Hardy's generalization of eᶻ and related analogs of cosine and sine

Hardy’s Generalization of e

and Related Analogs of Cosine and Sine

Iossif V. Ostrovskii

(Communicated by Richard S. Varga)

Abstract. In 1904, Hardy introduced an entire function depending on two

parameters being a generalization ofez. He had studied in detail its asymptotic properties and that of its zeros. We consider the two following non-asymptotic problems related to the zeros. (i) Determine values of the parameters such that all the zeros belong to the open left half-plane. For these values, the analogs of sine and cosine generated by Hardy’s function have real, simple and interlacing zeros. (ii) Determine the number of real zeros as a function of the parameters.

Keywords. Class P, integral representation, Levin’s generalization of the

Hermite-Biehler Theorem, logarithmic derivative, Rolle’s Theorem.

2000 MSC. 30D10, 30D15.

1. Introduction

In 1904, Hardy [1] had studied the entire function (1) E_s,a(z) = ∞  n=0 (n + a)szn n! , s∈ C, a ∈ C \ {0, −1, −2, . . . }.

This function is a generalization of ez. Observe that

(2) E_0,a(z) = ez, E_k,a(z) = T_k(z)ez, k = 1, 2, . . . ,

whereT_k is a polynomial of degree k. For other values of s, E_s,a does not admit a simple representation and has infinitely many zeros. In this case, Hardy [1] obtained asymptotic formulae forE_s,a and its zeros.

In the present paper, the case s∈ R, a > 0, is considered. In this case it suffices

to study E_s,a and its zeros only in the upper half-plane. Hardy’s results can be stated in the following way.

Received December 12, 2005.

Theorem A ([1]). For each δ > 0, 0 < δ < π/2, and s ∈ R \ {0, 1, 2, . . . }, the following asymptotic formulae hold as z → ∞ in the upper half-plane:

E_s,a(z) = zsez(1 +O(1)), 0≤ arg z ≤ π 2 − δ, E_s,a(z) = Γ(a) Γ(−s)(−z)a(log(−z))s+1(1 +O(1)), π 2 +δ≤ arg z ≤ π, E_s,a(z) = zsez(1 +O(1)) + Γ(a) Γ(−s)(−z)a(log(−z))s+1(1 +O(1)), | arg z − π 2| ≤ δ,

where the branches of zs, (−z)a and (log(−z))s+1 are so chosen as to be real when z or (−z) are positive and greater than 1.

Theorem B ([1]). Let s ∈ R \ {0, 1, 2, . . . }, and let zn, n = 1, 2, ..., be the zeros of E_s,a with Im(z_n)≥ 0. Then the following asymptotic formula holds:

(3)

z_n = −(s + a) log(2nπ) − (s + 1) log log n + log Γ(a) Γ(−s) +i[(2n + 1)π + 1 2(a− s)π] +O(1) as n→ +∞. Define (4) C_s,a(z) := 1

2(Es,a(iz) + Es,a(−iz)) ,

S_s,a(z) := 1

2i(Es,a(iz)− Es,a(−iz)) .

These functions can be viewed as generalizations of cosz and sin z. The zeros of

cosz and sin z are real, simple and interlace. We can ask the following question: for which values of (s, a) do the zeros of C_s,a and S_s,a inherit this important property? Evidently, this problem is of non-asymptotic character and cannot be

solved with help of asymptotic formulae for C_s,a,S_s,a derived from Theorem A. The generalization of the Hermite-Biehler Theorem due to Levin ([2, ch. 27], [3, ch. 7, sec. 4]) shows that the above problem is equivalent to the following, also non-asymptotic, one: for which values of (s, a) are all zeros of E_s,a located in the open left half-plane?

Evidently, Theorem B solves a similar but less subtle problem: for which values of (s, a) are all but finite number zeros of E_s,a located in the open left half-plane? By Theorem B, this is the case if and only if

One can conjecture that (5) also gives a complete solution to our problem. The-orem B only implies that this condition is necessary. In this paper we prove sufficiency of a more restrictive condition (Corollary 2.2(i)).

Theorem B also shows that if (s, a) belongs to the subset of{(s, a) : s ∈ R, a > 0}

where (5) does not hold, then all but finite number zeros of E_s,a lie in the open right half-plane. We find a sufficient condition for all zeros of E_s,a to be located in the open right half-plane (Corollary 2.2(ii)).

It turns out that condition 5 is also sufficient for E_s,a to be included into the known class P (Theorem 2.1(i)) introduced by Levin in the 1940s and playing an important role in generalizations of Hermite-Biehler Theorem and Bernstein inequalities for derivatives ([2, ch. 28], [3, ch. 7, ch. 9]). A theorem of Levin’s allows us to derive from E_s,a∈ P that zeros of Cs,a and S_s,a are real, simple and interlace.

We study also the following question of non-asymptotic character posed by Moshe Newman in a letter to the author: how many real zeros doesE_s,a have for s∈ R, a > 0? Clearly, E_s,a does not have non-negative zeros, so, only negative ze-ros are possible. He conjectured that E_s,a has exactly k + 1 negative zeros for k < s≤ k + 1, k = 0, 1, . . . . We confirm this conjecture (Theorem 2.5 below).

Finally we extend the above results (Theorems 2.6–2.10) to the “limiting case as

a→ +0” by considering the function ([1, p. 427])

(6) E_s(z) := ∞  n=1 nszn n! , s ∈ R,

and related analogs of sine and cosine.

2. Statement of results

According to the definition, the class P consists of entire functions ω of expo-nential type satisfying the conditions

ω(z) = 0 for Im(z) < 0,

(7)

2d(ω) := h_ω(−π/2) − hω(π/2) ≥ 0,

(8)

where h_ω is the indicator of ω defined as h_ω(θ) = lim sup

r→∞ r

−1_log|ω(reiθ_{)|, θ ∈ [0, 2π].}

For results related to the class P we refer to [2, ch. 27] and [3, ch. 9, sec. 4]. Let

G+ := {(s, a) : s ≥ 0, a > 0} ∪ {(s, a) = (−1, 1) : −1 ≤ s ≤ 0, a ≥ 1},

Theorem 2.1.

(i) If (s, a)∈ G+, then the function

(9) ω_s,a,γ(z) := e−iγzE_s,a(iz), γ ≤ 1

2,

belongs to P and, moreover, does not have real zeros.

(ii) If (s, a)∈ G−, then the function

(10) ω_s,a,γ(z) := eiγzE_s,a(−iz), γ ≥ 1

2,

belongs to P and, moreover, does not have real zeros.

The following corollary is immediate.

Corollary 2.2.

(i) If (s, a)∈ G+, then all zeros of E_s,a lie in the open left half-plane.

(ii) If (s, a)∈ G−, then all zeros of E_s,a lie in the open right half-plane.

The proof of Theorem 2.1 is based on an integral representation for E_s,a (see Lemma 3.3 below) having different forms fors > 0 and s < 0. In the case s < 0

it had been proved by Hardy in an equivalent form (cf. [1, p. 415]). For s > 0

Hardy’s representation ([1, p. 411], ) and its proof are different from ours. In the proof of Theorem 2.1, we also use some asymptotic properties of E_s,a

(Lemma 3.4 below). These properties are immediate corollaries of Theorem A. Since Hardy’s proof of this theorem is rather long and cumbersome, we present, for the reader’s convenience, a simple proof of Lemma 3.4 based on Lemma 3.3. Using Theorem 2.1 and the Hermite-Biehler phenomenon, we consider the zeros of functions a bit more general, than C_s,a and S_s,a.

Theorem 2.3.

(i) If (s, a)∈ G+, γ ≤ 1/2, then the zeros of functions

(11) C_s,a(z) cos γz + S_s,a(z) sin γz, S_s,a(z) cos γz − Cs,a(z) sin γz are real, simple, and interlace.

(ii) If (s, a) ∈ G−, γ ≥ 1/2, then also the zeros of functions (11) are real, simple, and interlace.

Using (i) with γ = 0, we derive the following corollary.

Corollary 2.4. If (s, a)∈ G+, then the zeros of C_s,a and S_s,a are real, simple, and interlace.

Theorem 2.3 is an immediate corollary of Theorem 2.1 and the following fact due to Levin ([2, pp. 222–223], [3, pp. 321–322]): if a function ω belongs to P and does not have real zeros, then the zeros of functions

(12) 1 2  ω(z) + ω(z)  , 1 2i  ω(z)− ω(z) 

are real, simple, and interlace.

Further we consider real zeros of E_s,a.

Theorem 2.5 (conjectured by Moshe Newman).

(i) For k < s ≤ k + 1, k = 0, 1, 2, . . . , the set of all real zeros of E_s,a consists of k + 1 negative simple zeros.

(ii) For s < 0 the function E_s,a does not have real zeros.

Proof of Theorem 2.5 also uses Lemmata 3.3 and 3.4.

The rest of results are analogues of the previous ones for function (6). Neverthe-less, they cannot be derived directly from them by taking limit as a→ +0. Theorem 2.6.

(i) For s > 0, the function ω_s,γ(z) := 1

ze −iγz_E

s(iz), γ ≤ 1₂, belongs to P and does not have real zeros.

(ii) For s < 0, the function ω_s,γ(z) := 1

ze iγz_E

s(−iz), γ ≥ 1

2,

belongs to P and does not have real zeros. Corollary 2.7.

(i) If s > 0, then all zeros of z−1E_s(z) are located in the open left half-plane.

(ii) If s < 0, then all zeros of z−1E_s(z) are located in the open right half-plane.

DefineC_s, S_s by (4) with E_s,a replaced by E_s.

Theorem 2.8.

(i) For s > 0, γ ≤ 1/2, the zeros of functions

1

z(Cs(z) cos γz + Ss(z) sin γz) ,

1

z (Ss(z) cos z− Cs(z) sin z) are real, simple, and interlace.

(ii) The above assertion remains true also for s < 0, γ ≥ 1/2.

Corollary 2.9. If s > 0, then the zeros of z−1C_s(z) and z−1S_s(z) are real, simple, and interlace.

Theorem 2.10.

(i) If k < s ≤ k + 1, k = 0, 1, . . . , then the set of all real zeros of z−1E_s(z) consists of k simple negative zeros.

(ii) If s < 0, then z−1E_s(z) does not have real zeros.

3. Integral representations and asymptotic properties

Lemma 3.1. The function E_s,a, a > 0, s ∈ C, defined by (1) is an entire function of exponential type 1. It satisfies the functional equations

E_s+1,a(z) = zE_s,a (z) + aE_s,a(z),

(13)

E_s,a+1(z) = E_s,a (z).

(14)

Proof. The first assertion follows from the well-known formulae (e.g. [2, ch. 1]) connecting order and type of an entire function with its Taylor coefficients. Equa-tions (13) and (14) are obtained by application of operator z(d/dz) and d/dz,

respectively, to (1).

Define entire functions Q_k, k = 0, 1, 2, . . . , by the power series expansion

(15) exp(ze−u− au) =

∞  k=0 (−1)kQ_k(z)u k k!. Lemma 3.2. The functions Q_k satisfy the recurrence relation

(16) Q_k+1(z) = zQ_k(z) + aQ_k(z), k = 0, 1, 2, . . . ,

with initial condition Q0(z) = ez. Moreover, the following formulae hold:

(17) Q_k(z) = E_k,a(z) = T_k(z)ez, k = 0, 1, 2, . . . , where T_k is a polynomial of degree k, normalized by T0(z) = 1. Proof. Differentiating (15) with respect to u and z yields

exp(ze−u− (a + 1)u) + a exp(ze−u− au) = ∞  k=0 (−1)kQ_k+1(z)u k k!, (18)

exp(ze−u− (a + 1)u) = ∞  k=0 (−1)kQ_k(z)u k k!.

Substituting the latter and (15) in the left hand side of (18) and then comparing coefficients of uk, we obtain (16).

Equations (13) and (16) imply that E_a,k satisfies the same recurrence relation and initial condition as Q_k. Hence the first of the equalities (17) follows. The second follows from (16) by induction.

Lemma 3.3. For k < s < k + 1, k = 0, 1, 2, . . . , the following representation holds: (19) E_s,a(z) = 1 Γ(−s)  _∞ 0 

exp(ze−u− au) − k  j=0 (−1)jQj(z) j! u j  du us+1. For s < 0, the following representation holds:

(20) E_s,a(z) = 1

Γ(−s)  _∞

0

exp(ze−u− au)u−s−1du.

Proof. Let k < s < k + 1, k = 0, 1, . . . . According to the Cauchy-Saalsch¨utz formula ([5, sec. 12.21]) one has

(21) Γ(−s) =  _∞ 0  e−t − k  j=0 (−1)jt j j!  dt ts+1.

Changing variable t = (n + a)u, we get

(n + a)s = 1 Γ(−s)  _∞ 0  e−(n+a)u− k  j=0 (−1)j(n + a) j_uj j!  du us+1. Hence E_s,a(z) = ∞  n=0 (n + a)szn n! = ∞  n=0 zn n! 1 Γ(−s)  _∞ 0  e−(n+a)u− k  j=0 (−1)j(n + a) j_uj j!  du us+1 = 1 Γ(−s)  _∞ 0  _∞  n=0 zn n!e −(n+a)u₋k j=0 (−1)ju j j! ∞  n=0 (n + a)jzn n!  du us+1.

The term-wise integration of the series can be readily justified by making use of the inequality   e−x− k  j=0 (−1)kx j j!   ≤ C min(xk, xk+1), x > 0. Using the first of equalities (17), we obtain (19).

The proof of (20) is simpler than that of (19) because instead of (21) we use the Euler representation

Γ(−s) =  _∞

0

Remark. By change of variableu = log(1/v) we obtain from (19) and (20) the

following representations respectively: (22) E_s,a(z) = 1 Γ(−s)  1 0  ezvva− k  j=0 (−1)jQj(z) j! log1 v _j log1 v _−s−1 dv v , for k < s < k + 1, k = 0, 1, 2, . . . , and (23) E_s,a(z) = 1 Γ(−s)  1 0 ezvva−1 log1 v _−s−1 dv,

for s < 0. The representation (23) had been obtained by Hardy (see [1, p. 415],

taking into account that our E_s,a corresponds to Hardy’s F_a,−s). Instead of (22) Hardy had obtained another representation for s > 0 ([1, p. 411]).

Now, we derive from Lemma 3.3 the asymptotic properties of E_s,a which will be used to prove the main results. As we have mentioned above, these properties are immediate corollaries of Theorem A whose proof is much more complicated.

Lemma 3.4. For k < s < k + 1, k = 0, 1, 2, . . . , the function E_s,a(z) possesses the following asymptotic properties:

(i) E_s,a(z) =O(|z|k+1) for Re(z)≤ 0, z → ∞;

(ii) E_s,a(z) =O(1) for | arg z + π| ≤ π/2 − ε, z → ∞ for all ε ∈ (0, π/2); (iii) lim_x→−∞|x|−1log|Es,a(x)| = 0 and limx→+∞x−1log|Es,a(x)| = 1.

(iv) Fors = 0, 1, . . . , all these assertions remain true with the one exception: the first formula in (iii) should be replaced by lim_x→−∞|x|−1log|Es,a(x)| = −1.

(v) For s < 0, (ii) and (iii) remain true and (i) is replaced by: E_s,a(z) is bounded in any half-plane {z : Re(z) ≤ b}, b ∈ R.

Proof. Note that in the case s = k = 0, 1, . . . , the lemma is trivial since E_k,a(z) = T_k(z)ez whereT_k is a polynomial of degree k. Moreover, for s < 0 the

lemma readily follows from representation (20). So, it remains to consider only the cases > 0, s= [s] (where [s] denotes the integer part of s).

(i): Lemma 3.3 and (15) imply that, fork < s < k + 1, k = 0, 1, 2, . . . ,

(24) E_s,a(z) =  _∞ 0 g_k(z, u)uk−sdu, where (25) g_k(z, u) = 1 uk+1 

exp(ze−u− au) − k  j=0 (−1)jTj(z)e z_uj j!  .

Taking into account that T_j(z)ez = Q_j(z), j = 0, 1, . . . , are generated by (15)

and using the formula for remainder of Taylor series, we can writeg_k(z, u) in the

form (26) g_k(z, u) = 1 uk+1 1 k!  _u 0 (u− t)kd k+1 dtk+1exp(ze −t_{− at) dt.}

Using (25), we get the estimate (27) |gk(z, u)| ≤

C uk+1

exp(Re(z)e−u− au)

+(1 +uk)(1 +|z|k)eRe(z), u > 0, Re(z)≤ 0.

Using (26) we get also the estimate

(28) |gk(z, u)| ≤ C(1 + |z|k+1) exp(Re(z)e−u), u > 0, Re(z)≤ 0,

where C > 0 is independent of z and u.

Taking any δ > 0 and using (28) for 0 < u < δ and (27) for u > δ, we obtain |Es,a(z)| ≤ C(1 + |z|k+1)

 _δ

0

exp(Re(z)e−u)uk−sdu

+C

 _∞

δ

exp(Re(z)e−u− au) du us+1 (29) +C(1 +|z|k)eRe(z)  _∞ δ (1 +uk) du us+1 Re(z)≤ 0.

Since δ can be taken arbitrarily small and Re(z)≤ 0, the assertion (i) follows.

(ii): Note that either of inequalities (28), (27) implies that for each u > 0 and p > 0

lim

z→∞, | arg z+π|<π/2−ε|z| p_g

k(z, u) = 0.

Moreover, (28) and (27) imply respectively that sup

| arg z+π|<(π/2)−ε, 0<u<1|gk(z, u)| < ∞,

sup

| arg z+π|<(π/2)−ε, u>1

u|gk(z, u)| < ∞.

Using (24) and the Dominated Convergence Theorem, we get (ii). (iii) Evidently, (29) implies that

lim sup

x→−∞ |x|

−1_{log|Es,a(x)| ≤ 0, lim sup} x→+∞ x

−1_{log|Es,a(x)| ≤ 1.}

Therefore it suffices to show that (30) lim inf

x→−∞ |x|

−1_{log|Es,a(x)| ≥ 0, lim inf}

x→+∞ log|Es,a(x)| ≥ 1.

The second of these inequalities is trivial because the inequality E_s,a(x) > asex, forx > 0, follows immediately from the definition of E_s,a. Let us prove the first. We divide the interval (0,∞) of integration in (24) into two parts: (0, R) and

(R,∞), R > 0. We obtain from (28) that

(31)   _R 0 g_k(z, s)uk−sdu ≤ C(1 + |z|k+1)  _R 0

Hence, for z = x < 0, we get  ₀Rg_k(x, s)uk−sdu ≤ CR s+1−k s + 1− k(1 +|x| k+1_{) exp(xe}−R_).

The integral along (R,∞) is estimated from below by



_R∞g_k(x, u)uk−sdu ≥

 _∞

R

exp(xe−u− au) du us+1 − e x ∞ R k  j=0 |Tj(x)|uj j! du us+1 ≥  _∞ 2R

exp(xe−u− au) du

us+1 +O(|x| k_ex_{), as x→ −∞.} This implies |Es,a(x)| ≥ exp(xe−2R)  _∞ 2R e−au du us+1 +O(|x| k_exp(xe−R_{)), as x→ −∞.} Hence lim inf x→−∞ |x|

−1_{log|Es,a(x)| ≥ −e}−2R_.

SinceR can be taken arbitrarily large, we get the first of the inequalities (30). Remark. It is easy to see that the estimate E_s,a(z) = O(|z|k+1) as z → ∞,

remains true in any half-plane {z : Re(z) ≤ b}, b ∈ R.

4. The function

E

and Levin’s class

P

Proof of Theorem 2.1(i). Firstly, we observe that, for the functions ω_s,a,γ de-fined by (9), condition (8) of the definition of the classP is satisfied. Indeed, by Lemma 3.4(iii), we have

h(−π/2, ωs,a,γ) = h(0, E_s,a)− γ = 1 − γ,

h(π/2, ω_s,a,γ) = h(π, E_s,a) +γ ≤ γ

and hence 2d(ω_s,a,γ)≥ 1 − 2γ ≥ 0.

It remains to prove that all zeros of ω_s,a,γ lie in the open upper half-plane. We divide the set G+ into three disjoint parts to be considered separately:

(a) G+1 :={(s, a) : 0 < s = [s], a > 0},

(b) G+2 :={(s, a) = (−1, 1) : −1 ≤ s < 0, a ≥ 1},

(c) G+3 :={(s, a) : s = 0, 1, 2, . . . , a > 0}.

The case (b) is the simplest. In this case, the result is an immediate consequence of Hardy’s representation (23) and the following theorem of P´olya ([4, part V, #177]): If f is a positive continuously differentiable function with positive (neg-ative) derivative on (0, 1), then the function

F (z) :=

 1 0

has no zeros in the closed right (left) half-plane. In our case (32) f (v) = va−1 log 1 v _−s−1 has positive derivative.

The case (c) is also easy. In this case we have E_k,a(z) = T_k(z)ez,k = 0, 1, 2, . . . .

It suffices to prove that all zeros of T_k lie in the open left half-plane. We prove more: all zeros of T_k, k = 1, 2, . . . , are negative and simple.

By (17), Q_k=ezT_k holds, so we may prove that all zeros ofQ_k are negative and simple. We use induction. For k = 1, Lemma 3.2 implies Q1(z) = (z + a)ez.

Assume that Q_k has k simple negative zeros. Equation (16) for z = −x < 0 can

be rewritten in the formx−a−1Q_k+1(−x) = −(x−aQ_k(−x)). Taking into account that Q_k tends to zero at −∞ and using Rolle’s Theorem, we get the desired assertion.

Now, consider the case (a). Letk < s < k + 1, k = 0, 1, 2, . . . . We use induction.

Letk = 0, 0 < s < 1. The representation (19) implies ω_s,a,γ(z) = e −iz(γ−1) Γ(−s) hs,a(iz), where h_s,a(z) =  _∞ 0

exp(z(e−u− 1) − au) − 1 du us+1.

Taking the real part, we get Reh_s,a(z) =

 _∞

0

exp(Re(z)(e−u− 1) − au) cos(Im(z)(e−u− 1) − 1) du us+1.

Since

Re(z)(e−u− 1) − au ≤ 0 for Re(z)≥ −a, u ≥ 0,

we see that Reh_a,s(z) < 0 for Re(z) ≥ −a and therefore ωs,a,γ(z) = 0 for

Im(z)≤ a and hence ωs,a,γ ∈ P for 0 < s < 1.

Further we need the following Levin’s formula for logarithmic derivative of a function ω∈ P ([2, p. 231, (9)]): Imω _(z) ω(z) =d(ω) +  q −(Im(z) − Im(aq)) |z − aq|2 ,

where a_q’s are zeros of ω.

Assume now that, for some positive integer k, k < s < k + 1, we have ω_s,a,γ ∈ P

and each its zero a_q satisfies Im(a_q)> 0. Using (9) and (13), we get ω_s+1,a,γ(z) zω_s,a,γ(z) = ω_s,a,γ (z) ω_s,a,γ(z)+ a z +iγ.

Since 2d(ω_s,a,γ)≥ 1 − 2γ and Im(a_q)> 0, Levin’s formula implies Imωs+1,a,γ(z) zω_s,a,γ(z) > d(ωs,a,γ) +γ + Im a z ≥ 1 2− a Im(z) |z|2 > 0, for Im(z)≤ 0.

Hence ω_s+1,a,γ ∈ P and ωs+1,a,γ(z) = 0 for Im(z) ≤ 0. This completes the proof

of (i).

Proof of Theorem 2.1(ii). Let (s, a) ∈ G−. By the last assertion of Lemma 3.4, its part (iii) remains valid for s < 0. Hence 2d(ω_s,a,γ) = 2γ − 1 ≥ 0, so,

condition (8) is satisfied.

To prove that all zeros of E_s,a lie in the open right half-plane, we use Hardy’s formula (23) and P´olya’s Theorem (quoted in the previous part of the proof) once again. In the case under consideration, the function f in (32) has negative

derivative on (0, 1).

5. Real zeros of

E

Proof of Theorem 2.5. For s < 0, the assertion readily follows from (20)

be-cause the integrand in the right hand side is positive for realz.

For non-integer s > 0 we divide the proof into two steps.

Step 1. We prove that for k < s < k + 1, k = 0, 1, 2, . . . , the number of all real

zeros is less than or equal to k + 1.

We rewrite the representation (19) in the form (33) E_s,a(z) = e z Γ(−s)ga,s(z), where (34) g_a,s(z) =  _∞ 0 

exp(z(e−u− 1) − au) − k  j=0 (−1)jTj(z)u j j!  du us+1.

Differentiating k + 1 times, we obtain dk+1

dzk+1ga,s(z) =

 _∞

0

(e−u− 1)k+1exp(z(e−u− 1) − au) du us+1.

For real values of z the integrand has a constant sign. Hence, the (k + 1)-th

derivative of g_a,s does not vanish on the real axis. Rolle’s Theorem implies that

g_a,s has at most k + 1 real zeros. By (33) the function E_s,a(z) also has at most k + 1 real zeros.

Step 2. Let us prove that for k < s < k + 1, k = 0, 1, 2, . . . , the number of

negative zeros of E_s,a, is greater than or equal to k + 1. We use induction with

Letk = 0, 0 < s < 1. Then, according to (34), we have g_a,s(z) =

 _∞

0

[exp(z(e−u− 1) − au) − 1] du us+1.

Evidently, g_a,s(x) steadily decreases along the real line. Since E_s,a(0) = as > 0,

we have g_a,s(0) =asΓ(−s) < 0. Noting that ga,s(x)→ +∞ as x → −∞, we see

that E_s,a(z) has just one real negative zero.

Now, assume that for some k ≥ 0, k < s < k + 1, the function Es,a(z) has m≥ k + 1 negative zeros xm < x_m−1 <· · · < x1 < 0. Define

f (x) := (−x)aE_s,a(x), x≤ 0.

This function has the same zeros on (−∞, 0) as E_s,a(x). By Rolle’s Theo-rem f has at least m − 1 zeros in (x_m, x1). Moreover, since f (0) = 0 and

by Lemma 3.4(ii) we have lim_x→−∞f (x) = 0, the function f also has at least one zero in each of the intervals (x1, 0) and (−∞, x_m). Thus, f has at least

m + 1 negative zeros. The functional equation (13) implies

(−x)a−1E_s+1,a(x) = −[(−x)aE_s,a(x)] =−f(x).

ThereforeE_s+1,a(x) has at least m + 1 ≥ k + 2 negative zeros.

For k < s < k + 1, k = 0, 1, 2, . . . , the desired assertion follows immediately

from the results of steps 1 and 2. For s = k + 1 the assertion was proved in the

previous section.

6. Limiting case

α

→ +0, zeros of the function E

The proofs of Theorems 2.6–2.10 are quite similar to those of the previous theo-rems, and even a bit simpler. Therefore we omit the proofs and restrict ourselves to statements of analogs of Lemmata 3.1–3.4.

Lemma 3.1. The function E_s defined by (6) is an entire function of exponential type 1. It satisfies the functional equation

E_s+1(z) = zE_s(z).

Define entire functions Q_k, k = 0, 1, 2, . . . , by

exp(ze−u) = ∞  k=0 (−1)kQ_k(z)u k k!.

Lemma 3.2. The functions Q_k satisfy the recurrence relation Q_k+1(z) = zQ_k(z), k = 0, 1, 2, . . . and the initial condition Q0(z) = ez. Moreover,

Q_k(z) = E_k(z) = R_k(z)ez, k = 1, 2, . . . , holds, where R_k is a polynomial of degree k and R1(z) = z.

Note that in contrast to the case a > 0, we now have Q0(z)= E0(z) = ez− 1. Lemma 3.3. For k < s < k + 1, k = 0, 1, 2, . . . , the following representation holds E_s(z) = 1 Γ(−s)  _∞ 0  exp(ze−u)− ∞  j=0 (−1)jQj(z) j! u j  du us+1. For s < 0 we have the representation

E_s(z) = 1

Γ(−s)  _∞

0

(exp(ze−u)− 1)u−s−1du.

The last representation (with change variable v = e−u) had been obtained by Hardy ([1, p. 427]).

Lemma 3.3implies that Lemma 3.4 remains true withE_s,areplaced byE_sexcept assertion (ii) which is replaced by

E_s(z) = O((log |z|)−s), z → ∞, Re(z) ≤ b, for any b ∈ R.

Note that the last estimate is an immediate corollary of Hardy’s asymptotic formula for E_s ([1, p. 428]).

Acknowledgement. I express my deep gratitude to Moshe Newman (Jerusalem) for drawing my attention to [1] and sending me the conjecture confirmed by The-orem 2.5.

My great thanks to the copy editor for his suggestions which improved the pre-sentation of the paper.

References

1. G. H. Hardy, On the zeros of certain class of integral Taylor series II, Proc. London Math. Soc. (2) 2 (1905), 401–431.

2. B. Ya. Levin, Lectures on Entire Functions, American Mathematical Society, Providence, RI, 1996.

3. _{, Distribution of Zeros of Entire Functions, American Mathematical Society,} Prov-idence, RI, 1980.

4. G. P´olya and G. Szeg¨_{o, Problems and Theorems in Analysis II, Springer, Berlin, 1998.} 5. E. T. Whittaker and G. N. Watson, A Course of Modern Analysis, Cambridge University

Press, Cambridge, 1962.

Iossif V. Ostrovskii E-mail: [email protected]

Address: Department of Mathematics, Bilkent University, 06800 Bilkent, Ankara, Turkey; In-stitute for Low Temperature Physics and Engineering, 47 Lenin ave, 61103, Kharkov, Ukraine.