)m y m
I (s) = ζ(1 +iy) 2 − +O (x−M−1) dyx−1−sdx
2 | 2 | m! x M
(2.13) Z1 Z0 mX=0 (cid:16) (cid:17) !
M ( 1)m 1
= − h +H (s),
m M
m! · m+s
m=0
X
say, where H (s) is a regular function of s for σ > M 1, and
M
− −
1
h := ζ(1 +iy) 2ymdy
m | 2 |
Z0
is a constant. Inserting (2.3) in I (s) in (2.12) we have, for σ > 1,
1
∞ N ∞ N
x
I (s) = (log +γ) a x−2n−sdx+ b x−1−2n−sdx+K (s)
1 n n N
2π
Z1 n=0 Z1 n=0
(2.14) X X
N
1 γ log2π
= a + − +K (s),
n (2n+s 1)2 2n+s 1 N
n=0 (cid:18) − − (cid:19)
X
say, where K (s) is regular for σ > 2N. Taking M = 2N it follows from (2.12)–(2.14)
N
−
that
N
1 γ log2π
(s)Γ(s) = a + −
Z1 n (2n+s 1)2 2n+s 1
n=0 (cid:18) − − (cid:19)
(2.15) X
2N ( 1)m 1
+ − h +R (s),
m N
m! · m+s
m=0
X
5
6 Aleksandar Ivi´c
say, where R (s) is a regular function of s for σ > 2N. This holds initially for σ > 1,
N
−
but by analytic continuation it holds for σ > 2N. Since N is arbitrary and Γ(s) has no
−
zeros, it follows that (2.15) provides meromorphic continuation of (s) to C. Taking into
1
Z (−1)m
account that Γ(s) has simple poles at s = m (m = 0,1,2,...) with residues , and
− m!
that near s = 1 we have the Taylor expansion
∞
1 f
= 1+γ(s 1)+ n(s 1)n
Γ(s) − n! −
n=2
X
with f = (1/Γ(s))(n) , it follows that
n
s=1
(cid:12)
(cid:12)
(cid:12) 1 2γ log(2π)
(s) = + −
Z1 (s 1)2 s 1
− −
N
1 1 γ log2π
+ a + −
(2.16) Γ(s) n (2n+s 1)2 2n+s 1
( )
n=1 (cid:18) − − (cid:19)
X
1 2N ( 1)m 1
+ − h +U (s),
m N
Γ(s) m! · m+s
!
m=0
X
say, where U (s) is a regular function of s for σ > 2N. Moreover, the function
N
−
1/(Γ(s)(2n + s 1)) is regular for n N and s C. Therefore (2.16) provides ana-
− ∈ ∈
lytic continuation to C, showing that besides s = 1 the only poles of (s) can be simple
1
Z
poles at s = 1 2n for n N, as asserted by Theorem 1. With more care the residues at
− ∈
these poles could be explicitly evaluated.
3. Proof Theorem 2
We suppose that X t and start from
≫
X ∞
(s) = + ζ(1 +ix) 2x−sdx (σ = es > 1).
Z1 | 2 | ℜ
Z1 ZX !
We use ζ(1 +ix) 2 = log(x/2π)+2γ +(E(x) π)′ and integrate by parts to obtain
| 2 | −
1−ε
t X
(s) = ζ(1 +ix) 2x−sdx+ ζ(1 +ix) 2x−sdx
Z1 | 2 | | 2 |
Z1 Zt1−ε
X1−s 1
+ +logX +2γ log2π
(3.1) s 1 s 1 −
− (cid:18) − (cid:19)
∞
(E(X) π)X−s +s (E(x) π)x−s−1dx
− − −
ZX
= I +I +O(X−σ +t−1X1−σlogX)+I ,
1 2 3
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The Mellintransform of the square of Riemann’s zeta-function 7
provided E(X) = 0. But as proved in [6], every interval [T,T + C√T] with sufficiently
large C > 0 contains a zero of E(t), hence we can assume that E(X) = 0 is fulfilled
with suitable X. Since E(x) is x1/4 in mean square (see e.g., [5] or [7]), it follows by
≍
analytic continuation that (3.1) is valid for σ > 1, t t > 0, t X tA(A 1). By
4 ≥ 0 ≤ ≤ ≥
repeated integration by parts it is found that I t−1. The integral I is split in O(logT)
1 2
≪
subintegrals of the form
5K/2
J = ϕ(x) ζ(1 +ix) 2x−sdx,
K | 2 |
ZK/2
where ϕ(x) C∞ is a nonnegative, smooth function supported in [K/2, 5K/2] that is
∈
equal to unity in [K, 2K]. To bound J one can start from (1.1) with k = 1 and use
K
the Mellin inversion formula (see [10] for a detailed discussion concerning the analogous
situation with the fourth moment of ζ(1 +it) ), which yields
| 2 |
1
ζ(1 +ix) 2 = (s)xs−1ds (x > 1).
| 2 | 2πi Z1
Z(1+ε)
Here we replace the line of integration by the contour , consisting of the same straight
L
line from which the segment [2 + ε i, 1 + ε + i] is removed and replaced by a circular
−
arc of unit radius, lying to the left of the line, which passes over the pole s = 1 of the
integrand. By the residue theorem we have
1 x
ζ(1 +ix) 2 = (s)xs−1ds+log +2γ (x > 1).
| 2 | 2πi Z1 2π
ZL (cid:16) (cid:17)
This gives
1 5K/2
J = (w) ϕ(x)xw−s−1dxdw+O(t−1).
K 1
2πi Z
ZL ZK/2
Repeated integration by parts in the x–integral shows that only the portion mw
|ℑ −
ms tε gives a non-negligible contribution. Hence replacing by the line ew = 1 we
ℑ | ≤ L ℜ 2
obtain
(3.2) J K12−σtε max (1 +iv) +t−1.
K ≪ε t−tε≤v≤t+tε|Z1 2 |
To estimate I in (3.1) we integrate by parts and use (2.7) and (2.8). We obtain
3
∞ ∞
(E(x) π)x−s−1dx = G(X)X−s−1 +(s+1) G(x)x−s−2dx
− −
ZX ZX
∞
= O(X−1/4−σ) (s+1)G (X)X−s−2 +(s+1)(s+2) G (x)x−s−3dx
1 1
−
ZX
X−1/4−σ +t2X−3/4−σ.
≪
7
8 Aleksandar Ivi´c
Thus it follows from (3.1) and (3.2) that we have, for 1 σ 1, t t ,
2 ≤ ≤ ≥ 0
(σ +it) t21−σ+ε max (1 +iv) +(X1−σt−1 +t−1)logt
(3.3) Z1 ≪ε t−tε≤v≤t+tε|Z1 2 |
+X−σ +tX−1/4−σ +t3X−3/4−σ.
In (3.3) we choose X to satisfy
X1−σt−1 = t3X−3/4−σ, i.e., X = t16/7.
Then tX−1/4−σ +X−σ t3X−3/4−σ, and (1.3) follows from (3.3).
≪
In conclusion, one may ask what should be the true order of magnitude of (σ+it).
1
Z
This is a difficult question, but it seems reasonable to expect that
(3.4) (σ +it) t12−σ+2µ(21)+ε (1 σ 1, t t > 0)
Z1 ≪ε 2 ≤ ≤ ≥ 0
holds, where
log ζ(σ +it)
µ(σ) := limsup | |
logt
t→∞
is the Lindel¨of exponent of ζ(s). One could try to obtain (3.4) from (1.3) by refining
the estimation of (s) in [11]. This procedure leads to an exponential sum with the
1
divisor function whZich is estimated as ε t2µ(21)+ε (this, of course, is non-trivial and needs
≪
elaboration).
References
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Math. Oxford 10(1939), 122-128.
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Math. Soc. 47(1941), 174-200.
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moments of L-functions, 71pp, ArXiv.math.NT/0206018.
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The Mellintransform of the square of Riemann’s zeta-function 9
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99(2001), 115-145.
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[12] M. Jutila, Mean values of Dirichlet series via Laplace transforms, in “Analytic Number
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[13] M. Jutila, The Mellin transform of the square of Riemann’s zeta-function, Periodica
Math. Hung. 42(2001), 179-190.
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[15] Y. Motohashi, A relation between the Riemann zeta-function and the hyperbolic
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Press, Oxford, 1986.
Aleksandar Ivic´
Katedra Matematike RGF-a
Universitet u Beogradu
D- uˇsina 7, 11000 Beograd
Serbia and Montenegro
e-mail: [email protected],
[email protected]
9