# New Inequalities for Gamma and Digamma Functions

Authors: M. R. Farhangdoost; M. Kargar Dolatabadi
Journal: Journal of Applied Mathematics
 (2014)

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## Abstract

By using the mean value theorem and logarithmic convexity, we obtain some new inequalities for gamma and digamma functions.


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## Body

## 1. Introduction

LetΓ(x), ψ(x), ψn(x), and ζ(x) denote the Euler gamma function, digamma function, polygamma functions, and Riemann zeta function, respectively, which are defined by
(1)Γx=∫0∞e-ttx-1dt,forx>0,ψx=Γ′xΓx,forx>0,(2)ψnx=-1n+1∫0∞tne-xt1-e-tdt,forx>0;n=1,2,3,…,(3)ζx=∑n=1∞1nx,forx>1.
In the past different papers appeared providing inequalities for the gamma, digamma, and polygamma functions (see [1–18]).By using the mean value theorem to the functionlog⁡Γ(x) on [u,u+1], with x>0 and u>0, Batir [19] presented the following inequalities for the gamma and digamma functions:
(4)ψx⩽log⁡x-1+e-γ,forx>0,log⁡x-ψx<12ψ′x,forx>1,ψ′x⩾π26eγe-ψx,forx⩾1.In Section2, by applying the mean value theorem on
(5)log⁡Γx′=ψx,forx>0,
we obtain some new inequalities on gamma and digamma functions.Section3 is devoted to some new inequalities on digamma function, by using convex properties of logarithm of this function.Note that in this paper byγ=lim⁡n→∞(∑k=1n1/k-log⁡(n))=0.5772156⋯ we mean Euler’s constant [5].


## 2. Inequalities for Gamma and Digamma Functions by the Mean Value Theorem

Lemma 1.
Fort>0, one has
(6)-ψ′′tψ′t2<1.Proof.
By [6, Proposition 1], we have
(7)ψ′tψ′′′t-2ψ′′t2<0,fort>0.
Thus the function ψ′′t/ψ′t2 is strictly decreasing on (0,∞).
By using asymptotic expansions [20, pages 253–256 and 364],
(8)ψ′t=1t+12t2+16t3+θ′30t5,0⩽θ′⩽1,(9)ψ′′t=-1t2-1t3-12t4+16t6-θ′′6t8,0⩽θ′′⩽1.
For t>0, we get
(10)lim⁡t→∞ψ′′tψ′t2=-1.
Now, the proof follows from the monotonicity of ψ′′t/ψ′(t)2 on (0,∞) and
(11)lim⁡t→∞ψ′′tψ′t2=-1.Theorem 2.
One has the following:(a)
x-1/2<1/ψ′x⩽x+(6/π2)-1 for x⩾1;(b)
1/x2<ψ′xψ′x+1<2/x2 for x>0;(c)
[ψ′(x)]2/ψ′′(x)⩾-π4/72ζ(3) for x⩾1 and x2ψ′x+1ψ′(x)<π4/72ζ(3) for x>2;(d)
[ψ′x+h]2-ψ′xψ′x+h/hψ′x>ψ′′x+h for x>0 and h>0;(e)
ψ′x+hψ′x-[ψ′x]2/hψ′x+h<ψ′′x for x>0 and h>0;(f)
-x2ψ′′x<ψ′x/ψ′x+1 and ψ′x+1/ψ′x<-x2ψ′′x+1 for x>0;(g)
π2x/6+1(x+6/π2)e-x(γ+1)⩽Γ(x+1)<(2x+1)(x+1/2)e-x(1+γ) for x⩾1;(h)
1/x-ψ′x<1/2ψ′′x+1/2 for x>0 and 1/x-ψ′x>ψ′-11-1ψ′′x for x>1;(i)
ψ(x+1)>log⁡(x+1/2)+ψ((ψ′)-1(1)) for x⩾1/2;(j)
(π4/72ζ3)log⁡(x-(ψ′)-1(1)+2)+ψ((ψ′)-1(1))⩾ψ(x+1) for x>(ψ′)-1(1)-1.Proof.
Letu be a positive real number and ψ(x) defined on the closed interval [u,u+1]. By using the mean value theorem for the function ψ(x) on [u,u+1] with u>0 and since ψ′ is a decreasing function, there is a unique θ depending on u such that 0⩽θ=θ(u)<1, for all u⩾0; then
(12)ψu+1-ψu=ψ′u+θu,
Since ψ(x+1)-ψ(x)=1/x and ψ′x+1-ψ′x=-1/x2, we have
(13)ψ′u+θu=1u,foru>0.
We show that the function θ(u) has the following properties:(1)
θ(u) is strictly increasing on (0,∞);(2)
lim⁡u→∞θ(u)=1/2;(3)
θ′u is strictly decreasing on (0,∞);(4)
lim⁡u→∞θ′u=0.
To prove these four properties, sinceψ′ is a decreasing function on (0,∞), we put u=1/ψ′(t), where t>0; by formula (13) we have
*ψ′1ψ′t+θ1ψ′t=ψ′t.
Since by formula (8) we have ψ′′t<0 and ψ′t>0, for all t>0, then the mapping t→ψ′t from (0,∞) into (0,∞) is injective since also ψ′t→0 and ψ′t→∞ when t→∞ and t→0+, respectively, then the mapping t→ψ′t from (0,∞) into (0,∞) is a bijective map. Clearly, by injectivity of ψ′, we find that
(14)θ1ψ′t=t-1ψ′t,fort>0.
Differentiating between both sides of this equation, we get
(15)θ′1ψ′t=-ψ′t2+ψ′′tψ′′t.
Since by formula (8), ψ′′t<0, where t>0, hence formula (15) gives θ′(1/ψ′t)>0, for all t>0. Since the mapping t→1/ψ′t from (0,∞) to (0,∞) is also bijective, then θ′t>0 for all t>0, and the proof of (1) is completed.
From (8) we have
(16)lim⁡u→∞θu=lim⁡t→∞θ1ψ′t=lim⁡t→∞t-1ψ′t=lim⁡t→∞t-11/t+1/2t2+1/6t3+1/3t5=12.
Differentiating between both sides of (15), we obtain
**θ′′1ψ′t=ψ′t3ψ′′t2ψ′′t2-ψ′tψ′′t.
Since ψ′t>0 and ψ′′t<0, where t>0, then θ′′1/ψ′t<0 for all t>0. Proceeding as above we conclude that θ′′t<0, for t>0. This proves (3).
For (4), from (8), (9), we conclude that
(17)lim⁡u→∞θ′u=lim⁡t→∞θ′1ψ′t=lim⁡t→∞-ψ′t2+ψ′′tψ′′t=-1-lim⁡t→∞ψ′tψ′′t2=0.
Now, we prove the theorem. To prove (a), let 1/ψ′1=6/π2⩽t<∞; then by (1) and (2) we have
(18)θ1ψ′1⩽θt<lim⁡t→∞θt.
Equation (13) and ψ′′t<0 for all t>0 give
(19)θt=ψ′-11t-t.
By substituting the value of θ(t) into (18), we get
(20)1-1ψ′1⩽ψ′-11t-t<lim⁡t→∞θt=12.
By substituting the value t=1/ψ′u into this inequality, we get
(21)u-12<1ψ′u⩽u+6π2-1,
where u⩾1.
In order to prove (b), by using the mean value theorem on the interval[1/ψ′t,1/ψ′t+1], and since θ is a decreasing function, there exists a unique δ such that
(22)0<δt<1,
for t>0 and
(23)θ1ψ′t+1-θ1ψ′t=1ψ′t+1-1ψ′tθ′1ψ′t+δt.
Now, by (14), we have
(24)1-1ψ′t+1+1ψ′t=1ψ′t+1-1ψ′tθ′1ψ′t+δt.
Since θ is strictly increasing on (0,∞), by (1), we have
(25)1+ψ′t+1-ψ′tψ′t+1ψ′t=θ1ψ′t+1-θ1ψ′t>0.
By using this inequality and the fact that ψ(x+1)-ψ(x)=1/x and
(26)ψ′x+1-ψ′x=-1x2,
we obtain
(27)ψ′t+1ψ′t>1t2,t>0.
Since θ is strictly increasing on (0,∞), by (1), it is clear that
(28)θ1ψ′t+1-θ1ψ′t<lim⁡t→∞θt-θ0+=12,t>0.
and then it is clear that (b) holds.
For (c), sincet>2, t+δ(t)>1+δ(1), and θ′ is strictly decreasing on (0,∞) by (3), then
(29)θ′1ψ′t+δt<θ′1ψ′1=-1-ψ′12ψ′′1,00000000000000000000000000.00000∀t>2.
Since ψ(x+1)-ψ(x)=1/x and ψ′x+1-ψ′x=-1/x2, by using (24), we obtain
(30)t2ψ′t+1ψ′t<π472ζ3,
where t>2.
Sinceθ′ is strictly decreasing on (0,∞) by (3) and ψ′′t<0, for all t>0, we have
(31)θ′1ψ′t⩽θ′1ψ′1,
where t⩾1.
Then it is clear that (c) is true.
Now we prove (d) and (e) by using the mean value theorem on[1/ψ′t,1/ψ′t+h](t>0,h>0), for θ, we conclude
(32)θ1ψ′t+h-θ1ψ′t=1ψ′t+h-1ψ′tθ′1ψ′t+a,
where 0<a<h.
After brief computation we have(33)θ′1ψ′t+a=hψ′t+hψ′tψ′t-ψ′t+h-1,t>0.
Since t+a>t for all a>0, t>0, and by the monotonicity of θ′ and ψ′ we have θ′(1/ψ′t+a)<θ′(1/ψ′t); then
(34)ψ′t+hψ′t-ψ′t2hψ′t+h<ψ′′t,t>0,h>0.
By monotonicity of  θ′ and ψ′, we have
(35)θ′1ψ′t+a>θ′1ψ′t+h.
After some simplification of this inequality (d) is proved.
For (f), we puth=1 in (e) and (d).
For (g), we integrate (a) on[1,t] for t>0; then we have
(36)log⁡t-1π26+1-γ⩽ψt<log⁡2t-1-γ,fort⩾1;
the proof is completed when we integrate these inequalities on [1,s], for s>0.
By using the mean value theorem for theψ′t on [t,t+θ(t)], there is a α(t) depending on t such that 0<α(t)<θ(t) for all t>0, and so
(37)ψ′t+θt=θtψ′′t+αt+ψ′t.
By formula (13) and (2), since ψ′′ is strictly increasing on (0,∞), we have
(38)ψ′′t+αtθt=1t-ψ′t<lim⁡t→∞θtψ′′t+lim⁡t→∞θt,fort>0,
or
(39)1t-ψ′t<12ψ′′t+12,fort>0;
since ψ′′ is strictly increasing on (0,∞), by (1), we have
(40)θtψ′′t+αt=1t-ψ′t>θ1ψ′′t,θtψ′′t+αt1.1tψ′t>θ1fort>1,
or
(41)1t-ψ′t>ψ′-11-1ψ′′t,fort>1.
In order to prove (i) and (j), we integrate both sides of (13) over 1⩽u⩽x to obtain
(42)∫1xψ′u+θudu=∫1x1udu.
Making the change of variable u=1/ψ′t on the left-hand side, by (14), we have
(43)∫ψ′-11x+θxψ′t-ψ′′tψ′t2dt=log⁡x;
since ψ′t>0 for all t>0 and ψ′xψ′′x-2[ψ′′x]2<0, we find that, for x>1,
(44)log⁡x<∫ψ′-11x+θxψ′tdt=ψx+θx-ψψ′-11
or
(45)log⁡x+ψψ′-11<ψx+θx.
Again using the monotonicity of θ and ψ, after some simplifications as for x⩾1/2, we can rewrite
(46)log⁡x+12+ψψ′-11<ψx+1.
This proves (i). By inequality (c) for x⩾1, we have
(47)log⁡x⩾72ζ3π4∫ψ′-11x+θxψ′tdt=72ζ3π4ψx+θx-ψψ′-11;
since for x⩾1, θ(x)⩾θ(1)=((ψ′)-1(1)-1))=(ψ′)-1(1)-1, from this inequality we find that
(48)π472ζ3log⁡x+ψψ′-11⩾ψx+ψ′-11-1;
replacing x by x-(ψ′)-1(1)+2, we get for x⩾(ψ′)-1(1)-1(49)π472ζ3log⁡x-ψ′-11+2+ψψ′-11⩾ψx+1,
which proves (j). Then the proof is completed.Example 3.
Consider the matrix(50)An=311⋯1141⋯1⋮11⋯1n+1.
By using inequalities (a), we obtain
(51)π2π2x+6-π2⩽ψ′x<22x-1,x⩾1.
Now, we integrate on [1,t] (for t>0) from both sides of (51) to obtain
(52)log⁡t-1π26+1-γ⩽ψt<log⁡2t-1-γ;
replacing t by n+1 (n is an integer number) and using the identity ψ(n+1)=Hn-γ [6] and det⁡An=n!Hn [21], where Hn=∑k=1n1/k is the nth harmonic number, then we have
(53)log⁡nπ26+1n!⩽n!Hn<log⁡2n+1n!.


## 3. New Inequalities for Digamma Function by Properties of Strictly Logarithmically Convex Functions

Definition 4.
A positive functionf is said to be logarithmically convex on an interval I if f has derivative of order two on I and
(54)log⁡fx′′⩾0
for all x∈I.
If inequality (54) is strict, for all x∈I, then f is said to be strictly logarithmically convex [22].Lemma 5.
The functionΓ is increasing on [c,∞), where c=1/46163⋯ is the only positive zero of ψ [1, 19].Lemma 6.
Ifx⩾c and k(x)=1/ψ(x), then k is strictly logarithmically convex on [c,∞).Proof.
By differentiation we have(55)log⁡kx′′=-ψ′xψx′=-ψ′′xψx+ψ′x2ψx2;
by Lemma 5, we obtain ψ(x)=Γ′x/Γx>0, for every x∈[c,∞) and since ψ′′x<0 on (0,∞), then we have log⁡kx′′>0, for x⩾c.
This implies that1/ψ(x) is strictly logarithmically convex on [c,∞).Theorem 7.
One has the following:(a)
[ψ(x+3)]a/ψ(ax+3)>(3/2-γ)a-1,  for a>1 and x>-3/a;(b)
[ψ(x+3)]a/ψ(ax+3)<(11/6-γ)a/ψ(3+a), for a>1 and x∈(0,1);(c)
[ψ(x+3)]a/ψ(ax+3)>(11/6-γ)a/ψ(3+a), for a>1 and x>1;(d)
[ψ(x+3)]a/ψ(ax+3)>(11/6-γ)a/ψ(3+a), for a∈(0,1) and x∈(0,1);(e)
[ψ(x+3)]a/ψ(ax+3)<(11/6-γ)a/ψ(3+a), for a∈(0,1) and x>1.Proof.
By Lemma6 we have, for a>1,
(56)ψup+vq>ψu1/pψv1/q,
where p>1, q>1, 1/p+1/q=1, u⩾c, and v⩾c.
Ifp=a and q=a/a-1, then
(57)ψ1au+1-1av>ψu1/aψv1-1/a
for u⩾c and v⩾c.
Letv=3 and u=ax+3. Note that ψ(3)=3/2-γ and 1/au+(1-1/a)v=x+3; also we obtain
(58)ψx+3aψax+3>32-γa-1Mforx=u-3a>-3a.
In order to prove (b), let
(59)fx=log⁡ψax+3-log⁡ψ3+a-alog⁡ψx+3;
since ψ(4)=11/6-γ, we have f(1)=log⁡(11/6-γ)-a. Also
(60)f′x=aψ′ax+3ψax+3-ψ′x+3ψx+3.
By Lemma 6, log⁡(1/ψ(t)) is strictly convex on [c,∞); then log⁡ψt′′<0 and so (ψ′t/ψ(t))′<0; this implies that (ψ′t/ψ(t)) is strictly decreasing on [c,∞). Since a>1 and x∈(0,1), we have ax+3>x+3. Then
(61)ψ′ax+3ψax+3<ψ′x+3ψx+3.
And then f′x<0; also f(1)=log⁡(11/6-γ)-a. Then
(62)fx>f1=log⁡116-γ-a
for a>1 and x∈(0,1) or
(63)ψx+3aψax+3<11/6-γaψ3+a.
So (b) is proved.
By(64)ax+3>x+3,fora>1,x>1,ax+3<x+3,fora∈0,1,x∈0,1,ax+3<x+3,fora∈0,1,x>1,
(c), (d), and (e) are clear.Corollary 8.
For allx∈(0,1) and all integers n>1, one has
(65)32-γn-1<ψx+3nψnx+3<11/6-γnHn+2-γ,
where Hn=∑k=1n(1/k) is the nth harmonic number.Proof.
By [6], for all integers n⩾1, we have
(66)ψn+1=Hn-γ,
and replacing a by n in Theorem 7, the proof is completed.Theorem 9.
Letf be a function defined by
(67)fx=ψ3+bxaψ3+axb;∀x>0,
where 3+ax⩾c and 3+bx⩾c; then for all a>b>0 or 0>a>b  (a>0 and b<0), f is strictly increasing (strictly decreasing) on (0,∞).Proof.
Letg be a function defined by
(68)gx=log⁡fx=alog⁡ψ3+bx-blog⁡ψ3+ax;
then
(69)g′x=abψ′3+bxψ3+bx-ψ′3+axψ3+ax.
By proof of Theorem 7, we have
(70)log⁡ψx′′<0,forx∈c,∞;
this implies that g′x>0 if a>b>0 or 0>a>b  (g′x<0 if a>0 and b<0); that is, g is strictly increasing on (0,∞) (strictly decreasing on (0,∞)). Hence f is strictly increasing on (0,∞), if a>b>0 or 0>a>b (strictly decreasing if a>0 and b<0).Corollary 10.
For allx∈(0,1) and all a>b>0 or 0>a>b, one has
(71)32-γa-b<ψ3+bxaψ3+axb<ψ3+baψ3+ab,
where 3+bx⩾c, 3+ax⩾c, 3+b⩾c, and 3+a⩾c.Proof.
To prove (71), applying Theorem 9 and taking account of ψ(3)=3/2-γ, we get f(0)<f(x)<f(1) for all x∈(0,1), and we obtain (71).Corollary 11.
For allx∈(0,1) and all a>0 and b<0, one has
(72)ψ3+baψ3+ab<ψ3+bxaψ3+axb<32-γa-b,
where 3+ax⩾c, 3+bx⩾c, 3+b⩾c, and 3+a⩾c.Proof.
Applying Theorem9, we get f(1)<f(x)<f(0) for all x∈(0,1), and we obtain (72).Corollary 12.
For allx∈(0,1) and all a>b>0 or 0>a>b, one has
(73)ψ3+byaψ3+ayb<ψ3+bxaψ3+axb,
where 3+ax⩾c, 3+bx⩾c, 3+ay⩾c, 3+by⩾c, and 0<y<x<1.Corollary 13.
For allx∈(0,1) and all a>0 and b<0, one has
(74)ψ3+bxaψ3+axb<ψ3+byaψ3+ayb,
where 3+ax⩾c, 3+bx⩾c, 3+ay⩾c, 3+by⩾c, and 0<y<x<1.Remark 14.
Takinga=n and b=1 in Corollary 10, we obtain inequalities of Corollary 8.



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