# Relation of the Cyclotomic Equation with the Harmonic and Derived Series

Authors: Luis J. Boya; Cristian Rivera
Journal: The Scientific World Journal
 (2015)

---

## Abstract

We associate some (old) convergent series related to definite integrals with the cyclotomic equationx
m
-
1
=
0, for several natural numbers m; for example, for m
=
3, x
3
-
1
=
(
x
-
1
)
(
1
+
x
+
x
2
) leads to ∫
0
1
d
x
(
1
/
(
1
+
x
+
x
2
)
)
=
π
/
(
3
3
)
=
(
1
-
1
/
2
)
+
(
1
/
4
-
1
/
5
)
+
(
1
/
7
-
1
/
8
)
+
⋯. In some cases, we express the results in terms of the Dirichlet characters. Generalizations for arbitrary m are well defined but do imply integrals and/or series summations rather involved.


---

## Body

## 1. Introduction

Nicola Oresmes proved ca. 1350, thedivergence of theharmonic series (e.g., see [1, page 183]). Indeed, today we know precisely how it doesdiverge ([2, page 14]; see also [3]):
(1)
Harm
N
∶
=
∑
n
=
1
N
‍
1
n
.ForN
≫
1
(2)
Harm
N
⟶
log
⁡
N
+
γ
+
1
2
N
-
1
12
N
N
+
1
-
⋯
,
where γ is the Euler-Mascheroni constant (written and named C at times; for a recent reference, see [4]):
(3)
γ
=
lim
⁡
N
→
∞
∑
n
=
1
N
‍
1
n
-
log
⁡
N
≈
.
577215
…
.We do not know, even today (winter 2015), whetherγ is rational or not.Later, in 1668 Mercator (= Kremer) proved [1, page 185] the convergence of the alternative series (of even/odd numbers) and summed it (i.e., taking the N
→
∞ limit):
(4)
1
-
1
2
+
1
3
-
1
4
+
⋯
+
1
2
N
-
1
-
1
2
N
+
⋯
⟶
∫
0
1
‍
d
x
1
+
x
=
log
⁡
2
.Apparently, some work from India preceded the (even later) so-called Gregory-Leibniz formula ([1] again, page 184) for another alternative series (of inverse of odd numbers):
(5)
1
-
1
3
+
1
5
-
1
7
+
⋯
+
1
4
N
-
3
-
1
4
N
-
1
+
⋯
⟶
∫
0
1
‍
d
x
1
+
x
2
=
π
4
.In this paper we interpret  (4) and (5) as arising from the cyclotomic equation of roots of the unity (Gauss; see, e.g., [5]):
(6)
x
m
-
1
=
0
m
∈
N
;
m
=
2,3
,
4
,
…
;
x
m
-
1
=
x
-
1
1
+
x
+
x
2
+
⋯
+
x
m
-
1
.
x
2
m
-
1
=
x
-
1
1
+
x
2
+
x
4
+
.
.
.
+
x
2
m
-
1
.In our cases,m
=
2,4, respectively, for (4) and (5), and then perhaps we can write anew the results in terms of some natural arithmetic functions. In (6), the roots ≠
±
1 are complex conjugate pairs, as (6) describes a real equation, so the roots have modulus 1, and all of them lie in the complex unit circle ≈
S
1.The main purpose of this paper is to generalize these results for other (generic) natural numbersm
∈
N. We will try to relate the series to some definite integrals.We anticipate already here part of the workings form
=
2 and 4.Form
=
2, x
2
-
1
=
(
x
-
1
)
(
x
+
1
); we always leave out the x
=
1 root. The inverse of (
x
+
1
) enters into (4) and we proceed to the four operations:(a)
inversion, (
x
+
1
)
=
(
1
+
x
)
→
(
1
+
x
)
-
1;
(b)
expansion, (
1
+
x
)
-
1
=
1
-
x
+
x
2
-
x
3
+
x
4
-
⋯ (|
x
|
<
1);
(c)
term-by-termintegration, ⇒
x
-
x
2
/
2
+
x
3
/
3
-
x
4
/
4
+
⋯;
(d)
taking thelimits
x
=
1 minus x
=
0, ⇒
(
1
-
1
/
2
)
+
(
1
/
3
-
1
/
4
)
+
(
1
/
5
-
1
/
6
)
⋯.Then we obtain the above result form
=
2:
(7)
x
+
1
=
1
+
x
⟹
1
1
+
x
⟹
1
-
x
+
x
2
-
⋯
⟹
x
-
x
2
2
+
x
3
3
-
x
4
4
+
⋯
⟹
x
=
1
minus
x
=
0
=
∑
n
=
1
∞
‍
-
1
n
+
1
n
=
log
⁡
(
2
)
in terms of the arithmetic modulated “sign” function (= even/odd) f
(
n
)
∶
=
(
-
1
)
n
+
1
/
n. Note thatafter the integration, the limit x
=
1 can be taken. Equation (7) involves the series
(8)
∑
n
=
1
∞
‍
1
2
n
-
1
-
1
2
n
=
∫
0
1
‍
d
x
1
1
+
x
=
log
⁡
(
2
)
expressed as a genuine definite integral.Similarly, form
=
4, it is x
4
-
1
=
(
x
2
-
1
)
(
x
2
+
1
), separating the roots ±
1 and ±
i; now (
1
+
x
2
)
-
1 enters into (5), which becomes, again after inversion, expansion, integration, and taking limits,
(9)
1
+
x
2
-
1
⟹
1
-
1
3
+
1
5
-
1
7
+
1
9
-
1
11
k
+
⋯
=
π
4
which can be written as the (alternative) difference between the two series and of course also as one integral
(10)
∑
n
=
1
∞
‍
1
4
n
-
3
-
1
4
n
-
1
=
∫
0
1
‍
d
x
1
1
+
x
2
=
π
4
.And it can be given now in terms of the so-called Dirichlet characters (see below).In this paper, as said, we will generalize the constructions above (form
=
2,4) for any integer m
∈
N (in principle) and include (when possible) the appropriate Dirichlet character.But first we want to fix our notation. For eachm
∈
N, m
>
1, we will consider first thefinite sums, up to m
N terms; so we define primarily
(11)
∑
m
N
≡
1
+
1
2
+
⋯
+
1
m
N
-
m
-
1
+
1
m
N
-
m
-
2
+
⋯
+
1
m
N
.For example,∑
3
(
N
)
≡
1
+
1
/
2
+
1
/
3
+
⋯
+
1
/
N
+
⋯
+
1
/
(
3
N
-
2
)
+
1
/
(
3
N
-
1
)
+
1
/
3
N; so these sumsdiverge, for N arbitrarily large, and indeed as (2) indicates,
(12)
∑
m
N
⟶
log
⁡
m
N
+
γ
+
O
1
m
N
=
log
⁡
N
+
γ
+
log
⁡
m
(from now on we will keeponly thedominant and theconstant terms in thedivergent expansion (2)).Also, we definepartial bounded sums
∑
m
i
=
∑
m
i
(
N
) for i
=
1,2
,
…
,
m as displaced sums:
(13)
∑
m
1
=
1
1
+
1
m
+
1
+
1
2
m
+
1
+
⋯
+
1
m
N
-
m
-
1
,
∑
m
2
=
1
2
+
1
m
+
2
+
⋯
+
1
m
N
-
m
-
2
,
⋮
∑
m
m
-
1
=
1
m
-
1
+
1
2
m
-
1
+
⋯
+
1
m
N
-
1
.And finally,∑
m
m
=
1
/
m
+
1
/
2
m
+
⋯
+
1
/
m
N
=
(
1
/
m
)
(
1
+
1
/
2
+
⋯
+
1
/
N
) so
(14)
for
N
≫
1
∑
m
m
N
⟶
log
⁡
N
m
+
γ
m
+
O
1
N
.Therefore,∑
m
(
N
) (in (11)) is the sum of all ∑
m
i
(
N
) with i
:
{
1,2
,
…
,
m
}:
(15)
∑
m
N
=
Sum
i
=
1
m
∑
m
i
N
=
∑
m
1
+
∑
m
2
+
⋯
+
∑
m
m
.Note that these sums make sense, spite diverging, because all have only positive terms and they are finite sums.The convergence or divergence of the above series is usually self-explained: convergence occurs always for the decreasing alternative series (Leibniz), but as the convergence is not absolute, but conditional, the ordering in the series should be maintained. The higher divergence behaves, if at all, with a factor∝
log
⁡
(
N
): divergences are no higher than logarithmic. We will try to be careful in subtracting two divergent series. We use a philosophy close to physics (in particular, to quantum electrodynamics or q. e. d.): we first truncate the series, taking afixed upper bound N
≫
1 (called thecut-off; in physics this process is calledregularization). Then we subtract other series, also divergent but with a similar type of divergence (this is calledrenormalization): the result should be convergent (radiative corrections); see, for example, the book by Schwinger [6].For a modern treatment of the Dirichlet series consult [7].


## 2. The Case form
=
2: Four Methods

Here we repeat thelog
⁡
(
2
) result for the case m
=
2  (7) byfour different methods, because eventually the four might be useful.(1) Redundance (Direct) Proof.
∑
2
‍
(
N
) and ∑
2
2
‍
(
N
), defined above, diverge in a known way, for fixed (and large) N and, as said, ((12), (14))
(16)
∑
2
1
=
∑
2
-
∑
2
2
⟶
log
⁡
(
N
)
+
log
⁡
2
+
γ
-
1
2
log
⁡
N
-
γ
2
kkkkkkkk
=
1
2
log
⁡
N
+
γ
2
+
log
⁡
2
.So(17)
∑
2
1
-
∑
2
2
⟶
N
⟶
∞
=
log
⁡
(
2
)
directly, without any integration or (infinite) series summation.(2) Integration Proof. Trivial here, as it is already (4):
(18)
∫
0
1
‍
d
x
1
+
x
=
log
⁡
2
.(3) Summation Proof. We have theconvergent series (so now N
=
∞)
(19)
∑
2
1
-
∑
2
2
=
1
-
1
2
+
1
3
-
1
4
+
⋯
=
∑
n
=
1
∞
‍
1
2
n
-
1
-
1
2
n
=
1
4
∑
n
=
1
∞
‍
1
n
n
-
1
/
2
which can be explicitly done using thedigamma function
Ψ
(
z
) (see, e.g., [8, page 258]):
(20)
Ψ
(
z
)
∶
=
d
log
⁡
Γ
z
d
z
=
Γ
′
z
Γ
z
z
∉
Z
-
∶
=
Z
∖
N
(for Γ
(
z
) is the ordinary, Euler’s gamma function), with help of the expression [8, page 259, formula 6.3.16]
(21)
∑
n
=
1
∞
‍
z
n
n
+
z
=
Ψ
(
z
)
+
γ
+
1
z
for z
=
-
1
/
2. In total we get, as expected,
(22)
∑
n
=
1
∞
‍
1
n
n
-
1
/
2
=
4
log
⁡
2
.
(4) Use of Hansen Formula. This formula, again depending on the Digamma function Ψ
(
z
), reads (cf. [9]; notice the sum starts in zero; x
,
y
,
and
z are some parameters)
(23)
∑
n
=
0
∞
‍
1
n
x
+
y
2
-
z
2
=
1
2
x
z
Ψ
y
+
z
x
-
Ψ
y
-
z
x
.We put ourn
′ above in ∑
1
‍
(
1
/
n
′
(
n
′
-
1
/
2
)
) as (n
′
=
n
+
1), because n runs now from zero; hence this is equivalent to (23) with the parameter values x
=
1, y
=
3
/
4, and z
=
1
/
4. So our sum is (with Ψ
(
1
)
=
-
γ and Ψ
(
1
/
2
)
=
-
γ
-
2
log
⁡
(
2
))
(24)
∑
=
2
Ψ
(
1
)
-
Ψ
1
2
=
4
log
⁡
(
2
)
as it should be.Of the four methods, the most common and “easy” is the second (integration); it can, in principle (i.e., if the integrand is known and the integration feasible), always be used. TheMathematica Computer Programs give many integrals and double sums also directly.Please note in this (initial)m
=
2 case that what appears as result is thesign function
f
(
n
)
∶
=
∑
n
=
1
∞
‍
(
(
-
1
)
n
+
1
/
n
); later, for some m
>
2, (including m
=
4) this will become a Dirichlet character: here we have, as résumé (to repeat),
(25)
∑
1
2
-
∑
2
2
=
log
⁡
2
=
∑
n
=
1
∞
‍
-
1
n
+
1
n
≡
∑
odd
-
∑
even
so NO Dirichlet character this time.


## 3. The Case form
=
3,4
,
6

We discussm
=
3 first. Now (
x
3
-
1
)
=
(
x
-
1
)
(
1
+
x
+
x
2
) as the roots are +
1, ω
=
exp
⁡
⁡
(
2
π
i
/
3
) and conjugate ω
¯; here ω is equivalent to a plane rotation by 120
∘. By direct integration, we obtain at once
(26)
∫
0
1
‍
d
x
1
+
x
+
x
2
=
π
3
3
which is a particular case of
(27)
∫
‍
1
A
d
x
for
A
=
x
-
ω
x
-
ω
¯
=
x
2
-
Tr
⁡
ω
x
+
1
,
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
i
Tr
⁡
ω
=
ω
+
ω
¯
.For a generalpositive integer m we obtain, after an easy calculation (ω is root of x
m
-
1
=
0),
(28)
∫
0
1
‍
d
x
1
-
Tr
⁡
ω
x
+
x
2
=
π
2
sin
(
2
π
/
m
)
·
m
-
2
m
.This formula (28) can be used at once for m
=
3,4, and 6, yielding (26) and
(29)
∫
0
1
‍
d
x
1
-
Tr
⁡
ω
·
x
+
x
2
=
π
4
m
=
4
2
π
3
3
m
=
6
.later, we shall need alsoI
=
∫
0
1
‍
d
x
(
1
/
1
-
Tr
⁡
ω
x
+
x
2
) for ω
=
exp
⁡
⁡
(
2
π
i
r
/
m
), for r natural between 1 and m
-
1. The integral is now
(30)
I
=
π
2
sin
2
π
r
/
m
·
m
-
2
r
m
1
≤
r
≤
m
-
1
.Coming back tom
=
3, the specific series is 1
-
S
+
S
2
-
S
3
+
S
4
-
⋯, with S
≡
x
+
x
2; after developing, we get (after term-by-term integration and taking limits 1, 0)
(31)
1
-
1
2
+
1
4
-
1
5
+
⋯
+
1
3
N
-
2
-
1
3
N
-
1
+
⋯
=
∑
n
=
1
∞
‍
1
3
n
-
2
-
1
3
n
-
1
≡
∑
3
1
-
∑
3
2
=
π
3
3
.Hence∑
3
1
-
∑
3
2 finishes the calculation for m
=
3 (as the third sum ∑
3
3 trivially computable).In terms of the [10] Dirichlet character mod 3, namely χ
2
(
3
), we have
(32)
∑
3
1
-
∑
3
2
=
π
3
3
=
∑
n
=
1
∞
‍
χ
2
3
n
n
,
where of course χ
2
(
3
)
(
1,2
,
3,4
,
5,6
)
=
(
1
,
-
1,0
,
1
,
-
1,0
) (so 3-periodic). We can even use Dirichlet’s L-functions, which will include the 1
/
n factors, but we refrain from doing that, as it does not illuminate the matter any further.Asrecapitulation, the series are obtained from the polynomial of roots ≠1:
(33)
m
=
3
P
2
(
x
)
=
x
2
+
x
+
1
after INVersion, EXPansion, INTegration, and TAKing x
=
1. In principle, the result of the series summation can be also obtained from the Hansen formula (22).Finally for thism
=
3 case, we quote the fourdiverging series for completeness (the signal → meaning just the limit for N
≫
1):
(34)
∑
3
1
-
∑
3
2
=
∑
n
=
1
∞
‍
1
3
n
-
2
-
1
3
n
-
1
=
π
3
3
=
∑
n
=
1
∞
‍
χ
2
(
3
)
(
n
)
n
as thecomplete solution for the m
=
3 case. We quote the following fourdivergent series for later use, still in this m
=
3 case (the limit → meaning just N
≫
1):
(35)
∑
3
=
1
+
1
2
+
⋯
+
1
3
N
⟶
log
⁡
N
+
γ
+
log
⁡
3
,
∑
3
3
=
1
3
+
1
6
+
⋯
+
1
3
N
⟶
1
3
log
⁡
N
+
γ
3
,
∑
3
1
=
1
4
+
1
7
+
⋯
+
1
3
N
-
2
⟶
1
3
log
⁡
N
+
γ
3
+
1
2
log
⁡
3
+
π
6
3
,
∑
3
2
=
1
2
+
1
5
+
⋯
+
1
3
N
-
1
⟶
1
3
log
⁡
(
N
)
+
γ
3
+
1
2
log
⁡
3
-
π
6
3
.Now, form
=
4, we have first the natural cyclotomic expression
(36)
I
x
4
-
1
=
(
x
2
-
1
)
(
x
2
+
1
)
(we write (
I
) because this is not the only possible factorization to be used). Repeating the steps as before for m
=
3, our first final result is here:
(37)
∑
4
1
-
∑
4
3
=
∑
n
=
1
∞
‍
1
4
n
-
3
-
1
4
n
-
1
=
∫
0
1
‍
d
x
x
2
+
1
=
arctan
1
=
π
4
with χ
2
m
=
4
(
1,2
,
3,4
,
5,6
,
7,8
)
=
(
1,0
,
-
1,0
,
1,0
,
-
1,0
) being periodic mod 4 (and restricted multiplicative). Note also why do we get i
=
1 and 3 in ∑
4
i (not 2!) mod 4: the expansion is for 1
/
(
1
+
x
2
)
≈
1
-
x
2
+
x
4
-
x
6, and so forth, so it is witheven powers only, so with onlyodd powers after integration!We are done with this, as∑
4, ∑
4
0, and ∑
4
2 are automatically obtainable.Thesecond factorization of x
4
-
1 is obtained by separating only the x
=
1 root:
(38)
II
x
4
-
1
=
x
-
1
1
+
x
+
x
2
+
x
3
.AsB
∶
=
(
1
+
x
+
x
2
+
x
3
) contains the x
=
-
1 root, one writes B
=
(
x
+
1
)
(
1
+
x
2
), where the integral can be computed at once (indeed, it is indicated above). The final result for factorization (
II
) is
(39)
∫
0
1
‍
d
x
1
+
x
+
x
2
+
x
3
=
1
4
log
⁡
(
2
)
+
π
8
which is aredundant result, as defines ∑
4
1
-
∑
4
2 computable from the above calculation (
I
). In fact, we end up this m
=
4 case by writing the analogy to (35):
(40)
∑
4
4
=
1
4
+
1
8
+
⋯
+
1
4
N
⟶
1
4
log
⁡
⁡
N
+
γ
4
,
∑
4
2
=
1
2
+
1
6
+
⋯
+
1
4
N
-
2
⟶
1
2
∑
2
1
=
1
4
log
⁡
⁡
N
+
γ
4
+
1
2
log
⁡
⁡
2
,
∑
4
1
=
1
2
+
1
5
+
⋯
+
1
3
N
-
1
⟶
1
4
log
⁡
⁡
(
N
)
+
γ
4
+
3
4
log
⁡
⁡
(
2
)
+
π
8
,
∑
4
3
⟶
1
4
log
⁡
(
N
)
+
γ
4
+
3
4
log
⁡
(
2
)
-
π
8
.The nonprime structure ofm
implies only a difference calculation, as it was also the case for m
=
3, prime.So the full solution for them
=
4 case has one redundancy (two factorizations), and it is done (solved) once a single calculation is made; for example, ∫
0
1
‍
d
x
(
1
/
1
+
x
2
)
=
π
/
4.Now it is the turn ofm
=
6. But here there are also several factorizations, as for the m
=
4 case above. The simplest is (perhaps)
(41)
I
Firstly
is
x
6
-
1
=
x
3
-
1
x
3
+
1
.As(
1
+
x
3
)
=
(
1
+
x
)
(
1
-
x
+
x
2
), the integral is easy, with this factoring:
(42)
∫
0
1
‍
d
x
1
+
x
3
=
1
3
log
⁡
2
+
π
3
3
.The operations INV, EXP, INT, and TAKx
=
1 applied to the polynomial P
3
(
x
)
=
1
+
x
3 yield the infinite but convergent sum
(43)
∑
6
1
-
∑
6
4
=
∑
n
=
1
∞
‍
1
6
n
-
5
-
1
6
n
-
2
=
log
⁡
⁡
2
3
+
π
3
3
.Notice again thejump, now by three: it is due to the cubic x
3
n terms in 1
/
(
1
+
x
3
).All other factorizations are thereforeredundant; we just write them:
(44)
II
x
6
-
1
=
x
-
1
1
+
x
+
x
2
+
x
3
+
x
4
+
x
5
⟹
∑
6
1
-
∑
6
2
III
x
6
-
1
=
x
2
-
1
1
+
x
2
+
x
4
⟹
∑
6
1
-
∑
6
3
.Redundancy arises because from (39) one obtains ∑
6
1, as ∑
6
4 is directly computable (Section 2); so, in (
II
), we know already the result, as ∑
6
2 is again trivial, the same in (
III
), as ∑
6
3 is again directly computable.It is remarkable here that in the original expansion(
x
6
-
1
)
=
(
x
3
-
1
)
(
x
3
+
1
) the simplest factorization (
x
3
+
1
)
=
(
x
+
1
)
(
1
-
x
+
x
2
) yields, for P
2
(
x
)
=
1
-
x
+
x
2, the double series expansion
(45)
1
+
1
2
-
1
4
+
1
5
+
1
7
+
1
8
-
1
10
+
1
11
±
⋯
=
∑
6
1
+
∑
6
2
-
∑
6
4
-
∑
6
5
=
∫
0
1
‍
d
x
1
-
x
+
x
2
=
2
π
3
3
which is again aredundant calculation, as included in the (summed) ∫
0
1
‍
d
x
(
1
/
1
+
x
3
).To finish these simple cases, we just write down the nontrivial summations:(46)
∑
6
1
N
=
1
6
log
⁡
N
+
γ
6
+
log
⁡
⁡
2
3
+
log
⁡
⁡
3
4
+
log
⁡
⁡
2
3
+
π
4
3
,
∑
6
4
=
log
⁡
(
N
)
6
+
γ
6
+
log
⁡
(
3
)
4
-
6
π
12
3
,
∑
6
1
-
∑
6
5
=
log
⁡
⁡
3
4
.And, in thism
=
6 case, we get also
(47)
∑
6
1
-
∑
6
4
=
π
3
3
,
computing ∑
6
2 and ∑
6
4
=
(
1
/
2
)
(
π
/
3
3
) as in Section 1.In terms of some Dirichlet functions we get(48)
∑
6
1
-
∑
6
5
=
∑
χ
2
(
6
)
n
,
as
χ
2
6
(
1,2
,
3,4
,
5,6
)
=
(
1,0
,
0,0
,
-
1,0
)
=
log
⁡
⁡
3
4
.For a modern treatment of sums involving harmonic numbers, see [4].


## 4. General Numbers

Here the cyclotomic equation is, form
=
5,
(49)
x
5
-
1
=
x
-
1
1
+
x
+
x
2
+
x
3
+
x
4
=
x
-
1
x
-
ω
x
-
ω
¯
x
-
ω
2
x
-
ω
2
¯
,
where ω
=
exp
⁡
⁡
(
2
π
i
/
5
) = Rotation by 72
∘ and ω
2
=
exp
⁡
⁡
(
4
π
i
/
5
) = Rotation by 144
∘.We obtain easily(50)
cos
⁡
72
∘
=
5
-
1
4
,
cos
⁡
144
∘
=
-
1
-
5
4
.The integralT
≡
∫
0
1
‍
(
d
x
/
1
+
x
+
x
2
+
x
3
+
x
4
) can be done as the real denominator splits in two real ones and quadratic, but we abstain to write explicitly (
ω
=
e
2
π
i
/
5
)
(51)
1
+
x
2
+
x
3
+
x
4
=
1
-
x
Tr
⁡
ω
+
x
2
1
-
x
Tr
⁡
ω
2
+
x
2
.The expression for the integral is too long to be written. In summation terms, it is1
/
(
1
+
x
+
x
2
+
x
3
+
x
4
)
=
⋯
=
(
1
-
1
/
2
)
+
(
1
/
6
-
1
/
7
)
+
⋯
=
∑
n
=
1
∞
‍
(
1
/
(
5
n
-
4
)
-
1
/
5
n
-
3
) still computable (e.g., with Mathematica) but still too long. The final result will be
(52)
∫
0
1
‍
d
x
1
1
+
x
+
x
2
+
x
3
+
x
4
=
∑
5
1
-
∑
5
2
.This does not correspond to none mod 5 Dirichlet characters, but one can always put a (Dirichlet) function(53)
f
2
5
1,2
,
3,4
,
5,6
,
7
,
…
=
1
,
-
1,0
,
0,0
,
1
,
-
1
periodic
mod
⁡
5
.The simplest summation (according toMathematica) is
(54)
∑
5
1
-
∑
5
4
=
1
+
5
π
5
2
-
5
.Notice that the factorization in (49) does not allow us to write it as difference between two series, because the coefficients are not integer numbers.For the next prime, namely,m
=
7, we have three couples of complex roots, plus the x
=
1 value: the sextic integral has not been attempted, but the summation can be again done; we refrain from elaborating.This is the general trend for prime numbersp; there are (
p
-
1
)
/
2 pairs of complex conjugate roots; “a priori,” the only integral versus series is the simplest case, generalizing the above result:
(55)
∫
0
1
‍
d
x
1
1
+
x
+
x
2
+
⋯
+
x
p
-
1
=
∑
p
1
-
∑
p
2
.The number(
p
-
1
)
/
2 coincides also with the Euler number.


## 5. General Case

We deal now with some composite numbers. Compound numbers are easier; we just add a calculation form
=
8, with ϕ
(
8
)
=
8
-
4
=
4:
(56)
x
8
-
1
=
(
x
4
-
1
)
(
x
4
+
1
)
,
x
4
+
1
=
x
2
-
2
x
+
1
x
2
+
2
x
+
1
,
and so
(57)
∫
0
1
‍
d
x
x
4
+
1
=
π
+
log
⁡
3
+
2
2
4
2
=
⋯
=
∑
8
1
-
∑
8
5
=
∑
n
=
1
∞
‍
1
8
n
-
7
-
1
8
n
-
3
=
π
4
2
+
log
⁡
3
+
2
2
4
2
.Another factorization is(
x
8
-
1
)
=
(
x
2
-
1
)
(
1
+
x
2
+
x
4
+
x
6
). It is equivalent to
(58)
∫
0
1
‍
d
x
1
1
+
x
2
+
x
4
+
x
6
=
∑
8
1
-
∑
8
3
=
∑
n
=
1
∞
‍
1
8
n
-
7
-
1
8
n
-
5
.Still, a third factorization is(
x
8
-
1
)
=
(
x
-
1
)
(
1
+
x
+
x
2
+
⋯
+
x
7
), which yields ∑
8
1
-
∑
8
2, so the remaining ∑
8
7 is obtained by difference (with ∑
8
2,4
,
6,8 inmediate). The case m
=
8 is potentially resolved. Again, we refrain from elaborating.Form
=
9, ∑
9
3 and ∑
9
6 can be computed directly, whereas the odd cases ∑
9
1,3
,
5,7 require further work, but it is again feasible, similarly for m
=
10.Asgeneral conclusion, we have shown a remarkable relation between the cyclotomic equation x
m
-
1
=
0 and some series and definite integrals; they go from the simplest integrals (and series) in the literature (like ∫
‍
d
x
(
1
/
(
1
+
x
)
)
=
log
⁡
(
2
)) to very complicated cases, still feasible: the integrals have denominators factoring in quadratic ones, and the series are of the type ∑
‍
(
1
/
a
n
2
+
b
n
+
c
), computable, in principle, by means of the Hansen’s formula.There are, however, some questions left in our work: for example, the series for1
/
(
1
+
x
+
x
2
+
x
3
+
⋯
+
x
q
) we identify it with the series ∑
q
+
1
1; this is correct, but we have checked it “case by case,” offering no general proof, and so forth. Also we feel that some new series might perhaps appear, whenever the quadratic components offer an integer expansion: those are two questions for the future.



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*Source: 950521-2015-01-22.xml*