Volume 10 (2009), Issue 1, Article 12, 8 pp.
SOME INEQUALITIES FOR THE q-DIGAMMA FUNCTION
TOUFIK MANSOUR AND ARMEND SH. SHABANI
D EPARTMENT OF M ATHEMATICS
U NIVERSITY OF H AIFA
31905 H AIFA , I SRAEL
toufik@math.haifa.ac.il
D EPARTMENT OF M ATHEMATICS
U NIVERSITY OF P RISHTINA
AVENUE "M OTHER T HERESA "
5 P RISHTINE 10000, R EPUBLIC OF KOSOVA
armend_shabani@hotmail.com
Received 17 June, 2008; accepted 21 February, 2009
Communicated by A. Laforgia
A BSTRACT. For the q-digamma function and it’s derivatives are established the functional inequalities of the types:
f 2 (x · y) ≶ f (x) · f (y),
f (x + y) ≶ f (x) + f (y).
Key words and phrases: q-digamma function, inequalities.
2000 Mathematics Subject Classification. 33D05.
1. I NTRODUCTION
The Euler gamma function Γ(x) is defined for x > 0 by
Z ∞
Γ(x) =
tx−1 e−t dt.
0
The digamma (or psi) function is defined for positive real numbers x as the logarithmic derivative of Euler’s gamma function, ψ(x) = Γ0 (x)/Γ(x). The following integral and series
representations are valid (see [1]):
Z ∞ −t
e − e−xt
1 X
x
(1.1)
ψ(x) = −γ +
dt
=
−γ
−
+
,
−t
1−e
x n≥1 n(n + x)
0
where γ = 0.57721 . . . denotes Euler’s constant. Another interesting series representation for
ψ, which is “more rapidly convergent" than the one given in (1.1), was discovered by Ramanujan
[3, page 374].
177-08
2
T OUFIK M ANSOUR AND A RMEND S H . S HABANI
Jackson (see [5, 6, 7, 8]) defined the q-analogue of the gamma function as
(q; q)∞
(1 − q)1−x , 0 < q < 1,
(1.2)
Γq (x) = x
(q ; q)∞
and
x
(q −1 ; q −1 )∞
(1.3)
Γq (x) = −x −1 (q − 1)1−x q (2) , q > 1,
(q ; q )∞
Q
j
where (a; q)∞ = j≥0 (1 − aq ).
The q-analogue of the psi function is defined for 0 < q < 1 as the logarithmic derivative of
the q-gamma function, that is,
d
ψq (x) =
log Γq (x).
dx
Many properties of the q-gamma function were derived by Askey [2]. It is well known that
Γq (x) → Γ(x) and ψq (x) → ψ(x) as q → 1− . From (1.2), for 0 < q < 1 and x > 0 we get
X q n+x
(1.4)
ψq (x) = − log(1 − q) + log q
1 − q n+x
n≥0
X q nx
= − log(1 − q) + log q
1 − qn
n≥1
and from (1.3) for q > 1 and x > 0 we obtain
(1.5)
!
1 X q −n−x
ψq (x) = − log(q − 1) + log q x − −
2 n≥0 1 − q −n−x
!
1 X q −nx
= − log(q − 1) + log q x − −
.
2 n≥1 1 − q −n
A Stieltjes integral representation for ψq (x) with 0 < q < 1 is given in [4]. It is well-known
that ψ 0 is strictly completely monotonic on (0, ∞), that is,
(−1)n (ψ 0 (x))(n) > 0 for x > 0 and n ≥ 0,
see [1, Page 260]. From (1.4) and (1.5) we conclude that ψq0 has the same property for any q > 0
(−1)n (ψq0 (x))(n) > 0 for x > 0 and n ≥ 0.
If q ∈ (0, 1), using the second representation of ψq (x) given in (1.4), it can be shown that
X nk · q nx
(1.6)
ψq(k) (x) = logk+1 q
1 − qn
n≥1
(k)
and hence (−1)k−1 ψq (x) > 0 with x > 1, for all k ≥ 1. If q > 1, from the second representation of ψq (x) given in (1.5), we obtain
!
X nq −nx
(1.7)
ψq0 (x) = log q 1 +
1 − q −nx
n≥1
and for k ≥ 2,
ψq(k) (x) = (−1)k−1 logk+1 q
(1.8)
X nk q −nx
1 − q −nx
n≥1
(k)
and hence (−1)k−1 ψq (x) > 0 with x > 0, for all q > 1.
J. Inequal. Pure and Appl. Math., 10(1) (2009), Art. 12, 8 pp.
http://jipam.vu.edu.au/
S OME I NEQUALITIES FOR THE q-D IGAMMA F UNCTION
3
In this paper we derive several inequalities for ψ (k) (x), where k ≥ 0.
2. I NEQUALITIES OF THE TYPE f 2 (x · y) ≶ f (x) · f (y)
We start with the following lemma.
Lemma 2.1. For 0 < q <
1
2
and 0 < x < 1 we have that ψq (x) < 0.
Proof. At first let us prove that ψq (x) < 0 for all x > 0. From (1.4) we get that
ψq (x) =
X q nx
qx
log q − log(1 − q) + log q
.
1−q
1 − qn
n≥2
In order to see that ψq (x) < 0, we need to show that the function
qx
g(x) =
log q − log(1 − q)
1−q
is a negative for all 0 < x < 1 and 0 < q < 12 . Indeed g 0 (x) =
that g(x) is an increasing function on 0 < x < 1, hence
qx
1−q
log2 q > 0, which implies
q
log q − log(1 − q)
1−q
qq
1
=
log
< 0,
1−q
(1 − q)1−q
g(x) < g(1) =
for all 0 < q < 21 .
Theorem 2.2. Let 0 < q <
1
2
and 0 < x, y < 1. Let k ≥ 0 be an integer. Then
ψq(k) (x)ψq(k) (y) < (ψq(k) (xy))2 .
Proof. We will consider two different cases: (1) k = 0 and (2) k ≥ 1.
(1) Let f (x) = ψq2 (x) defined on 0 < x < 1. By Lemma 2.1 we have that
f 0 (x) = 2ψq (x)ψq0 (x) < 0
for all 0 < x < 1, which gives that f (x) is a decreasing function on 0 < x < 1. Hence, for all
0 < x, y < 1 we have
ψq2 (xy) > ψq2 (x)
and
ψq2 (xy) > ψq2 (y),
which gives that
ψq4 (xy) > ψq2 (x)ψq2 (y).
Since ψq (x)ψq (y) > 0 for all 0 < x, y < 1, see Lemma 2.1, we obtain that
ψq2 (xy) > ψq (x)ψq (y),
as claimed.
J. Inequal. Pure and Appl. Math., 10(1) (2009), Art. 12, 8 pp.
http://jipam.vu.edu.au/
4
T OUFIK M ANSOUR AND A RMEND S H . S HABANI
(2) From (1.6) we have that
ψq(k) (x)ψq(k) (y) − (ψq(k) (xy))2
!
!
X nk q ny
X nk q nx
= logk+1 q
logk+1 q
−
n
1
−
q
1 − qn
n≥1
n≥1
= (logk+1 q)2
= (log
k+1
logk+1 q
X nk q nxy
n≥1
!2
1 − qn
X (nm)k q (n+m)xy
X nk q nx mk q my
k+1
2
·
−
(log
q)
1 − qn 1 − qm
(1 − q n )(1 − q m )
n,m≥1
n,m≥1
X (nm)k (q nx+my − q (n+m)xy )
q)
.
(1 − q n )(1 − q m )
n,m≥1
2
For 0 < x, y < 1 , q nx+my − q (n+m)xy < 0 and for x, y > 1, q nx+my − q (n+m)xy > 0 and the
results follow.
1 Note that the above theorem for k ≥ 1 remains true also for q ∈ 2 , 1 . Also, if x, y > 1,
k ≥ 1 and 0 < q < 1 then
2
ψq(k) (x)ψq(k) (y) > ψq(k) (xy) .
Now we extend Lemma 2.1 to the case q > 1. In order to do that we denote the zero of the
q−3
function f (q) = 2(q−1)
log(q) − log(q − 1), q > 1, by q ∗ . The numerical solution shows that
q ∗ ≈ 1.56683201 . . . as shown on Figure 2.1.
1
0.5
0
1.5
2
2.5
3
3.5
4
4.5
5
q
–0.5
–1
Figure 2.1: Graph of the function
q−3
2(q−1)
log q − log(q − 1).
Lemma 2.3. For q > q ∗ and 0 < x < 1 we have that ψq (x) < 0.
Proof. From (1.5) we get that
X q −nx
q −x
1
ψq (x) = −
log q − log(q − 1) + log q x −
− log q
.
1 − q −1
2
1 − q −n
n≥2
J. Inequal. Pure and Appl. Math., 10(1) (2009), Art. 12, 8 pp.
http://jipam.vu.edu.au/
S OME I NEQUALITIES FOR THE q-D IGAMMA F UNCTION
5
In order to show our claim, we need to prove that
1
q −x
g(x) = −
log q − log(q − 1) + log q x −
<0
1 − q −1
2
q −x
1−q −1
on 0 < x < 1. Since g 0 (x) =
function on 0 < x < 1. Hence
log2 q + log q > 0, it implies that g(x) is an increasing
g(x) < g(1) =
q−3
log q − log(q − 1) < 0,
2(q − 1)
for all q > q ∗ , see Figure 2.1.
Theorem 2.4. Let q > 2 and 0 < x, y < 1. Let k ≥ 0 be an integer. Then
2
ψq(k) (x)ψq(k) (y) < ψq(k) (xy) .
Proof. As in the previous theorem we will consider two different cases: (1) k = 0 and (2)
k ≥ 1.
(1) As shown in the introduction the function ψq0 (x) is an increasing function on 0 < x < 1.
Therefore, for all 0 < x, y < 1 we have that
ψq (xy) < ψq (x)
Hence, Lemma 2.3 gives that
ψq2 (xy)
and
ψq (xy) < ψq (y).
> ψq (x)ψq (y), as claimed.
(2) Analogous to the second case of Theorem 2.2.
Note that Theorem 2.4 for k ≥ 1 remains true also for q > 1. Also, if x, y > 1, k ≥ 1 and
q > 1 then
2
ψq(k) (x)ψq(k) (y) > ψq(k) (xy) .
3. I NEQUALITIES OF THE T YPE f (x + y) ≶ f (x) + f (y)
The main goal of this section is to show that ψq (x + y) ≥ ψq (x) + ψq (y), for all 0 < x, y < 1
and 0 < q < 1. In order to do that we define
X q j (q j − 2)
ρ(q) = log(1 − q) + log q
.
j
1
−
q
j≥1
Lemma 3.1. For all 0 < q < 1, ρ(q) > 0.
P
q j (q j −2)
with constant c > 0 for m ≥ 2. Then
Proof. Let 0 < q < 1 and let gm (q) = c + m−1
j=1
1−q j
gm (0) = c, limq→1− gm (q) < 0 and gm (q) is a decreasing function since
0
gm
(q)
=−
m−1
X
j=1
jq j−1 (1 + (1 − q j )2 )
< 0.
(1 − q j )2
On the other hand
q m (q m − 2)
<0
1 − qm
for all 0 < q < 1. Hence, for all m ≥ 2 we have that
gm+1 (q) − gm (q) =
gm+1 (q) < gm (q),
0 < q < 1.
Thus, if bm is the positive zero of the function gm (q) (because gn (q) is decreasing) on 0 <
q < 1 (by Maple or any mathematical programming we can see that b1 = 0.38196601 . . .,
b2 = 0.3184588966 and b3 = 0.3055970874), then gm (q) > 0 for all 0 < q < bm and gm (q) < 0
J. Inequal. Pure and Appl. Math., 10(1) (2009), Art. 12, 8 pp.
http://jipam.vu.edu.au/
6
T OUFIK M ANSOUR AND A RMEND S H . S HABANI
for all bm < q < 1. Furthermore, the sequence {bm }m≥0 is a strictly decreasing sequence of
positive real numbers, that is 0 < bm+1 < bm , and bounded by zero, which implies that
X q j (q j − 2)
lim gm (q) = c +
< 0,
m→∞
1 − qj
j≥1
for all 0 < q < 1. Hence, if we choose c = 2 log(1−q)
(c is positive since 0 < q < 1), then we
log q
have that
X q j (q j − 2)
2 log(1 − q)
<−
,
j
1
−
q
log
q
j≥1
which implies that
ρ(q) = log(1 − q) + log q
X q j (q j − 2)
1 − qj
j≥1
> −2 log(1 − q) + log(1 − q)
= − log(1 − q) > 0,
as requested.
Theorem 3.2. For all 0 < q < 1 and 0 < x, y < 1,
ψq (x + y) > ψq (x) + ψq (y).
Proof. From the definitions we have that
ψq (x + y) − ψq (x) − ψq (y) = log(1 − q) + log q
X q n(x+y) − q nx − q ny
n≥1
1 − qn
.
Since 0 < x, y, q < 1, we have that
q n(x+y) − q nx − q ny = (1 − q nx )(1 − q ny ) − 1
< (1 − q n )2 − 1
= q n (q n − 2).
Hence, by Lemma 3.1
ψq (x + y) − ψq (x) − ψq (y) > ρ(q) > 0,
which completes the proof.
The above theorem is not true for x, y > 1, for example
ψ1/10 (4) = 0.1051046497 . . . ,
ψ1/10 (5) = 0.1053349312 . . . ,
ψ1/10 (9) = 0.1053605131 . . . .
Theorem 3.3. For all q > 1 and 0 < x, y < 1,
ψq (x + y) > ψq (x) + ψq (y).
Proof. From the definitions we have that
X Qn(x+y) − Qnx − Qny
1
ψq (x + y) − ψq (x) − ψq (y) = log(q − 1) + log q + log Q
,
2
1 − Qn
n≥1
J. Inequal. Pure and Appl. Math., 10(1) (2009), Art. 12, 8 pp.
http://jipam.vu.edu.au/
S OME I NEQUALITIES FOR THE q-D IGAMMA F UNCTION
7
where Q = 1/q. Thus
ψq (x + y) − ψq (x) − ψq (y)
= log(q − 1) +
1
log q + ψQ (x + y) − ψQ (x) − ψQ (y) − log(1 − Q).
2
Using Theorem 3.2 we get that
ψq (x + y) − ψq (x) − ψq (y) > log(q − 1) +
1
log q − log(q − 1) + log q > 0,
2
which completes the proof.
Note that the above theorem holds for q > 2 and x, y > 1, since
ψq (x + y) − ψq (x) − ψq (y)
= log(q − 1) +
X q −nx (1 − q −ny ) + q −ny
1
log q + log q
> 0.
−n
2
1
−
q
n≥1
The above theorem is not true for x, y > 1 when 1 < q < 2, for example
ψ3/2 (4) = 1.83813910 . . . ,
ψ3/2 (5) = 2.34341101 . . . ,
ψ3/2 (9) = 4.10745515 . . . .
Theorem 3.4. Let q ∈ (0, 1). Let k ≥ 1 be an integer.
(1) If k is even then
ψq(k) (x + y) ≥ ψq(k) (x) + ψq(k) (y).
(2) If k is odd then
ψq(k) (x + y) ≤ ψq(k) (x) + ψq(k) (y).
Proof. From (1.6) we have
ψqk (x + y) − ψqk (x) − ψqk (x) = logk+1 q
X
n≥1
Since the function f (z) = q
nz
nk
(q n(x+y) − q nx − q ny ).
1 − qn
is convex from
x+y
1
f
≤ (f (x) + f (y)),
2
2
we obtain that
2 · qn
(3.1)
x+y
2
≤ q nx + q ny .
On the other hand it is clear that
2 · qn
(3.2)
x+y
2
> q n(x+y) .
From (3.1) and (3.2) we have that
q n(x+y) − q nx − q ny < 0.
(1) Since for q ∈ (0, 1) and k even we have logk+1 q < 0, hence
ψq(k) (x + y) − ψq(k) (x) − ψq(k) (x) ≥ 0.
(2) The other case can be proved in a similar manner.
Using a similar approach one may prove analogue results for q > 1.
J. Inequal. Pure and Appl. Math., 10(1) (2009), Art. 12, 8 pp.
http://jipam.vu.edu.au/
8
T OUFIK M ANSOUR AND A RMEND S H . S HABANI
R EFERENCES
[1] M. ABRAMOWITZ AND I.A. STEGUN, Handbook of Mathematical Functions with Formulas and
Mathematical Tables, Dover, NewYork, 1965.
[2] R. ASKEY, The q-gamma and q-beta functions, Applicable Anal., 8(2) (1978/79) 125–141.
[3] B.C. BERNDT, Ramanujan’s Notebook, Part IV. Springer, New York 1994.
[4] M.E.H. ISMAIL AND M.E. MULDOON, Inequalities and monotonicity properties for gamma and
q-gamma functions, in: R.V.M. Zahar (Ed.), Approximation and Computation, International Series
of Numerical Mathematics, vol. 119, Birkhäuser, Boston, MA, 1994, pp. 309–323.
[5] T. KIM, On a q-analogue of the p-adic log gamma functions and related integrals, J. Number Theory,
76 (1999), 320–329.
[6] T. KIM, A note on the q-multiple zeta functions, Advan. Stud. Contemp. Math., 8 (2004), 111–113.
[7] T. KIM AND S.H. RIM, A note on the q-integral and q-series, Advanced Stud. Contemp. Math., 2
(2000), 37–45.
[8] H.M. SRIVASTAVA, T. KIM AND Y. SIMSEK, q-Bernoulli numbers and polynomials associated
with multiple q-zeta functions and basic L-series, Russian J. Math. Phys., 12 (2005), 241–268.
J. Inequal. Pure and Appl. Math., 10(1) (2009), Art. 12, 8 pp.
http://jipam.vu.edu.au/