# Relations among Sums of Reciprocal Powers—Part II

Authors: José María Amigó
Journal: International Journal of Mathematics and Mathematical Sciences
 (2008)

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## Abstract

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.


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## Body

## 1. Introduction

In [1], we studied some arithmetic relations among the
classical numbers:(1.1)λ(n)=∑ν=0∞1(2ν+1)n(n≥2),L(n)=∑ν=0∞(−1)ν(2ν+1)n(n≥1).In this paper, we extend this
analysis to the remaining(1.2)ζ(n)=∑ν=1∞1νn(n≥2),η(n)=∑ν=1∞(−1)ν−1νn(n≥1).Although the numbers λ(n),ζ(n), and η(n) are related to each other through the
identities(1.3)λ(n)=(1−2−n)ζ(n),η(n)=(1−21−n)ζ(n),(n≥2)(η(1)=ln2) and thus(1.4)η(n)+ζ(n)=2λ(n)(n≥2),we will use all of them in order
to keep the algebraic expressions as simple as possible.Fork≥2, define(1.5)D(k)=2πk−1∑j=1∞λ(2j)2j⋯(2j+k−1).It will be shown below that these
numbers can be alternatively expressed as(1.6)D(k)=12(k−1)!∫0πtk−1cott2dt=2k−1(k−1)!∫0π/2tk−1cottdt.Indefinite integrals of this type
were considered by Ramanujan [2, page 260]. The constants D(k) have the property of relating the values of ζ-numbers (eventually, η- or λ-numbers) with odd argument to the elementary
values ζ(2n)=(2π)2n|B2n|/2(2n)! (where B2n are the Bernoullian numbers) as, for example,
in (1.7)∑j=0n(−1)n−jπ2n−2j(2n−2j)!η(2j+1)=2π∑j=1n(−1)jD(2j)ζ(2n−2j+2)+(−1)nπD(2n+2)
(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 D(k) by means of sums of reciprocal powers with odd
arguments and, conversely, η(2n+1),ζ(2n+1),  and λ(2n+1) via D(k) 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 λ(n), η(n), ζ(n), and D(n). 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.Sections3, 4  deal with the numbers D(k) and with the classical sums of reciprocal
powers, respectively. In particular, Section 3 is mainly devoted to the
calculation of several series containing D(k).
In Section 4, the focus changes onto some particular series whose terms contain λ-,η-,ζ-, and L-numbers like, for example, (1.8)∑n=1∞(ζ(2n)−ζ(2n+1))=12,∑n=1∞(ζ(2n+1)−λ(2n))=0.Finally, in the brief Section5, the generating
function of the numbers D(k) is expressed using the Psi (Digamma) function ψ(x)=Γ'(x)/Γ(x).


## 2. Main Statements

Define the generating functionsΛ(x),ℰ(x), and 𝒵(x) by(2.1)Λ(x)=∑n=2∞λ(n)xn−1,ℰ(x)=∑n=1∞η(n)xn−1,𝒵(x)=∑n=2∞ζ(n)xn−1,where, in principle, x∈ℂ  (since limn→∞λ(n)=limn→∞η(n)=limn→∞ζ(n)=1, these formal power series converge only for |x|<1). Furthermore, denote by Λ+(x) and Λ−(x) the even and odd parts, respectively, of Λ(x) and similarly for ℰ±(x) and 𝒵±(x). Then, the following identities hold [3, 4.3.67/68/70]: (2.2)tanπx2=4π∑n=1∞λ(2n)x2n−1=4πΛ−(x)(|x|<1),(2.3)cscπx=1πx+2π∑n=1∞η(2n)x2n−1=1πx+2πℰ−(x)(0<|x|<1),(2.4)cotπx=1πx−2π∑n=1∞ζ(2n)x2n−1=1πx−2π𝒵−(x)(0<|x|<1).Owing to (1.3) and (1.4), the above generating
functions fulfill the trivial relations(2.5)ℰ(x)=𝒵(x)−𝒵(x2)+ln2,Λ(x)=𝒵(x)−12𝒵(x2),ℰ(x)+𝒵(x)=2Λ(x)+ln2,respectively.On substituting the definitions ofλ(n),η(n),  and ζ(n) into the corresponding generating functions,
we find the following expansions in partial fractions:(2.6)Λ(x)=∑ν=0∞(12ν+1−x−12ν+1),ℰ(x)=∑ν=1∞(−1)ν−1ν−x,𝒵(x)=∑ν=1∞(1ν−x−1ν).In particular, the
expansions(2.7)ℰ+(x)=12(ℰ(x)+ℰ(−x))=∑ν=1∞(−1)ν−1νν2−x2,𝒵+(x)=12(𝒵(x)+𝒵(−x))=∑ν=1∞(νν2−x2−1ν)will be used below.Proposition 2.1.
The
integral representation (1.6) holds.Proof.
Letf(t)=tan(πt/2). By repeated partial integration, one
gets(2.8)∫01tk−1f(1−t)dt=f(0)k+f'(0)k(k+1)+f′′(0)k(k+1)(k+2)+⋯=∑ν=0∞f(ν)(0)k(k+1)⋯(k+ν),where according to the Taylor
expansion (2.2),(2.9)f(ν)(0)={0,ifν=2j,j∈ℕ,4(2j−1)!λ(2j)/π,ifν=2j−1,j∈ℕ.Hence,(2.10)2k−1(k−1)!∫0π/2tk−1tan(π2−t)dt=πk2(k−1)!∫01tk−1tanπ2(1−t)dt=πk2(k−1)!∑j=1∞4(2j−1)!λ(2j)/πk(k+1)⋯(k+2j−1)=2πk−1∑j=1∞λ(2j)2j⋯(2j+k−1).=D(k).Since, for0≤x≤π/2 and n∈ℕ, (2.11)xncotx≲12n(π2)n+1,it follows that limn→∞D(n+1)=0 faster (in fact, much faster) than πn+2/8nn!.Proposition 2.2.
The numbersD(k)(k≥2) can be evaluated in terms of η(1),η(3),…,η(2⌊k/2⌋−1),  supplemented with λ(k) if k is odd, by the formula(2.12)D(k)=∑j=1⌊k/2⌋(−1)j−1πk−2j+1(k−2j+1)!η(2j−1)+{0,(keven),(−1)(k−1)/22λ(k),(kodd).Proof.
For eachν∈ℕ,
we have upon k−1 integrations by parts(2.13)2k(k−1)!∫0π/2tk−1sin2νtdt=∑j=1⌊k/2⌋(−1)jπk−2j+1(k−2j+1)!(−1)νν2j−1+{0,(keven),(−1)(k−1)/21−(−1)ννk,(kodd).Sum now both sides on  ν,1≤ν≤N, and use the identity [4, 1.342(1)](2.14)∑ν=1Nsin2νt=sin(N+1)tsinNtcsct=12(cost−cos(2N+1)t)csct,to get on the left
side(2.15)2k(k−1)!∑ν=1N∫0π/2tk−1sin2νtdt=2k−1(k−1)!∫0π/2(tk−1cott−tk−1sintcos(2N+1)t)dt.
To finish the proof, letN→∞ and use the Riemann-Lebesgue lemma.In particular,(2.16)D(2)=π1!η(1)=πln2as in [4, 3.747(7)],
and(2.17)D(3)=π22ln2−2λ(3)=π22ln2−74ζ(3)as in [5, 4.2(3)]. Other expressions
for D(k) different  from (2.12) can be found in [4, 3.748(2)].Due to the definition (1.5) and
formula (2.12), each
number D(k) embodies a relation among all elementary
values λ(2j) and a finite number of their nonelementary
counterparts λ(2j+1),
namely,(2.18)∑j=1∞λ(2j)2j⋯(2j+2k−1)=12∑j=1k(−1)j−1(1/π)2j−2(2k−2j+1)!η(2j−1),∑j=1∞λ(2j)2j⋯(2j+2k)=12∑j=1k(−1)j−1(1/π)2j−2(2k−2j+2)!η(2j−1)+(−1)kλ(2k+1)π2k,for k≥1.Define next the generating function(2.19)𝒟(x)=∑n=2∞D(n)(ix)n−1.Owing to the vanishing rate of
the coefficients D(n), this power series is convergent for all x∈ℂ. Then,(2.20)𝒟(x)=∫0π/2dtcott∑n=2∞(i2xt)n−1(n−1)!=∫0π/2(ei2xt−1)cottdt.Furthermore, let 𝒟+(x) and 𝒟−(x) be the even and odd parts of 𝒟(x), that is,(2.21)𝒟+(x)=∑n=1∞D(2n+1)(ix)2n=∑n=1∞(−1)nD(2n+1)x2n=∫0π/2(cos2xt−1)cottdt,𝒟−(x)=∑n=1∞D(2n)(ix)2n−1=i∑n=1∞(−1)n−1D(2n)x2n−1=i∫0π/2sin2xtcottdt.
Thus, if x is meant to be real, 𝒟+(x) and (1/i)𝒟−(x) are the real and imaginary part, respectively,
of 𝒟(x).Proposition 2.3.
The following relations hold:(2.22)𝒟+(x)=cos(πx)ℰ+(x)+𝒵+(x)−ln2,𝒟−(x)=isin(πx)ℰ+(x).Proof.
We claim
that(2.23)𝒟(x)=eiπxℰ+(x)+𝒵+(x)−ln2.In fact, from(2.24)𝒟(x)=∫0π/2(ei2xt−1)cottdt=2∑ν=1∞∫0π/2(ei2xt−1)sin2νtdt,∫0π/2ei2xtsin2νtdt=(−1)ν−1νeiπx+ν2(ν2−x2),∫0π/2sin2νtdt=1+(−1)ν−12ν,we obtain(2.25)𝒟(x)=eiπx∑ν=1∞(−1)ν−1νν2−x2+∑ν=1∞(νν2−x2−1ν)−∑ν=1∞(−1)ν−1ν.Comparison with (2.7)
proves the claim. Take now even and odd parts.Solving forℰ+(x) and 𝒵+(x) in (2.22), we get the
identities (2.26)ℰ+(x)=1icsc(πx)𝒟−(x)(0<|x|<1),(2.27)𝒵+(x)=𝒟+(x)+icot(πx)𝒟−(x)+ln2(0<|x|<1),which lead to a kind of converse
of Proposition 2.2.Proposition 2.4.
The numbersη(2n+1) and ζ(2n+1),n≥1, can be expressed in terms of the D(k) and the elementary values η(2k) and ζ(2k) by(2.28)η(2n+1)=2π∑j=1n(−1)j−1η(2n−2j+2)D(2j)+(−1)nπD(2n+2),ζ(2n+1)=2π∑j=1n(−1)j−1ζ(2n−2j+2)D(2j)+(−1)nD(2n+1)+(−1)n+1πD(2n+2).Proof.
(1)  Equation
(2.26) reads explicitly (use η(0)=1/2 in (2.3)),(2.29)∑n=0∞η(2n+1)x2n=1icsc(πx)𝒟−(x)=2iπ∑n=0∞η(2n)x2n−1∑j=1∞i(−1)j−1D(2j)x2j−1=2π∑n=0∞∑j=1n+1(−1)j−1η(2n−2j+2)D(2j)x2n=1πD(2)+2π∑n=1∞(∑j=1n(−1)j−1η(2n−2j+2)D(2j)+(−1)n2D(2n+2))x2n.
(2) The
second identity follows from (2.27) since (use ζ(0)=−1/2 in (2.4)),(2.30)cot(πx)𝒟−(x)=−2π∑n=0∞ζ(2n)x2n−1∑j=1∞i(−1)j−1D(2j)x2j−1=−2iπ∑n=0∞(∑j=1n+1(−1)j−1ζ(2n−2j+2)D(2j))x2n=iπD(2)−2iπ∑n=1∞(∑j=1n(−1)j−1ζ(2n−2j+2)D(2j)+(−1)n+12D(2n+2))x2n,Remark 2.5.
Equation (1.7) is nothing else but the formula(2.31)cos(πx)ℰ+(x)=1icot(πx)𝒟−(x)written explicitly.
Indeed,(2.32)cos(πx)ℰ+(x)=∑n=0∞(−1)nπ2n(2n)!x2n∑j=1∞η(2j−1)x2j−2=∑n=0∞(∑j=0n(−1)n−jπ2n−2j(2n−2j)!η(2j+1))x2n=η(1)+∑n=1∞(∑j=0n(−1)n−jπ2n−2j(2n−2j)!η(2j+1))x2n,and cot(πx)𝒟−(x) was calculated in the last proof. For n=1 one gets the representation(2.33)ζ(3)=29π2ln2−43πD(4)=29π2ln2−169π∫0π/2t3cottdt.Equation (1.7) and also the first formula of
Proposition 2.4 show that η(2n+1) (and, for that case, ζ(2n+1) and λ(2n+1)) can be expressed in terms of only D(2),D(4),…,D(2n+2). Furthermore, adding the two formulas of
Proposition 2.4 and using (1.4), we obtain the following result.Corollary 2.6.
Forn≥1,(2.34)λ(2n+1)=2π∑j=1n(−1)j−1λ(2n−2j+2)D(2j)+(−1)n2D(2n+1).Forn=1, we recover (2.17).


## 3. Summation Formulas for theD(k)s

Before deriving more relations involving the sums of
reciprocal powers, we obtain next some “summation" formulas for the
numbers D(k). Two (integral) summation formulas follow
trivially from the very definition  of the generating function 𝒟(x) and (2.20), namely,(3.1)∑n=2∞D(n)=𝒟(1i)=∫0π/2(e2t−1)cottdt,∑n=2∞(−1)nD(n)=−𝒟(i)=∫0π/2(1−e−2t)cottdt.Also, (3.2)∑n=1∞(−1)nD(2n+1)=𝒟+(1)=∫0π/2(cos2t−1)cottdt=−1,(3.3)∑n=1∞(−1)nD(2n+1)22n=𝒟+(12)=12∫0π/2(cost−1)cottdt=ln2−1,(3.4)∑n=1∞(−1)n+1D(2n)=1i𝒟−(1)=∫0π/2sin2tcottdt=π2,(3.5)∑n=1∞(−1)n+1D(2n)22n−1=1i𝒟−(12)=∫0π/2sintcottdt=1.Other similar series can be also
straightforwardly deduced after differentiating (2.21),(3.6)∑n=1∞(−1)n+1nD(2n+1)x2n−1=12𝒟+'(x)=∫0π/2tsin2xtcottdt,∑n=1∞(−1)n+1(2n−1)D(2n)x2n−2=1i𝒟−'(x)=2∫0π/2tcos2xtcottdt,and substituting fixed values for x.
In particular, the series(3.7)∑n=1∞(−1)n+1(2n−1)D(2n)=1i𝒟−'(1)=2∫0π/2tcos2tcottdt=D(2)−π2=π(ln2−12)(where cos2t=1−2sin2t,
(1.6) and (2.16) were used) will be needed below. Note that from
(3.7) and
(3.4), it follows(3.8)D(2)=2∑n=2∞(−1)nnD(2n).Furthermore, from (2.20), we have(3.9)𝒟(x+12)−𝒟(x−12)=2i∫0π/2ei2xtcostdt=2ieiπx−i2x1−4x2,so, after separating real and
imaginary parts, the equations (3.10)𝒟+(x+12)−𝒟+(x−12)=21−4x2(2x−sinπx),(3.11)𝒟−(x+12)−𝒟−(x−12)=i21−4x2cosπxhold for x∈ℝ.
Letting x→1/2 one recovers (3.2) and (3.4).Proposition 3.1.
The
following identities hold:
(1)
Forn=0,1,2,…,(3.12)∑k=1∞(−1)k+1(2n+2k2n+1)D(2n+1+2k)22n+1+2k=(−1)n4(∑k=0n(−1)kπ2k+1(2k+1)!−2).
In particular, forn=0,(3.13)∑k=1∞(−1)k+1kD(2k+1)22k+1=18(π−2).(2)
Forn=0,1,2,…,(3.14)∑k=1∞(−1)k+1(2n+2k−12n)D(2n+2k)22n+2k=(−1)n2∑k=0n(−1)kπ2k(2k)!.Proof.
(1) In
fact,(3.15)𝒟+(x+12)−𝒟+(x−12)=∑k=1∞(−1)kD(2k+1)((x+12)2k−(x−12)2k)=2∑k=1∞(−1)kD(2k+1)∑n=0k−1(2k2n+1)x2n+12−(2k−2n−1)=22n+3∑n=0∞(∑k=1∞(−1)n+k(2n+2k2n+1)D(2n+1+2k)22n+1+2k)x2n+1.Comparison with (3.10), that
is,(3.16)11−4x2(2x−sinπx)=((2−π)x+∑n=1∞(−1)n+1π2n+1(2n+1)!x2n+1)∑j=0∞(2x)2j=∑n=0∞(2−π)(2x)2n+1+∑n=1∞(∑k=1n(−1)k+1π2k+1(2k+1)!)(2x)2n+1=22n+1∑n=0∞(2+∑k=0n(−1)k+1π2k+1(2k+1)!)x2n+1,where |x|<1 yields the result.
(2) Analogously to (1), the summation formula follows comparing(3.17)𝒟−(x+12)−𝒟−(x−12)=22n+2i∑n=0∞(∑k=1∞(−1)n+k+1(2n+2k−12n)D(2n+2k)22n+2k)x2nwith (3.11), that
is,(3.18)2i1−4x2cosπx=2i∑n=0∞(−1)nπ2n(2n)!x2n∑j=0∞(2x)2j=22n+1i∑n=0∞(∑k=0n(−1)kπ2k(2k)!)x2n,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,(3.19)∫0π/2[(π2−t)n−(π2)n]tan(π2−t)dt=∫0π/2[∑k=1n(nk)(π2)n−k(−t)k]tan(π2−t)dt=∑k=1n(−1)k(nk)(π2)n−k∫0π/2tkcottdt=∑k=1n(−1)k(nk)(π2)n−kk!2kD(k+1).Therefore, changing the variable
in the integral, we get(3.20)∑k=1n(−1)k−1πn−k(n−k)!D(k+1)=2nn!∫0π/2((π2)n−tn)tantdt.


## 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 seriesζ(n),λ(n),  or η(n).
For completeness, we will also include the series L(n),
what requires the consideration of a new generating function 𝒞(x) defined in [1]. As a first example, we will prove the following
proposition.Proposition 4.1.
The following identities  hold
true:(4.1)∑n=1∞ζ(2n)22n=12,∑n=1∞ζ(2n+1)22n+1=ln2−12,∑n=1∞λ(2n)22n=π8,∑n=1∞λ(2n+1)22n+1=14ln2,∑n=1∞η(2n)22n=12(π2−1),∑n=0∞η(2n+1)22n+1=12,∑n=1∞L(2n)22n=24ln(1+2),∑n=0∞L(2n+1)22n+1=π28.Note, in particular, the series(4.2)∑n=2∞ζ(n)2n=ln2,∑n=1∞η(n)2n=π4.Proof.
(1) Settingx=1/2 in (2.27), we get(4.3)𝒵+(12)=𝒟+(12)+ln2.Thus, by (3.3),(4.4)∑n=1∞ζ(2n+1)22n+1≡12𝒵+(12)=12𝒟+(12)+12ln2=ln2−12.On the other hand (see
(2.4)),(4.5)∑n=1∞ζ(2n)22n≡12𝒵−(12)=12.Adding up the even and odd parts,
we get(4.6)∑n=2∞ζ(n)2n=ln2,∑n=2∞(−1)nζ(n)2n=1−ln2.
(2) Settingx=1/2 in (2.26), we get(4.7)ℰ+(12)=1i𝒟−(12).Hence, by (3.5),(4.8)∑n=0∞η(2n+1)22n+1≡12ℰ+(12)=12i𝒟−(12)=12.On the other hand, by
(2.3),(4.9)∑n=1∞η(2n)22n≡12ℰ−(12)=12(π2−1).Adding up the even and odd parts,
we get(4.10)∑n=1∞η(n)2n=π4,∑n=1∞(−1)n+1η(n)2n=1−π4.
(3) In
[1], it is proved
that, for |x|<1,(4.11)∑n=1∞λ(2n+1)x2n≡Λ+(x)=𝒞+(x)+1itanπx2𝒞−(x)holds, where(4.12)𝒞+(x)=12∫0π/2cosxt−1sintdt,𝒞−(x)=i2∫0π/2sinxtsintdt.Hence,(4.13)∑n=1∞λ(2n)22n≡12Λ−(12)=π8by (2.2), and(4.14)∑n=1∞λ(2n+1)22n+1≡12Λ+(12)=12𝒞+(12)+12i𝒞−(12)=14ln2,since(4.15)𝒞+(12)=12∫0π/2cos(t/2)−1sintdt=12ln21+2,𝒞−(12)=i2∫0π/4sectdt=i2lntan3π8=i2ln(1+2).Adding up the even and odd
parts,(4.16)∑n=2∞λ(n)2n=14(π2+ln2),∑n=2∞(−1)nλ(n)2n=14(π2−ln2).
(4) In
[1], it is proved
that (4.17)∑n=0∞L(2n+1)x2n≡ℒ+(x)=π4secπx2,(4.18)∑n=1∞L(2n)x2n−1≡ℒ−(x)=1isecπx2𝒞−(x),for |x|<1. Hence,(4.19)∑n=1∞L(2n)22n≡12ℒ−(12)=12isecπ4𝒞−(12)=122ln(1+2),∑n=0∞L(2n+1)22n+1≡12ℒ+(12)=π8secπ4=π28.Adding up the even and odd
parts,(4.20)∑n=1∞L(n)2n=π28+122ln(1+2),∑n=1∞(−1)n+1L(n)2n=π28−122ln(1+2).The next theorem shows that more
challenging results can be achieved by being slightly more sophisticated.Theorem 4.2.
The following equalities hold
true:(4.21)∑(ζ(2n)−ζ(2n+1))=12,∑(η(2n)−η(2n−1))=ln2−12,∑(λ(2n)−λ(2n+1))=ln2−1,∑(L(2n)−L(2n−1))=12(1−ln2),∑(ζ(2n+1)−λ(2n))=0,∑(ζ(2n+1)−η(2n))=12,∑(ζ(2n+1)−L(2n+1))=12(π2−1),∑(ζ(2n)−λ(2n+1))=ln2,∑(λ(2n+1)−η(2n))=1−ln2,∑(λ(2n+1)−L(2n+1))=π4−ln2,∑(ζ(2n)−η(2n−1))=12+ln2,∑(λ(2n)−η(2n−1))=ln2,∑(L(2n−1)−η(2n−1))=ln2−12,∑(ζ(2n)−L(2n))=12(1+ln2),∑(ζ(2n+1)−L(2n))=12ln2,∑(λ(2n)−L(2n))=12ln2,∑(λ(2n+1)−L(2n))=12(1−ln2),∑(L(2n)−η(2n))=12(1−ln2),∑(L(2n)−η(2n−1))=12ln2,where all the series start with n=1.Proof.
We will proceed
left to right and top to bottom.
(1) From
(2.4) and (2.27), it follows(4.22)∑n=1∞(ζ(2n)−ζ(2n+1))x2n=x𝒵−(x)−𝒵+(x)=12−π2xcot(πx)−𝒟+(x)−icot(πx)𝒟−(x)−ln2for |x|<1. Thus, by the Tauber theorem [6], (3.2) and
(3.7),(4.23)∑n=1∞(ζ(2n)−ζ(2n+1))=limx→1−∑n=1∞(ζ(2n)−ζ(2n+1))x2n=12−𝒟+(1)−ln2−limx→1−cot(πx)(π2x+i𝒟−(x))=12+1−ln2−(12+iπ𝒟−'(1))=12,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(4.24)∑n=1∞(η(2n)−η(2n−1))x2n−1=ℰ−(x)−xℰ+(x)=−12x+π2csc(πx)+ixcsc(πx)𝒟−(x)for 0<|x|<1. Thus, by Tauber's theorem, (3.4) and
(3.7),(4.25)∑n=1∞(η(2n)−η(2n−1))=limx→1−∑n=1∞(η(2n)−η(2n−1))x2n−1=−12+limx→1−csc(πx)(π2+ix𝒟−(x))=−12−iπ(𝒟−(1)+𝒟−'(1))=ln2−12.
(3) Analogously, from (2.2) and
(4.11),(4.26)∑n=1∞(λ(2n)−λ(2n+1))=limx→1−(xΛ−(x)−Λ+(x))=limx→1−(π4xtanπx2−𝒞+(x)+itanπx2𝒞−(x))=−𝒞+(1)+limx→1−tanπx2(π4x+i𝒞−(x))=−𝒞+(1)−(12+2iπ𝒞−'(1))=ln2−1,since (see (4.12))(4.27)𝒞+(1)=12∫0π/2(cott−csct)dt=−12ln2,(4.28)𝒞−'(1)=i12∫0π/2tcottdt=iD(2)=iπ4ln2.
(4) Using
(4.18), (4.17), and (4.28), it follows(4.29)∑n=1∞(L(2n)−L(2n−1))=limx→1−(ℒ−(x)−xℒ+(x))=limx→1−(1isecπx2𝒞−(x)−π4xsecπx2)=limx→1−secπx2(1i𝒞−(x)−π4x)=−2iπ𝒞−'(1)+12=12(1−ln2).
(5) From
(2.27), (2.2), and
(3.7), we get(4.30)∑n=1∞(ζ(2n+1)−λ(2n))=limx→1−(𝒵+(x)−xΛ−(x))=limx→1−(𝒟+(x)+icot(πx)𝒟−(x)+ln2−π4xtanπx2)=𝒟+(1)+ln2+12limx→1−tanπx2(i(cot2πx2−1)𝒟−(x)−π2x)=−1+ln2+12(i2π𝒟−'(1)+1)=0.
(6) From
(2.27), (2.3), and
(3.7), we get(4.31)∑n=1∞(ζ(2n+1)−η(2n))=limx→1−(𝒵+(x)−xℰ−(x))=limx→1−(𝒟+(x)+icot(πx)𝒟−(x)+ln2+12−π2xcscπx)=𝒟+(1)+ln2+12+limx→1−cscπx(icosπx𝒟−(x)−π2x)=𝒟+(1)+ln2+12+iπ𝒟−'(1)+12=12.
(7) From
(2.27), (4.17), and (3.7), we get(4.32)∑n=1∞(ζ(2n+1)−L(2n+1))=limx→1−(𝒵+(x)−ℒ+(x)+L(1))=limx→1−(𝒟+(x)+icot(πx)𝒟−(x)+ln2−π4secπx2+π4)=𝒟+(1)+ln2+π4+12limx→1−secπx2(isinπx2(cot2πx2−1)𝒟−(x)−π2)=−1+ln2+π4+iπ𝒟−'(1)=12(π2−1).
(8) From
(2.4), (4.11), and (4.28), we get(4.33)∑n=1∞(ζ(2n)−λ(2n+1))=limx→1−(x𝒵−(x)−Λ+(x))=limx→1−(12−π2xcotπx−𝒞+(x)−1itanπx2𝒞−(x))=12+12ln2−limx→1−tanπx2(π4x(cot2πx2−1)+1i𝒞−(x))=12(1+ln2)−(12−2iπ𝒞−'(1))=ln2.
(9) From (4.11),
(2.3), and (4.28), we get(4.34)∑n=1∞(λ(2n+1)−η(2n))=limx→1−(Λ+(x)−xℰ−(x))=limx→1−(𝒞+(x)+1itanπx2𝒞−(x)+12−π2xcscπx)=𝒞+(1)+12+limx→1−secπx2(1isinπx2𝒞−(x)−π4xcscπx2)=−12ln2+12−2iπ𝒞−'(1)+12=1−ln2.
(10) From
(4.11), (4.17), and (4.28), we get(4.35)∑n=1∞(λ(2n+1)−L(2n+1))=limx→1−(Λ+(x)−ℒ+(x)+L(1))=limx→1−(𝒞+(x)+1itanπx2𝒞−(x)−π4secπx2+π4)=𝒞+(1)+π4+limx→1−secπx2(1isinπx2𝒞−(x)−π4)=−12ln2+π4−2iπ𝒞−'(1)=π4−ln2.
(11) From
(2.4), (2.26), (3.4), and
(3.7), we get(4.36)∑n=1∞(ζ(2n)−η(2n−1))=limx→1−(𝒵−(x)−xℰ+(x))=limx→1−(12x−π2cotπx−1ixcscπx𝒟−(x))=12−limx→1−cscπx(π2cosπx+1ix𝒟−(x))=12+1iπ(𝒟−(1)+𝒟−'(1))=12+ln2.
(12) From
(2.2), (2.26),
(3.4), and (3.7), we get(4.37)∑n=1∞(λ(2n)−η(2n−1))=limx→1−(Λ−(x)−xℰ+(x))=limx→1−(π4tanπx2−1ixcscπx𝒟−(x))=limx→1−secπx2(π4sinπx2−12ixcscπx2𝒟−(x))=1iπ(𝒟−(1)+𝒟−'(1))=ln2.
(13) From
(4.17), (2.26), and (3.7), we get(4.38)∑n=0∞(L(2n+1)−η(2n+1))=limx→1−(ℒ+(x)−ℰ+(x))=limx→1−(π4secπx2−1icsc(πx)𝒟−(x))=12limx→1−secπx2(π2+icscπx2𝒟−(x))=1iπ𝒟−'(1)=ln2−12.
(14) From
(2.4), (4.18), and (4.28), we get(4.39)∑n=1∞(ζ(2n)−L(2n))=limx→1−(𝒵−(x)−ℒ−(x))=limx→1−(12x−π2cotπx+isecπx2𝒞−(x))=12−limx→1−secπx2(π4cosπxcscπx2−i𝒞−(x))=12−2iπ𝒞−'(1)=12(1+ln2).
(15) From
(2.27), (4.18), (3.7), (4.12), and (4.28), we get(4.40)∑n=1∞(ζ(2n+1)−L(2n))=limx→1−(𝒵+(x)−xℒ−(x))=limx→1−(𝒟+(x)+icot(πx)𝒟−(x)+ln2−1ixsecπx2𝒞−(x))=𝒟+(1)+ln2+ilimx→1−secπx2(sinπx2(cot2πx2−1)𝒟−(x)+x𝒞−(x))=−1+ln2+i2π(𝒟−'(1)−𝒞−(1)−𝒞−'(1))=12ln2.
(16) From
(2.2), (4.18), and (4.28), we get(4.41)∑n=1∞(λ(2n)−L(2n))=limx→1−(Λ−(x)−ℒ−(x))=limx→1−(π4tanπx2+isecπx2𝒞−(x))=limx→1−secπx2(π4sinπx2+i𝒞−(x))=2iπ𝒞−'(1)=12ln2.
(17) From
(4.11), (4.18), (4.27), and (4.12), we get(4.42)∑n=1∞(λ(2n+1)−L(2n))=limx→1−(Λ+(x)−xℒ−(x))=limx→1−(𝒞+(x)+1itanπx2𝒞−(x)−1ixsecπx2𝒞−(x))=𝒞+(1)+1i𝒞−(1)limx→1−secπx2(sinπx2−x)=12(1−ln2).
(18) From
(4.18), (2.3), and
(4.28), we get(4.43)∑n=1∞(L(2n)−η(2n))=limx→1−(ℒ−(x)−ℰ−(x))=limx→1−(1isecπx2𝒞−(x)+12x−π2cscπx)=12−limx→1−secπx2(i𝒞−(x)+π4cscπx2)=12−2iπ𝒞−'(1)=12(1−ln2).
(19) From
(4.18), (2.26), (4.28), (3.4), and (3.7), we get(4.44)∑n=1∞(L(2n)−η(2n−1))=limx→1−(ℒ−(x)−xℰ+(x))=limx→1−(1isecπx2𝒞−(x)−1ixcscπx𝒟−(x))=1ilimx→1−secπx2(𝒞−(x)−x2cscπx2𝒟−(x))=−2iπ𝒞−'(1)+1iπ𝒟−(1)+1iπ𝒟−'(1)=12ln2.


## 5. Enterψ(x)

Letψ(x),x∈ℂ, be the logarithmic derivative of the Gamma
function. Then, if |x|<1 [7, 1.17(5)](5.1)ψ(1+x)=−γ+∑n=2∞(−1)nζ(n)xn−1,hence [7, 1.7(11)](5.2)𝒵(x)=−ψ(1−x)−γ=−ψ(x)−πcotπx−γ,ℰ(x)=𝒵(x)−𝒵(x2)+ln2=−ψ(x)+ψ(x2)+πcscπx+ln2.It follows(5.3)𝒵+(x)=−ψ+(x)−γ,ℰ+(x)=−ψ+(x)+ψ+(x2)+ln2,where ψ+(x)≡(ψ(x)+ψ(−x))/2.Substitution of these expressions forℰ+(x) and 𝒵+(x) in (2.22) and comparison with
(2.21)  yields(5.4)𝒟+(x)=∫0π/2(cos2xt−1)cottdt=−2cos2πx2ψ+(x)+cosπx(ψ+(x2)+ln2)−γ−ln2,1i𝒟−(x)=∫0π/2sin2xtcottdt=sinπx(−ψ+(x)+ψ+(x2)+ln2) respectively.



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*Source: 421478-2008-10-27.xml*