Bulletin of Mathematical Analysis and Applications
ISSN: 1821-1291, URL: http://www.bmathaa.org
Volume 2 Issue 3(2010), Pages 40-52.
SUMS IN TERMS OF POLYGAMMA FUNCTIONS
ANTHONY SOFO
Abstract. We consider sums involving the product of reciprocal binomial
coefficient and polynomial terms and develop some double integral identities.
In particular cases it is possible to express the sums in closed form, give some
general results, recover some known results in Coffey [8] and produce new
identities.
1. Introduction
In a recent paper Coffey [8], considers summations over digamma and polygamma
functions and develops many results, namely two of his propositions are respectively
equations (58) and (66b)
𝑝
𝑛
∑
( −𝑚
)
(−1)
=
1
+
𝑝
−
2
ln
2
+
2
− 1 𝜁 (𝑚 + 1)
𝑝+1
𝑛(𝑛
+
1)
𝑛=1
𝑚=1
(1.1)
𝑝
∑
1
=1+𝑝−
𝜁 (𝑚 + 1) .
𝑛(𝑛 + 1)𝑝+1
𝑛=1
𝑚=1
(1.2)
∞
∑
and
∞
∑
Coffey [8] also constructs new integral representations for these sums. The major
aim of this paper is to investigate general binomial sums with various parameters
that then enables one to give more general representations of (1.1) and (1.2), thereby
generalizing the propositions of Coffey, both in closed form in terms of zeta functions
and digamma functions at possible rational values of the argument, and in double
integral form. The following definitions will be used throughout this paper. The
generalized hypergeometric representation 𝑝 𝐹𝑞 [⋅, ⋅], is defined as
]
[
∞
∑
𝑎1 , 𝑎2 , ..., 𝑎𝑝 (𝑎1 )𝑛 (𝑎2 )𝑛 ... (𝑎𝑝 )𝑛 𝑧 𝑛
𝑧 =
𝑝 𝐹𝑞
(𝑏1 )𝑛 (𝑏2 )𝑛 ... (𝑏𝑞 )𝑛 𝑛!
𝑏1 , 𝑏2 , ..., 𝑏𝑞 𝑛=0
⎞
⎛
𝑝, 𝑞 ∈ {0, 1, 2, 3...} ; 𝑝 ≤ 𝑞 + 1; 𝑝 ≤ 𝑞 and ∣𝑧∣ < ∞;
⎟
⎜
𝑞 + 1 and ∣𝑧∣ <
⎜
⎟
{𝑝 =
} 1; 𝑝 = 𝑞 + 1, ∣𝑧∣ = 1 and
𝑞
𝑝
⎝
⎠
∑
∑
Re
𝑏𝑚 −
𝑎𝑚 > 0, 𝑏𝑚 ∈
/ {0, −1, −2, −3, ...}
𝑚=1
𝑚=1
2000 Mathematics Subject Classification. Primary 05A10, 11B65, 05A19, 33C20.
Key words and phrases. Double integrals, combinatorial identities, Harmonic numbers,
Polygamma functions, recurrences, hypergeometric functions.
c
⃝2010
Universiteti i Prishtinës, Prishtinë, Kosovë.
Submitted April, 2010. Published June, 2010.
40
POLYGAMMA FUNCTIONS
{
41
Γ(𝑤+𝛼)
Γ(𝑤) ,
for 𝛼 > 0
is Pochhammer’s symbol. The Gamma and
1, for 𝛼 = 0
Beta functions are defined respectively as
∫ ∞
Γ (𝑧) =
𝑤𝑧−1 𝑒−𝑤 𝑑𝑤, for Re (𝑧) > 0,
where (𝑤)𝛼 =
0
and
∫
1
Γ (𝑠) Γ (𝑤)
Γ (𝑠 + 𝑤)
0
for Re (𝑠) > 0 and Re (𝑧) > 0. The well known Riemann zeta function is defined as
∞
∑
1
, Re (𝑧) > 1,
𝜁 (𝑧) =
𝑟𝑧
𝑟=1
𝑤𝑠−1 (1 − 𝑤)
𝐵 (𝑠, 𝑧) =
𝑧−1
𝑑𝑤 =
and we have
∞
∑
(−1)𝑟
𝑟=1
𝑟𝑧
(
)
= 21−𝑧 − 1 𝜁 (𝑧) , Re (𝑧) > 0, 𝑧 ∕= 1.
The generalized harmonic numbers of order 𝛼 are given by
𝑛
∑
1
𝐻𝑛(𝛼) =
for (𝛼, 𝑛) ∈ ℕ := {1, 2, 3, ...} .
𝑟𝛼
𝑟=1
In the case of non integer values we may write the generalized harmonic numbers
in terms of polygamma functions
𝛼
𝐻𝑧(𝛼) = 𝜁 (𝛼) −
(−1) (𝛼−1)
𝜓
(𝑧 + 1) , 𝑧 ∕= {−1, −2, −3, ...}
Γ (𝛼)
(1.3)
and for 𝛼 = 1,
𝐻𝑛(1)
∫1
= 𝐻𝑛 =
0
𝑛
∑
1 − 𝑡𝑛
1
𝑑𝑡 =
= 𝛾 + 𝜓 (𝑛 + 1) ,
1−𝑡
𝑟
𝑟=1
where 𝛾 denotes the Euler-Mascheroni constant, defined by
( 𝑛
)
∑1
𝛾 = lim
− log (𝑛) = −𝜓 (1) ≈ 0.5772156649015.....
𝑛→∞
𝑟
𝑟=1
and where 𝜓 (𝑧) denotes the Psi, or digamma function, defined by
)
∞ (
𝑑
Γ′ (𝑧) ∑
1
1
𝜓 (𝑧) =
log Γ (𝑧) =
=
−
− 𝛾.
𝑑𝑧
Γ (𝑧)
𝑛+1 𝑛+𝑧
𝑛=0
Furthermore
∞ (
∑
1
)
1
𝜓 (𝑧 + 1) =
−
− 𝛾,
𝑛 𝑛+𝑧
𝑛=1
(
)
1
2𝜓 (2𝑧) = 𝜓 (𝑧) + 𝜓 𝑧 +
+ 2 ln 2,
2
and the polygamma functions
∫∞
∞
𝑘+1
𝑘+1 𝑘 −(𝑧+1)𝑡
∑
(−1)
𝑘!
(−1)
𝑡 𝑒
(𝑘)
𝜓 (𝑧 + 1) =
=
𝑑𝑡,
𝑘+1
1 − 𝑒−𝑡
𝑛=1 (𝑛 + 𝑧)
0
(1.4)
(1.5)
(1.6)
42
A. SOFO
see [3]. The Lerch transcedent Φ (𝑧, 𝑠, 𝑏) is defined as
Φ (𝑧, 𝑠, 𝑏) =
∞
∑
𝑧𝑛
𝑠 , ∣𝑧∣ < 1, 𝑏 ∕= 0, −1, −2, ...
(𝑛 + 𝑏)
𝑛=0
and the Hurwitz zeta function
𝜁 (𝑠, 𝑏) = Φ (1, 𝑠, 𝑏) =
∞
∑
1
𝑠 , Re (𝑠) > 1.
(𝑛
+
𝑏)
𝑛=0
The Polylogarithmic function, see [25],
𝐿𝑖𝑘 (𝑧) = 𝑃 𝑜𝑙𝑦𝐿𝑜𝑔 (𝑘, 𝑧) =
∞
∑
𝑧𝑛
𝑛𝑘
𝑛=1
Sums of reciprocals of binomial coefficients appear in the calculation of massive Feynman diagrams [12] within several different approaches: for instance, as
solutions of differential equations for Feynman amplitudes, through a naive 𝜀expansion of hypergeometric functions within Mellin-Barnes technique or in the
framework of recently proposed algebraic approach [11]. There has recently been
renewed interest in the study of series involving binomial coefficients and a number
of authors have obtained either closed form representation or integral representation for some particular cases of these series. The interested reader is referred to
([1, 2, 4, 5, 6, 9, 10, 15, 16, 17, 18, 19, 20, 21, 22, 23, 25, 26, 27]).
The following Lemma and Theorem are the main results presented in this paper.
2. the main results
The following Lemma will be useful in the proof of the main theorem.
Lemma 2.1. Let ∣𝑧∣ ≤ 1, 𝑚 > 1 and 𝑞 ≥ 0. Then
∞
∑
𝑟=1
𝑚−1
𝑧𝑟
(−1)
𝑚 =
(𝑚 − 1)!
(𝑞 + 𝑟)
∫1
𝑧 𝑦 𝑞 (ln (𝑦))
1 − 𝑧𝑦
𝑚−1
𝑑𝑦.
(2.1)
0
1
1−𝑧𝑦
Proof. First we expand
as a power series about 𝑦 = 0, then interchange the
order of integration and summation so that
𝑚−1
(−1)
(𝑚 − 1)!
∫1
𝑧 𝑦 𝑞 (ln (𝑦))
1 − 𝑧𝑦
0
𝑚−1
1
∫
∞
𝑚−1 ∑
(−1)
𝑚−1
𝑑𝑦 =
𝑧 𝑠+1
𝑦 𝑞+𝑠 (ln (𝑦))
𝑑𝑦.
(𝑚 − 1)! 𝑠=0
0
Now we successively integrate (𝑚 − 1) times so that
∫1
𝑚−1
𝑦 𝑞+𝑠 (ln (𝑦))
𝑚−1
𝑑𝑦 =
(−1)
(𝑚 − 1)!
𝑚
(𝑞 + 𝑠 + 1)
0
and replacing the counter 𝑠 + 1 = 𝑟 we obtain (2.1).
The next Lemma deals with four infinite sums.
□
POLYGAMMA FUNCTIONS
43
Lemma 2.2. Let 𝑎 and 𝑟 be positive real numbers. Then
(1)
∞
∑
𝐻𝑟
1
= 𝑎
𝑛 (𝑎𝑛 + 𝑟)
𝑟
𝑛=1
∞
∑
,
(2.2)
𝑛
}
1 { (1)
(−1)
(1)
𝐻 𝑟 − 𝐻𝑟
=
,
2𝑎
𝑎
𝑛 (𝑎𝑛 + 𝑟)
𝑟
𝑛=1
∞
∑
𝑛=1
∞
∑
𝑛=1
1
𝑛 (𝑎𝑛 + 𝑏)
(−1)
𝑚+1
1
= 𝑚
𝑏
(𝑚+1 ( )
∑ 𝑏 𝑠−1
𝑎
𝑠=1
(𝑠)
𝐻𝑏 −
𝑚+1
∑(
𝑎
𝑠=2
𝑏
𝑎
(2.3)
)
)𝑠−1
𝜁 (𝑠)
and
(2.4)
)
(2.5)
𝑛
𝑛 (𝑎𝑛 + 𝑏)
=
𝑚+1
1
𝑏𝑚+1
{
} 𝑚+1
∑
(1)
(1)
𝐻 𝑏 − 𝐻𝑏
+
2𝑎
𝑎
𝑝=2
(
𝑎
𝑏𝑚+2
𝑏
2𝑎
)𝑝 (
𝐻
(𝑝)
𝑏
2𝑎
−𝐻
(𝑝)
𝑏
1
2𝑎 − 2
Proof.
)
∞ (
1∑ 1
1
1
=
−
𝑛 (𝑎𝑛 + 𝑟)
𝑟 𝑛=1 𝑛 𝑛 + 𝑎𝑟
𝑛=1
(𝑟
)]
1[
=
𝛾+𝜓
+ 1 , using (1.4)
𝑟
𝑎
(1)
𝐻𝑟
= 𝑎 , hence (2.2) is attained.
𝑟
∞
∑
∞
∑
[∞ (
]
) ∑
) ∑
∞ (
∞
𝑛
(−1)
1 ∑ 1
1
1
1
1
=
−
−
−
−
𝑟
𝑛 (𝑎𝑛 + 𝑟)
2𝑟 𝑛=1 𝑛 𝑛 + 2𝑎
𝑛 𝑛 + 𝑟−𝑎
𝑛 (2𝑛 − 1)
2𝑎
𝑛=1
𝑛=1
𝑛=1
[
(
)
]
(
)
𝑟
𝑟
1
1
−𝛾 + 𝜓
+1 +𝛾−𝜓
+
− 2 ln 2 ,
=
2𝑟
2𝑎
2𝑎 2
(𝑟
)
from (1.5) we may write after substituting for 𝜓 2𝑎
+ 12
∞
∑
𝑛
)
(𝑟
)]
(−1)
1 [ (𝑟
=
2𝜓
+ 1 − 2𝜓
+1
𝑛 (𝑎𝑛 + 𝑟)
2𝑟
2𝑎
𝑎
𝑛=1
[
]
1
(1)
(1)
=
𝐻 𝑟 − 𝛾 − 𝐻 𝑟 + 𝛾 , therefore (2.3) follows.
2𝑎
𝑎
𝑟
The partial fraction decomposition
(
1
1
1
1
−
𝑚+1 = 𝑏𝑚+1
𝑛 𝑛+
𝑛 (𝑎𝑛 + 𝑏)
𝑏
𝑎
−
𝑚+1
∑(
𝑠=2
𝑏
𝑎
)𝑠−1
)
1
(
𝑛+
)
𝑏 𝑠
𝑎
44
A. SOFO
will be useful in the expansion of (2.4). Consider the sum (2.4)
(
(
) 𝑚+1
)
∞
∞
∑
∑
∑ 𝑎 ( 𝑏 )𝑠
1
1
1
1
1
−
−
(
)𝑠
𝑚+1 =
𝑏𝑚+1 𝑛 𝑛 + 𝑎𝑏
𝑏𝑚+2 𝑎
𝑛 + 𝑎𝑏
𝑛=1 𝑛 (𝑎𝑛 + 𝑏)
𝑛=1
𝑠=2
=
∞
∑
1
𝑏𝑚+1
(
𝑛=1
1
1
−
𝑛 𝑛+
)
−
𝑏
𝑎
𝑚+1
∑
𝑎
𝑏𝑚+2
𝑠=2
( )𝑠 ∑
∞
𝑏
1
(
)
𝑎 𝑛=1 𝑛 + 𝑎𝑏 𝑠
using (1.6) and (2.3) we have
∞
∑
𝑛=1
1
𝑛 (𝑎𝑛 + 𝑏)
=
𝑚+1
1
(1)
𝑏𝑚+1
𝐻𝑏 −
𝑎
𝑚+1
∑
𝑠=2
𝑠
𝑎 (−1)
𝑚+2
𝑏
(𝑠 − 1)!
(
( )𝑠
)
𝑏
𝑏
(𝑠−1)
𝜓
+1 .
𝑎
𝑎
From (1.3)
∞
∑
1
𝑚+1
𝑛 (𝑎𝑛 + 𝑏)
𝑛=1
1
(1)
𝐻𝑏 −
𝑏𝑚+1 𝑎
=
𝑚+1
∑
=
𝑠=2
( )𝑠 (
)
𝑏
(𝑠)
𝐻 𝑏 − 𝜁 (𝑠)
𝑎
𝑎
𝑠
𝑎 (−1)
𝑏𝑚+2 (𝑠 − 1)!
( )𝑠
𝑚+1
∑ 𝑎 ( 𝑏 )𝑠
𝑏
(𝑠)
𝐻𝑏 −
𝜁 (𝑠)
𝑎
𝑎
𝑏𝑚+2 𝑎
𝑠=2
𝑎
𝑏𝑚+2
𝑠=1
𝑚+1
∑
and rearranging we obtain (2.4). For the last identity (2.5) let
∞
∑
𝑛=1
𝑛
(−1)
𝑛 (𝑎𝑛 + 𝑏)
𝑚+1
=
∞
∑
(
(−1)
(
1
𝑛
𝑏𝑚+1
𝑛=1
∞
𝑛
∑
(−1)
=
𝑏𝑚+1
𝑛=1
(
1
1
−
𝑛 𝑛+
1
1
−
𝑛 𝑛+
)
𝑏
𝑎
−
)
−
𝑏
𝑎
𝑚+1
∑
𝑝=2
𝑚+1
∑
𝑝=2
𝑎
𝑏𝑚+2
𝑎
𝑏𝑚+2
)
( )𝑝
𝑏
1
(
)𝑝
𝑎
𝑛 + 𝑎𝑏
( )𝑝 ∑
∞
𝑛
(−1)
𝑏
(
)
𝑎 𝑛=1 𝑛 + 𝑎𝑏 𝑝
from (2.3), and (1.6)
∞
∑
𝑛=1
𝑛
(−1)
𝑛 (𝑎𝑛 + 𝑏)
𝑚+1
=
1
(
𝑏𝑚+1
−
𝐻
𝑚+1
∑
𝑝=2
=
−
𝑚+1
∑
𝑝=2
𝑝
1
𝑏𝑚+1
(−1) 𝑎
(𝑝 − 1)!𝑏𝑚+2
Applying (1.3)
(
(
𝑏
2𝑎
(1)
− 𝐻𝑏
(
𝑏𝑚+2
(1)
𝑏
2𝑎
)
𝑎
𝑎
𝐻
𝑏
2𝑎
(1)
𝑏
2𝑎
(1)
− 𝐻𝑏
)𝑝 (
(
)𝑝 ∑
∞
1
(
𝑛=1
𝑛+
)
𝑏 𝑝
2𝑎
−(
)
1
𝑛+
𝑏
2𝑎
−
)
1 𝑝
2
)
𝑎
𝜓 (𝑝−1)
(
)
(
))
𝑏
𝑏
1
+ 1 − 𝜓 (𝑝−1)
+
.
2𝑎
2𝑎 2
POLYGAMMA FUNCTIONS
∞
∑
(−1)
𝑛
𝑚+1
𝑛=1
𝑛 (𝑎𝑛 + 𝑏)
=
1
(
𝑏𝑚+1
𝐻
(1)
𝑏
2𝑎
(1)
− 𝐻𝑏
)
−
𝑎
𝑚+1
∑
𝑝=2
𝑎
𝑏𝑚+2
45
(
𝑏
2𝑎
)𝑝 (
−𝐻
(𝑝)
𝑏
2𝑎
+𝐻
(𝑝)
1
𝑏
2𝑎 − 2
and rearranging we obtain (2.5).
)
□
Remark. In the following Corollaries and remarks we encounter harmonic num(𝛼)
bers at possible rational values of the argument, of the form 𝐻 𝑟 where 𝑟 =
𝑎
1, 2, 3, ..., 𝑘, 𝛼 = 1, 2, 3, ... and 𝑘 ∈ ℕ. The polygamma function 𝜓 (𝛼) (𝑧) is defined as:
𝜓 (𝛼) (𝑧) =
(𝛼)
To evaluate 𝐻 𝑟
𝑎
𝑑𝛼
𝑑𝛼+1
[log
Γ
(𝑧)]
=
[𝜓 (𝑧)] , 𝑧 ∕= {0, −1, −2, −3, ...} .
𝑑𝑧 𝛼+1
𝑑𝑧 𝛼
we have available a relation in terms of the polygamma function
𝜓 (𝛼) (𝑧) , (1.3), or, for rational arguments 𝑧 = 𝑎𝑟 ,
𝛼
(𝛼+1)
𝐻𝑟
𝑎
= 𝜁 (𝛼 + 1) +
)
(−1) (𝛼) ( 𝑟
𝜓
+1
𝛼!
𝑎
we also define
(𝑟
)
(𝛼)
+ 1 , and 𝐻0 = 0.
𝑎
𝑎
( )
The evaluation of the polygamma function 𝜓 (𝛼) 𝑎𝑟 at rational values of the argument can be explicitly done via a formula as given by Kölbig [14], (see also [13]),
or Choi and Cvijovic [7] in terms of the Polylogarithmic or other special functions.
Some specific values are given as, many others are listed in the book [24]:
( )
(
)
1
𝑛
(𝑛)
𝜓
= (−1) 𝑛! 2𝑛+1 − 1 𝜁 (𝑛 + 1)
2
(1)
𝐻𝑟 = 𝛾 + 𝜓
(4)
(2)
2
4
16
+ 8𝐺 − 5𝜁 (2) ,
9
√
6
3𝜋 3 ln (3)
= +
−
− 2 ln (2) .
5
2
2
𝐻 1 = 16 − 4𝜁 (2) , 𝐻 3 =
(1)
𝐻1 =
3
𝜋
3 ln 3
3
(1)
− √ −
, and 𝐻 5
6
2 2 3
2
The main result of this paper is embodied in the following theorem.
Theorem 2.3. Let 𝑎 be a positive real number, ∣𝑡∣ ≤ 1 , 𝑗 ≥ 0, and 𝑘 ∈ ℕ ∪ {0} .
Then
∞
∑
𝑡𝑛
(
)
𝑆𝑘+1 (𝑎, 𝑗, 𝑡) =
(2.6)
𝑎𝑛
+
𝑗
+
1
𝑘
𝑛=1 𝑛 (𝑎𝑛 + 𝑗 + 1)
𝑗+1
=
⎧


⎨


⎩
(𝑗+1)𝑡(−1)𝑘
𝑎𝑘 𝑘!
𝑎𝑡
∫1
0
∫1 ∫1
0 0
𝑗+1
(1−𝑥)𝑗 𝑥𝑎−1 𝑦 𝑎 (ln(𝑦))𝑘
𝑑𝑥𝑑𝑦,
1−𝑡𝑥𝑎 𝑦
(1−𝑥)𝑗+1 𝑥𝑎−1
1−𝑡𝑥𝑎
for 𝑘 ≥ 1
.
𝑑𝑥,
for 𝑘 = 0
46
A. SOFO
⎡
= 𝑇0
(𝑎−1)−𝑡𝑒𝑟𝑚𝑠
𝑘−𝑡𝑒𝑟𝑚𝑠
z
}|
{ z
}|
{
𝑎+𝑗+1
𝑎+𝑗+1 𝑎+1
2𝑎 − 1
1, 1,
, ......,
,
, .. . . . .,
𝑎
𝑎
𝑎
𝑎
2𝑎 + 𝑗 + 1
2𝑎 + 𝑗 + 1 𝑎 + 𝑗 + 2
𝑎+𝑗+𝑎
,
, ......,
, . . . ...,
𝑎
𝑎
𝑎 }
|
{z
} | 𝑎
{z
⎢
⎢
⎢
⎢
𝑎+𝑘+1 𝐹𝑎+𝑘 ⎢
⎢
⎣
(𝑘+1)−𝑡𝑒𝑟𝑚𝑠
(𝑎−1)−𝑡𝑒𝑟𝑚𝑠
⎤
⎥
⎥
⎥
𝑡⎥
⎥
⎥
⎦
(2.7)
where
𝑇0 =
𝑡 (𝑗 + 1) 𝐵 (𝑗 + 1, 𝑎 + 1)
(𝑎 + 𝑗 + 1)
𝑘
.
Proof. Consider
∞
∑
𝑛=1
𝑡𝑛
(
𝑛 (𝑎𝑛 + 𝑗 + 1)
𝑘
𝑎𝑛 + 𝑗 + 1
𝑗+1
)=
∞
∑
(𝑗 + 1) 𝑡𝑛 Γ (𝑗 + 1) 𝑎𝑛 Γ (𝑎𝑛)
𝑛=1
𝑛 (𝑎𝑛 + 𝑗 + 1)
= 𝑎 (𝑗 + 1)
𝑘+1
∞
∑
𝑛=1
Γ (𝑎𝑛 + 𝑗 + 1)
𝑡𝑛
(𝑎𝑛 + 𝑗 + 1)
𝑘+1
𝐵 (𝑎𝑛, 𝑗 + 1)
now replacing the Beta function with its integral representation, we have
𝑎 (𝑗 + 1)
∞
∑
𝑛=1
𝑡𝑛
(𝑎𝑛 + 𝑗 + 1)
𝐵 (𝑎𝑛, 𝑗 + 1) =
𝑘+1
∞
∑
𝑛=1
𝑎 (𝑗 + 1)
=
𝑎𝑘+1
∫1
0
𝑎 (𝑗 + 1) 𝑡𝑛
(𝑎𝑛 + 𝑗 + 1)
∞
𝑗
∫1
𝑘+1
𝑗
𝑥𝑎𝑛−1 (1 − 𝑥) 𝑑𝑥
0
𝑛
∑
(𝑡𝑥𝑎 )
(1 − 𝑥)
.
𝑑𝑥
(
)
𝑗+1 𝑘+1
𝑥
𝑛=1 𝑛 +
𝑎
By a justified changing the order of integration and summation, by the dominated
convergence theorem, we have,
∞
∑
𝑛=1
𝑡𝑛
(
𝑛 (𝑎𝑛 + 𝑗 + 1)
𝑘
∫1
∞
𝑛
𝑗
(𝑡𝑥𝑎 )
𝑎 (𝑗 + 1) ∑
(1 − 𝑥)
)=
𝑑𝑥
𝑎𝑘+1 𝑛=1 (𝑛 + 𝑗+1 )𝑘+1
𝑥
𝑎𝑛 + 𝑗 + 1
𝑎
0
𝑗+1
𝑘
=
(𝑗 + 1) 𝑡 (−1)
𝑎𝑘 𝑘!
∫1 ∫1
0
𝑗+1
𝑗
𝑘
(1 − 𝑥) 𝑥𝑎−1 𝑦 𝑎 (ln (𝑦))
𝑑𝑥𝑑𝑦, for 𝑘 ≥ 1
1 − 𝑡𝑥𝑎 𝑦
0
upon utilizing Lemma 2.1. The case of 𝑘 = 0 follows in a similar way so that
𝑆1 (𝑎, 𝑗, 𝑡) =
∞
∑
(
𝑛=1
𝑛
∫1
= 𝑎𝑡
0
𝑡𝑛
)
𝑎𝑛 + 𝑗 + 1
𝑗+1
𝑗+1
(1 − 𝑥)
𝑥𝑎−1
𝑑𝑥,
1 − 𝑡𝑥𝑎
POLYGAMMA FUNCTIONS
47
hence the integrals in (2.6) are attained. By the consideration of the ratio of
successive terms 𝑈𝑈𝑛+1
where
𝑛
𝑡𝑛
(
𝑈𝑛 =
𝑛 (𝑎𝑛 + 𝑗 + 1)
𝑘
𝑎𝑛 + 𝑗 + 1
𝑗+1
)
we obtain the result (2.7).
□
The following interesting corollaries follow from Theorem 2.3.
Corollary 2.4. Let 𝑡 = 1 and 𝑎 > 0. Also let 𝑗 ≥ 0 and 𝑘 ≥ 1 be integers. Then
∞
∑
1
(
)
𝑆𝑘+1 (𝑎, 𝑗, 1) =
𝑎𝑛
+
𝑗
+
1
𝑘
𝑛=1 𝑛 (𝑎𝑛 + 𝑗 + 1)
𝑗+1
(𝑗 + 1) (−1)
=
𝑎𝑘 𝑘!
𝑘
∫1 ∫1
0
𝑗+1
𝑗
𝑘
(1 − 𝑥) 𝑥𝑎−1 𝑦 𝑎 (ln (𝑦))
𝑑𝑥𝑑𝑦, for 𝑘 ≥ 1
1 − 𝑥𝑎 𝑦
0
(2.8)
=
𝑘
∑
𝐴𝑠 (𝑗 + 1)!
𝑝=1
𝑠=0
+
(𝑝)
𝐻 𝑗+1
𝑘+1−𝑠
∑
𝑎
𝑎𝑠+𝑝−1 (𝑗 + 1)
(2.9)
𝑘+2−𝑠−𝑝
( )
𝑗
𝑘
𝑘+1−𝑠
𝑟+1
∑
∑
∑
(−1)
(𝑗 + 1)
𝜁 (𝑝)
𝑗
(1)
𝐻
−
𝐴
(𝑗
+
1)!
𝑟
𝑠
𝑘+1
𝑘+2−𝑠−𝑝
𝑎
𝑟
(𝑗 + 1 − 𝑟)
𝑎𝑠+𝑝−1 (𝑗 + 1)
𝑟=1
𝑠=0
𝑝=2
where
[
1 𝑑𝑠
𝐴𝑠 =
lim
𝑠! 𝑑𝑛𝑠
𝑛→(− 𝑗+1
𝑎 )
{
𝑘+1
(𝑎𝑛 + 𝑗 + 1)
𝑘+1 ∏𝑗
(𝑎𝑛 + 𝑗 + 1)
𝑟=1 (𝑎𝑛 + 𝑟)
}]
, 𝑠 = 0, 1, 2, ...𝑘.
(2.10)
Proof. By expansion,
∞
∞
∑
∑
1
(𝑗 + 1)!
(
)=
𝑘
𝑎𝑛 + 𝑗 + 1
𝑛=1 𝑛 (𝑎𝑛 + 𝑗 + 1)𝑘
𝑛=1 𝑛 (𝑎𝑛 + 𝑗 + 1) (𝑎𝑛 + 1)𝑗+1
𝑗+1
∞
∑
(𝑗 + 1)!
=
𝑘 ∏𝑗+1
𝑛=1 𝑛 (𝑎𝑛 + 𝑗 + 1)
𝑟=1 (𝑎𝑛 + 𝑟)
∞
∑
(𝑗 + 1)!
=
𝑘+1 ∏𝑗
𝑛=1 𝑛 (𝑎𝑛 + 𝑗 + 1)
𝑟=1 (𝑎𝑛 + 𝑟)
[
]
𝑗
∞
𝑘
∑
∑
(𝑗 + 1)! ∑ 𝐵𝑟
𝐴𝑠
=
+
,
𝑛
𝑎𝑛 + 𝑟 𝑠=0 (𝑎𝑛 + 𝑗 + 1)𝑘+1−𝑠
𝑛=1
𝑟=1
where
{
𝐵𝑟 =
lim
𝑟
𝑛→(− 𝑎
)
𝑎𝑛 + 𝑟
𝑘+1 ∏𝑗
(𝑎𝑛 + 𝑗 + 1)
𝑟=1 (𝑎𝑛 + 𝑟)
}
𝑟+1
(−1)
=
𝑗!
𝑟
𝑘+1
(𝑗 + 1 − 𝑟)
( )
𝑗
,
𝑟
48
A. SOFO
𝐴𝑠 is defined by (2.10). Hence, after interchanging the sums, we have
∞
∑
𝑛=1
1
(
𝑛 (𝑎𝑛 + 𝑗 + 1)
𝑗
∑
𝑘
𝑎𝑛 + 𝑗 + 1
𝑗+1
)
∞
∑
𝑘
∞
∑
∑
1
1
+
(𝑗 + 1)!𝐴𝑠
𝑘+1−𝑠
𝑛
(𝑎𝑛
+
𝑟)
𝑟=1
𝑛=1
𝑠=0
𝑛=1 𝑛 (𝑎𝑛 + 𝑗 + 1)
( )
𝑗
∑
𝑗
𝑗+1
(1)
𝑟+1
𝐻𝑟
=
(−1)
𝑘+1
𝑎
𝑟
(𝑗 + 1 − 𝑟)
𝑟=1
⎞
⎛
(𝑝)
𝑘
𝑘+1−𝑠
𝑘+1−𝑠
𝐻 𝑗+1
∑
∑
∑
𝜁
(𝑝)
𝑎
⎠
+
−
(𝑗 + 1)!𝐴𝑠 ⎝
𝑠+𝑝−1 (𝑗 + 1)𝑘+2−𝑠−𝑝
𝑠+𝑝−1 (𝑗 + 1)𝑘+2−𝑠−𝑝
𝑎
𝑎
𝑠=0
𝑝=2
𝑝=1
=
=
𝑗
∑
𝑟=1
+
𝑘
∑
(𝑗 + 1)!𝐵𝑟
(−1)
𝑟+1
( )
𝑗
𝑗+1
(1)
𝐻𝑟
𝑟 (𝑗 + 1 − 𝑟)𝑘+1 𝑎
(𝑗 + 1)!𝐴𝑠
(𝑝)
𝐻 𝑗+1
𝑘+1−𝑠
∑
𝑠=0
𝑝=1
𝑎
𝑎𝑠+𝑝−1
(𝑗 + 1)
𝑘+2−𝑠−𝑝
−
𝑘
∑
(𝑗 + 1)!𝐴𝑠
𝑠=0
𝑘+1−𝑠
∑
𝑝=2
𝜁 (𝑝)
𝑎𝑠+𝑝−1
(𝑗 + 1)
𝑘+2−𝑠−𝑝
upon utilizing Lemma 2.2, which is the result (2.9). The degenerate case, for
𝑗 = −1, gives the known result
∞
∑
1
1
= 𝑘 𝜁 (𝑘 + 1) .
𝑘 𝑛𝑘+1
𝑎
𝑎
𝑛=1
The integral (2.8) follows from the integral in (2.6).
□
A similar result is evident for the case 𝑡 = −1.
Corollary 2.5. Let 𝑡 = −1 and 𝑎 > 0. Also let 𝑗 ≥ 0 and 𝑘 ≥ 1 be integers. Then
𝑆𝑘+1 (𝑎, 𝑗, −1) =
∞
∑
𝑛
(−1)
(
)
𝑎𝑛 + 𝑗 + 1
𝑛=1 𝑛 (𝑎𝑛 + 𝑗 + 1)𝑘
𝑗+1
𝑘+1
(𝑗 + 1) (−1)
=
𝑎𝑘 𝑘!
∫1 ∫1
0
=
𝑗
∑
𝑟=1
+
𝑘
∑
𝑠=0
+
𝑟+1
(−1)
𝑎𝑠 (𝑗 + 1)
(
𝑘+1
𝐴𝑠 (𝑗 + 1)!
(
𝑘+1−𝑠
𝑘
∑
𝐴𝑠 (𝑗 + 1)! 𝑎1−𝑠
𝑠=0
(𝑗 + 1)
𝑘
0
(𝑗 + 1)
(𝑗 + 1 − 𝑟)
𝑗+1
𝑗
(1 − 𝑥) 𝑥𝑎−1 𝑦 𝑎 (ln (𝑦))
𝑑𝑥𝑑𝑦, for 𝑘 ≥ 1
1 + 𝑥𝑎 𝑦
𝑗
𝑟
)(
𝐻
(1)
𝑟
2𝑎
(1)
(1)
2𝑎
𝑎
𝐻 𝑗+1 − 𝐻 𝑗+1
𝑘+1−𝑠
∑ (
𝑘+2−𝑠
𝑝=2
(1)
− 𝐻𝑟
𝑗+1
2𝑎
)
(2.11)
𝑎
)
)𝑝 (
(𝑝)
(𝑝)
2𝑎
2𝑎
𝐻 𝑗+1 − 𝐻 𝑗+1 − 1
)
2
Proof. The proof, uses (2.3) and (2.5) and follows the same details as that of Corollary 2.4, and will not be given here.
POLYGAMMA FUNCTIONS
49
The degenerate case, for 𝑗 = −1, gives the known result
∞
𝑛
∑
)
(−1)
1 (
=
1 − 2𝑘 𝜁 (𝑘 + 1) .
𝑘
𝑘
𝑘+1
𝑎 𝑛
(2𝑎)
𝑛=1
□
We give the following example to illustrate some of the above identities.
Example. From (2.11)
𝑆3 (6, 2, −1)
=
∞
∑
(
)
𝑛
√
(−1)
139
5 11
𝜋
(
) = − 𝐺−
8 3−
3
9
8
9
6𝑛 + 3
𝑛=1 𝑛 (6𝑛 + 3)2
3
(
√ )
√
)
26 3 3
𝜋3
3 3 (√
+
+
ln
3+1 ,
−
ln 2 −
144
9
4
2
here 𝐺 is Catalan’s constant, defined by
𝐺=
∞
𝑟
∑
(−1)
𝑟=0
(2𝑟 + 1)
2
≈ 0.915965....
Remark. The very special case of 𝑎 = 1 and 𝑗 = 0 allows one to evaluate (1.1)
and (1.2).
A recurrence relation for a degenerate case, 𝑗 = 0, of Theorem 2.3 is embodied
in the following corollary.
Corollary 2.6. Let the conditions of Theorem 2.3 hold with 𝑗
∑
𝑡𝑛
0
0
,
𝑆𝑘+1
: = 𝑆𝑘+1
(𝑎, 𝑡) =
𝑘+1
𝑛≥1 𝑛 (𝑎𝑛 + 1)
⎡ (𝑘+2)−𝑡𝑒𝑟𝑚𝑠
z }| {
⎢ 1, 1, ....., 1 , 𝑎+1
𝑡
𝑎
⎢
=
𝑘+3 𝐹𝑘+2 ⎢
𝑎+1
⎣ 2, 2, ....., 2 , 2𝑎+1
𝑎
| {z }
(𝑘+1)−𝑡𝑒𝑟𝑚𝑠
= 0 and put
⎤
⎥
⎥
𝑡⎥ ,
⎦
then
0
𝑆𝑘+1
− 𝑆𝑘0 +
)
(
1
1
Φ
𝑡,
𝑘
+
1,
= 𝑎, for 𝑘 ≥ 1
𝑎𝑘
𝑎
with solution
0
𝑆𝑘+1
=
𝑆10
(
)
𝑘
∑
1
1
+ 𝑎𝑘 −
Φ 𝑡, 𝑟 + 1,
𝑎𝑟
𝑎
𝑟=1
where
𝑆10
=
∑
𝑛≥1
𝑡𝑛
𝑡
=
3 𝐹2
𝑛 (𝑎𝑛 + 1)
𝑎+1
and Φ is the Lerch transcendent.
[
]
1, 1, 𝑎+1
𝑎
𝑡
2𝑎+1 2, 𝑎
50
A. SOFO
Proof. We notice that
⎡
0
𝑆𝑘+1
= 𝑆𝑘0 − 𝑎 ⎣
𝑡
∑
⎤
𝑛
𝑛≥0 (𝑎𝑛 + 1)
𝑘+1
− 1⎦
hence the solution follows by iteration.
□
Some examples are:
∙ For 𝑡 = −1
0
𝑆𝑘+1
=
=
(𝑎, −1) =
𝑆10
(
)
𝑘
∑
1
1
Φ −1, 𝑟 + 1,
+ 𝑎𝑘 −
, for 𝑘 ≥ 1
𝑎𝑟
𝑎
𝑟=1
(
)
(
)
[ ( )
(
)] ∑
𝑘
− 𝜁 𝑟 + 1, 2𝑎+1
𝜁 𝑟 + 1, 𝑎+1
1
𝑎+1
1
2𝑎
2𝑎
𝜓
−𝜓
+
,
𝑎𝑘 − ln (2) +
2
2𝑎
2𝑎
2𝑟+1 𝑎𝑟
𝑟=1
)
(
)
(
( )
( ) ∑
𝑘
− 𝜁 𝑟 + 1, 2𝑎+1
𝜁 𝑟 + 1, 𝑎+1
1
1
2𝑎
2𝑎
, using (1.5).
𝑎𝑘 + 𝜓
−𝜓
+
2𝑎
𝑎
2𝑟+1 𝑎𝑟
𝑟=1
When 𝑎 = 1, we obtain Coffey’s [8], result (1.1)
0
𝑆𝑘+1
(1, −1) = 1 − 2 ln (2) + 𝑘 +
𝑘
∑
(
)
2−𝑟 − 1 𝜁 (𝑟 + 1) .
𝑟=1
When 𝑎 = 2,
0
𝑆𝑘+1
(2, −1) = 2 −
(
(
)
)
𝑘
∑
𝜁 𝑟 + 1, 43 − 𝜁 𝑟 + 1, 14
𝜋
− ln (2) + 2𝑘 +
.
2
22𝑟+1
𝑟=1
∙ For 𝑡 = 1
(
)
𝑘
∑
1
1
Φ
1,
𝑟
+
1,
, for 𝑘 ≥ 1
𝑎𝑟
𝑎
𝑟=1
(
)
( ) ∑
𝑘
𝜁 𝑟 + 1, 𝑎1
1
= 𝛾 + 𝑎 (𝑘 + 1) − 𝜓
−
.
𝑎
𝑎𝑟
𝑟=1
0
𝑆𝑘+1
(𝑎, 1) = 𝑆10 + 𝑎𝑘 −
When 𝑎 = 1, we obtain Coffey’s [8] result (1.2), by noting that 𝜓 (1) = −𝛾
0
𝑆𝑘+1
(1, 1) = 1 + 𝑘 −
𝑘
∑
𝜁 (𝑟 + 1) .
𝑟=1
When 𝑎 = 2,
0
𝑆𝑘+1
(2, 1) = 2 − 2 ln (2) + 2𝑘 −
)
𝑘 ( 𝑟+1
∑
2
−1
𝑟=1
2𝑟
𝜁 (𝑟 + 1) ,
similarly for 𝑎 = 8,
0
𝑆𝑘+1
𝜋
(8, 1) = 8 (1 + 𝑘)−
2
√
(
)
𝑘
√ ) ∑
1
2𝑟
3 + 2 2−4 ln (2)+ 2 ln 3 − 2 2 −
2 𝜁 𝑟 + 1,
.
8
𝑟=1
√
√
(
POLYGAMMA FUNCTIONS
51
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52
A. SOFO
Anthony Sofo
School of Engineering and Science, Victoria University, PO Box 14428, Melbourne
City, VIC 8001, Australia.
E-mail address: anthony.sofo@vu.edu.au; asofo60@gmail.com