CC BY
11
УДК 517.55
On the Zeta-Function of Zeros of Some Class of Entire Functions
Vyacheslav I. Kuzovatov*
Institute of Mathematics and Computer Science Siberian Federal University Svobodny, 79, Krasnoyarsk, 660041
Russia
Alexey A. Kytmanov^
Institute of Space and Information Technology Siberian Federal University Kirenskogo, 26, Krasnoyarsk, 660074
Russia
Received 25.07.2014, received in revised form 20.08.2014, accepted 26.09.2014 Using the residue theory, we give an integral representation for the zeta-function that enables us to construct its analytic continuation.
Keywords: zeta-function, integral representation, entire function.
Introduction
The question of the integral representation for the zeta-function associated with some entire function was studied by V.B.Lidskii and V.A. Sadovnichii in [1], where an entire function f (z) of a certain type was considered.
In [2] A.M.Kytmanov and S.G. Myslivets introduced the concept of the zeta-function associated with a system of meromorphic functions f = (fi,...,fn) in Cn. Using the residue theory, these authors gave an integral representation for the zeta-function, but the system of functions f 1,..., fn was subject to rigid constraints.
Let f (z) be an entire function of order p with zeros z1, z2,..., such that f (0) = 0. Then, according to Hadamard's theorem on factorization (see, for example, [3, Chapter VIII, S. 8.2.4]), the function f (z) is represented in the form
f (z) = eQ(z)P (z),
where P (z) is the canonical product constructed by the zero set of the function f (z), and Q (z) is a polynomial with the degree not higher than p. In this case, the canonical product P (z) has the form
^ / \ ^ / \
P (z) = П Д ( = П ( 1 - - ) e zn+(zn )2/2+'"+( zn )p/p
-I \ zn J V zn J
m — 1 4 ' m — 1 4 '
* [email protected] [email protected] © Siberian Federal University. All rights reserved
and p < p.
Thus, under f (z) = 0 we have locally (in accordance with [3, Chapter VIII, S. 8.2.4])
ln f (z) = ln(eQ(z)P (z)) = ln eQ(z) + lnP (z) =
= Q (z) + ln «
= Q (z) + £ln
n=1 «
= Q (z) + E
n=1 «
= Q (z) + E
« / \
J] h - — jezn+( zn T/2+-+( znr/p
n = 1 ^ n /
1 - — ^ ezn+(zn)2/2+-+(zn)P/P z„ J
zn + ( zn )2/2+- + ( zn r/p
ln ( 1--J + ln e
zn
n=1
ln(1 - -) + - +( J^y/2 + ... + ( JlV/,"
zn / zn V zn / V zn /
Differentiate this relation once. We obtain
f ' (z) f (z)
Q' (z) + E
n=1 «
Q' (z) + E
1 - zn
M 1 z 1 ( z
— + — +---+... + —
zn / zn zn zn \ zn
1
zn z z
z
zp-1
p-1
+ - + - + ... + ~JT
We rewrite the expression under the summation sign to get
— + — + •
zn zn
+
1
zn z zn
+ — + + ... +
zp-1
zp
zp-1
zp
1- ^
1- ^
1- ^
1 -
1 1 - —
-1 + V zn
zn z
(z„ - z) z.
p n
Thus
f ' (z) z^
f (z)
= Q' (z) -
n= (zn - z) z
The resulting series converges absolutely and locally uniformly for z = zn since p < p1 < p ([3, Chapter VIII, S. 8.2.3]). Here p1 is the index of convergence of zeros.
Recall ([3, Chapter VIII, S. 8.2.2]) that the lower boundary of positive numbers a for which the series J2 |zn| a converges is called the index of convergence of zeros. Denote it by p1.
Henceforth we assume |z1| < |z2| < ... < |zn| < .... Let some number of zeros of the function f (z) lies inside the circle |z| = R, and the rest lies out. Then
« zp
-, (zn z) zn
n=1
no
E
n=1
(zn z) zP
+
« E
n=no + 1
(zn - z) z;
p
Suppose that for the zeros of the function f (z) the following estimates hold:
|z - z„| >£ |z|, when |z„| < |z|,
|z - z„| >£ |z„|, when |z„| > |z|,
(1) (2)
1
1
z
n
1
p
p
p
1
1
1
z
zn =
z
z
n
n
n
n
p
p
p
z
z
zz
zz
n
p
p
z
z
where for convenience only one constant S > 0 is introduced. We discuss the conditions under which the estimates (1) and (2) would hold true below.
Estimate the first sum using (1) and the fact that -r—r < -.—r. We have
N M
E
""^ (zn z) zn n=1 v /
no n=1
izp n0
< 4-s
( zn z ) zn
û-
1
n=1
|Zn - z| |Zn|P
< z
npE
1
= S |z||zn|p
n=1
<
S ^ \z P+1 n=1 | zn|
< |z|p • £1.
Estimate the second sum using (2). We get
E
n=no + 1
( zn z ) zP
œ
E
n=no + 1
<
(zn - z) z;
< |z|
œ
p E
n=no + 1
|zn - z| |zn|
s ^
1
ip+1
< |z|P • £2.
n=no + 1 |zn|î
Let us now discuss the conditions (1) and (2). Let
|z - zn| > |z| - |zn| > S |z|,
p < |z|p E
n=no + 1
S |zn| |zn|p
i. e.
|z| S |z| > |zn| , |z| >
1S
Setting z = zn+1, we obtain the following system of inequalities:
|zn+1| > ^ >
1 |zn-1|
1 - S 1 - S 1 - S
|zn-1|
(1 - S)2
>... >
1 < (1 - S)r
|zn+1|
|z11
Then
œ .. .. œ ..
E^ < Aid -s)n = A • s,
| z | | z | | z1 |
^ |zn+1| |z11
n=1 1 1 1 11 n=1
E
|zn+1|a |z1|
<ûVE ((1 - s)c
|z11
(1 - S)n '
|z1 r
• Sa
a > 0,
where S and Sa denote the corresponding sums of the series. Thus, we have shown that the conditions (1) and (2) would be true for an entire function with p1 = 0.
In what follows we consider entire functions f (z) of the zero order with the index of convergence of zeros p1 = 0. The conditions (1) and (2) hold for the zeros of such functions.
Given the above calculations and reasoning, as well as the fact that p < p1 < p (see [3, Chapter VIII, S. 8.2.3]), we obtain
f ' (z) f (z)
Ezr
1
n=1
if z = zn.
P
P
z
z
1
œ
P
P
1
1
z
z
z
n
1
1
1
1
œ
n
1. The first integral representation
Let f (z) be an entire function of the zero order in C. Consider the equation
f (z) = 0. (4)
Denote by Nf = f-1 (0) the set of all roots of (4) with the multiplicity counted. The number of zeros is at most countable.
Further assume that the following asymptotic representation holds true on a positive part of the real axis is a non-negative integer)
- f -, z - i.e. fM - f «V = J . (5)
f (z) ¿0 zV f (z) V=0 zV V|z|V0+V
If v0 =0 then
f'(z) - «0 = o(. (6)
f (z) V |z|
Our goal is to obtain an integral representation for the zeta-function Zf (s) of (4) that was defined in [2] as
Zf (s)= £ (-a)-S ,
where s G C. We choose the minus sign in the definition of the zeta-function only for convenience in writing the integral formulas; below we explain what value is taken for the multivalued function (-z)-s.
Let z = x + iy. Suppose that the function f is not equal to zero at any point of R+ := {z G C : x > 0, y = 0}. This means that Nf n R+ = 0. Consider a domain D C C of the form
D = {z G C : r < |z| < R} \ {z G C : r < Re z < R, Im z = 0}
and 0 < r < R.
Observe that D is a simply connected domain. Its boundary 7 = dD consists of the intervals [r, R] on the real axis, the circle SR of radius R centered at the origin with positive (counterclockwise) orientation, the interval [R, r] on R obtained from [r, R] by the change of orientation, and the circle -Sr obtained from Sr by the change of orientation. Choose the radii r and R so that 7 n Nf = 0. Consider the integral
The functions (-z) s = e-sln(-z) are holomorphic in D, where lnZ denotes the principal branch of the logarithm, i. e. the holomorphic branch ln Z on C \ {Z G C : Re Z ^ 0, Im Z = 0} equal to zero for Z = 1. It is obvious that 1 (s) is an entire function in s G C.
Note that 1 (s) can be written in the form (in accordance with the logarithmic residue theorem)
Thus
1 (s) = (-a)-
aewf nD
Choose a sequence Rk so that SRk does not contain the zeros of f. f f'(z)
Lemma 1. The integral ( —z) s ^ N dz ^ 0 as Rk ^ +to and Re s > 1.
JSr,_
Proof. We have
(-z)
f (z)
-s(ln Rk n))
g—Re s ln Rfc+Im s(<^-n) _ o (R-Re
where ^ = arg z.
Then the estimates hold for the module of the integral
(-z)
SRk
f' (z) f (z)
dz
SRk
f' (z)
f (z)
|dz| < R—Re s•C•Rk = C•R1-Re s
C
R
Re s-1
as Rk ^ +to and Re s > 1.
0,
□
Denote by r0 = (TO,r] U [r, to), r0 = Sr U r0. Now, taking R = Rk and letting k tend to infinity in (7), we obtain
1 (s) = 2ni L (—z)~
f' (z) f (z)
dz.
Lemma 2. The integral 1 (s) can be continued analytically into the half plane Re s > —v0 when the condition (5) holds.
Proof. We argue as in [1]. For the proof of this lemma we write the integral 1 (s) as the sum of four parts:
1 ^ = ¿1 (—z)-S fl dz (—z)-S(
f' (z) f (z)
vo \ £ £) d-z+
v=0 /
1
+ /no ( z)
Vo
-s
zV
v=0
^ dz — S (—z)-^ ^ dz = /1 (s) + 12 (s) + 13 (s)+ 14 (s) .
v=0
We easily see that 11 (s) and 14 (s) are entire functions in s, and 12 (s), in view of the asymptotic representation (5), can be continued analytically into Re s > — v0. Finally, 13 (s) is zero for Re s > 1 and so its analytic continuation is equal to zero in the whole plane. □
Suppose further that v0 = 0. Then, by Lemma 2,1 (s) = 11 (s)+12 (s)+14 (s) and for Re s < 1 (as in Lemma 1) the integrals 11 (s) and 14 (s) tend to zero when r ^ 0. Thus, we have Corollary 1. For 0 < Re s < 1
f (s)=iL( —(—z)-s dz
f (z)
where r0' = (to, 0] U [0, to) .
Consider the integrals over the intervals with opposite orientation. Since
(—z)-s = (—xe2n^ = e-2nis (—x)-s ,
s
e
s
s
k
s
we obtain
/'- ^ (-z)" dz = /7fM - wo ) e"*™ (-xp dx =
— wo (— x) s dx.
/(z) У У™ V/ (x)
/' (x)
= - e"2nis
f (x)
Summing all integrals over the intervals, we obtain
iff' (x) A ( )-s d _2nis rff' (x) A ( )-s d
L Ifx) - "V(-x) dx - e J0 ITm - (-x) dx
= (1 - e-2niS) r (fx) - *») (-x)-s dx.
By obvious calculations we have
i\ t3 ^ k /I О
1 - e"2nis = e"nis (enis - e"nis) = e"nis2i-—- = e"nis2i sinns = (-1)" 2i sinns.
2г
Summing all integrals over the intervals, we arrive at the integral
^ f' (x)
2i sin ns / ———— wo x s dx.
Jo \/ (x) /
Summarizing the above, we obtain the integral representation for the zeta-function Zf (s) in the strip 0 < Re s < 1.
Theorem 1.1 Let / (z) be an entire function of the zero order in C and satisfy the condition (6). Suppose that 0 < Re s < 1. Then
/' (x)
л . . sin ns f° f/' (x) \ s ,
0 w = — Jo (ж- wo)dx
where «0 is the limit value of f. ( at infinity.
f ( x)
The method of proof shows that if the asymptotic condition (5) holds, we have the following result.
Corollary 2. Suppose that the asymptotic condition (5) holds. Then for -v0 < Re s < 1 the following holds
sin ns (/' (x) wA
Zf (s) = ~ Jo - V=o x7/
v 1 dx.
, . / , x
(x) z—' xv
v 7 v=0
To conclude this section we compare the obtained integral representation with the integral representation for the classical Riemann zeta-function Z (s) (see, for example, [4, Chapter 2, S. 9]) in the strip 0 < Re s < 1. Namely,
sinns ff r' (1 + x) . _s ,
' J - ln x > x s dx, 0 < Re s < 1.
Z<•> = -^f - 'nx}
o
OO
2. The second integral representation
Consider an entire function f (z) of order p. In this section we obtain another integral representation for the zeta-function Zf (s) of the zeros zn of f that have the form
-qn + is», qn > 0.
(8)
For this purpose we consider the integral I xs 1eZnX dx, in which we make the change of
variables
Thus, by (8)
zn • x y x
dx =
dy.
cs 1eZnX dx =
s -1
= y-dy = -
ys - 1e-у dy =
г (-z„)s -1 (-*«)" (-zn)
n) J0
(-zn)S '
where l is a ray (corresponding to the change of variables zn • x = —y) from the origin, and r (s) is the Euler gamma-function defined by the formula
f(s) =
cs 1e x dx.
Further, we consider the product
f(s) • Zf (s)=f(s)^(-z„) —s = f(s)¿
(-zn)
E
r(s)s
(-zn)S
E
cs 1 ez-x dx =
E:
s-1.
s — 1
ez-x dx.
Denoting
we obtain
or
F (/,x) = £ eZnX,
n=1 oo
Г (s) • Zf (s) = / xs—1F (f,x) dx,
o
1 ío
Zf (s) = f(Sy /„ xs—1F (f,x) dx.
(9)
r(s^ 0
It is necessary to justify the change of the order of summation and integration and explain why the series (9) converges.
To prove that the series (9) is convergent we use the Cauchy criterion. Consider
|ez-x|
|e qnx , eisnx I — |e xl — e x
Then for the convergence of the series (9) it is necessary and sufficient that
lim —e q-x = lim e
i.e.
= lim < 1,
n—ю n—ю n—Ю e -
lim ^ > 0.
z
n
oo
0
1
y
z
z
n
n
oo
oo
y
s
0
o
0
1
0
1
1
1
1
o
x
0
0
1
1
o
n
n—
To justify the change of the order of summation and integration it is necessary to prove
o
the uniform convergence of the series J2 xs—1eZnX on the set [0;+^). We enumerate the ze-
n=1
ros z1,..., zn,... in the order of increasing absolute values of the real parts, i.e., q1 < q2 < ... < qn < ..., and let Re s = a > 1. Consider the series of modules
o O
ZnX I _ \ ■ o—1 _qnX
|xs—1ez"x | = ^^ xo—1e_= ^^ xo—1e_9ixe(9l_= e_91 x ^^ xo—1e(9l_
n=1 n=1 n=1 n=1
Consider the function gn (x) = xo—1e(qi — 9n)x and find its extremums. Consider the equation
gn(x) =
For root x0 of this equation we have the relation
a - 1 + x (q1 - q„) = 0, a - 1 = x (qn - q1) , a-1
x0
qn - q1
It is easy to see that for x > x0 the function gn (x) is decreasing, and for 0 < x < x0 the function gn (x) is increasing. Thus, the point x0 is a local maximum of the function gn (x). Then
OO OO OO / 1 \ o —1
V |xs—1 | = e_9qix V xo—1e(9l—< e_9qix V ( —— ) e(qi—qn-1 =
1 n=1 n=1 V qn- qj
O o—1 O o—1 o—1
-9i*W a - M e_(o —1) = e_'a - n ' 1
Elf-qrJ e_(0_1) = e_qiXE
O
E
1 vqn- q1j e / \qn- q1
' \ o —1 OO / \ o—1
-qi* ' a - ^ 1
1 qn - q1
For the uniform convergence of the series under study it is necessary that the seO / X o—1
O 1 o—1
ries (- I converges. The convergence of this series is equivalent to the conver-
V qn- qJ
O 1 o—1
gence of the series ^^ (—J . Thus, to change the order of summation and integration it is necessary that
O 1 o—1
the series ^^ I — I is convergent. (11)
n=A qn/
n=1
We have proved the following result. Theorem 2.1 Suppose that the conditions (10) and (11) are satisfied and Re s > 1. Then
r (s) Jo
where F (f, x) is defined by formula (9).
1O
Zf (s)=F^ J0 xs_1F (f,x) dx, (12)
O
e
Corollary 3. Suppose that the conditions of Theorem 2.1 are satisfied. Then for 0 < Re s < 1 the following formula holds
Zf (s)r(s) = J (f (f,x) - X) xs-1 dx,
where F (f, x) is defined by formula (9).
Proof. We write the expression (12) in the form
Zf (s) r (s) = / F (f,x) xs-1 dx = / (f (f,x) - xs-1dx+/ 1 xs-1dx+/ F (f, x) xs-1dx. Jo Jo V x/ Jo x ./1
In the last equation we calculate the second integral. We have
f1 1 xs-1 dx = f1 xs-2 dx = — 'o x Jo s - 1
s1
since Re s > 1 .
Thus, for Re s > 1 the following equalities hold
Zf (s)r(s)
^ ^F (f,x) - xs-1 dx + —+ ^ F (f,x) xs-1 dx.
According to the principle of analytic continuation, this formula holds for Re s > 0. Moreover, for 0 < Re s < 1 we have
rœ xs-1
dx =
s-2
s — 1
dx =
s1
1
1
s — 1 x
1-s
1 (0 -1)= 1
s1
s1
Hence we obtain
Zf (s) r (s) = / F (f,x) -- xs-1 dx -
œ xs-1
dx + / F (f, x) xs 1 dx.
Simplifying the expression, we obtain the statement of the corollary. □
In the conclusion of this section we give (see, for example, [4, Chapter 2, S. 4]) one more integral representation for the classical Riemann zeta-function Z (s). Namely,
Z (s) =
1
œ xs-1
r (s) Jo ex - 1
dx, Re s > 1 .
(13)
If 0 < Re s < 1 the integral representation (13) can be written (see, for example, [4, Chapter 2, S. 7]) in the form
Z (s) r (s) = f°
o
1 -- ] xs 1 dx, 0 < Re s < 1.
ex — 1 x
The difference between the classical integral representation (13) and the obtained integral representation (12) is that in the classical case it is possible to calculate explicitly the series (9),
since the Riemann zeta-function is defined by the zeros of
1
r(1+ x)
Therefore, this classical formula follows from the formula of Corollary 3.
, i.e., zn = -1, -2, -3,....
1
1
o
œ
œ
œ
x
x
x
1
1
1
œ
x
o
3. Examples
In this section we consider the examples of entire functions f (z) of zero order, constructed by the zero set zn, for which the following relation holds
f' (x) lim -TTY = ^
x^+to f (x)
f' (z)
where the ratio is defined by formula (3).
f (z)
It is well-known that the limit of the sum of a series is equal to the sum of the series consisting of the limits of its terms, when there is the uniform convergence, i. e.
œ œ
limy; Un (x) = y] { lim u„ (x) f . (14)
x—>a *—' *—' l^x—>a J
= 1 n=1
Example 1. Let zn = -2n. Then, in accordance with formula (3)
f' (x) = ^ 1
f (x) x + 2n .
11 O 1 Since-— < —, where x > 0, the series V^-— converges uniformly on the set [0,
x + 2n 2n x + 2n
n=1
in accordance with the Weierstrass criterion of a uniform convergence of functional series. Given formula (14), we have
f' (x) O 1 O 1 lim = lim -= lim -= 0,
x^+TO f (x) x^+TO ^—' x + 2n Z—' x^+TO x + 2n
v ' n=1 n=1
i. e. «0 =0 and f, (x.) — 0 as x — f (x)
Example 2. Let zn = -qn + isn, qn > 0. Then, in accordance with formula (3)
f' (x) = O 1
f (x) ^ x - zn .
Estimate the terms:
11 1 11
< , = = -—r, x > 0.
|x - z„| |x + qn - is„| /(x + q )2
V(x + qn)2 + sn v/qJ+4 |zn|:
1 O 1
Since the series ^^ |—^ converges ([3, Chapter VIII, S. 8.2.2]), when a > p, the series ^^ |—|
O1
converges. Then the series ^^-converges uniformly on the set [0, in accordance with
x - zn
n=1
the Weierstrass criterion of a uniform convergence of functional series. Given formula (14), we have
f' (x) ^ 1 ^ , 1
lim „, x = lim > --— = > lim —n
+œ f (x) x—+œ ; x + qn - isn ^ x—+œ x + qn - is^
1 n=1
i. e. = 0 and f, ,(x) ~ 0 as x ^ f (x)
This work was supported by the Russian Foundation of Basic Research (grants 12-01-00007-a, 14-01-00283-a), first author; the state order of the Ministry of Education and Science of the Russian Federation for Siberian Federal University (task 1.1462.2014/K, second author).
References
[1] V.B.Lidskii, V.A.Sadovnichii, Regularized Sums of Zeros of a Class of Entire Functions, Functional Analysis and Its Applications, 1(1967), no. 2, 133-139.
[2] A.M.Kytmanov, S.G.Myslivets, On the Zeta-Function of Systems of Nonlinear Equations, Siberian Math. J, 48(2007), no. 5, 863-870.
[3] E.C.Titchmarsh, The Theory of Functions, Oxford University Press, Oxford, 1939.
[4] E.C.Titchmarsh, The Theory of the Riemann Zeta-Function, Oxford University Press, Oxford, 1951.
О дзета-функции корней одного класса целых функций
Вячеслав И. Кузоватов Алексей А. Кытманов
С использованием теории вычетов дается интегральное представление для дзета-функции, которое позволяет построить аналитическое продолжение дзета-функции.
Ключевые слова: дзета-функция, интегральное представление, целая функция.
