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ASYMPTOTIC BEHAVIOUR AND ZEROS OF THE BERNOULLI
POLYNOMIALS OF THE SECOND KIND
FRANTISEK STAMPACH
Abstract. The main aim of this article is a careful investigation of the asymptotic
behavior of zeros of Bernoulli polynomials of the second kind. It is shown that the zerosare all real and simple. The asymptotic expansions for the small, large, and the middle
zeros are computed in more detail. The analysis is based on the asymptotic expansions
of the Bernoulli polynomials of the second kind in various regimes.
1. Introduction
The Bernoulli polynomials of the second kind bn are defined by the generating function∞∑n=0
bn(x)tn
n!=
t
ln(1 + t)(1 + t)x, |t| < 1. (1)
Up to a shift, they coincide with the generalized Bernoulli polynomials of order n, precisely,
bn(x) = B(n)n (x+ 1), (2)
where the generalized Bernoulli polynomials of order a ∈ C are defined by the generatingfunction
∞∑n=0
B(a)n (x)
tn
n!=
(t
et − 1
)aext, |t| < 2π. (3)
Among the numerous branches of mathematics where the generalized Bernoulli polynomialsplay a significant role, we emphasize the theory of finite integration and interpolation whichis nicely exposed in the classical Norlund’s treatise [21] (unfortunately still not translated toEnglish).
In contrast to the Bernoulli polynomials of the first kind Bn ≡ B(1)n , the second kind
Bernoulli polynomials appear less frequently. Still a great account of the research focuses onthe study of their various generalizations and their combinatorial, algebraic, and analyticalproperties; let us mention at least [3, 12, 13, 15, 23, 25, 26]. Concerning the Bernoullipolynomials of the first kind Bn, a significant part of the research is devoted to the study ofthe asymptotic properties of their zeros that exhibit a fascinating complexity and structure;the asymptotic zero distribution of Bn, for n → ∞, was obtained in [1], for the results onthe asymptotic behavior of real zeros of Bn for n large, see [5, 6, 11, 14, 28], the proof thatthe zeros of Bn are actually all simple is in [2, 8], and further results on the number of realzeros of Bn can be found in [10, 16].
On the other hand, it seems that the asymptotic behavior of zeros of the Bernoulli poly-nomials of the second kind has not been studied yet. This contrast was a motivation forthe present work whose main goal is to fill this gap. To this end, let us mention that the
asymptotic behavior of the polynomials B(a)n , for n large and the order a not depending on n,
was studied in [17, 18].
Throughout the paper, we prefer to work with the polynomials B(n)n rather than the
Bernoulli polynomials of the second kind bn themselves which, however, makes no differencedue to the simple relation (2). First, we prove that, unlike zeros of the Bernoulli polyno-mials of the first kind, the zeros of the second kind Bernoulli polynomials can be only real.
Date: February 4, 2019.
2010 Mathematics Subject Classification. 11B68, 30C15, 41A60.Key words and phrases. Bernoulli polynomials of the second kind, asymptotic behavior, zeros, integral
representation.
1
2 FRANTISEK STAMPACH
Moreover, all zeros are simple, located in the interval [0, n], and interlace with the integers1, 2, . . . n (Theorem 1).
Next, we focus on the small zeros of B(n)n , i.e., the zeros that are located in a fixed distance
from the origin. The proof of their asymptotic behavior (Theorem 10) is based on a complete
local uniform asymptotic formula for B(n)n (Theorem 6). It turns out that the zeros of B
(n)n
are distributed symmetrically around the point n/2 and hence we obtain also the asymptoticformulas for the large zeros, i.e., the zeros located in a fixed distance from n.
Further, the asymptotic behavior of zeros of B(n)n located in a fixed distance from a
point αn, where α ∈ (0, 1), is examined. The analysis uses an interesting integral represen-
tation of B(n)n which can be a formula of independent interest (Theorem 13). With the aid of
Laplace’s method, the leading term of the asymptotic expansion of B(n)n (z+αn), as n→∞,
is deduced (Theorem 18) and limit formulas for the zeros located around αn then follow(Corollary 20). A particular attention is paid to the middle zeros, i.e., the case α = 1/2.In this case, more detailed results are obtained. First, a complete local uniform asymptotic
expansion for B(n)n (z + n/2), as n → ∞, is derived (Theorem 23). As a consequence, we
obtain several terms in the asymptotic expansion of the middle zeros (Theorem 25).
The asymptotic formulas for B(n)n that are used to analyze the zeros can be viewed as the
asymptotic expansion of the scaled polynomials B(n)n (nz) in the oscilatory region z ∈ (0, 1)
and close to the edge points z = 0 and z = 1. To complete the picture, the leading term
of the asymptotic expansion of B(n)n (nz) in the non-oscilatory regime or the zero-free region
z ∈ C \ [0, 1] is derived (Theorem 32) using the saddle-point method. Finally, we formulateseveral open research problems in the end of the paper.
2. Asymptotic behavior and zeros of the Bernoulli polynomials of thesecond kind
First, we recall several basic properties of the polynomials B(a)n that are used in this
section. For n ∈ N and x, a ∈ C, the identities
B(a)n (x+ 1) = B(a)
n (x) + nB(a−1)n−1 (x),
∂
∂xB(a)n (x) = nB
(a)n−1(x) (4)
and
B(a)n (−x) = (−1)nB(a)
n (x+ a) (5)
can be derived readily from (3), see also [23, Chp. 4, Sec. 2]. Next, by making use of thespecial value
B(n)n−1(x) =
n−1∏k=1
(x− k) (6)
together with the identities from (4), one easily deduces the well-known integral formula
B(n)n (x) =
∫ 1
0
n∏k=1
(x+ y − k) dy. (7)
2.1. Real and simple zeros. An elementary proof of the reality and simplicity of the zerosof the Bernoulli polynomials of the second kind is based on an inspection of the integer valuesof these polynomials.
Theorem 1. Zeros of B(n)n are real and simple. In addition, if x
(n)1 < x
(n)2 < . . . x
(n)n denotes
the zeros of B(n)n , then
k − 1 < x(n)k < k, (8)
for 1 ≤ k ≤ n. Further, the zeros of B(n)n are distributed symmetrically around the value
n/2, i.e.,
x(n)k = n− x(n)
n−k+1, (9)
for 1 ≤ k ≤ n. In particular, x(2n−1)n = n− 1/2.
3
Proof. We start with the integral representation (7) which implies that
B(n)n (k) =
∫ 1
0
n∏j=1
(y + k − j) dy = (−1)n+k
∫ 1
0
[k−1∏i=0
(y + i)
]n−k∏j=1
(j − y)
dy.
for any integer 0 ≤ k ≤ n. Since each factor in the above integral is positive for y ∈ (0, 1),the whole integral has to be positive and therefore
(−1)n+kB(n)n (k) > 0, (10)
for 0 ≤ k ≤ n.
Consequently, the signs of the values B(n)n (k) alternate for 0 ≤ k ≤ n and hence there has
to be at least one root of B(n)n in each interval (k−1, k) for 1 ≤ k ≤ n. Since the polynomial
B(n)n is of degree n, there has to be exactly one zero in each interval (k−1, k), for 1 ≤ k ≤ n.
Thus, B(n)n has n distinct roots located in the intervals (k − 1, k), 1 ≤ k ≤ n. These roots
are necessarily simple.The symmetry of the distribution of the roots around n/2 follows readily from the iden-
tity (5) which implies
B(n)n
(n2− x)
= (−1)nB(n)n
(n2
+ x), (11)
for n ∈ N0.
According to the Gauss–Lucas theorem, the zeros of
B(n)k (x) =
k!
n!
dn−k
dxn−kB(n)n (x),
where 0 ≤ k ≤ n, are located in the convex hull of the zeros of B(n)n which is, by Theorem 1,
a subset of the interval (0, n).
Corollary 2. For all n ∈ N0 and 0 ≤ k ≤ n, the zeros of B(n)k are located in (0, n).
2.2. The asymptotic expansion of B(n)n (z) and small and large zeros. First, we derive
a complete locally uniform asymptotic expansion of B(n)n in negative powers of log n. As an
application, this expansion allows us to derive asymptotic formulas for the zeros of B(n)n that
are located in a fixed distance from the origin or the point n for n large.
In the proof of the asymptotic expansion of B(n)n we will make use of a particular case
of Watson’s lemma given below; for the more general version and its proof, see, e.g., [22,Chp. 3, Thm. 3.1] or [29, Sec. I.5].
Lemma 3 (Watson). Let f(t) be a function of positive variable t, such that
f(t) ∼∞∑m=0
amtm+λ−1, as t→ 0+, (12)
where λ > 0. Then one has the complete asymptotic expansion∫ ∞0
e−xtf(t)dt ∼∞∑m=0
Γ(m+ λ)amxm+λ
, as x→∞, (13)
provided that the integral converges absolutely for all sufficiently large x.
Remark 4. If additionally the coefficients am = am(ξ) in (12) depend continuously on aparameter ξ ∈ K where K is a compact subset of C and the asymptotic expansion (12)is uniform in ξ ∈ K, then the expansion (13) holds uniformly in ξ ∈ K provided that theintegral converges uniformly in ξ ∈ K for all sufficiently large x. This variant of Watson’slemma can be easily verified by a slight modification of the proof given, for example, in [22,Chp. 3, Thm. 3.1].
Yet another auxiliary statement, this time an inequality for the Gamma function with acomplex argument, will be needed to obtain the desired asymptotic expansion of Bnn(z) forn→∞.
4 FRANTISEK STAMPACH
Lemma 5. For all s ∈ [0, 1] and z ∈ C such that Re z > 1, it holds∣∣∣∣zsΓ(z − s)Γ(z)
− 1
∣∣∣∣ ≤ 2s
Re z − 1.
Proof. Let fz(s) := zsΓ(z − s), Re z > 1, and s ∈ [0, 1]. Then, by the Lagrange theorem,
|fz(s)− fz(0)| ≤ |f ′z(s∗)|s, (14)
for some s∗ ∈ (0, s). The differentiation of fz(s) with respect to s yields
f ′z(s) = zsΓ(z − s) (log z − ψ(z − s)), (15)
where ψ = Γ′/Γ is the Digamma function. Recall that [9, Eq. 5.9.13]
ψ(z) = log z +
∫ ∞0
(1
t− 1
1− e−t
)e−tzdt,
for Re z > 0. From the above formula, one deduces that, for Re z > 1 and s ∈ [0, 1],
|ψ(z − s)− log(z − s)| ≤∫ ∞
0
e−t(Re z−1)dt =1
Re z − 1,
where we used that ∣∣∣∣1t − 1
1− e−t
∣∣∣∣ < 1, ∀t > 0.
Taking also into account that [9, Eq. 5.6.8]∣∣∣∣Γ(z − s)Γ(z)
∣∣∣∣ ≤ 1
|z|s
and
| log(z − s)− log(z)| ≤∞∑n=1
1
n
sn
|z|n≤ 1
|z| − 1≤ 1
Re z − 1,
for Re z > 1 and s ∈ [0, 1], we get from (15) the estimate∣∣∣∣f ′z(s)Γ(z)
∣∣∣∣ ≤ 2
Re z − 1.
The above inequality together with (14) yields the statement.
Now, we are at the position to deduce the complete asymptotic expansion of the polyno-
mials B(n)n (z) for n→∞.
Theorem 6. The asymptotic expansion
(−1)nnz
n!B(n)n (z) ∼
∞∑k=0
ck(z)
logk+1 n, n→∞, (16)
holds locally uniformly in z ∈ C, where
ck(z) =dk
dzk1
Γ(1− z). (17)
Proof. The integral formula (7) can be rewritten in terms of the Gamma functions as
B(n)n (z) = (−1)n
∫ 1
0
Γ(1− s− z + n)
Γ(1− s− z)ds.
Using this together with Lemma 5, we obtain∣∣∣∣∣ B(n)n (z)
Γ(1− z + n)− (−1)n
∫ 1
0
1
(1− z + n)sΓ(1− s− z)ds
∣∣∣∣∣≤ 2
n− Re z
∫ 1
0
s
|(1− z + n)sΓ(1− s− z)|ds,
for Re z < n. Since 1/Γ is an entire function, for K ⊂ C a compact set, there is a constantC > 0 such that
sups∈[0,1]
supz∈K
1
|Γ(1− z − s)|≤ C.
5
Hence∣∣∣∣∣ B(n)n (z)
Γ(1− z + n)− (−1)n
∫ 1
0
ds
(1− z + n)sΓ(1− s− z)
∣∣∣∣∣ ≤ 2C
n− Re z
∫ 1
0
sds
|1− z + n|s
=2C
n− Re z
|1− z + n| − log |1− z + n| − 1
|1− z + n| log2 |1− z + n|≤ 2C
n− Re z
1
log2 |1− z + n|,
for Re z < n. In total, we conclude that
B(n)n (z)
Γ(1− z + n)= (−1)n
∫ 1
0
ds
(1− z + n)sΓ(1− s− z)+O
(1
n log2 n
)= (−1)n
∫ 1
0
ds
nsΓ(1− s− z)+O
(1
n
), (18)
as n→∞, locally uniformly in z ∈ C.The analyticity of the reciprocal Gamma function implies
1
Γ(1− s− z)=
∞∑k=0
ck(z)sk
k!,
where
ck(z) =dk
dsk
∣∣∣∣s=0
1
Γ(1− s− z)=
dk
dzk1
Γ(1− z).
Moreover, ck is an entire function for any k ∈ N0. Consequently, if χ(0,1) denotes theindicator function of the interval (0, 1), then
fz(s) :=χ(0,1)(s)
Γ(1− s− z)∼∞∑k=0
ck(z)sk
k!, s→ 0+,
and the application of Lemma 3 yields∫ 1
0
ds
nsΓ(1− s− z)=
∫ ∞0
e−s lognfz(s)ds ∼∞∑k=0
ck(z)
logk+1 n, n→∞. (19)
This asymptotic formula is again local uniform in z ∈ C. To see this, one may proceed asfollows. For m ∈ N0, we have∣∣∣∣∣ 1
Γ(1− s− z)−
m∑k=0
ck(z)sk
k!
∣∣∣∣∣ ≤∣∣∣∣ dk
dsk
∣∣∣∣s=s∗
1
Γ(1− s− z)
∣∣∣∣ sm+1
(m+ 1)!= |cm+1(z + s∗)| sm+1
(m+ 1)!,
where s∗ ∈ (0, s). Now, if K ⊂ C is compact, then, by the analyticity of cm+1, there is aconstant C ′ > 0 such that
sups∈[0,1]
supz∈K|cm+1(z + s)| ≤ C ′.
Consequently, the error term in the expansion (19) is majorized by
C ′
(m+ 1)!
∫ ∞0
sm+1
nsds =
C ′
logm+2 n,
for all z ∈ K; cf. also Remark 4. Thus, (18) together with (19) imply that
B(n)n (z)
Γ(1− z + n)∼ (−1)n
∞∑k=0
ck(z)
logk+1 n, n→∞, (20)
locally uniformly in z ∈ C.Finally, it follows from the well-known Stirling asymptotic expansion that
1
Γ(1− z + n)=nz
n!
(1 +O
(1
n
)), n→∞,
locally uniformly in z ∈ C. By applying the above formula in (20), we arrive at the statement.
6 FRANTISEK STAMPACH
Remark 7. By putting z = 1 in (16), one gets the complete asymptotic expansion of the
Bernoulli numbers of the second kind B(n)n (1)/n! which was obtained by Van Veen in [27]
and recently rediscovered by Nemes [19].
By using Theorem 6, we immediately get the following corollary, where the second state-ment follows from the Hurwitz theorem, see [4, Thm. 2.5, p. 152], and the fact that the zerosof 1/Γ are located at non-positive integers and are simple.
Corollary 8. It holds
limn→∞
(−1)nnz log n
n!B(n)n (z) =
1
Γ(1− z)uniformly in compact subsets of C. Consequently, for k ∈ N, one has
limn→∞
x(n)k = k.
Remark 9. Both sequences above converge quite slowly. The error terms turn out to decayas 1/ log n, for n→∞.
Since we known all the coefficients in the asymptotic expansion (16) by any power of
1/ log n, we can compute also the asymptotic expansions of zeros x(n)k for any fixed k ∈ N,
as n → ∞, to an arbitrary order, in principle. However, the coefficients by the powers of1/ log n become quickly complicated and no closed formula for the coefficients was found.We provide the first three terms of the asymptotic expansions.
Theorem 10. For k ∈ N, we have the asymptotic expansion
x(n)k = k − 1
log n− ψ(k)
log2 n+O
(1
log3 n
), as n→∞,
where ψ = Γ′/Γ is the Digamma function.
Remark 11. Theorem 10 gives the asymptotic behavior of zeros located in a fixed distancefrom 0. As a consequence of the symmetry (9), we know also the asymptotic behavior of thezeros located in a fixed distance from n. Namely,
x(n)n−k+1 = n− k +
1
log n+
ψ(k)
log2 n+O
(1
log3 n
), as n→∞,
for any fixed k ∈ N.
Proof. We fix k ∈ N and introduce ε(k)n := k− x(n)
k . If we substitute for z = k− ε(k)n in (17),
we get
cj
(k − ε(k)
n
)= (−1)j
dj
dzj
∣∣∣∣z=ε
(k)n
1
Γ(1− k + z)= (−1)j
dj
dzj
∣∣∣∣z=ε
(k)n
(z + 1− k)kΓ(z + 1)
, (21)
where (a)k := a(a+ 1) . . . (a+ k − 1) is the Pochhammer symbol.Recall that, for k ∈ N,
(z + 1− k)k =
k∑l=1
s(k, l)zl, (22)
where s(k, l) are the Stirling numbers of the first kind; see [9, Sec. 26.8]. The Stirling numberscan be defined recursively but here we will make use only of the special values
s(k, 1) = (−1)k−1(k − 1)! and s(k, 2) = (−1)k−1(k − 1)! (γ + ψ(k)), (23)
for k ∈ N. We will also need the Maclaurin series for the reciprocal Gamma function [9,Eqs. (5.7.1) and (5.7.2)]
1
Γ(z + 1)=
∞∑m=0
πmzm, (24)
where
π0 = 1 and mπm = γπm−1 +
m∑j=2
(−1)j+1ζ(j)πm−j , for m ∈ N,
7
and ζ stands for the Riemann zeta function. In particular,
π0 = 1 and π1 = γ. (25)
By using (22) and (24) in (21), we obtain
cj
(k − ε(k)
n
)= (−1)jj!
∞∑m=j
(m
j
)X(k)m
(ε(k)n
)m−j, (26)
where
X(k)m :=
min(k,m)∑i=1
s(k, i)πm−i. (27)
It follows from Theorem 6 that ε(k)n = O(1/ log n). Therefore we may write
ε(k)n =
µ(k)n
log n, (28)
where µ(k)n is a bounded sequence. If we take the first two terms from the expansion (16)
and use that the left-hand side of (16) vanishes at z = x(n)k , we arrive at the equation
c0
(k − ε(k)
n
)log n+ c1
(k − ε(k)
n
)+O
(1
log n
)= 0, as n→∞.
With the aid of (26) and (28), the above equation can be written as
X(k)1 µ(k)
n −X(k)1 +O
(1
log n
)= 0, as n→∞,
which implies
µ(k)n = 1 +O
(1
log n
), as n→∞,
since X(k)1 6= 0. Consequently, we get
x(n)k = k − ε(k)
n = k − 1
log n+O
(1
log2 n
), as n→∞.
Similarly, by writing
ε(k)n =
1
log n+
ν(k)n
log2 n, (29)
where ν(k)n is a bounded sequence, repeating the same procedure that uses the first three
terms of the asymptotic expansion (16), we compute another terms in the expansion of ε(k)n ,
for n→∞. More precisely, it follows from (16) that
c0
(k − ε(k)
n
)log2 n+ c1
(k − ε(k)
n
)log n+ c2
(k − ε(k)
n
)+O
(1
log n
)= 0, as n→∞.
By using (26) and (29), the above equation implies that
X(k)1 ν(k)
n +X(k)2 +O
(1
log n
)= 0, as n→∞.
Since
X(k)1 = (−1)k−1(k − 1)! and X
(k)2 = (−1)k(k − 1)!ψ(k)
as one computes from (27) by using the special values (23), (25), and the identity ψ(1) = −γ,see [9, Eq. (5.4.12)], we conclude that
ν(k)n = ψ(k) +O
(1
log n
), as n→∞.
The above formula used in (29) imply the asymptotic expansion of x(n)k from the statement.
8 FRANTISEK STAMPACH
Remark 12. Note that for k = 1, the coefficients (27) simplify. Namely, X(1)0 = 0 and
X(1)m = πm−1, for m ∈ N. Without going into details, we write down several other terms in
the asymptotic expansion of the smallest zero:
x(n)1 = 1− 1
log n+
γ
log2 n− γ2 − π2/6
log3 n+γ3 − γπ2/2 + 3ζ(3)
log4 n
− γ4 − γ2π2 + 12γζ(3)− π4/90
log5 n+O
(1
log6 n
), as n→∞.
2.3. Another integral representation and more precise localization of zeros. Fur-ther asymptotic analysis as well as an improvement of the localization (8) will rely on an
integral representation for the polynomials B(n)n derived below. This integral representation,
however, can be a formula of an independent interest. All fractional powers appearing belowhave their principal values.
Theorem 13. One has
B(n)n (z) = (−1)n
n!
π
∫ ∞0
uz−1
(1 + u)nπ cos(πz)− log(u) sin(πz)
π2 + log2 udu, (30)
for n ∈ N and 0 < Re z < n.
Proof. We again starts with the integral formula (7) expressing the integrand as a ratio ofthe Gamma functions:
B(n)n (z) =
∫ 1
0
n∏j=1
(s+ z − j)
ds =
∫ 1
0
Γ(s+ z)
Γ(s+ z − n)ds.
By using the well-know identity
Γ(z)Γ(1− z) =π
sinπz,
we may rewrite the integral formula for B(n)n (z) as
B(n)n (z) =
1
π
∫ 1
0
sin (π(s+ z)− πn) Γ (s+ z) Γ (1− s− z + n) ds. (31)
Recall that for u, v ∈ C, Reu > 0, Re v > 0, it holds
Γ(u)Γ(v)
Γ(u+ v)= 2
∫ π/2
0
sin2u−1 θ cos2v−1 θdθ,
see [9, Eqs. 5.12.2 and 5.12.1]. Using the above formula in (31), we get
B(n)n (z) = (−1)n
2n!
π
∫ 1
0
sin (π(s+ z))
∫ π/2
0
sin2s+2z−1 θ cos2n−2s−2z+1 θdθds, (32)
where z has to be restricted such that 0 < Re z < n to guarantee that both arguments ofthe Gamma functions in (31) are of positive real part. By using Fubini’s theorem, we maychange the order of integration in (32). Doing also some elementary manipulations with thetrigonometric functions, we arrive at the expression
B(n)n (z) = (−1)n
2n!
π
∫ π/2
0
cos2n(θ) tan2z θ
∫ 1
0
sin (π(s+ z)) tan2s−1 θdsdθ, (33)
for 0 < Re z < n.As the last step, we evaluate the inner integral in (33). We can make use of the elementary
integral ∫eax sin(bx)dx =
eax
a2 + b2(a sin bx− b cos bx),
to compute that∫ 1
0
sinπ(s+ z) tan2s−1 θds =π cosπz − 2 log(tan θ) sinπz
sin θ cos θ(π2 + 4 (log tan θ)
2) , (34)
9
for θ ∈ (0, π/2). By using (34) in (33), we arrive at the integral representation
B(n)n (z) = 2(−1)n
n!
π
∫ π/2
0
cos2n−2(θ) tan2z−1(θ)π cos(πz)− 2 log(tan θ) sin(πz)
π2 + 4 log2(tan θ)dθ, (35)
for n ∈ N and 0 < Re z < n. Substituting for tan2 θ = u in (35), we arrive at the formula (30).
As a first application of the integral formula (30), we improve the localization of the zeros
of B(n)n given by the inequalities (8). It turns out that the zeros are located in a half of the
respective intervals between two integers.To do so, we rewrite the integral in (30) to a slightly different form. By writing the integral
in (30) as the sum of two integrals integrating from 0 to 1 and from 1 to ∞, respectively,substituting u = 1/u in the second one (and omitting the tilde notation afterwards), oneobtains the formula
B(n)n (z) = (−1)n
n!
π
∫ 1
0
1
(1 + u)n[ρz(u) + unρ−z(u)] du, (36)
for n ∈ N and 0 < Re z < n, where
ρz(u) := uz−1π cos(πz)− log(u) sin(πz)
π2 + log2 u.
Below, bxc denotes the integer part of x ∈ R.
Theorem 14. For n ∈ N, one has
k − 1
2< x
(n)k < k, if 1 ≤ k ≤
⌊n
2
⌋,
and
k < x(n)k+1 < k +
1
2, if
⌊n+ 1
2
⌋≤ k ≤ n− 1.
Recall that x(2n−1)n = n− 1/2.
Remark 15. The global localization of the zeros in the intervals of fixed lengths given above
is the best possible. Indeed, as it is shown below, the zeros of B(n)n located around n/2 cluster
at half-integers as n→∞, while the zeros located in a left neighborhood of the point n (orin a right neighborhood of 0) cluster at integers as n→∞.
Proof. Clearly, if the first set of inequalities is established then the second one follows readilyfrom the symmetry (9).
For 1 ≤ k ≤ bn/2c, write z = k − 1/2 in (36). Since
ρk−1/2(u) = −u2k−1ρ−k+1/2(u) = (−1)kuk−3/2 log u
π2 + log2 u,
we have
B(n)n
(k − 1
2
)= (−1)n+k n!
π
∫ 1
0
uk−3/2(1− un−2k+1
)(1 + u)n
log u
π2 + log2 udu
The integrand is obviously a negative function on (0, 1) for any 1 ≤ k ≤ bn/2c. Hence
(−1)n+k+1B(n)n
(k − 1
2
)> 0, for 1 ≤ k ≤
⌊n
2
⌋.
Taking also (10) into account, we observe that the values B(n)n (k) and B
(n)n (k − 1/2) differ
in sign for 1 ≤ k ≤ bn/2c. Hence
x(n)k ∈
(k − 1
2, k
), for 1 ≤ k ≤
⌊n
2
⌋.
10 FRANTISEK STAMPACH
2.4. Asymptotic expansion of B(n)n (z + αn) and consequences for zeros. Theorem 10
and Remark 11 give an information on the asymptotic behavior of the small or large zeros ofthe Bernoulli polynomials of the second kind. Bearing in mind the symmetry (11), the zeros
of B(n)n located “in the middle”, i.e., around the point n/2 are of interest, too. More generally,
we will study the asymptotic behavior of zeros of B(n)n that are traced along the positive
real line at a speed α ∈ (0, 1) by investigating the asymptotic behavior of B(n)n (z + αn), for
n→∞, focusing particularly on the case α = 1/2.
In order to derive the asymptotic behavior B(n)n (z + αn) for n large, we apply Laplace’s
method to the integral representation obtained in Theorem 13. We refer reader to [22,Sec. 3.7] for a general description of Laplace’s method. Here we use a particular case ofLaplace’s method adjusted to the situation which appears below. In particular, we need thevariant where the extreme point is an inner point of the integration interval which is an easymodification of the standard form of Laplace’s method where the extreme point is assumedto be one of the endpoints of the integration interval.
Lemma 16 (Laplace’s method with an interior extreme point). Let f be real-valued andg complex-valued continuous functions on (0,∞) independent of n. Assume further that fand g are analytic functions at a point a ∈ (0,∞) where f has a unique global minimum in(0,∞). Let fk and gk be the coefficients from the Taylor series
f(x) = f(a) +
∞∑k=0
fk(x− a)k+2 and g(x) =
∞∑k=0
gk(x− a)k+`, (37)
where ` ∈ 0, 1 and g0 6= 0. Suppose moreover that f0 6= 0. Then, for n→∞, one has∫ ∞0
e−nf(u)g(u)du =2√π
`+ 1e−nf(a)
[c`
n`+1/2+O
(1
n`+3/2
)](38)
provided that the integral converges absolutely for all n sufficiently large. The coefficients ckare expressible in terms of fk and gk as follows:
c0 =g0
2f1/20
and c1 =2f0g1 − 3f1g0
4f5/20
.
Remark 17. Suppose that the coefficients gk = gk(ξ) in (37) depend continuously on anadditional parameter ξ ∈ K where K is a compact subset of C and the power series for gin (37) converges uniformly in ξ ∈ K. Then the asymptotic expansion (38) holds uniformlyin ξ ∈ K as well provided that the integral converges uniformly in ξ ∈ K for all n sufficientlylarge.
Now, we are ready to deduce an asymptotic expansion of B(n)n (z + αn) for n→∞.
Theorem 18. For α ∈ (0, 1), the asymptotic expansion
(−1)n√n
n!ααn(1− α)(1−α)nB(n)n (z + αn) =
√2
π
αz−1/2(1− α)−z−1/2
π2 + τ2α
× [π cos(πz + παn)− τα sin(πz + παn)] +O
(1√n
)(39)
holds locally uniformly in z ∈ C as n→∞, where
τα := log
(α
1− α
). (40)
Proof. By writing z + αn instead of z in (30), one obtains
(−1)nπ
n!B(n)n (z + αn) = I1(n) cos(πz + παn)− I2(n) sin(πz + παn), (41)
for −αn < Re z < (1− α)n, where
Ii(n) :=
∫ ∞0
e−nf(u)gi(u)du, i ∈ 1, 2, (42)
11
and
f(u) := log(1 + u)− α log u, (43)
g1(u) :=πuz−1
π2 + log2(u)and g2(u) :=
uz−1 log u
π2 + log2(u). (44)
The integrals (42) are in the suitable form for the application of Laplace’s method.One easily verifies that the function f defined by (43) has the simple global minimum at
the point
uα :=α
1− α.
Further, the functions f and g1, g2 from (44) are analytic in a neighborhood of uα havingthe expansions
f(u) = f(uα) +(1− α)3
2α(u− uα)2 +O
((u− uα)3
),
and
gi(u) = gi(uα) +O (u− uα) , (45)
for u→ uα. Moreover, the expansions for gi in (45) are local uniform in z ∈ C as one readilychecks by elementary means.
Suppose first that α 6= 1/2. Then g1(uα) 6= 0 as well as g2(uα) 6= 0 and Lemma 16 appliesto both I1(n) and I2(n) with ` = 0 resulting in the asymptotic formulas
I1(n) = ααn(1− α)(1−α)n
[√2π3αz−1/2(1− α)−z−1/2
π2 + log2 (α/(1− α))
1√n
+O
(1
n
)](46)
and
I2(n) = ααn(1− α)(1−α)n
[√2παz−1/2(1− α)−z−1/2
π2 + log2 (α/(1− α))log
(α
1− α
)1√n
+O
(1
n
)],
for n → ∞. By plugging the above expressions for Ii(n), i ∈ 1, 2, into (41) one gets theexpansion (39).
If α = 1/2, g2(uα) = 0 and hence Lemma 16 applies to I2(n) with ` = 1. It follows that2nI2(n) = O(n−3/2), as n → ∞, for α = 1/2. The asymptotic expansion (46) for I1(n)remains unchanged even if α = 1/2. In total, taking again (41) into account, we see that theexpansion (39) remains valid also for α = 1/2 since τα vanishes in this case.
In order to conclude that the expansion (39) is local uniform in z ∈ C, it suffices to checkthat the integrals Ii(n), i ∈ 1, 2, converge locally uniformly in z ∈ C for all n sufficientlylarge; see Remark 17. Let K ∈ C be a compact set. Then if z ∈ K, |Re z| < C for someC > 0. Concerning for instance I1(n), it holds that∫ ∞
0
∣∣∣∣uαn+z−1
(1 + u)n1
π2 + log2 u
∣∣∣∣ du ≤ ∫ 1
0
uαn−C−1du+
∫ ∞1
uαn+C−1
(1 + u)ndu.
For n sufficiently large, the first integral on the right-hand side above can be majorized by 1and the second integral converges at infinity because α < 1. Consequently, the integral I1(n)converges uniformly in z ∈ K for all n large enough. A similar reasoning shows that thesame is true for I2(n) which concludes the proof.
In the particular case when α = 1/2, Theorem 18 yields the following limit formulasthat can be compared with Dilcher’s limit formulas for the Bernoulli polynomials of the firstkind [7, Cor. 1].
Corollary 19. One has
limn→∞
(−1)n22n−1
√n
(2n)!β2n(z) =
cos(πz)
π3/2
and
limn→∞
(−1)n22n√n
(2n+ 1)!β2n+1(z) =
sin(πz)
π3/2
12 FRANTISEK STAMPACH
locally uniformly in C, where
βn(z) := B(n)n
(z +
n
2
).
We can combine Theorem 18 and the Hurwitz theorem [4, Thm. 2.5, p. 152] in order to
deduce the asymptotic behavior of the zeros of B(n)n located around the point αn for n large.
As it is seen from (39), the situation with general α is slightly different from the one with
α = 0 treated in Corollary 8 since the normalized function B(n)n (z + αn) does not converge
to any function of z for n → ∞ if α ∈ (0, 1). Nevertheless, passing to certain subsequenceswe get the desired result. Recall definition (40) of τα.
Corollary 20. Let α ∈ (0, 1) and nk be an increasing sequence of integers such that
limk→∞
(αnk − bαnkc) = ω, (47)
where bxc denotes the integer part of a real number x. Then, for any ` ∈ Z, one has
limk→∞
(x
(nk)bαnkc+` − αnk
)= `− 1− ω +
1
πarccot
(ταπ
).
Proof. It follows from the assumption (47) that
limk→∞
(−1)bαnkc cos(απnk) = cos(πω)
and
limk→∞
(−1)bαnkc sin(απnk) = sin(πω).
Thus, Theorem 18 tells us that
limk→∞
(−1)nk+bαnkc√nknk!ααnk(1− α)(1−α)nk
B(nk)nk
(z + αnk) = Cα(z) [π cos(π(z + ω))− τα sin(π(z + ω))],
where Cα(z) 6= 0 and the convergence is local uniform in z ∈ C. By the Hurwitz theorem,the zeros of the polynomial
z 7→ B(nk)nk
(z + αnk) (48)
cluster at the zeros of the function
z 7→ π cos(π(z + ω))− τα sin(π(z + ω))
which coincide with the solutions of the secular equation
cot(π(z + ω)) =ταπ. (49)
Further, one deduces from (8) that the zeros x(nk)bαnkc+` − αnk of (48) satisfy
bαnkc − αnk + `− 1 < x(nk)bαnkc+` − αnk < bαnkc − αnk + `.
Taking also (47) into account, one concludes that
limk→∞
(x
(nk)bαnkc+` − αnk
)= ζ`,
where ζ` is the unique solution of (49) such that ζ` ∈ (` − 1 − ω, ` − ω). Finally, it sufficesto note that ζ` = `− 1− ω + ζ, where ζ fulfills
cot(πζ) =ταπ
and ζ ∈ (0, 1).
Example 21. If we put α = 1/3, then τ1/3 = − log 2. Further, by taking the subsequencesnk = 3k, 3k + 1, 3k + 2, respectively, in Corollary 20, one obtains
limk→∞
(x
(3k)k+` − k
)= limk→∞
(x
(3k+1)k+` − k
)= limk→∞
(x
(3k+2)k+` − k
)= `− 1 +
1
πarccot
(− log 2
π
)≈ `− 1.430877,
for any ` ∈ Z.
13
By taking α = 1/2 and either nk = 2k or nk = 2k + 1 in Corollary 20, we get
limk→∞
(x
(2k)k+` − k
)= limk→∞
(x
(2k+1)k+` − k
)= `− 1
2,
for ` ∈ Z fixed. Thus, in contrast to the small or large zeros of B(n)n that cluster at integers as
shown in Theorem 10 and Remark 11, the zeros of B(n)n around the middle point n/2 cluster
at half-integers as n → ∞. Our next goal is to deduce more precise asymptotic expansions
for the middle zeros of B(n)n . To do so, we need to investigate the asymptotic behavior of
B(n)n (z + n/2), for n→∞, more closely. A complete asymptotic expansion will be obtained
by using the classical form of the Laplace method, see [22, Sec. 3.7], applied together withPerron’s formula for the expansion coefficients [29, p. 103] adjusted slightly to our needs.
Lemma 22 (Laplace’s method and Perron’s formula). Let f be real-valued and g complex-valued continous funtions on (0,∞) independent of n. Assume further that f and g areanalytic functions at the origin where f has a unique global minimum in (0,∞). Let theMaclaurin expansions of f and g are of the form
f(x) =
∞∑k=0
fkxk+2 and g(x) =
∞∑k=0
gkxk+`,
where ` ∈ N and f0 6= 0 as well as g0 6= 0. Then, for n→∞, one has∫ ∞0
e−nf(u)g(u)du ∼∞∑k=0
Γ
(k + `+ 1
2
)ck
n(k+`+1)/2
provided that the integral converges absolutely for all n sufficiently large. Perron’s formulafor the coefficients ck yields
ck =1
2k!
dk
dxk
∣∣∣∣x=0
g(x)xk+1
(f(x))(k+`+1)/2
, k ∈ N0.
Theorem 23. For n→∞, the complete asymptotic expansion
(−2)n√n
n!B(n)n
(z +
n
2
)∼ π1/2 cos
(πz +
πn
2
) ∞∑k=0
pk(z)
22kk!
1
nk
− π−1/2 sin(πz +
πn
2
) ∞∑k=0
qk(z)
22k+1k!
1
nk+1(50)
holds locally uniformly in z ∈ C. The coefficients pk and qk are polynomials given by theformulas
pk(z) =
k∑j=0
(2k
2j
)ω
(k)j z2k−2j and qk(z) =
k∑j=0
(2k + 1
2j
)ω
(k+1)j z2k+1−2j ,
where
ω(k)j =
d2j
dx2j
∣∣∣∣x=0
x2k+1
(π2 + x2) (log cosh(x/2))k+1/2
.
Remark 24. Several first coefficients pk and qk read
p0(z) =2√
2
π2, p1(z) =
2√
2
π4
(8π2z2 + π2 − 16
)p2(z) =
2√
2
π6
(64π4z4 + 16(5π2 − 48)π2z2 + π4 − 160π2 + 1536
),
and
q0(z) =16√
2z
π2, q1(z) =
16√
2z
π4
(8π2z2 + 5π2 − 48
)q2(z) =
16√
2z
3π6
(192π4z4 + 80(7π2 − 48)π2z2 + 91π4 − 3360π2 + 23040
).
14 FRANTISEK STAMPACH
Proof. The starting point is the equation (41) with α = 1/2:
(−1)nπ
n!B(n)n
(z +
n
2
)= I1(n) cos
(πz +
πn
2
)− I2(n) sin
(πz +
πn
2
), (51)
which holds true if |Re z| < n/2 and where
I1(n) =
∫ ∞0
(u1/2
1 + u
)nπuz−1du
π2 + log2 uand I2(n) =
∫ ∞0
(u1/2
1 + u
)nuz−1 log(u)du
π2 + log2 u.
First we split the integral I1(n) into two integrals integrating from 0 to 1 in the first oneand from 1 to ∞ in the second one. Next, we substitute for u = e−x in the first integral andu = ex in the second one. This results in the formula
I1(n) =π
2n−1
∫ ∞0
cosh(xz)
π2 + x2
dx
coshn(x/2). (52)
Similarly one shows that
I2(n) =1
2n−1
∫ ∞0
x sinh(xz)
π2 + x2
dx
coshn(x/2). (53)
To the integral in (52), we may apply Lemma 22 with
f(x) = log cosh(x
2
), g(x) =
cosh(xz)
π2 + x2,
and ` = 0 getting the expansion
I1(n) ∼ π
2n−1
∞∑k=0
Γ
(k + 1
2
)ck
n(k+1)/2, (54)
where
ck =1
2k!
dk
dxk
∣∣∣∣x=0
cosh(xz)
π2 + x2
xk+1
(log cosh(x/2))(k+1)/2
.
Notice that c2k−1 = 0 for k ∈ N. Next, by using the Leibnitz rule, one gets
c2k =1
2(2k)!
k∑j=0
(2k
2j
)d2k−2j
dx2k−2j
∣∣∣∣x=0
(cosh(xz))d2j
dx2j
∣∣∣∣x=0
x2k+1
(π2 + x2) (log cosh(x/2))k+1/2
which yields
c2k =1
2(2k)!
2k∑j=0
(2k
2j
)z2k−2jω
(k)j =
pk(z)
2(2k)!, (55)
where the notation from the statement has been used.By substituting from (55) and (54) in the equation (51), one arrives at the first asymptotic
series on the right-hand side of (50). In order to deduce the second expansion on the right-hand side of (50), one proceeds in a similar fashion applying Lemma 22 to the integral (53)this time with ` = 2. The local uniformity of the expansion can be justified using the analyticdependence of the integrands in (52) and (53) on z.
Theorem 23 allows to compute coefficients in the asymptotic expansion of the zeros of B(n)n
located in a fixed distance from n/2, for n→∞, similarly as it was done in Theorem 10 forthe small zeros based on the asymptotic expansion from Theorem 6. Since the proof of thestatement below is completely analogous to the proof of Theorem 10 with the only exceptionthat the asymptotic formula of Theorem 23 is used, it is omitted.
Theorem 25. For any k ∈ Z, one has
x(2n)n+k = n+ k − 1
2− 2k − 1
π2n− (2k − 1)(π2 − 12)
2π4n2+O
(1
n3
), n→∞,
and
x(2n+1)n+k+1 = n+ k +
1
2− 2k
π2n+
12k
π4n2+O
(1
n3
), n→∞.
15
Remark 26. Note the difference between the polynomial decay in the asymptotic expansion
of the middle zeros of B(n)n and the logarithmic decay in the asymptotic formulas for the
small and large zeros (Theorem 10 and Remark 11).
Remark 27. A more detailed expansions for the zeros located most closely to n/2 read
x(2n)n = n− 1
2+
1
π2n+π2 − 12
2π4n2+
3π4 − 100π2 + 720
12π6n3
+3π6 − 216π4 + 3856π2 − 20160
24π8n4+O
(1
n5
),
and
x(2n+1)n+2 = n+
3
2− 2
π2n+
12
π4n2− 3π4 − 40π2 + 720
6π6n3+
14(3π4 − 44π2 + 360)
3π8n4+O
(1
n5
),
for n→∞.
Remark 28. The numbers D(2n)2n := 4nB
(2n)2n (n) are known as the Norlund D-numbers and
appear in formulas for a numerical integration, see [21, Chp. 8, § 7]. Theorem 23 impliesthat
(−1)n√
2n
(2n)!D
(2n)2n ∼
√π
∞∑k=0
ω(k)k
23kk!
1
nk, n→∞.
Explicitly, the first three terms read
(−1)n√
2n
(2n)!D
(2n)2n =
2√
2
π3/2+
π2 − 16
2√
2π7/2n+π4 − 160π2 + 1536
32√
2π11/2n2+O
(1
n3
).
2.5. The asymptotic behavior outside the oscilatory region. We can scale the argu-
ment of B(n)n by n and consider the polynomials B
(n)n (nz). Their zeros are located in the
interval (0, 1) for all n ∈ N which follows from (8). Consequently, the function z 7→ B(n)n (nz)
oscillates in (0, 1) and the formula (39) shows the asymptotic behavior in the oscillatoryregion
B(n)n (nx) =(−1)nn!
√2
πn
xxn−1/2(1− x)(1−x)n−1/2
π2 + log2 (x/(1− x))
×[π cos(πxn)− log
(x
1− x
)sin(πxn) +O
(1√n
)],
for x ∈ (0, 1) and n→∞. In addition, the asymptotic behavior by the edges x = 0 and = 1can be obtained from Theorem 6 and the symmetry relation
B(n)n (z) = (−1)nB(n)
n (n− z), (56)
which follows from (5).
To complete the picture, it remains to deduce the asymptotic behavior of B(n)n (nz) for z
outside the interval [0, 1]. The asymptotic analysis is based on the following variant of thesaddle-point method taken from [22, Thm. 7.1, Chp. 4]; see also Perron’s method in [29,Sec. II.5].
Theorem 29 (the saddle-point method). Let the following assumptions hold:
i) Functions f and g are independent of n, single valued, and analytic in a regionM ⊂ C.
ii) The integration path γ is independent of n and its range is located in M with apossible exception of the end-points.
iii) There is a point ξ0 located on the path γ which is not an end-point and is such thatf ′(ξ0) = 0 and f ′′(ξ0) 6= 0 (i.e., ξ0 is the saddle point of Re f).
iv) The integral ∫γ
g(ξ)e−nf(ξ)dξ
converges absolutely for all n sufficiently large.
16 FRANTISEK STAMPACH
v) One hasRe (f(ξ)− f(ξ0)) > 0,
for all ξ 6= ξ0 that lie on the range of γ.
Then ∫γ
g(ξ)e−nf(ξ)dξ = g(ξ0)e−nf(ξ0)
√2π
nf ′′(ξ0)
(1 +O
(1√n
)), as n→∞. (57)
Remark 30. A uniform version of the saddle-point method will be used again. In our case,the function f = f(·, z) from Theorem 29 will depend analytically on an additional variablez ∈ K where K is a compact subset of C. In order to conclude that the expansion (57) holdsuniformly in z ∈ K, it suffices to require the assumptions (i)-(iii) and (v) to remain valid forall z ∈ K and the integral from (iv) to converge absolutely and uniformly in z ∈ K for alln sufficiently large. The reader is referred to [20] for a more general uniform version of thesaddle-point method and the proof.
Our starting point is the contour integral representation
B(n)n (z) =
n!
2πi
∮γ
(1 + ξ)z−1
ξn log(1 + ξ)dξ, z ∈ C, (58)
which follows readily from the generating function formula (1) and (2). The curve γ can beany Jordan curve with 0 in its interior not crossing the branch cut (−∞,−1] of the integrand,i.e., the range of γ has to be located in C \ ((−∞,−1] ∪ 0). The principal branches of themulti-valued functions like the logarithm are chosen if not stated otherwise. Writing nzinstead of z in (58), the contour integral can be written in the form
B(n)n (nz) =
n!
2πi
∮γ
g(ξ)e−nf(ξ,z)dξ, z ∈ C, (59)
where
f(ξ, z) = log ξ − z log(1 + ξ) and g(ξ) =1
(1 + ξ) log(1 + ξ),
which is suitable for the application of the saddle-point method. Without loss of generality,we may restrict z ∈ C \ [0, 1] to the half-plane Re z ≤ 1/2 due to the symmetry (56).
The assumptions (i),(ii), and (iv) of Theorem 29 can be readily checked. Concerning theassumption (iii), one finds that the point
ξ0 :=1
z − 1
is the only solution of ∂ξf(ξ, z) = 0 and is simple.The most difficult part is the justification of the assumption (v) of Theorem 29. The idea
is to investigate the level curves in the ξ-plane determined by the equation
Re f(ξ, z) = Re f(ξ0, z)
with z ∈ C\ [0, 1] and Re z ≤ 1/2 being fixed. This level curve, denoted as Ω0, is the commonboundary of the two open sets
Ω± := ξ ∈ C \ ((−∞,−1] ∪ 0) | Re f(ξ, z) ≷ Re f(ξ0, z) .Since Re f(·, z) is harmonic in C \ ((−∞,−1] ∪ 0), the curves of Ω0 have no end-point inC\((−∞,−1] ∪ 0) and, moreover, they cannot form a loop located in C\((−∞,−1] ∪ 0)with its interior; see, for instance, [24, Lemma 24 and 26]. Hence the only possible loop of Ω0
encircles the origin (the singularity of f(·, z)) and the level curves of Ω0 either goes to ∞or end at the cut (−∞,−1). In general, Ω0 need not be connected but the curves of Ω0
intersect at exactly one point ξ0 since it is the only stacionary point of f(·, z).To fulfill the assumption (v) of Theorem 29, we have to show that the Jordan curve γ can
be homotopically deform to a Jordan curve which crosses ξ0 and is located entirely in Ω+
with the only exception of the saddle point ξ0. The following lemma will be used to justifythe assumption (v).
Lemma 31. Let z ∈ C \ [0, 1], Re z ≤ 1/2, be fixed. For any θ ∈ (−π, π], there existsξ ∈ C \ ((−∞,−1] ∪ 0) with arg ξ = θ such that ξ ∈ Ω0.
17
Proof. We show that Ω0 has an non-empty intersection with any complex ray in C\ (−∞, 0].First, assume θ ∈ (−π, π). Since
limr→0−
Re f(reiθ, z
)= −∞ and lim
r→∞Re f
(reiθ, z
)=∞,
and Re f(reiθ, z
)is continuous in r ∈ (0,∞), there exists r = r(θ) > 0 such that r(θ)eiθ ∈ Ω0.
Second, we verify that Ω0 intersects the interval (−1, 0). If Re z > 0, the task is easy since
limx→−1+
Re f(x, z) =∞ and limx→0−
Re f(x, z) = −∞,
and Re f(x, z) is continuous in x ∈ (−1, 0).Suppose Re z < 0. Then 1/(Re z − 1) ∈ (−1, 0). We show that 1/(Re z − 1) ∈ Ω+, i.e.,
Re f
(1
Re z − 1, z
)> Re f
(1
z − 1, z
). (60)
Then Ω0 ∩ (−1, 0) 6= ∅ because Ω− contains a neighborhood of 0.To verify the inequality (60), we introduce the auxiliary function
χ(z) := Re
(f
(1
Re z − 1, z
)− f
(1
z − 1, z
))and show that χ(z) > 0 for Re z < 0. Noticing that
∂
∂ξ
∣∣∣∣ξ=1/(Re z−1)
Re f(ξ, z) = 0
one computes that∂χ
∂ Re z(z) = log
∣∣∣∣ z
z − 1
∣∣∣∣− logRe z
Re z − 1.
It is easy to check that the above expression is positive if Re z < 0 and Im z 6= 0. Hence, ifIm z 6= 0, then χ is a strictly increasing function of Re z ∈ (−∞, 0). Taking also into accountthat
limRe z→−∞
χ(z) = 0,
one infers that χ(z) > 0 whenever Re z < 0 and Im z 6= 0.If Re z < 0 and Im z = 0, then 1/(Re z − 1) coincides with the saddle point ξ0 and hence
Ω0 intersects the interval (−1, 0), too. At last, if Re z = 0 and Im z 6= 0, it suffices to checkthat
Re f
(1
z − 1, z
)< 0,
sincelim
x→−1+Re f(x, z) = 0 and lim
x→0−Re f(x, z) = −∞.
The verification of the above inequality is a matter of an elementary analysis.
Now, we are at the position to deduce the asymptotic formula for B(n)n (nz), as n → ∞,
in the non-oscilatory regime when z ∈ C \ [0, 1].
Theorem 32. One has
B(n)n (nz) =
n!√2πn
(z − 1)n
z log (z/(z − 1))
(z
z − 1
)nz+1/2(1 +O
(1√n
)), (61)
for n→∞ locally uniformly in z ∈ C \ [0, 1].
Remark 33. Note that z/(z − 1) ∈ (−∞, 0) if and only if z ∈ (0, 1). Hence the leading termin the asymptotic formula (61) is an analytic function of z in C\[0, 1] with a branch cut [0, 1].
Proof. Assume first that z ∈ C \ [0, 1] and Re z ≤ 1/2. For the application of Theorem 29to the contour integral in (59), it remains to justify the crucial assumption (v). Lemma 31implies that the level curve set Ω0 includes a loop which encircles the origin, intersects theinterval (−1, 0) and the saddle point ξ0. Moreover, the interior of this loop is a subset ofΩ−. Indeed, no other loop can separate the interior because in such a case there would be adomain in which f(·; z) is analytic and Re f(·; z) constant on the boundary of this domain.This would imply that f is a constant by general principles.
18 FRANTISEK STAMPACH
The only crossing of curves in Ω0 occurs at the point ξ0 where exactly two curves crossesat the angle π/2 since the saddle point ξ0 is simple. Two of the out-going arcs enclosesinto the loop around the origin. The remaining two arcs either continue to ∞ or end atthe cut (−∞,−1]. These curves cannot cross the loop at another point different from ξ0since ξ0 is the only stationary point of f(·, z). Thus, with the only exception of the point ξ0,a right neighborhood of the loop, if transversed in the counter-clockwise orientation, is asubset of Ω+; see Figure 1.
As a result, one observes that there exists a Jordan curve with 0 in its interior, crossingthe interval (−1, 0) and entirely located in the set Ω+ except the only point ξ0 which belongto the image of this curve. This is the possible choice for the descent path satisfying theassumption (v) of Theorem 29 into which the curve γ in (59) can be homotopically deformed.
The asymptotic formula (61) follows from the application of Theorem 29 and is determinedup to a sign since the branch of the square root in the asymptotic formula (57) has not beenspecified. The local uniformity of the expansion can be justified using the fact that theintegrand of (59) depends analytically on z, see Remark 30.
Using (10) and the fact that all zeros of B(n)n are positive, we get (−1)nB
(n)n (x) > 0 if
x < 0. By inspection of the obtained asymptotic formula (61), one shows the correct choiceof the sign is plus that results in (61). Finally, if z ∈ C \ [0, 1] with Re z > 1/2, one usesthe symmetry (56) together with the already obtained asymptotic formula extending thevalidity of (61) to all z ∈ C \ [0, 1].
Figure 1. An illustration of the level curves Ω0 (blue solid line) and a possiblechoice of the curve γ that fulfill the assumption (v) of Theorem 29 (red dashedline) for z = 1/2 + i/6.
19
Remark 34. It is not very surprising that the sequence of zero-counting measures of B(n)n (nz),
i.e., the uniform probability measures supported on the roots of B(n)n (nz):
µn =1
n
n∑k=1
δx(n)k /n
,
converges weakly to the uniform probability measure supported on the interval [0, 1]. Thiscan be verified by using the Cauchy transform of µn and the asymptotic formula (61). Indeed,for z ∈ C \ [0, 1], one has
limn→∞
∫C
dµn(ξ)
z − ξ= limn→∞
∂
∂zlogB(n)
n (nz) = log
(z
z − 1
)=: Cµ(z).
Now, the Stieltjes–Perron inversion formula implies that the sequence µn converges weaklyto the absolutely continuous measure supported on [0, 1] whose density reads
dµ
dx(x) = lim
ε→0+
1
πImCµ(x− iε) = 1,
for x ∈ [0, 1].
3. Final remarks and open problems
In the end, after a short remark concerning the Euler polynomials of the second kind, weindicate several research problems related to the Bernoulli polynomials of the second kind.These interesting problems appeared during the work on this paper and remained unsolved.
3.1. A remark on the Euler polynomials of the second kind. The Bernoulli poly-nomials are often studied jointly with the Euler polynomials since they share many similarproperties [21]. The Euler polynomials of higher order are defined by the generating function
∞∑n=0
E(a)n (x)
tn
n!=
(2
1 + et
)aext.
The second identity in (4) remains valid at the same form even if B(a)n is replaced by E
(a)n .
On the other hand, there is no simple expression for E(n)n−1 comparable with (6). Hence,
following the same steps as in the case of polynomials B(n)n that resulted in (7), one cannot
deduce a simple integral expression for E(n)n . The simple integral formula (7) for B
(n)n was
the crucial ingredience that allowed to obtain the most important result of this paper. No
similar integral formula for E(n)n is known to the best knowledge of the author. Moreover,
numerical experiments indicate that the zeros of E(n)n are not all real.
3.2. Open problem: Alternative proofs of the reality of zeros. Althogh the proof of
the reality of zeros of B(n)n used in Theorem 1 is elementary, it would be interesting to find
another proof that would not be based on the particular values of B(n)n .
One way of proving the reality of zeros of a polynomial is based on finding a Hermitianmatrix whose characteristic polynomial coincides with the studied polynomial. This is afamiliar fact, for example, for orthogonal polynomials that are characteristic polynomials ofJacobi matrices. Since various recurrence formulas are known for the generalized Bernoullipolynomials, one may believe that there exists a Hermitian matrix An with explicitly ex-
pressible elements such that B(n)n (x) = det(x − An). We would like to stress that no such
matrix was found.Concerning this problem, one can, for instance, show the linear recursion
B(n+1)n+1 (x) = (x− n)B(n)
n (x)−n∑k=0
(n
k
)bn−k+1
n− k + 1B
(k)k (x), n ∈ N0.
20 FRANTISEK STAMPACH
by making use of (1), where bn := bn(0) = B(n)n (1). As a consequence, B
(n+1)n+1 is the
characteristic polynomials of the lower Hessenberg matrix
An =
b1 1 0 0 . . . 0(
10
)12b2 b1 + 1 1 0 . . . 0(
20
)13b3
(21
)12b2 b1 + 2 1 . . . 0
......
...... . . .
...(n0
)1
n+1bn+1
(n1
)1nbn
(n2
)1
n−1bn−1
(n3
)1
n−2bn−2 . . . b1 + n
.However, An is clearly not Hermitian and the reality (and simplicity) of its eigenvalues is byno means obvious.
3.3. Open problem: Beyond the reality of the zeros, a positivity. A certain posi-tivity property of a convolution-like sums with generalized Bernoulli polynomials seems to
hold. This positivity would imply the reality of zeros of B(n)n .
It follows from the operational (umbral) calculus that [23, p. 94]
B(a)n (x+ y) =
n∑k=0
(n
k
)B
(a)n−k(x)yk,
for any x, y, a ∈ C and n ∈ N0. With the aid of the above formula, one can verify the identity
|B(n)n (x+ iy)|2 = |B(n)
n (x)|2 +
bn2 c∑k=1
αn,k(x)y2k +
bn−12 c∑
k=0
βn,k(x)y2n−2k, (62)
where
αn,k(x) := (−1)k2k∑l=0
(−1)l(n
l
)(n
2k − l
)B
(n)n−l(x)B
(n)n−2k+l(x)
and
βn,k(x) := (−1)k2k∑l=0
(−1)l(n
l
)(n
2k − l
)B
(n)l (x)B
(n)2k−l(x).
Conjecture 1. For all x ∈ R and n ∈ N, one has
αn,k(x) ≥ 0, for 1 ≤ 2k ≤ n, and βn,k(x) ≥ 0, for 0 ≤ 2k ≤ n− 1.
If the above conjecture holds true, one would obtain from (62), for example, the inequality
|B(n)n (x+ iy)|2 ≥
(B(n)n (x)
)2+ y2n,
where we used that βn,0(x) = 1. From this inequality, the reality of zeros of B(n)n immediately
follows.
3.4. Open problem: A transition between the asymptotic zero distributions ofthe Bernoulli polynomials of the first and second kind. The asymptotic zero distri-
bution µ0 for Bernoulli polynomials of the first kind Bn = B(1)n was found by Boyer and Goh
in [1]. It is a weak limit of the sequence of the zero-counting measures associated with poly-nomials Bn. The measure µ0 is absolutely continuous, supported on certain arcs of analyticcurves in C, and its density is also described in [1]. On the other hand, the asymptotic zero
distribution µ1 of the polynomials B(n)n is simply the uniform probability measure supported
on the interval [0, 1], see Remark 34.The two measures µ0 and µ1 can be viewed as two extreme points of the asymptotic zero
distribution µλ of the polynomials
z 7→ B(1−λ+λn)n (nz), (63)
where λ ∈ [0, 1]. The order of the above generalized Bernoulli polynomial is nothing butthe convex combination of 1 and n. The measures µλ seem to be absolutely continuous andcontinuously dependent on λ. An interesting research problem would be to describe thesupport (the zero attractor) as well as the density of µλ for λ ∈ (0, 1). For an illustration ofapproximate supports of µλ for several values of λ, see Figure 2.
21
Figure 2. Plots of the zeros of polynomials (63) in the complex plane for n = 500and λ ∈ 0, 1/4, 1/2, 3/4 illustrating how the zero attractor of the Bernoullipolynomials (case λ = 0) deforms into the interval [0, 1] (case λ = 1) as λ changesfrom 0 to 1.
Acknowledgement
The author acknowledges financial support by the Ministry of Education, Youth andSports of the Czech Republic project no. CZ.02.1.01/0.0/0.0/16 019/0000778.
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(Frantisek Stampach) Department of Applied Mathematics, Faculty of Information Technology,
Czech Technical University in Prague, Thakurova 9, 160 00 Praha, Czech RepublicEmail address: [email protected]
