Abstract

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

1. Introduction

Let , , , and denote the Euler gamma function, digamma function, polygamma functions, and Riemann zeta function, respectively, which are defined by 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 function on , with and , Batir [19] presented the following inequalities for the gamma and digamma functions:

In Section 2, by applying the mean value theorem on we obtain some new inequalities on gamma and digamma functions.

Section 3 is devoted to some new inequalities on digamma function, by using convex properties of logarithm of this function.

Note that in this paper by we mean Euler’s constant [5].

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

Lemma 1. For , one has

Proof. By [6, Proposition 1], we have Thus the function is strictly decreasing on .
By using asymptotic expansions [20, pages 253–256 and 364], For , we get Now, the proof follows from the monotonicity of on and

Theorem 2. One has the following:(a) for ;(b) for ;(c) for and for ;(d) for and ;(e) for and ;(f) and for ;(g) for ;(h) for and for ;(i) for ;(j) for .

Proof. Let be a positive real number and defined on the closed interval . By using the mean value theorem for the function on with and since is a decreasing function, there is a unique depending on such that , for all ; then Since and , we have We show that the function has the following properties:(1) is strictly increasing on ;(2);(3) is strictly decreasing on ;(4).
To prove these four properties, since is a decreasing function on , we put , where ; by formula (13) we have Since by formula (8) we have and , for all , then the mapping from into is injective since also and when and , respectively, then the mapping from (0,) into (0,) is a bijective map. Clearly, by injectivity of , we find that Differentiating between both sides of this equation, we get Since by formula (8), , where , hence formula (15) gives , for all . Since the mapping from to is also bijective, then for all , and the proof of (1) is completed.
From (8) we have Differentiating between both sides of (15), we obtain Since and , where , then for all . Proceeding as above we conclude that , for . This proves (3).
For (4), from (8), (9), we conclude that Now, we prove the theorem. To prove (a), let ; then by (1) and (2) we have Equation (13) and for all give By substituting the value of into (18), we get By substituting the value into this inequality, we get where .
In order to prove (b), by using the mean value theorem on the interval , and since is a decreasing function, there exists a unique such that for and Now, by (14), we have Since is strictly increasing on , by (1), we have By using this inequality and the fact that and we obtain Since is strictly increasing on , by , it is clear that and then it is clear that (b) holds.
For (c), since , , and is strictly decreasing on by (3), then Since and , by using (24), we obtain where .
Since is strictly decreasing on by (3) and , for all , we have where .
Then it is clear that (c) is true.
Now we prove (d) and (e) by using the mean value theorem on , for , we conclude where .
After brief computation we have Since for all , , and by the monotonicity of and we have ; then By monotonicity of   and , we have After some simplification of this inequality (d) is proved.
For (f), we put in (e) and (d).
For (g), we integrate (a) on for ; then we have the proof is completed when we integrate these inequalities on , for .
By using the mean value theorem for the on , there is a depending on such that for all , and so By formula (13) and (2), since is strictly increasing on , we have or since is strictly increasing on , by (1), we have or In order to prove (i) and (j), we integrate both sides of (13) over to obtain Making the change of variable on the left-hand side, by (14), we have since for all and , we find that, for , or Again using the monotonicity of and , after some simplifications as for , we can rewrite This proves (i). By inequality (c) for , we have since for , , from this inequality we find that replacing by , we get for which proves (j). Then the proof is completed.

Example 3. Consider the matrix By using inequalities (a), we obtain Now, we integrate on (for ) from both sides of (51) to obtain replacing by ( is an integer number) and using the identity [6] and [21], where is the th harmonic number, then we have

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

Definition 4. A positive function is said to be logarithmically convex on an interval if has derivative of order two on and for all .
If inequality (54) is strict, for all , then is said to be strictly logarithmically convex [22].

Lemma 5. The function is increasing on , where is the only positive zero of [1, 19].

Lemma 6. If and , then is strictly logarithmically convex on .

Proof. By differentiation we have by Lemma 5, we obtain , for every and since on , then we have , for .
This implies that is strictly logarithmically convex on .

Theorem 7. One has the following:(a),  for and ;(b), for and ;(c), for and ;(d), for and ;(e), for and .

Proof. By Lemma 6 we have, for , where , , , , and .
If and , then for and .
Let and . Note that and ; also we obtain In order to prove (b), let since , we have . Also By Lemma 6, is strictly convex on ; then and so ; this implies that is strictly decreasing on . Since and , we have . Then And then ; also . Then for and or So (b) is proved.
By (c), (d), and (e) are clear.

Corollary 8. For all and all integers , one has where is the th harmonic number.

Proof. By [6], for all integers , we have and replacing by in Theorem 7, the proof is completed.

Theorem 9. Let be a function defined by where and ; then for all or    and , is strictly increasing (strictly decreasing) on .

Proof. Let be a function defined by then By proof of Theorem 7, we have this implies that if or    if and ; that is, is strictly increasing on (strictly decreasing on . Hence is strictly increasing on , if or (strictly decreasing if and ).

Corollary 10. For all and all or , one has where , , , and .

Proof. To prove (71), applying Theorem 9 and taking account of , we get for all , and we obtain (71).

Corollary 11. For all and all and , one has where , , , and .

Proof. Applying Theorem 9, we get for all , and we obtain (72).

Corollary 12. For all and all or , one has where , , , , and .

Corollary 13. For all and all and , one has where , , , , and .

Remark 14. Taking and in Corollary 10, we obtain inequalities of Corollary 8.

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.