Table Of ContentUNIQUENESS OF NONTRIVIALLY COMPLETE
MONOTONICITY FOR A CLASS OF FUNCTIONS INVOLVING
POLYGAMMA FUNCTIONS
0
1
0
FENGQIANDBAI-NIGUO
2
n
a Abstract. Form,n∈N,letfm,n(x)=(cid:2)ψ(m)(x)(cid:3)2+ψ(n)(x)on(0,∞). Inthe
J
presentpaper,weproveusingtwomethodsthat,amongallfm,n(x)form,n∈
6
2 N,onlyf1,2(x)isnontriviallycompletelymonotonicon(0,∞). Accurately,the
functionsf1,2(x)andfm,2n−1(x)arecompletelymonotonicon(0,∞),butthe
]
A functions fm,2n(x) for (m,n) 6= (1,1) are not monotonic and does not keep
C thesamesignon(0,∞).
.
h
t
a
m
1. Introduction
[
Recall [4, Chapter XIII] and [10, Chapter IV] that a function f is said to be
2
v
completely monotonic on an interval I if f has derivatives of all orders on I and
4
0
1 (−1)nf(n)(x)≥0 (1)
1
. for x ∈ I and n ≥ 0. The celebrated Bernstein-Widder’s Theorem [10, p. 160,
4
0
Theorem12a] states that a function f(x) is completely monotonic on [0,∞) if and
9
0 only if there exists a bounded and non-decreasing function α(t) such that
:
v ∞
i f(x)= e−xtdα(t) (2)
X
Z
0
r
a converges for x ∈ [0,∞). This expresses that a completely monotonic function f
on [0,∞) is a Laplace transform of the measure α.
It is general knowledge that the classical Euler gamma function
∞
Γ(x)= tx−1e−tdt, x>0, (3)
Z
0
2000 Mathematics Subject Classification. Primary26A48,33B15;Secondary 26A51,66R10.
Keywords and phrases. uniqueness,completemonotonicity,polygammafunction,squareofa
polygammafunction.
TheauthorswerepartiallysupportedbytheChinaScholarshipCouncilandtheScienceFoun-
dationofTianjinPolytechnic University.
ThispaperwastypesetusingAMS-LATEX.
1
2 F.QIANDB.-N.GUO
the psifunction ψ(x)= Γ′(x), and the polygammafunctions ψ(i)(x) for i∈N are a
Γ(x)
seriesofimportantspecialfunctions,sincetheyhavemuchextensiveapplicationsin
manybranchessuchasstatistics,probability,numbertheory,theoryof0-1matrices,
graphtheory,combinatorics,physics,engineering,andothermathematicalsciences.
For m,n∈N, let
f (x)=ψ(n)(x)+ ψ(m)(x) 2, x>0. (4)
m,n
(cid:2) (cid:3)
The positivity and complete monotonicity of f (x) have been proved, recovered,
1,2
generalized,andappliedinanumberofliteratureinrecentyears. Formuchdetailed
information on these, please refer to [7, 8] and closely-relatedreferences therein.
The aim of this paper is to demonstrate using two methods that, among all
f (x) for m,n∈N, only f (x) is nontrivially completely monotonic on (0,∞).
m,n 1,2
Our main results may be stated as follows.
Theorem 1. For m,n ∈ N, the functions f1,2(x) and fm,2n−1(x) are completely
monotonic on (0,∞), but the functions f (x) for (m,n)6=(1,1) are not mono-
m,2n
tonic and does not keep the same sign on (0,∞).
Remark 1. As shown in the proof of Theorem 1 below, the complete monotonicity
of the functions fm,2n−1(x) is trivial, so f1,2(x) is the only nontrivial completely
monotonic function on (0,∞) among all f (x) for m,n∈N.
m,n
2. First proof of Theorem 1
The following formulas are listed in [1]: For r >0, x>0 and n∈N, we have
∞
1 1
= tr−1e−xtdt, (5)
xr Γ(r)Z
0
∞ tn
ψ(n)(x)=(−1)n+1 e−xtdt, (6)
Z 1−e−t
0
(−1)n−1(n−1)!
ψ(n−1)(x+1)=ψ(n−1)(x)+ . (7)
xn
For n>m≥1, direct calculation and utilization of the above three formulas give
f (x)−f (x+1)= ψ(m)(x) 2− ψ(m)(x+1) 2+ψ(n)(x)−ψ(n)(x+1)
m,n m,n
(cid:2) (cid:3) (cid:2) (cid:3)
= ψ(m)(x)−ψ(m)(x+1) ψ(m)(x)+ψ(m)(x+1) +ψ(n)(x)−ψ(n)(x+1)
(cid:2) (cid:3)(cid:2) (cid:3)
COMPLETE MONOTONICITY OF FUNCTIONS INVOLVING POLYGAMMA FUNCTIONS 3
(−1)m+1m! (−1)mm! (−1)n+1n!
= 2ψ(m)(x)+ +
xm+1 (cid:20) xm+1 (cid:21) xn+1
(−1)m+12·m! (−1)mm! (−1)n−mn!
= ψ(m)(x)+ +
xm+1 (cid:20) 2xm+1 2·m!xn−m(cid:21)
(−1)m+12·m! ∞ tm
= (−1)m+1 e−xtdt
xm+1 (cid:20) Z 1−e−t
0
(−1)m ∞ (−1)n−mn! ∞
+ tme−xtdt+ tn−m−1e−xtdt
2 Z 2·m!(n−m−1)!Z (cid:21)
0 0
2·m! ∞ 1 n−1 (−1)n−1ntn−2m−1 1
= + − tme−xtdt
xm+1 Z (cid:20)1−e−t (cid:18) m (cid:19) 2 2(cid:21)
0
∞
2·m! 1 1 1 n n−1
= − −(−1)n tn−m−1e−xtdt
xm+1 Z (cid:20)tn−2m−1(cid:18)1−e−t 2(cid:19) 2(cid:18) m (cid:19)(cid:21)
0
∞
2·m! n n−1
, xm+1 Z (cid:20)hn−2m−1(t)−(−1)n2(cid:18) m (cid:19)(cid:21)tn−m−1e−xtdt
0
and
1−e2t k−2tet/ 1−e2t 1−e2t [k−ω(t)]
h′(t)= , ,
k (cid:0) 2(cid:1)t(cid:2)k+1 et−1(cid:0)2 (cid:1)(cid:3) (cid:0) 2tk+1(cid:1)et−1 2
(cid:0) (cid:1) (cid:0) (cid:1)
where k is any given integer and ω(t) is increasing on (0,∞) with the limits
lim ω(t)=−1 and lim ω(t)=0.
t→0+ t→∞
Hence, if k ≥ 0 then the function h′(t) is negative on (0,∞), and if k ≤ −1 then
k
the derivative h′(t) is positive on (0,∞). This means that the function h (t) is
k k
decreasing for k ≥0 and increasing for k ≤−1 on (0,∞). Furthermore, from
0, k ≤−2;
t→lim0+hk(t)=1, k =−1;
∞, k ≥0
and
∞, k ≤−1;
tl→im∞hk(t)=12, k =0;
0, k ≥1;
it follows that
(1) if k ≤−2 or k ≥1, then 0<h (t)<∞;
k
(2) if k =−1, then 1<h (t)<∞;
k
(3) if k =0, then 1 <h (t)<∞.
2 k
4 F.QIANDB.-N.GUO
Therefore,byarecoursetoBernstein-Widder’sTheorem[10,p.160,Theorem12a],
asmentionedin(2),andafactthataproductoffinitecompletemonotonicfunctions
is also completely monotonic, it follows that
(1) if n=2i−1 for i∈N, the function
fm,n(x)−fm,n(x+1)=fm,2i−1(x)−fm,2i−1(x+1)
is completely monotonic on (0,∞);
(2) if n=2i for i∈N, the quantity
=1, i=m=1;
n n−1 2i−1
2(cid:18) m (cid:19)=i(cid:18) m (cid:19)=0, 2i−1<m;
≥2, i≥2 and 2i−1≥m;
which implies that the function
f (x)−f (x+1)=f (x)−f (x+1)
m,n m,n m,2i m,2i
is completely monotonic on (0,∞) for i=m=1 or 2i−1<m;
(3) if i ≥ 2 and 2i−1 ≥ m, then the function f (x)−f (x+1) is not
m,2i m,2i
completely monotonic on (0,∞).
If the function f (x)−f (x+1) is completely monotonic on (0,∞), then
m,n m,n
(−1)ℓ[f (x)−f (x+1)](ℓ) =(−1)ℓf(ℓ) (x)−(−1)ℓf(ℓ) (x+1)≥0
m,n m,n m,n m,n
which can be rewritten as
(−1)ℓf(ℓ) (x)≥(−1)ℓf(ℓ) (x+1)≥···≥(−1)ℓf(ℓ) (x+i)→0
m,n m,n m,n
asi→∞,whereℓ≥0. Thus,thefunctionf (x)isalsocompletelymonotonicon
m,n
(0,∞). As a result, the functions fm,2i−1(x) for 2i−1 > m, f1,2(x), and fm,2i(x)
for 2i−1 < m and 2i > m are completely monotonic on (0,∞). In a word, the
functionsf1,2(x)andfm,2i−1(x)for2i−1>marecompletelymonotonicon(0,∞).
The formula (6) shows that the functions (−1)m+1ψ(m)(x) and ψ(2n−1)(x) for
m,n∈N are completely monotonic on (0,∞), so the function
(−1)m+1ψ(m)(x) 2+ψ(2n−1)(x)= ψ(m)(x) 2+ψ(2n−1)(x)=fm,2n−1(x)
(cid:2) (cid:3) (cid:2) (cid:3)
for m,n∈N, the result of a square and an additive operation of completely mono-
tonic functions, is completely monotonic on (0,∞).
COMPLETE MONOTONICITY OF FUNCTIONS INVOLVING POLYGAMMA FUNCTIONS 5
In [3, Theorem 2.1], [5, Theorem 2.1] and [6, Lemma 1.3], the function ψ(x)−
lnx+ α was proved to be completely monotonic on (0,∞), i.e.,
x
α (i)
(−1)i ψ(x)−lnx+ ≥0 (8)
x
h i
for i≥0, if and only if α≥1, so is its negative, i.e., the inequality (8) is reversed,
if and only if α ≤ 1. In [2], the function exΓ(x) was proved to be logarithmically
2 xx−α
completely monotonic on (0,∞), i.e.,
exΓ(x) (k)
(−1)k ln ≥0 (9)
(cid:20) xx−α (cid:21)
fork ∈N,ifandonlyifα≥1,soisitsreciprocal,i.e.,theinequality(9)isreversed,
if and only if α ≤ 1. Considering the fact (see, for example, [9, p. 82]) that a
2
completely monotonic function which is non-identically zero cannot vanish at any
point on (0,∞) and rearranging either (8) or (9) leads to the double inequalities:
For x∈(0,∞) and k ∈N,
1 1
lnx− <ψ(x)<lnx− (10)
x 2x
and
(k−1)! k! (k−1)! k!
+ <(−1)k+1ψ(k)(x)< + . (11)
xk 2xk+1 xk xk+1
Using the inequality (11) in the equation
f′ (x)=2ψ(m)(x)ψ(m+1)(x)+ψ(2n+1)(x)
m,2n
yields that
p (x)
′ m,n
f (x)≤
m,2n 4x2m+2n+3
and
q (x)
′ m,n
f (x)≥ ,
m,2n 2x2m+2n+3
where
p (x)=4(2n)!x2m+2+4(2n+1)!x2m+1−m!(m+1)!x2n
m,n
−2[(m!)2+(m−1)!(m+1)!]x2n+1−4(m−1)!m!x2n+2
and
q (x)=2(2n)!x2m+2+(2n+1)!x2m+1−2m!(m+1)!x2n
m,n
−2[(m!)2+(m−1)!(m+1)!]x2n+1−2(m−1)!m!x2n+2.
6 F.QIANDB.-N.GUO
If m>n, then the powers of x in q (x) have the relations
m,n
2m+2>2n+2>2n+1>2n and 2m+2>2m+1, (12)
so the function q (x) tends to ∞ as x→∞, and so f′ (x) is positive when x
m,n m,2n
islargeenough,thatis,the functionf (x)isincreasingonsomeinfinite interval
m,2n
whose left end point is large enough. If m < n, then the powers of x in q (x)
m,n
have the relations
2m+1<2n<2n+1<2n+2 and 2m+1<2m+2, (13)
thus the function q (x) tends to 0+ as x→0+, and so f′ (x) is positive when
m,n m,2n
x is enough close to 0 from the right-hand side, that is, the function f (x) is
m,2n
increasingonsomeintervalwherethenumber0isitsleftendpoint. Ifm=n,then
q (x)=2[(2m)!−(m−1)!m!]x2m+2+[(2m+1)!−2(m!)2
m,m
−2(m−1)!(m+1)!]x2m+1−2m!(m+1)!x2m;
From (2m)!−(m−1)!m! ≥ 1, it follows as above that the function f (x) is
m,2m
increasingonsomeinfiniteintervalwhoseleftendpointislargeenough. Inaword,
the function f (x) is not decreasing on (0,∞).
m,2n
Now consider the function p (x). If m > n, then the relations in (12) still
m,n
holds and the quantity 2n is the smallest power of x in the polynomial p (x);
m,n
andso, by the similar argumentas above,the function p (x) is negative and the
m,n
functionf (x)isdecreasingonsomeintervalwhoseleftendpointis0. Ifm<n,
m,2n
the relations in (13) is also valid and the number 2n+2 is the greatest power of
x in the polynomial p (x), consequently the function p (x) is negative and
m,n m,n
the function f (x) is decreasingon some infinite intervalwhose left end point is
m,2n
large enough. If m=n, then
p (x)=4[(2m)!−(m−1)!m!]x2m+2+2[2(2m+1)!−(m!)2
m,m
−(m−1)!(m+1)!]x2m+1−m!(m+1)!x2m;
Therefore,thefunctionp (x)isnegativeandthefunctionf (x)isdecreasing
m,m m,2m
on some interval whose left end point is 0. In a word, the function f (x) is not
m,2n
increasing on (0,∞).
COMPLETE MONOTONICITY OF FUNCTIONS INVOLVING POLYGAMMA FUNCTIONS 7
In conclusion, the functions f (x) for (m,n) 6= (1,1) are not monotonic on
m,2n
(0,∞).
′
Bythesameargumentasdiscussingthesignoff (x)above,wecanalsoturn
m,2n
out that the function f (x) for (m,n) 6= (1,1) does not keep the same sign on
m,2n
(0,∞). For the sake of shortening the length of this paper, we omit the detailed
argument. The proof of Theorem 1 is complete.
3. Second proof of Theorem 1
AftergoingthroughtheproofofTheorem1inSection2,wemayoutlineasketch
of a short proof for Theorem 1 as follows.
The complete monotonicity of f (x) has been proved in different manners and
1,2
in more general forms in [7, 8] and related references therein. For the convenience
of the reader, it would have been a good idea to include a short proof for it, to
avoid that the reader has to study either or both of the whole papers [7, 8].
Utilizing the formulas (5), (6) and (7) to compute straightforwardlygives
′ ′ ′ ′
f (x)−f (x+1)= ψ (x)−ψ (x+1) ψ (x)+ψ (x+1)
1,2 1,2
(cid:2) (cid:3)(cid:2) (cid:3)
′′ ′′
+ ψ (x)−ψ (x+1)
(cid:2) (cid:3)
1 1 2
′
= 2ψ (x)− −
x2(cid:20) x2(cid:21) x3
2 1 1
′
= ψ (x)− −
x2(cid:20) 2x2 x(cid:21)
∞
2 t t
= −1− e−xtdt
x2 Z (cid:18)1−e−t 2(cid:19)
0
∞
2 t/2
= −1 e−xtdt,
x2 Z (cid:18)tanh(t/2) (cid:19)
0
which shows that the function discussed is completely monotonic on (0,∞). So is
then the sum
N
[f (x+k)−f (x+k+1)]=f (x)−f (x+N +1)
1,2 1,2 1,2 1,2
kX=0
whichconvergestof (x)pointwiseasN →∞,andthereforef (x)iscompletely
1,2 1,2
monotonic on (0,∞).
8 F.QIANDB.-N.GUO
Since (−1)n+1ψ(n)(x) is a completely monotonic function for all n ∈ N, adding
thecompletelymonotonicfunctionψ(2n−1)(x)tothecompletelymonotonicfunction
ψ(m)(x) 2leadstothecompletemonotonicityoffm,2n−1(x)form,n∈Non(0,∞).
(cid:2) (cid:3)
Using the substitution xt=u in (6) yields
(n−1)!
ψ(n)(x)∼(−1)n+1 , x→∞ (14)
xn
and
n!
ψ(n)(x)∼(−1)n+1 , x→0+. (15)
xn+1
From these, it follows that
(1) for large x the behaviour of f (x) is like
m,2n
[(m−1)!]2 (2n−1)!
1− x2(m−n)
x2m (cid:26) [(m−1)!]2 (cid:27)
and this is clearly
(a) negative when m>n and x is large, or
(b) positive when m<n and x is large,
and if m=n the sign is determined by
2m−1
1−m
(cid:18)m−1(cid:19)
which is negative unless m=1,
(2) for small x the behaviour of f (x) is like
m,2n
(m!)2 (2n)!
1− x2(m−n)+1
x2(m+1)(cid:26) (m!)2 (cid:27)
and this is clearly
(a) negative when m<n and x is small, or
(b) positive when m≥n and x is small.
Asaresult,thefunctionf (x)for(m,n)6=(1,1)oritsnegativeisnotcompletely
m,2n
monotonic and does not keep the same sign on (0,∞). The proof of Theorem 1 is
complete.
Acknowledgements. The originalversionofthis manuscriptwascompleteddur-
ing the first author’s visit to the RGMIA, Victoria University, Australia between
March2008andFebruary2009. Thefirstauthorwouldliketoexpressmanythanks
to Professors Pietro Cerone and Server S. Dragomir and other local colleagues for
their invitation and hospitality throughout this period.
COMPLETE MONOTONICITY OF FUNCTIONS INVOLVING POLYGAMMA FUNCTIONS 9
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(F.Qi)DepartmentofMathematics,CollegeofScience,TianjinPolytechnicUniver-
sity,Tianjin City, 300160,China
E-mail address: [email protected], [email protected], [email protected]
URL:http://qifeng618.spaces.live.com
(B.-N.Guo) School of Mathematicsand Informatics,Henan Polytechnic University,
JiaozuoCity, HenanProvince,454010,China
E-mail address: [email protected], [email protected]
URL:http://guobaini.spaces.live.com
