Hindawi Publishing Corporation
International Journal of Mathematics and Mathematical Sciences
Volume 2008, Article ID 421478, 20 pages
doi:10.1155/2008/421478
Research Article
Relations among Sums of Reciprocal
Powers—Part II
José Marı́a Amigó
Centro de Investigación Operativa, Universidad Miguel Hernández, Avenida de la Universidad s/n,
03202 Elche (Alicante), Spain
Correspondence should be addressed to José Marı́a Amigó, jm.amigo@umh.es
Received 18 July 2008; Accepted 10 September 2008
Recommended by Pentti Haukkanen
Some formulas relating the classical sums of reciprocal powers are derived in a compact way
by using generating functions. These relations can be conveniently written by means of certain
numbers which satisfy simple summation formulas. The properties of the generating functions can
be further used to easily calculate several series involving the classical sums of reciprocal powers.
Copyright q 2008 José Marı́a Amigó. This is an open access article distributed under the Creative
Commons Attribution License, which permits unrestricted use, distribution, and reproduction in
any medium, provided the original work is properly cited.
1. Introduction
In 1, we studied some arithmetic relations among the classical numbers:
λn ∞
1
2ν
1n
ν0
n ≥ 2,
Ln ∞
−1ν
n
ν0 2ν 1
n ≥ 1.
1.1
In this paper, we extend this analysis to the remaining
ζn ∞
1
n
ν
ν1
n ≥ 2,
ηn ∞
−1ν−1
ν1
νn
n ≥ 1.
1.2
Although the numbers λn, ζn, and ηn are related to each other through the identities
λn 1 − 2−n ζn,
ηn 1 − 21−n ζn,
n ≥ 2
1.3
η1 ln 2 and thus
ηn ζn 2λn
n ≥ 2,
we will use all of them in order to keep the algebraic expressions as simple as possible.
1.4
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International Journal of Mathematics and Mathematical Sciences
For k ≥ 2, define
Dk 2π k−1
∞
λ2j
.
2j · · · 2j k − 1
j1
1.5
It will be shown below that these numbers can be alternatively expressed as
Dk 1
2k − 1!
π
0
2k−1
t
tk−1 cot dt 2
k − 1!
π/2
tk−1 cot t dt.
1.6
0
Indefinite integrals of this type were considered by Ramanujan 2, page 260. The constants
Dk have the property of relating the values of ζ-numbers eventually, η- or λ-numbers
with odd argument to the elementary values ζ2n 2π2n |B2n |/22n! where B2n are the
Bernoullian numbers as, for example, in
n
n
2
π 2n−2j
−1n
η2j 1 D2n 2
−1n−j
−1j D2jζ2n − 2j 2 2n − 2j!
π j1
π
j0
1.7
if n 0, the first term on the right hand side of 1.7 has to be dropped. This and other
formulas expressing the numbers Dk by means of sums of reciprocal powers with odd
arguments and, conversely, η2n 1, ζ2n 1, and λ2n 1 via Dk will be proved
in Section 2 see Propositions 2.2 and 2.4, and Corollary 2.6 below using some generating
functions defined by the numbers λn, ηn, ζn, and Dn. The properties of these
generating functions, which are given as expansions both in powers and in partial fractions,
will be instrumental for most of the subsequent results.
Sections 3, 4 deal with the numbers Dk and with the classical sums of reciprocal
powers, respectively. In particular, Section 3 is mainly devoted to the calculation of several
series containing Dk. In Section 4, the focus changes onto some particular series whose
terms contain λ-, η-, ζ-, and L-numbers like, for example,
∞
1
ζ2n − ζ2n 1 ,
2
n1
∞
ζ2n 1 − λ2n 0.
1.8
n1
Finally, in the brief Section 5, the generating function of the numbers Dk is expressed
using the Psi Digamma function ψx Γ x/Γx.
2. Main statements
Define the generating functions Λx, Ex, and Zx by
Λx ∞
n2
λnxn−1 ,
Ex ∞
n1
ηnxn−1 ,
Zx ∞
ζnxn−1 ,
2.1
n2
where, in principle, x ∈ C since limn→∞ λn limn→∞ ηn limn→∞ ζn 1, these formal
power series converge only for |x| < 1. Furthermore, denote by Λ x and Λ− x the even
José Marı́a Amigó
3
and odd parts, respectively, of Λx and similarly for E± x and Z± x. Then, the following
identities hold 3, 4.3.67/68/70:
tan
∞
4
4
πx
λ2nx2n−1 Λ− x
2
π n1
π
|x| < 1,
2.2
csc πx ∞
2
2
1
1
E− x 0 < |x| < 1,
η2nx2n−1 πx π n1
πx π
2.3
cot πx ∞
2
2
1
1
−
− Z− x
ζ2nx2n−1 πx π n1
πx π
2.4
0 < |x| < 1.
Owing to 1.3 and 1.4, the above generating functions fulfill the trivial relations
Ex Zx − Z
x
ln 2,
2
1
x
Λx Zx − Z
,
2
2
2.5
Ex Zx 2Λx ln 2,
respectively.
On substituting the definitions of λn, ηn, and ζn into the corresponding
generating functions, we find the following expansions in partial fractions:
Λx ∞ ν0
∞
1
−1ν−1
1
−
,
Ex ,
2ν 1 − x 2ν 1
ν−x
ν1
∞ 1
1
−
.
Zx ν−x ν
ν1
2.6
In particular, the expansions
E x ∞
ν
1
Ex E−x −1ν−1 2
,
2
ν
−
x2
ν1
∞ 1
1
ν
−
Z x Zx Z−x 2
ν 2 − x2 ν
ν1
2.7
will be used below.
Proposition 2.1. The integral representation 1.6 holds.
Proof. Let ft tanπt/2. By repeated partial integration, one gets
1
0
tk−1 f1 − tdt f 0
f 0
f0
···
k
kk 1 kk 1k 2
∞
f ν 0
,
kk 1 · · · k ν
ν0
2.8
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International Journal of Mathematics and Mathematical Sciences
where according to the Taylor expansion 2.2,
0,
ν
f 0 42j − 1!λ2j/π,
Hence,
2k−1
k − 1!
π/2
tk−1 tan
0
if ν 2j, j ∈ N,
if ν 2j − 1, j ∈ N.
2.9
1
πk
π
π
− t dt tk−1 tan 1 − tdt
2
2k − 1! 0
2
∞
42j − 1!λ2j/π
πk 2k − 1! j1 kk 1 · · · k 2j − 1
2π k−1
2.10
∞
λ2j
2j
·
·
·
2j
k − 1
j1
Dk.
Since, for 0 ≤ x ≤ π/2 and n ∈ N,
xn cot x 1 π n1
,
2n 2
2.11
it follows that limn→∞ Dn 1 0 faster in fact, much faster than π n2 /8nn!.
Proposition 2.2. The numbers Dk k ≥ 2 can be evaluated in terms of η1, η3, . . . , η2k/2−
1, supplemented with λk if k is odd, by the formula
k/2
0,
k even,
π k−2j1
j−1
η2j − 1 2.12
−1
Dk k−1/2
k
−
2j
1!
−1
2λk, k odd.
j1
Proof. For each ν ∈ N, we have upon k − 1 integrations by parts
π/2
2k
tk−1 sin 2νt dt
k − 1! 0
⎧
k/2
ν
⎨0,
k even,
k−2j1
π
−1
ν
−1j
1
−
−1
k − 2j 1! ν2j−1 ⎩−1k−1/2
, k odd.
j1
νk
Sum now both sides on ν, 1 ≤ ν ≤ N, and use the identity 4, 1.3421
N
sin 2νt sinN 1t sin Nt csc t ν1
1
cos t − cos2N 1tcsc t,
2
2.13
2.14
to get on the left side
N π/2
2k tk−1 sin 2νt dt
k − 1! ν1 0
2k−1
k − 1!
π/2 0
tk−1
cos2N 1t dt.
tk−1 cot t −
sin t
To finish the proof, let N → ∞ and use the Riemann-Lebesgue lemma.
2.15
José Marı́a Amigó
5
In particular,
π
η1 π ln 2
1!
2.16
π2
π2
7
ln 2 − 2λ3 ln 2 − ζ3
2
2
4
2.17
D2 as in 4, 3.7477, and
D3 as in 5, 4.23. Other expressions for Dk different from 2.12 can be found in 4, 3.7482.
Due to the definition 1.5 and formula 2.12, each number Dk embodies a relation
among all elementary values λ2j and a finite number of their nonelementary counterparts
λ2j 1, namely,
∞
k
λ2j
1
1/π2j−2
η2j − 1,
−1j−1
2j · · · 2j 2k − 1 2 j1
2k − 2j 1!
j1
∞
k
λ2j
1/π2j−2
λ2k 1
1
η2j − 1 −1k
−1j−1
,
2j
·
·
·
2j
2k
2
2k
−
2j
2!
π 2k
j1
j1
2.18
for k ≥ 1.
Define next the generating function
Dx ∞
Dnixn−1 .
2.19
n2
Owing to the vanishing rate of the coefficients Dn, this power series is convergent for all
x ∈ C. Then,
Dx π/2
dt cot t
0
∞
i2xtn−1
n2
n − 1!
π/2
ei2xt − 1 cot t dt.
2.20
0
Furthermore, let D x and D− x be the even and odd parts of Dx, that is,
D x ∞
∞
D2n 1ix2n −1n D2n 1x2n
n1
π/2
n1
cos 2xt − 1 cot t dt,
0
∞
∞
D2nix2n−1 i −1n−1 D2nx2n−1
D− x n1
i
2.21
n1
π/2
sin 2xt cot t dt.
0
Thus, if x is meant to be real, D x and 1/iD− x are the real and imaginary part,
respectively, of Dx.
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International Journal of Mathematics and Mathematical Sciences
Proposition 2.3. The following relations hold:
D x cosπxE x Z x − ln 2,
2.22
D− x i sinπxE x.
Proof. We claim that
Dx eiπx E x Z x − ln 2.
In fact, from
Dx π/2
∞ π/2
ei2xt − 1 cot t dt 2
ei2xt − 1 sin 2ν t dt,
0
π/2
ei2xt sin 2νt dt 0
Dx e
ν1
ν−1
π/2
−1 νe ν
,
2ν2 − x2 we obtain
iπx
iπx
∞
−1ν−1 ν
ν1
2.23
ν 2 − x2
∞ ν1
0
sin 2νt dt 0
1
ν
−
ν 2 − x2 ν
−
1 −1
2ν
∞
−1ν−1
ν1
ν
2.24
ν−1
,
.
2.25
Comparison with 2.7 proves the claim. Take now even and odd parts.
Solving for E x and Z x in 2.22, we get the identities
1
E x csc πxD− x 0 < |x| < 1,
i
Z x D x i cotπxD− x ln 2 0 < |x| < 1,
2.26
2.27
which lead to a kind of converse of Proposition 2.2.
Proposition 2.4. The numbers η2n 1 and ζ2n 1, n ≥ 1, can be expressed in terms of the Dk
and the elementary values η2k and ζ2k by
n
−1n
2
D2n 2,
−1j−1 η2n − 2j 2D2j η2n 1 π j1
π
2.28
n
2
−1n1
j−1
n
D2n 2.
ζ2n 1 −1 ζ2n − 2j 2D2j −1 D2n 1 π j1
π
Proof. 1 Equation 2.26 reads explicitly use η0 1/2 in 2.3,
∞
η2n 1x2n
n0
1
csc πxD− x
i
∞
∞
2
η2nx2n−1 i−1j−1 D2jx2j−1
iπ n0
j1
∞ n1
2
−1j−1 η2n − 2j 2D2jx2n
π n0 j1
∞
n
2
1
−1n
j−1
D2n 2 x2n .
D2 −1 η2n − 2j 2D2j π
π n1 j1
2
2.29
José Marı́a Amigó
7
2 The second identity follows from 2.27 since use ζ0 −1/2 in 2.4,
cotπxD− x
−
∞
∞
2
ζ2nx2n−1 i−1j−1 D2jx2j−1
π n0
j1
∞
n1
2i j−1
−
−1 ζ2n − 2j 2D2j x2n
π n0 j1
2.30
∞
n
2i i
−1n1
j−1
D2n 2 x2n ,
D2 −
−1 ζ2n − 2j 2D2j π
π n1 j1
2
Remark 2.5. Equation 1.7 is nothing else but the formula
cosπxE x 1
cotπxD− x
i
2.31
written explicitly. Indeed,
cosπxE x ∞
−1n
n0
∞
π 2n 2n x
η2j − 1x2j−2
2n!
j1
∞
n
2n−2j
n−j π
η2j 1 x2n
−1
2n
−
2j!
n0 j0
η1 ∞
n
n1
j0
−1
n−j
2.32
π 2n−2j
η2j 1 x2n ,
2n − 2j!
and cotπxD− x was calculated in the last proof. For n 1 one gets the representation
ζ3 2
4
16
2 2
π ln 2 −
D4 π 2 ln 2 −
9
3π
9
9π
π/2
t3 cot t dt.
2.33
0
Equation 1.7 and also the first formula of Proposition 2.4 show that η2n 1 and,
for that case, ζ2n 1 and λ2n 1 can be expressed in terms of only D2, D4, . . . , D2n 2. Furthermore, adding the two formulas of Proposition 2.4 and using 1.4, we obtain the
following result.
Corollary 2.6. For n ≥ 1,
λ2n 1 n
−1n
2
D2n 1.
−1j−1 λ2n − 2j 2D2j π j1
2
For n 1, we recover 2.17.
2.34
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International Journal of Mathematics and Mathematical Sciences
3. Summation formulas for the Dks
Before deriving more relations involving the sums of reciprocal powers, we obtain next some
“summation” formulas for the numbers Dk. Two integral summation formulas follow
trivially from the very definition of the generating function Dx and 2.20, namely,
π/2
∞
1
Dn D
e2t − 1 cot t dt,
i
0
n2
∞
−1n Dn −Di π/2
3.1
1 − e−2t cot t dt.
0
n2
Also,
∞
−1n D2n 1 D 1 π/2
cos 2t − 1 cot t dt −1,
3.2
0
n1
∞
1 π/2
1
n D2n 1
−1
D
cos t − 1 cot t dt ln 2 − 1,
2
2 0
22n
n1
3.3
π/2
∞
1
π
n1
−1 D2n D− 1 sin 2t cot t dt ,
i
2
0
n1
3.4
π/2
D2n 1
1
D
sin t cot t dt 1.
−
2n−1
i
2
2
0
3.5
∞
−1n1
n1
Other similar series can be also straightforwardly deduced after differentiating 2.21,
∞
−1
n1
nD2n 1x
2n−1
n1
∞
−1
n1
2n − 1D2nx
2n−2
n1
1 D x 2
π/2
1 D− x 2
i
t sin 2xt cot t dt,
0
3.6
π/2
t cos 2xt cot t dt,
0
and substituting fixed values for x. In particular, the series
∞
n1
−1
n1
1 2n − 1D2n D− 1 2
i
π/2
t cos 2t cot t dt D2 −
0
π
π
2
ln 2 −
1
2
3.7
where cos 2t 1 − 2 sin2 t, 1.6 and 2.16 were used will be needed below. Note that from
3.7 and 3.4, it follows
∞
D2 2 −1n nD2n.
3.8
n2
Furthermore, from 2.20, we have
π/2
1
1
eiπx − i2x
−D x−
2i
D x
ei2xt cos t dt 2i
,
2
2
1 − 4x2
0
3.9
José Marı́a Amigó
9
so, after separating real and imaginary parts, the equations
2
1
1
− D x −
D x 2x − sin πx,
2
2
1 − 4x2
2
1
1
D− x − D− x −
i
cos πx
2
2
1 − 4x2
3.10
3.11
hold for x ∈ R. Letting x → 1/2 one recovers 3.2 and 3.4.
Proposition 3.1. The following identities hold:
1 For n 0, 1, 2, . . . ,
∞
n
2k1
2n 2k D2n 1 2k −1n k1
k π
−2 .
−1
−1
4
2k 1!
22n12k
2n 1
k1
k0
3.12
In particular, for n 0,
∞
−1k1 k
k1
D2k 1 1
π − 2.
8
22k1
3.13
2 For n 0, 1, 2, . . . ,
∞
n
2k
2n 2k − 1 D2n 2k −1n k1
k π
.
−1
−1
2 k0
2k!
22n2k
2n
k1
3.14
Proof. 1 In fact,
∞
1 2k
1 2k
1
1
D x −1k D2k 1
x
− x−
− D x −
2
2
2
2
k1
∞
k−1
2 −1k D2k 1
k1
2
2n3
∞
∞
−1
n0
k1
2k
x2n1 2−2k−2n−1
2n 1
2n 2k D2n 1 2k
nk
n0
2n 1
22n12k
x2n1 .
3.15
Comparison with 3.10, that is,
∞
∞
2n1
1
n1 π
2n1
x
2x
−
sin
πx
2
−
πx
−1
2x2j
2n
1!
1 − 4x2
j0
n1
∞
∞
n
2k1
2n1
k1 π
2x2n1
2 − π2x
−1
2k
1!
n0
n1 k1
n
∞
2k1
k1 π
2n1
2 −1
x2n1 ,
2
2k
1!
n0
k0
where |x| < 1 yields the result.
3.16
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International Journal of Mathematics and Mathematical Sciences
2 Analogously to 1, the summation formula follows comparing
1
1
− D− x −
D− x 2
2
∞
∞
2n 2k − 1 D2n 2k
nk1
2n2
2
x2n
i
−1
2n2k
2
2n
n0 k1
3.17
with 3.11, that is,
∞
∞
n
∞
2n
2k
2i
n π
2j
k π
2n
2n1
x
x2n ,
cos πx 2i −1
2x 2
i
−1
2n!
2k!
1 − 4x2
n0
n0 k0
j0
3.18
where |x| < 1.
Of course, all these summation formulas can be also checked using the integral
representation 1.6. Finally, a finite summation formula can be derived in the following way.
For n ≥ 1,
π/2 0
π
−t
2
n
n−k
π/2 n n
n
π
π
π
π
k
− t dt − t dt
−
−t tan
tan
2
2
2
2
k
0
k1
n−k π/2
n
n
π
k
−1
tk cot t dt
2
k
0
k1
n−k
n
n
k!
π
k
−1
Dk 1.
2
2k
k
k1
3.19
Therefore, changing the variable in the integral, we get
n
−1k−1
k1
2n
π n−k
Dk 1 n − k!
n!
π/2 n
π
− tn tan t dt.
2
0
3.20
4. Further relations
From the close-form expressions obtained in the previous sections, we can derive a number
of interesting results for series containing, in turn, the series ζn, λn, or ηn. For
completeness, we will also include the series Ln, what requires the consideration of a new
generating function Cx defined in 1. As a first example, we will prove the following
proposition.
José Marı́a Amigó
11
Proposition 4.1. The following identities hold true:
∞
ζ2n
n1
22n
∞
λ2n
n1
22n
∞
ζ2n 1
1
,
2
n1
1
ln 2 − ,
2
∞
λ2n 1
π
,
8
22n1
n1
22n1
1
ln 2,
4
∞
η2n 1 1
1 π
−
1
,
,
2n
2
2
2
2
22n1
n0
n1
√
√
∞
∞
√ L2n
2 L2n 1 π 2
ln 1 2 ,
.
4
8
22n
22n1
n0
n1
∞
η2n
4.1
Note, in particular, the series
∞
ζn
n2
2n
ln 2,
∞
ηn
n1
2n
π
.
4
4.2
Proof. 1 Setting x 1/2 in 2.27, we get
Z
1
1
D
ln 2.
2
2
4.3
Thus, by 3.3,
∞
ζ2n 1
n1
22n1
1
1
1
1
1
1
≡ Z
D
ln 2 ln 2 − .
2
2
2
2
2
2
4.4
1
1
1
Z−
.
2
2
2
4.5
On the other hand see 2.4,
∞
ζ2n
n1
22n
≡
Adding up the even and odd parts, we get
∞
ζn
n2
2n
∞
ζn
−1n n 1 − ln 2.
2
n2
ln 2,
4.6
2 Setting x 1/2 in 2.26, we get
1
1
1
D−
.
E
2
i
2
4.7
Hence, by 3.5,
∞
η2n 1
n0
22n1
≡
1
1
1
1
1
E
D−
.
2
2
2i
2
2
4.8
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International Journal of Mathematics and Mathematical Sciences
On the other hand, by 2.3,
∞
η2n
22n
n1
≡
1
1
1 π
E−
−1 .
2
2
2 2
4.9
Adding up the even and odd parts, we get
∞
ηn
n1
2n
∞
π
,
4
−1n1
n1
ηn
π
1− .
n
2
4
4.10
3 In 1, it is proved that, for |x| < 1,
∞
λ2n 1x2n ≡ Λ x C x n1
πx
1
tan
C− x
i
2
4.11
holds, where
C x 1
2
π/2
cos xt − 1
dt,
sin t
0
C− x i
2
π/2
0
sin xt
dt.
sin t
4.12
Hence,
∞
λ2n
n1
22n
≡
π
1
1
Λ−
2
2
8
4.13
by 2.2, and
∞
λ2n 1
n1
22n1
≡
1
1
1
1
1
1
1
Λ
C
C−
ln 2,
2
2
2
2
2i
2
4
since
C
4.14
1
1 π/2 cost/2 − 1
dt
2
2 0
sin t
1
2
ln
√ ,
2 1 2
i π/4
1
sec t dt
C−
2
2 0
4.15
3π
i
ln tan
2
8
√ i ln 1 2 .
2
Adding up the even and odd parts,
∞
λn
n2
∞
n2
2n
λn
−1n n
2
1 π
ln 2 ,
4 2
1 π
− ln 2 .
4 2
4.16
José Marı́a Amigó
13
4 In 1, it is proved that
∞
πx
π
,
L2n 1x2n ≡ L x sec
4
2
n0
∞
L2nx2n−1 ≡ L− x n1
πx
1
sec
C− x,
i
2
4.17
4.18
for |x| < 1. Hence,
√ 1
π
1
1
1
1
L
sec
C
√ ln 1 2 ,
−
−
2n
2
2
2i
4
2
2
2 2
n1
√
∞
π
π π 2
L2n 1 1
1
L
sec
.
≡
2
2
8
4
8
22n1
n0
∞
L2n
≡
4.19
Adding up the even and odd parts,
√
√ 1
π 2
ln
1
2 ,
√
2n
8
2 2
n1
√
∞
√ 1
π 2
n1 Ln
−
ln
1
−1
2 .
√
2n
8
2 2
n1
∞
Ln
4.20
The next theorem shows that more challenging results can be achieved by being
slightly more sophisticated.
Theorem 4.2. The following equalities hold true:
ζ2n − ζ2n 1 1
,
2
λ2n − λ2n 1 ln 2 − 1,
ζ2n 1 − λ2n 0,
1 π
ζ2n 1 − L2n 1 −1 ,
2 2
λ2n 1 − η2n 1 − ln 2,
ζ2n − η2n − 1 1
ln 2,
2
1
L2n − 1 − η2n − 1 ln 2 − ,
2
1
ζ2n 1 − L2n ln 2,
2
1
λ2n 1 − L2n 1 − ln 2,
2
1
L2n − η2n − 1 ln 2,
2
where all the series start with n 1.
1
η2n − η2n − 1 ln 2 − ,
2
1
L2n − L2n − 1 1 − ln 2,
2
1
ζ2n 1 − η2n ,
2
ζ2n − λ2n 1 ln 2,
π
λ2n 1 − L2n 1 − ln 2,
4
λ2n − η2n − 1 ln 2,
1
1 ln 2,
2
1
λ2n − L2n ln 2,
2
1
L2n − η2n 1 − ln 2,
2
ζ2n − L2n 4.21
14
International Journal of Mathematics and Mathematical Sciences
Proof. We will proceed left to right and top to bottom.
1 From 2.4 and 2.27, it follows
∞
ζ2n − ζ2n 1x2n xZ− x − Z x
n1
4.22
1 π
− x cotπx − D x − i cotπxD− x − ln 2
2 2
for |x| < 1. Thus, by the Tauber theorem 6, 3.2 and 3.7,
∞
ζ2n − ζ2n 1 lim
x→1−
n1
∞
ζ2n − ζ2n 1x2n
n1
1
π
− D 1 − ln 2 − lim cotπx
x iD− x
x→1−
2
2
1
i 1
1
D− 1 ,
1 − ln 2 −
2
2 π
2
4.23
where we have used the L’Hopital rule in the last line. We will use L’Hopital’s rule also in the
sequel to resolve indeterminacies.
2 From 2.3 and 2.26, it follows
∞
π
1
csc πx ix csc πxD− x
η2n − η2n − 1x2n−1 E− x − xE x −
2x
2
n1
4.24
for 0 < |x| < 1. Thus, by Tauber’s theorem, 3.4 and 3.7,
∞
n1
η2n − η2n − 1 lim
x→1−
∞
η2n − η2n − 1x2n−1
n1
1
π
ixD− x
− lim csc πx
2 x→1−
2
4.25
i
1
1
− − D− 1 D− 1 ln 2 − .
2 π
2
3 Analogously, from 2.2 and 4.11,
∞
n1
λ2n − λ2n 1 lim xΛ− x − Λ x
x→1−
πx
π
πx
x tan
− C x i tan
C− x
x→1− 4
2
2
πx π
x iC− x
−C 1 lim tan
x→1−
2
4
1 2i C 1
−C 1 −
2 π −
lim
ln 2 − 1,
4.26
José Marı́a Amigó
15
since see 4.12
1
C 1 2
C− 1 i
π/2
1
2
0
1
cot t − csc tdt − ln 2,
2
π/2
t cot t dt iD2 i
0
4.27
π
ln 2.
4
4.28
4 Using 4.18, 4.17, and 4.28, it follows
∞
L2n − L2n − 1 lim L− x − xL x
x→1−
n1
πx
1
sec
C− x −
x→1− i
2
πx 1
lim sec
C− x −
x→1−
2
i
lim
−
πx
π
x sec
4
2
π
x
4
4.29
2 1 1
C− 1 1 − ln 2.
iπ
2 2
5 From 2.27, 2.2, and 3.7, we get
∞
ζ2n 1 − λ2n
n1
lim Z x − xΛ− x
x→1−
πx
π
lim D x i cotπxD− x ln 2 − x tan
x→1−
4
2
πx
1
πx
π
D 1 ln 2 lim tan
i cot2
− 1 D− x − x
2 x→1−
2
2
2
1 2 −1 ln 2 i D 1 1 0.
2 π −
4.30
6 From 2.27, 2.3, and 3.7, we get
∞
ζ2n 1 − η2n
n1
lim Z x − xE− x
x→1−
1 π
lim D x i cotπxD− x ln 2 − x csc πx
x→1−
2 2
1
π
D 1 ln 2 lim csc πx i cos πxD− x − x
2 x→1−
2
D 1 ln 2 i 1
1 1
D− 1 .
2 π
2 2
4.31
16
International Journal of Mathematics and Mathematical Sciences
7 From 2.27, 4.17, and 3.7, we get
∞
ζ2n 1 − L2n 1
n1
lim Z x − L x L1
x→1−
πx π
π
lim D x i cotπxD− x ln 2 − sec
x→1−
4
2
4
πx
π 1
πx
π
2 πx
D 1 ln 2 lim sec
i sin
cot
− 1 D− x −
4 2 x→1−
2
2
2
2
π
i 1 π
−1 ln 2 D− 1 −1 .
4 π
2 2
4.32
8 From 2.4, 4.11, and 4.28, we get
∞
ζ2n − λ2n 1
n1
lim xZ− x − Λ x
x→1−
πx
1
1 π
− x cot πx − C x − tan
C− x
lim
x→1− 2
2
i
2
1
πx
1 1
πx π
x cot2
− 1 C− x
ln 2 − lim tan
x→1−
2 2
2
4
2
i
2 1
1
−
C 1 ln 2.
1 ln 2 −
2
2 iπ −
4.33
9 From 4.11, 2.3, and 4.28, we get
∞
λ2n 1 − η2n
n1
lim Λ x − xE− x
x→1−
πx
1
1 π
C− x − x csc πx
lim C x tan
x→1−
i
2
2 2
πx
πx
1
πx 1
π
C 1 lim sec
sin
C− x − x csc
2 x→1−
2
i
2
4
2
1
2 1
1
C− 1 1 − ln 2.
− ln 2 −
2
2 iπ
2
4.34
José Marı́a Amigó
17
10 From 4.11, 4.17, and 4.28, we get
∞
λ2n 1 − L2n 1
n1
lim Λ x − L x L1
x→1−
πx
πx π
1
π
C− x − sec
lim C x tan
x→1−
i
2
4
2
4
πx
π
πx 1
π
C 1 lim sec
sin
C− x −
4 x→1−
2
i
2
4
4.35
1
π
2 π
− ln 2 −
C− 1 − ln 2.
2
4 iπ
4
11 From 2.4, 2.26, 3.4, and 3.7, we get
∞
ζ2n − η2n − 1 lim Z− x − xE x
x→1−
n1
π
1
1
− cot πx − x csc πxD− x
x→1− 2x
2
i
1
1
π
cos πx xD− x
− lim csc πx
2 x→1−
2
i
lim
4.36
1
1
1
D− 1 D− 1 ln 2.
2 iπ
2
12 From 2.2, 2.26, 3.4, and 3.7, we get
∞
λ2n − η2n − 1 lim Λ− x − xE x
n1
x→1−
πx 1
π
tan
− x csc πxD− x
x→1− 4
2
i
πx 1
πx
πx π
sin
− x csc
D− x
lim sec
x→1−
2
4
2
2i
2
lim
4.37
1
D− 1 D− 1 ln 2.
iπ
13 From 4.17, 2.26, and 3.7, we get
∞
L2n 1 − η2n 1 lim L x − E x
n0
x→1−
πx 1
π
sec
− csc πxD− x
x→1− 4
2
i
πx
1
πx π
i csc
D− x
lim sec
2 x→1−
2
2
2
lim
1 1
D− 1 ln 2 − .
iπ
2
4.38
18
International Journal of Mathematics and Mathematical Sciences
14 From 2.4, 4.18, and 4.28, we get
∞
ζ2n − L2n lim Z− x − L− x
x→1−
n1
π
πx
1
− cot πx i sec
C− x
x→1− 2x
2
2
πx
1
πx π
cos πx csc
− iC− x
− lim sec
2 x→1−
2
4
2
lim
4.39
1 2i 1
− C− 1 1 ln 2.
2 π
2
15 From 2.27, 4.18, 3.7, 4.12, and 4.28, we get
∞
ζ2n 1 − L2n
n1
lim Z x − xL− x
x→1−
πx
1
C− x
lim D x i cotπxD− x ln 2 − x sec
x→1−
i
2
πx
πx
2 πx
− 1 D− x xC− x
sin
cot
D 1 ln 2 i lim sec
x→1−
2
2
2
4.40
2 1
−1 ln 2 i D− 1 − C− 1 − C− 1 ln 2.
π
2
16 From 2.2, 4.18, and 4.28, we get
∞
λ2n − L2n lim Λ− x − L− x
x→1−
n1
πx
πx
π
tan
i sec
C− x
x→1− 4
2
2
πx
πx π
sin
iC− x
lim sec
x→1−
2
4
2
lim
4.41
2 1
C− 1 ln 2.
iπ
2
17 From 4.11, 4.18, 4.27, and 4.12, we get
∞
λ2n 1 − L2n
n1
lim Λ x − xL− x
x→1−
πx
πx
1
1
C− x − x sec
C− x
lim C x tan
x→1−
i
2
i
2
πx
1
1
πx
sin
− x 1 − ln 2.
C 1 C− 1 lim sec
x→1−
i
2
2
2
4.42
José Marı́a Amigó
19
18 From 4.18, 2.3, and 4.28, we get
∞
L2n − η2n lim L− x − E− x
x→1−
n1
πx
π
1
1
sec
C− x − csc πx
x→1− i
2
2x 2
1
πx
πx
π
− lim sec
iC− x csc
2 x→1−
2
4
2
lim
4.43
1
2 1
−
C− 1 1 − ln 2.
2 iπ
2
19 From 4.18, 2.26, 4.28, 3.4, and 3.7, we get
∞
L2n − η2n − 1 lim L− x − xE x
x→1−
n1
πx
1
sec
C− x −
x→1− i
2
1
πx
lim sec
C− x −
i x→1−
2
lim
−
1
x csc πxD− x
i
πx
x
csc
D− x
2
2
4.44
2 1
1 1
C− 1 D− 1 D− 1 ln 2.
iπ
iπ
iπ
2
5. Enter ψx
Let ψx, x ∈ C, be the logarithmic derivative of the Gamma function. Then, if |x| < 1 7,
1.175
ψ1 x −γ ∞
−1n ζnxn−1 ,
5.1
n2
hence 7, 1.711
Zx −ψ1 − x − γ −ψx − π cot πx − γ,
x
x
Ex Zx − Z
ln 2 −ψx ψ
π csc πx ln 2.
2
2
5.2
It follows
Z x −ψ x − γ,
x
E x −ψ x ψ ln 2,
2
where ψ x ≡ ψx ψ−x/2.
5.3
20
International Journal of Mathematics and Mathematical Sciences
Substitution of these expressions for E x and Z x in 2.22 and comparison with
2.21 yields
D x π/2
cos2xt − 1 cot t dt
0
−2cos2
1
D− x i
πx
x
ψ x cos πx ψ
ln 2 − γ − ln 2,
2
2
π/2
5.4
sin 2xt cot t dt
0
x
sin πx − ψ x ψ
ln 2
2
respectively.
Acknowledgments
This work was completed during a research stay at the Centre de Recerca Matemàtica in
Barcelona. The author would like to thank very much Professor Manuel Castellet for making
this stay possible. The kind hospitality of the Centre is also acknowledged.
References
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1, pp. 177–184, 2001.
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Mathematical Tables, Dover, New York, NY, USA, 1972.
4 I. S. Gradshteyn and I. M. Ryzhik, Table of Integrals, Series and Products, Academic Press, New York, NY,
USA, 1994.
5 H. M. Srivastava and J. Choi, Series Associated with the Zeta and Related Functions, Kluwer Academic
Publishers, Dordrecht, The Netherlands, 2001.
6 T. Apostol, Mathematical Analysis, Addison-Wesley, Reading, Mass, USA, 1974.
7 H. Bateman and I. M. Erdelyi, Higher Transcendental Functions. Vol. I, McGraw-Hill, New York, NY,
USA, 1953.