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# Lerch's Φ and the Polylogarithm at the Negative Integers

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## Abstract

At the negative integers, there is a simple relation between the Lerch Φ function andthe polylogarithms. The literature has a formula for the polylogarithm at the negativeintegers, which utilizes the Stirling numbers of the second kind. Starting from thatformula, we deduce a simple closed formula for the Lerch Φ function at the negativeintegers, where the Stirling numbers are not needed. Leveraging that finding, we alsoproduce alternative formulae for thek-th derivatives of the cotangent and cosecant (ditto,tangent and secant), as simple functions of the negative polylogarithm and the Lerch Φ,respectively, which is a testament to the importance of these functions (they are lessexotic than they seem). Lastly, we rewrite the Hurwitz zeta function at the positiveintegers using this new fact.
Lerch’s Φ and the Polylogarithm at the Negative Integers
Jose Risomar Sousa
September 17, 2021
Abstract
At the negative integers, there is a simple relation between the Lerch Φ function and
the polylogarithm. The literature has a formula for the polylogarithm at the negative
integers, which utilizes the Stirling numbers of the second kind. Starting from that
formula, we can deduce a simple closed formula for the Lerch Φ function at the negative
integers, where the Stirling numbers are not needed. Leveraging that ﬁnding, we also
produce alternative formulae for the k-th derivatives of the cotangent and cosecant (ditto,
tangent and secant), as simple functions of the negative polylogarithm and Lerch Φ,
respectively, which is evidence of the importance of these functions (they are less exotic
than they seem). Lastly, we present a new formula for the Hurwitz zeta function at the
positive integers using this novelty.
1 Introduction
As seen in reference [4], it’s possible to derive a formula for the Hurwitz zeta function at
the positive integers with results from both [2] and [3]. To greatly summarize the reasoning
presented there, if kis a positive integer greater than one, then:
Z1
0
uk(1 cos 2π n u) cot πu du H(n)
πk!
π
b(k1)/2c
X
j=1
(1)j(2π)2jζ(2j+ 1)
(k2j)! ,
which implies the following approximation (with the equality only valid for integer k):
Z1
0
(uku)(1cos 2π n u) cot πu du k!
π
b(k1)/2c
X
j=1
(1)j(2π)2jζ(2j+ 1)
(k2j)! =Z1
0
(uku) cot πu du,
which in turn justiﬁes the formula1shown next.
1The formulae that can be derived with this method are not unique and the one shown may be the simplest.
1
For every integer kgreater than one and every non-integer complex b:
ζ(k, b) = 1
2bk+(2πi)k
4 Lik+1 e2πib
(k1)! +e2πib
k
X
j=1
δ1j+ Lij+1(e2πib)
(j1)!(kj)! !
i(2πi)k
2Z1
0
k
X
j=1 δ1j+ Lij+1(e2πib)ukje2πib u e2πib
(j1)!(kj)! cot πu du
The discovery of a new, possibly ﬁrst, closed formula for the Lerch transcendent function
at the negative integers was made possible through the analysis of the above formula.
The big breakthrough is really in the following straightforward identity, which only works
for the analytic continuation of the Lerch Φ and the polylogarithm functions at the negative
integers, and makes it possible to obtain the former as a sum of the latter:
k
X
j=0
Lijebukj
j!(kj)! =eb
k!Φ(eb,k, u + 1) (1)
In fact, this identity also applies to the sum of Lerches:
k
X
j=0
Φeb,j, vukj
j!(kj)! =1
k!Φ(eb,k, u +v)(2)
Last but not least, from two known facts from the literature, namely, a recurrence for the
Bernoulli polynomials and a relation between them and the Hurwitz zeta function:
Bk(u+v) = k!
k
X
j=0
Bj(v)ukj
j!(kj)! and Bj(v) = j ζ(1 j, v),
one may conclude that,
ζ(k, u +v) = uk+1
k+ 1 +k!
k
X
j=0
ζ(j, v)ukj
j!(kj)! ,
and from this recurrence, one can obtain a natural expression relating the Hurwitz and the
Riemann zeta functions, which completes the picture:
ζ(k, u + 1) = uk+1
k+ 1 +k!
k
X
j=0
ζ(j)ukj
j!(kj)! (3)
Now, back to the Hurwitz formula, from relation (1) we obtain,
k
X
j=1
Lij+1 e2πibukj
(j1)!(kj)! =e2πib
(k1)!Φ(e2πib,k+ 1, u + 1),
2
which leads to the below analytic continuation for the Hurwitz zeta, which then holds for
<(k)>1:
ζ(k, b) = 1
2bk+(2πi)k
4(k1)! e2πib+e4πibΦ(e2πib,k+ 1,2) + Lik+1 e2πib
i(2πi)k
2(k1)! Z1
0uk1e2πib u e2πib
+e2πib(u+1) Φ(e2πib,k+ 1, u + 1) e4πibΦ(e2πib,k+ 1,2)cot πu du (4)
Since this formula depends on the Lerch’s Φ analytic continuation at the negative integers,
we can rewrite (4), if can we ﬁgure out Lerch’s Φ at the negative integers.
Relation (1) may be the counterpart to the functional equation that relates the Lerch Φ
to the Hurwitz zeta at the positive integers, as demonstrated in a previous paper5. This new
relation therefore presupposes the existence of another relation between the polylogarithm and
the Hurwitz zeta at the negative integers.
2 Stirling numbers of the second kind
The existing formula for the polylogarithm Lij+1(z), available in the literature, makes use
of the Stirling numbers of the second kind (the S(j, q) in the curly brackets):
Lij+1(z) =
j
X
q=1
(q1)!j
q z
1zq
(5)
Let’s see how to obtain the Lerch Φ from it. Making z=e2πib, we have:
z
1z=1 + icot π b
2
Therefore, back to equation (1):
k
X
j=1
Lij+1 e2πibuj
(j1)!(kj)! =
k
X
j=1
j
X
q=1
(q1)!j
q1 + icot π b
2q
k
X
j=1
j
X
q=1
(q1)!uj
(j1)!(kj)!j
q1 + icot π b
2q
k
X
q=1
(q1)! 1 + icot π b
2qk
X
j=q
uj
(j1)!(kj)!j
q(6)
3
Now we need to study the second sum from equation (6) and see if it’s possible to rewrite
it. An expression for a similar sum exists in the literature,
k
X
j=q
1
j!(kj)!j
q=1
k!k+ 1
q+ 1,
but this new one is much more complicated.
3 Binomial formula for Stirling numbers
Let’s try and rewrite the below ﬁnite sum,
k
X
j=q
uj
(j1)!(kj)!j
q(7)
First oﬀ, the literature provides us with the below relation:
k
X
j=qj
qxj
j!=(ex1)q
q!,
therefore,
X
k=q
k
X
j=qj
qxj
j!
ykj
(kj)! =ey(ex1)q
q!,
and diﬀerentiating with respect to x,
X
k=q
k
X
j=qj
qxj1
(j1)!
ykj
(kj)! =eyex(ex1)q1
(q1)!
Now, making x=u z and y=z, we have:
X
k=q
zk1
k
X
j=qj
quj1
(j1)!(kj)! =e(u+1)z(eu z 1)q1
(q1)! =
q1
X
j=0
(1)q1je(u+1+j u)z
j!(q1j)! ,
and hence diﬀerentiating k1 times with respect to z, we conclude that,
k
X
j=q
uj
(j1)!(kj)!j
q=u
(k1)!
q
X
j=1
(1)qj(j u + 1)k1
(j1)!(qj)! ,(8)
To obtain a more proper version of this formula, we integrate as below,
Zu
0
k
X
j=q
xj1
(j1)!(kj)!j
qdx =1
(k1)! Zu
0
q
X
j=1
(1)qj(j x + 1)k1
(j1)!(qj)! dx,
4
which gives us the neat expression below, which holds for every non-negative integer qand
every u:
k
X
j=q
uj
j!(kj)!j
q=1
k!
q
X
j=0
(1)qj(j u + 1)k
j!(qj)! (9)
Though it’s not going to be used here, another pattern similar to the binomial theorem
emerges in the factorial power of the sum of two numbers, xand y:
(x+y)(k)=k!
k
X
j=0
x(j)y(kj)
j!(kj)!, where x(j)=x!
(xj)!
4 Lerch’s Φat the negative integers
Now, if we combine equations (1), (6) and (8), we obtain:
Φ(e2πib,k+ 1, u + 1) = e2πib
k
X
q=1
(q1)! 1 + icot π b
2qq
X
j=1
(1)j(j+u)k1
(j1)!(qj)! (10)
For integer bthe formula is not deﬁned as the cotangent is inﬁnity, so we can’t extract the
Hurwitz zeta at the negative integers from it. But from the relation,
Φ(e2πib,k+ 1,1) = e2πibLik+1 e2πib,
we can derive the polylogarithm, which however is just a rewrite of equation (5) with an
expression for S(k, q) known from the literature, which nonetheless conﬁrms equation (10):
Lik+1 e2πib=
k
X
q=1
(q1)! 1 + icot π b
2qq
X
j=1
(1)jjk1
(j1)!(qj)! (11)
Looking at (10) and (11) now, it might look simple to go from the latter straight to the
former without having to ﬁgure (8), but that’s misleading.
Finally, we can turn (10) into another form, which holds for every non-negative integer k:
Φ(z, k, u) = 1
z1
k
X
q=0
q!z
z1qq
X
j=0
(1)j(j+u)k
j!(qj)! (12)
It’s interesting to note how much simpler the analytic continuation of the Lerch Φ at the
negative integers is than the proper function at the positive integers. And also how strikingly
similar it is to the power series for the Lerch Φ at the positive integers available in the literature:
Φ(z, k, u) = 1
z1
X
q=0
q!z
z1qq
X
j=0
(1)j(j+u)k
j!(qj)!
5
5 Derivatives of trigonometric functions
In his paper On the Hurwitz function for rational arguments 1, Victor Adamchik provides
the ﬁrst ever formula for the intricate patterns of the k-th derivatives of the cotangent. It
looks like this:
dk(cot a x)
dxk= (2 ia)k(i+ cot a x)
k
X
q=1
q!k
q1 + icot a x
2q
It’s possible to express this formula as a simple function of the polylogarithm. First, we
rewrite it as:
dk(cot a x)
dxk= (2 ia)k(i+ cot a x)
k
X
q=1
q!1icot a x
2qq
X
j=1
(1)jjk1
(j1)!(qj)! (13)
Secondly, we note how similar it looks to the polylogarithm from (11):
Lik+1 e2ia x=
k
X
q=1
(q1)! 1icot a x
2qq
X
j=1
(1)jjk1
(j1)!(qj)!
If the above polylog is diﬀerentiated once with respect to xand transformed,
Like2ia x=1
1e2ia x
k
X
q=1
q!1icot a x
2qq
X
j=1
(1)jjk1
(j1)!(qj)!,(14)
an alternative expression is obtained for the polylogarithm of order k, which, however, is not
exactly equal to form (11). That stems from a property of polylogs, that when diﬀerentiated
they yield the next order polylog. Finally, comparing the two expressions, (13) and (14), we
conclude that:
dk(cot a x)
dxk=iδ0k2i(2 ia)kLike2ia x, where δ0k= 1 iﬀ k= 0 (15)
To obtain the cosecant, we can resort to a simple logic:
cos a x +isin a x
sin a x =eia x
sin a x =i+ cot a x 1
sin a x =eia x (i+ cot a x)
Then the Leibniz rule for the derivative of a product of two functions leads to:
dk
dxk1
sin a x=2ieia x
k
X
q=0
k! (ia)kq
q!(kq)! (2 ia)qLiqe2ia x
Lastly, formula (1) allows us to rewrite the above expression as:
dk
dxk1
sin a x=2i(2 ia)keia x Φe2ia x ,k, 1
2,(16)
which holds for every non-negative integer k.
6
5.1 Tangent and secant
To be able to obtain the tangent and secant, ﬁrst we need to produce a formula for the
cotangent and cosecant of a translated arc. Adamchik’s formula1becomes:
dk(cot (a x +b))
dxk= (2 ia)k(i+ cot (a x +b))
k
X
q=1
q!1icot (a x +b)
2qq
X
j=1
(1)jjk1
(j1)!(qj)!
The polylog formula then changes to:
Like2i(a x+b)=1
1e2i(a x+b)
k
X
q=1
q!1icot (a x +b)
2qq
X
j=1
(1)jjk1
(j1)!(qj)!,
and then the ﬁnal formula is not diﬀerent from the simple case:
dk(cot (a x +b))
dxk=iδ0k2i(2 ia)kLike2i(a x+b), where δ0k= 1 iﬀ k= 0
Similarly, the cosecant for a translated arc is:
dk
dxk1
sin (a x +b)=2i(2 ia)kei(a x+b)Φe2i(a x+b),k, 1
2
Finally, to obtain the tangent and secant, we just need to set bto π/2. And since the
formulae for the translated arc are not very diﬀerent from b= 0, for the tangent we have:
dk(tan (a x +b))
dxk=iδ0k+ 2 i(2 ia)kLike2i(a x+b), where δ0k= 1 iﬀ k= 0, (17)
and for the secant:
dk
dxk1
cos (a x +b)= 2 (2 ia)kei(a x+b)Φe2i(a x+b),k, 1
2(18)
It’s surprising that these derivatives can be expressed by means of negative Lerch and
polylogs. For example, the negative polylog is known to yield the derivatives of a simple
exponential function at a point, but not the derivative itself:
dk
dxkx
ea x+b1x=0
=kδ1k+ Lik+1 ebak1
7
6 Hurwitz zeta at the positive integers
Now that fomula (10) is known, the Hurwitz zeta formula at the positive integers greater
than one can be rewritten with only elementary functions references:
ζ(k, b) = 1
2bk+(2πi)k
4(k1)!
e2πib+
k
X
q=1
(q1)! 1 + icot π b
2qq
X
j=1
(1)je2πib(j+ 1)k1+jk1
(j1)!(qj)!
i(2πi)k
2(k1)! Z1
0 uk1e2πib u e2πib+
k
X
q=1
(q1)! 1 + icot π b
2qq
X
j=1
(1)je2πib u(j+u)k1e2πib(j+ 1)k1
(j1)!(qj)! !cot πu du
The pattern of the above formula becomes clearer and looks better if we include the outside
terms into the summation symbol, assuming the convention that (1)!/(1)! = 1:
ζ(k, b) = 1
2bk+(2πi)k
4(k1)!
k
X
q=0
(q1)! 1 + icot π b
2qq
X
j=0
(1)je2πib(j+ 1)k1+jk1
(j1)!(qj)!
i(2πi)k
2(k1)! Z1
0
k
X
q=0
(q1)! 1 + icot π b
2qq
X
j=0
(1)je2πib u(j+u)k1e2πib(j+ 1)k1
(j1)!(qj)! cot πu du
One of the advantages of this new formula is the fact it allows one to get rid of its non-real
parts more easily, though the resulting formula is inevitably more complicated.
6.1 When the parameter bis a half-integer
The below result stems from e2πib=1 and cot π b = 0 when bis a half-integer:
ζ(k, b) = 1
2bk(2πi)k
4(k1)!
1 +
k
X
q=1
(q1)! 1
2qq
X
j=1
(1)j(j+ 1)k1jk1
(j1)!(qj)!
i(2πi)k
2(k1)! Z1
0
uk1e2πib u +1+
k
X
q=1
(q1)! 1
2qq
X
j=1
(1)je2πib u(j+u)k1+ (j+ 1)k1
(j1)!(qj)!
cot πu du
7 A new formula for the Hurwitz zeta
In [4], we had created a generating function for the Hurwitz zeta function, f(x). When b
is not a half-integer, the expression is:
f(x) =
X
k=2
xkζ(k, b) = x2
2b(xb)1
2 sin πb
πx sin πx
sin π(xb)
πx Z1
0sin 2πu(xb)
sin 2π(xb)sin 2πbu
sin 2πb cot πu du (19)
8
The k-th derivative of f(x) yields the Hurwitz zeta function of order k:
ζ(k, b) = f(k)(0)
k!
And now that we know how to diﬀerentiate the cosecant successively, it’s possible to pro-
duce an explicit formula from f(x), again through the Leibniz rule. However, to make this
process simpler, we resort to two artiﬁces. First, to get rid of the extra xfactor in the integral,
we divide f(x) by xand take the (k1)-th derivative instead of the k-th. Second, to avoid the
complications of diﬀerentiating the sine, we replace it with an equivalent sum of exponential
functions.
The ﬁrst and second parts of (19) are straightforward, they coincide with the terms outside
of the integral from (4), that is:
1
k!
dk
dxk1
2 sin πb
πx sin πx
sin π(xb)x=0
=(2πi)k
4(k1)! e2πib+e4πibΦ(e2πib,k+ 1,2) + Lik+1 e2πib
The same isn’t true for the third part of (19), since f(x) was created using a diﬀerent
process than (4) (see [4] for details). The integrals evaluate to the same number, but the
integrands are not the same.
After all is put together, the ﬁnal formula holds for any positive integer kgreater than one,
and any 2bthat’s not an integer:
ζ(k, b) = 1
2bk+(2πi)k
4(k1)! e2πib+e4πibΦ(e2πib,k+ 1,2) + Lik+1 e2πib+
i(2πi)ke2πib
4Z1
0
k
X
j=1
2jukje2πib u (1)kje2πib u
(j1)!(kj)! Φe4πib,j+ 1,1
2cot πu du (20)
Finally, by using the relation (2), an analytic continuation can be produced:
ζ(k, b) = 1
2bk+(2πi)k
4(k1)! e2πib+e4πibΦ(e2πib,k+ 1,2) + Lik+1 e2πib+
i(4πi)ke2πib
4(k1)! Z1
0e2πib uΦe4πib,k+ 1,u+ 1
2e2πib uΦe4πib,k+ 1,u+ 1
2cot πu du
9
References
[1] Adamchik, Victor S. On the Hurwitz function for rational arguments, Applied Mathematics
and Computation, Volume 187, Issue 1, 2007, Pages 3–12.
[2] Risomar Sousa, Jose Generalized Harmonic Numbers, eprint arXiv:1810.07877, 2018.
[3] Risomar Sousa, Jose Generalized Harmonic Progression Part II, eprint arXiv:1902.01008,
2019.
[4] Risomar Sousa, Jose The Hurwitz Zeta Function at the Positive Integers, eprint
arXiv:1902.06885, 2019.
[5] Risomar Sousa, Jose Lerch’s Φand the Polylogarithm at the Positive Integers, eprint
arXiv:2006.08406, 2020.
10
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Article
Using functional properties of the Hurwitz zeta function and symbolic derivatives of the trigonometric functions, the function ζ(2n + 1, p/q) is expressed in several ways in terms of other mathematical functions and numbers, including in particular the Glaisher numbers.
• Risomar Sousa
Risomar Sousa, Jose Generalized Harmonic Numbers, eprint arXiv:1810.07877, 2018.
• Risomar Sousa
Risomar Sousa, Jose Generalized Harmonic Progression Part II, eprint arXiv:1902.01008, 2019.
• Risomar Sousa
Risomar Sousa, Jose The Hurwitz Zeta Function at the Positive Integers, eprint arXiv:1902.06885, 2019.
• Risomar Sousa
Risomar Sousa, Jose Lerch's Φ and the Polylogarithm at the Positive Integers, eprint arXiv:2006.08406, 2020.