Article Short Remarks on Complete Monotonicity of Some Functions
Ladislav Matejíˇcka
Faculty of Industrial Technologies in Púchov, TrenˇcínUniversity of Alexander Dubˇcekin Trenˇcín,I. Krasku 491/30, 02001 Púchov, Slovakia; [email protected]
Received: 25 February 2020; Accepted: 28 March 2020; Published: 5 April 2020
Abstract: In this paper, we show that the functions xm|β(m)(x)| are not completely monotonic on − ( ) (0, ∞) for all m ∈ N, where β(x) is the Nielsen’s β-function and we prove the functions xm 1|β m (x)| − ( ) and xm 1|ψ m (x)| are completely monotonic on (0, ∞) for all m ∈ N, m > 2, where ψ(x) denotes the logarithmic derivative of Euler’s gamma function .
Keywords: completely monotonic functions; laplace transform; inequality; Nielsen’s β-function; polygamma functions
1. Introduction Completely monotonic functions have attracted the attention of many authors. Mathematicians have proved many interesting results on this topic. For example, Koumandos [1] obtained upper and lower polynomial bounds for the function x/(ex − 1) x > 0, with coefficients of the Bernoulli numbers Bk. This enabled him to give simpler proofs of some results of H. Alzer and F. Qi et al., concerning complete monotonicity of certain functions involving the functions Γ(x), ψ(x) and the polygamma functions ψ(n), n = 1, 2, .... [2]. For example, he proved by simpler way the following theorem [3].
Theorem 1. The functions
1 1 ψ(x) − log(x) + + , 2x 12x2
1 log(x) − − ψ(x), 2x
1 1 1 1 ψ0(x) − − − + , x 2x2 6x3 30x5
1 1 1 + + − ψ0(x) x 2x2 6x3 are strictly completely monotonic on (0, ∞).
Qi and Agarwal [4] surveyed some results related to the function [ψ0(x)]2 + ψ00(x), its q-analogous, variants, and divided difference forms; several ratios of gamma functions; and so on. Their results include the origins, positivity, inequalities, generalizations, completely monotonic degrees, (logarithmically) complete monotonicity, necessary and sufficient conditions, equivalences to inequalities for sums, applications, etc. Finally, the authors listed several remarks and posed several
Mathematics 2020, 8, 537; doi:10.3390/math8040537 www.mdpi.com/journal/mathematics Mathematics 2020, 8, 537 2 of 14 open problems. We note that the seventh open problem ([4] p. 39) was solved by Matejicka [5] and the fourth open problem ([4] p. 38) was solved by Matejicka [6]. H. Alzer et al. [7] disproved the following conjecture:
m (m) Conjecture 1. Let Φm(x) = −x ψ (x), where ψ(x) denotes the logarithmic derivative of Euler’s gamma (m) function [2]. Then, the function Φm (x) is completely monotonic on (0, ∞) for each m ∈ N.
(m) Clark and Ismail [2] proved that, if m ∈ 1, 2, ..., 16, then Φm (x) is completely monotonic on (0, ∞), and thus they conjectured that this is true for all natural numbers m. Alzer, Berg and Koumandos [7] disproved Conjecture1 by showing that there exists an integer m0 such that for all m ≥ m0 the (m) functions Φm (x) are not completely monotonic on (0, ∞). They defined the function
α (m) ∆α,m(x) = x | ψ (x) | f or x > 0, m ∈ N, α ∈ R,
+ and noted that it remains an open problem to determine all (α, m) ∈ R × N such that ∆α,m is completely monotonic. We note that this problem is still open. Nantomah ([8] p. 92) posed a similar open problem: Find all values of a ∈ R such that the function a (m) Ha(x) = x | β (x) | is completely monotonic on (0, ∞), where β(x) is the Nielsen’s β-function. Alzer, Berg and Koumandos [7] introduced new function
1 1 x s(x) = + H 2 π 2π where
∞ 1 x H(x) = ∑ sin k=1 k k
(m) is the Hardy–Littlewood function [9–13] defined for x ∈ C. They showed that the functions Φm (x) are all completely monotonic on (0, ∞), m ∈ N if and only if s(x) ≥ 0 for x > 0. Their main result was that for each K > 0 there is xK > 0 such that H(xK) < −K. It implies Conjecture1 is not valid. m (m) The goal of this paper is to show that the functions Υm(x) = x | β (x) | are not completely − ( ) − ( ) monotonic on (0, ∞) for all m ∈ N, but the functions xm 1 | ψ m (x) |, xm 1 | β m (x) | are completely monotonic on (0, ∞) for all m ∈ N, m > 2. A detailed list of references on completely monotonic functions can be found in [1–39]. Now, recall some useful definitions and theorems. The classical Nielsen’s β-function [8,31,32] is defined as
Definition 1.
∞ 1 Z e−xt Z tx−1 ∞ (−1)m ( ) = = = = β x −t dt dt ∑ 1 + e 1 + t m= m + x 0 0 0 1 1 + x x = ψ − ψ 2 2 2 for x > 0, where ψ(x) = d ln Γ(x)/dx is a digamma function and Γ(x) is the Euler’s Gamma function [4,24]. Mathematics 2020, 8, 537 3 of 14
It is generally known (see [8]) that the special function β(x) is related to the Euler’s beta function B(x, y) and to the Gauss hypergeometric function 2F1(a, b; c; d) by
d x 1 β(x) = − ln B , dx 2 2 1 β(x) = ( F (1, x; 1 + x; −1)) x 2 1 for x > 0. For additional information on the Nielsen’s β-function, one may refer to [8,32] and the related references therein.
Definition 2 ([4,24]). We say that a function f is a completely monotonic on the interval I, if f (x) has n (n) derivatives of all orders on I and the inequality (−1) f (x) ≥ 0 holds for x ∈ I and n ∈ N0.
A characterization of completely monotonic function can be given by the Bernsten–Widder theorem [36,37], which reads that a function f (x) on (0, ∞) is completely monotonic if and only if there exists bounded and non-decreasing function α(t) such that the integral
∞ Z f (x) = e−xtdα(t) 0 converges for x ∈ (0, ∞).
Definition 3 ([24]). Let h(t) be a completely monotonic function on (0, ∞) and let h(∞) = lim h(t) ≥ 0. If t→∞ the function tα [h(t) − h(∞)] is a completely monotonic on (0, ∞) when and only when 0 ≤ α ≤ r ∈ R, then we say h(t) is of completely monotonic degree r; if tα [h(t) − h(∞)] is a completely monotonic on (0, ∞) for all α ∈ R, then we say that the completely monotonic degree of h(t) is ∞.
t For convenience, Guo [21] designed a notation degcmh(t) to denote the completely monotonic degree r of h(t) with respect to t ∈ (0, ∞).
m (m) Matejicka [31] s showed that the function x β (x) is completely monotonic on (0, ∞) for m = 1, 2, 3.
Definition 4 ([35]). A function f has exponential order α if there exist constants M > 0 and α such that for some t0 ≥ 0
αt | f (t)| ≤ Me , t ≥ t0.
Definition 5. We say
∗ αt α = inf α; there are constants M, t0 such that | f (t)| ≤ Me , t ≥ t0 is a lower exponential order of function f .
Theorem 2 ([19,20]). (Weierstrass’s criterion for uniform convergence) Suppose f (x, t) is a continuous on [a, b] × [0 , ∞) and g(t) is integrable on [0 , ∞) . If | f (x, t)| ≤ g(t) for all a ≤ x ≤ b and all t ≥ 0 then the ∞ integral R f (x, t)dt is uniformly convergent on [a, b] . 0 Mathematics 2020, 8, 537 4 of 14
Theorem 3 ([19,20]). Suppose f (x, t) is continuous on [a, b] × [0 , ∞) together with its partial derivative ∂ f ∂x (x, t). In this case,
∞ ∞ d Z Z ∂ f (x, t)dt = f (x, t)dt, dx ∂x 0 0 when the first integral is convergent and the second is uniformly convergent on [a, b] .
2. Main Results
0 (m) ∗ ∗ ∗ Theorem 4. Let m ∈ N and ϕ, ϕ , ... ϕ be continuous functions of lower exponential orders L0, L1, ... Lm,, respectively, on (0, ∞). Let ∞ ∗ ∗ ∗ R −xt (m) L = max L0, L1, ..., Lm ≥ 0;F(x) = ϕ(t)e dt ≥ 0 for x > L; ϕ (t) ≥ 0 on (0, ∞). Let 0
m−1−k (k) ∂ n −xt p1 : lim ϕ (t) t e = 0 t→+∞ ∂tm−1−k and
m−1−k (k) ∂ n −xt p2 : lim ϕ (t) − − t e = 0 t→0+ ∂tm 1 k for x > L, n ∈ N, k = 0, ..., m − 1. Then, xmF(x) is a completely monotonic function on (L, +∞).
Remark 1. We note that, if the following conditions
lim ϕ(k)(t) = 0, lim ϕ(k)(t)e−xt = 0 t→0+ t→∞ are fulfilled for k = 0, ..., m − 1, and x > L where m ∈ N, then p1, p2 are also valid.
Proof. Let a, b be fixed and such that L < a < b. To prove Theorem4, it suffices to show
n m (n) Lm,n = (−1) (x F(x)) ≥ 0 (0) for n ≥ 1 and a ≤ x ≤ b. Some computation gives
∞ (n) Z m n+m ∂ −xt Lm,n = (−1) ϕ(t) e dt . ∂tm 0
Denote
f (x, t) = ϕ(t)xme−xt.
∗ ∗ (L+ε∗)t Let ε > 0 such that ε < a − L. Then, there are M, t0 such that | ϕ(t) |≤ Me for t ≥ t0. This implies
(L+ε∗)t m −xt (L+ε∗−a)t | f (x, t)| ≤ Me (max{|a|, |b|}) e ≤ g(t) = C(a, b, m, M)e f or t ≥ t0.
It is evident that g(t) is an integrable function on (0, +∞). A straightforward differentiation yields
∂ f (x, t) = ϕ(t)mxm−1e−xt − tϕ(t)xme−xt. ∂x Mathematics 2020, 8, 537 5 of 14
It follows immediately that
∂ ( + ∗− ) m f (x, t) ≤ C(a, b, m, M)e L ε a t + t f or t ≥ t . ∂x a 0
Thus, according to Theorems2 and3, the integral
∞ Z ∂ f (x, t)dt ∂x 0 is uniformly convergent on [a, b] and we observe that
∞ ∞ d Z Z ∂ f (x, t)dt = f (x, t)dt. dx ∂x 0 0
− ∗ It is easy to see that the function h(t) = tke εt, where k ∈ N, ε = a − L − ε is integrable on (0, +∞). Using mathematical induction and considering the formulas
∞ ∞ ∞ Z 1 Z k Z te−εtdt = , tke−εtdt = tk−1e−εtdt ε2 ε 0 0 0 for k ∈ N, ε > 0 bring the desired result. If we repeat the process n times, we obtain
∞ Z ∂n ∂m L = (−1)n+m ϕ(t) e−xtdt . m,n ∂xn ∂tm 0
Using Schwartz theorem (mixed partial derivatives) [19] yields
∞ ∞ Z ∂m ∂n Z ∂m L = (−1)n+m ϕ(t) e−xtdt = (−1)m ϕ(t) tne−xtdt dt. m,n ∂tm ∂xn ∂tm 0 0
Applying to Lm,n integration by parts m times and using p1, p2 leads to
∞ Z (m) n −xt Lm,n = ϕ (t)t e dt. 0
This completes the proof of our theorem.
Lemma 1. Let tm tm ϕ (t) = and ϕ (t) = 1 1 − e−t 2 1 + e−t for t > 0 and m > 2, m ∈ N. Then, m−1−k (k) ∂ − + (1) lim ϕ (t) (tne xt) = 0 for a = 0 and a = +∞, k = 0, ..., m − 1, n ∈ N, x > 0, t→a 2 ∂tm−1−k m−2−k (k) ∂ − + (2) lim ϕ (t) (tne xt) = 0 for i = 1, 2, a = 0 and a = +∞, k = 0, ..., m − 2, n ∈ N, x > 0. t→a i ∂tm−2−k
Proof. Differentiating k times the following equations
−t m −t m 1 − e ϕ1(t) = t and 1 + e ϕ2(t) = t Mathematics 2020, 8, 537 6 of 14 leads to
" k−1 # (k) ( ) k (j) 1 ϕ (t) = (tm) k + (−1)k−j ϕ (t)e−t (1) 1 ∑ 1 − −t j=0 j 1 e and
" k−1 # (k) ( ) k (j) 1 ϕ (t) = (tm) k − (−1)k−j ϕ (t)e−t . (2) 2 ∑ 2 + −t j=0 j 1 e
Similarly, we get
m−1−k m−1−k ∂ m − 1 − k ( ) α = (tne−xt) = e−xt (− )m−1−k−i (tn) i xm−1−k−i m−1−k ∑ 1 (3) ∂t i=0 i and
m−2−k m−2−k ∂ m − 2 − k ( ) γ = (tne−xt) = e−xt (− )m−2−k−i (tn) i xm−2−k−i m−2−k ∑ 1 . (4) ∂t i=0 i
H. Alzer et al. [7] (p. 108) demonstrated that the generating function for Bernoulli numbers Bk yields for |t| < 2π
t t ∞ B = 1 + + k tk, (5) − −t ∑ 1 e 2 k=2 k! the series on the right side of Equation (5) is uniformly convergent on each [−c, c] where 0 < c < 2π. This implies
m ∞ m−1 t Bj j+m−1 ϕ1(t) = t + + ∑ t 2 j=2 j! and
(k) (tm)(k) ∞ B (k) (k)( ) = m−1 + + j j+m−1 ϕ1 t t ∑ t . (6) 2 j=2 j!
(k) From Equation (6), we get lim ϕ1 (t) = 0 for k = 0, ..., m − 2. Using mathematical induction t→0+ (k) and Equation (2), we deduce that lim ϕ2 (t) = 0 for k = 0, ..., m − 1. It is evident that lim α = 0 or t→0+ t→0+ lim α = c where c ∈ R. Thus, if a = 0, then (1) is valid. Next, we have t→0+ x x m−1−k m−1−k (k) ∂ (k) − t − t m − 1 − k ( ) α∗ = ϕ (t) (tne−xt) = ϕ (t)e 2 e 2 (− )m−1−k−i (tn) i xm−1−k−i 2 m−1−k 2 ∑ 1 . (7) ∂t i=0 i x (k) − t By using mathematical induction it is easy to show that lim ϕ (t)e 2 = 0 for k = 0, ..., m − 1. t→∞ 2 Thus, lim α∗ = 0 for k = 0, ..., m − 1. The proof of (1) is complete. t→∞ Mathematics 2020, 8, 537 7 of 14
Similarly, it can be shown that
m−2−k (k) ∂ lim ϕ (t) (tne−xt) = 0 t→∞ 1 ∂tm−2−k for k = 0, ..., m − 2 and
m−2−k (k) ∂ lim ϕ (t) (tne−xt) = 0. t→∞ 2 ∂tm−2−k The proof of the Lemma1 would be done if we show
m−2−k (k) ∂ n −xt lim ϕ1 (t) − − (t e ) = 0 t→0+ ∂tm 2 k for k = 0, ..., m − 2, n ∈ N, x > 0. However, it follows from Equation (6). This completes our proof.
m (m) Theorem 5. There exists an integer m0 such that for each m > m0 the function x |β (x)| is not completely monotonic on (0, ∞).
Proof. Using Theorem4 and Lemma1 yields
∞ (n) Z dm tm L = (−1)n xm|β(m)(x)| = tne−xtdt. m,n dtm 1 + e−t 0
Denote
dm tm dm tm f (t) = and g (t) = . m dtm 1 − e−t m dtm 1 + e−t
H. Alzer et al. [7] (p. 109 (3.4)) proved that
1 x lim fm = s(x) f or x > 0, m→∞ m! m where ([7] p. 109 (3.7))
1 1 x s(x) = + H 2 π 2π and
∞ 1 x H(x) = ∑ sin f or x ∈ C. k=1 k k
Elementary calculation yields
1 1 2 + = . (8) 1 + e−t 1 − e−t 1 − e−2t
A straightforward computation gives
dm tm dm tm 1 dm (2t)m + = . (9) dtm 1 + e−t dtm 1 − e−t 2m−1 dtm 1 − e−2t Mathematics 2020, 8, 537 8 of 14
This establishes
gm(x) = 2 fm(2x) − fm(x). (10)
By using (see ([7] p. 109 (3.4)))
f (x) lim m = s(x), m→∞ m! we obtain
g (x) g(x) = lim m = 2s(2x) − s(x). m→∞ m!
Let m be any natural number and xm|β(m)(x)| be completely monotonic on (0, +∞). m (m) The Bernstein–Widder theorem [36] implies gm(x) ≥ 0 on (0, +∞). Thus, if x |β (x)| is completely monotonic on (0, +∞) for all m ∈ N then g(x) ≥ 0 on (0, +∞). It is easy to see that s(0) = 1/2. Continuity of s(x) at 0 gives that there is ε > 0 such that s(x) > 0 for x ∈ (0, ε). From Theorem 2.1 (see ([7] p. 105)), we can derive that there is x0 > 0 such that s(x0) < 0. Denote
M = {x0; such that s(x0) = 0}.
It is obvious that M 6= ∅. Then, there is x∗ = inf M. It is easy to see that 0 < ε ≤ x∗ and s(x∗) = 0. From the definition of x∗, we deduce
x∗ x∗ x∗ g = 2s(x∗) − s = −s < 0. 2 2 2
∗ Thus, there is m0 ∈ N such that, if m ≥ m0, m ∈ N, then gm(x /2) < 0. The Bernstein–Widder theorem [36] implies xm|β(m)(x)| is not completely monotonic on (0, ∞). This completes the proof.
Lemma 2. Let y > 0. Then,
∞ y ∞ y V(y) = ∑ sin2 − ∑ sin2 > 0. n=1 n n=1 2n
Proof. Straightforward calculation gives that, for m ∈ N,
V(y) = Am(y) − Bm(y) + Cm(y) where
m+1 y A (y) = sin2 , m ∑ − n=1 2n 1
m+1 y B (y) = sin2 m ∑ + n=1 2m 2n Mathematics 2020, 8, 537 9 of 14 and
∞ 2 y 2 y Cm = ∑ sin − sin . n=2m+2 n 2n
It is clear that, if 0 < y < π(2m + 2)/2, then sin2(y/(2m + 2)) < y2/(2m + 2)2. Thus, we obtain that ! m+1 1 −B (y) > −y2 . m ∑ ( + )2 n=1 2m 2n
There are two cases:
• (α) 0 < y 6= kπ for all k ∈ N. • (β) y = k0π for some k0 ∈ N.
2 In the case (α), there is j0 ∈ N such that y < π(2j0 + 2)/2. Put e = sin (y) > 0. It is obvious that there is m0 ∈ N, m0 > j0 such that |Cm(y)| < e/2 and Bm(y) < e/2 for m > m0. This implies V(y) > 0. 2 Consider the case β) y = k0π for some k0 ∈ N. One can easily see that e = sin (k0π/(2k0 + 1)) > 0. It is obvious again that there is m0 > k0 such that |Cm(y)| < e/2 and Bm(y) < e/2 for m > m0. This implies V(y) > 0. The proof is complete.
m−1 (m) Theorem 6. Let m ∈ N, m > 2. Then, x ψ (x) is completely monotonic function on (0, +∞).
Proof. We need to prove
(n) n m−1 (m) Km,n = (−1) x ψ (x) ≥ 0 for x > 0, n ∈ N. Theorem4 and Lemma1 imply that the proof will be done if we show
tm (m−1) r (t) = ≥ 0 m 1 − e−t for t > 0 and m ∈ N, m > 2. It is clear that
t Z rm(t) = fm(u)du + rm(0), 0 where
tm (m) f (t) = . m 1 − e−t
The formula ([7] p. 108)
xm xm ∞ B = xm−1 + + k xk+m−1 − −x ∑ 1 e 2 k=2 k! Mathematics 2020, 8, 537 10 of 14
for 0 < x < 2π, where Bk are Bernoulli numbers, implies rm(0) = (m − 1)!. The proof would be done if we prove rm(x) ≥ 0 for x > 0. H. Alzer et al. [7] (p.113) showed that
∞ Z −t m fm(x) = e t s(xt)dt 0 for |x| < 2π, where
1 1 y s(y) = + H 2 π 2π and
∞ 1 y H(y) = ∑ sin . n=1 n n
This implies
u ∞ Z Z −t m rm(u) = e t s(vt)dtdv + (m − 1)! 0 0 for |u| < 2π which can be rewritten as
∞ u Z Z −t m rm(u) = e t s(vt)dv dt + (m − 1)!. 0 0
Direct computation yields
ut u u 2π Z 1 1 Z vt 1 2 Z s(vt)dv = u + H dv = u + H(p)dp > 0 2 π 2π 2 t 0 0 0 because of (see ([13] p. 1))
x∗ Z ∞ x∗ H(p)dp = 2 ∑ sin2 ≥ 0. = 2n 0 n 1
Thus, rm(u) > 0 for 0 < |u| < 2π. H. Alzer et al. [7] (p. 112) showed that fm(x) ≥ 0 for x ≥ 2 log 2, m ∈ N. Thus, if x ≥ 2π, then
x 2 log 2 x Z Z Z rm(x) = fm(p)dp + (m − 1)! = fm(p)dp + fm(p)dp + (m − 1)! 0 0 2 log 2 x Z = rm(2 log 2) + fm(p)dp ≥ 0. 2 log 2
This completes the proof.
m−1 (m) Theorem 7. Let m ∈ N, m > 2. Then, x β (x) is completely monotonic function on (0, +∞). Mathematics 2020, 8, 537 11 of 14
Proof. We need to show
(n) n m−1 (m) Jm,n = (−1) x β (x) ≥ 0 for n ∈ N, x > 0. Theorem4 and Lemma1 imply that the proof would be done if we show
tm (m−1) h (t) = ≥ 0. m 1 + e−t
It is clear that
x Z hm(x) = gm(t)dt + hm(0). 0
From Equation (8), we get
dm−1 tm dm−1 tm 1 dm−1 (2t)m + = dtm−1 1 + e−t dtm−1 1 − e−t 2m−1 dtm−1 1 − e−2t which can be rewritten as hm(t) = rm(2t) − rm(t). This implies hm(0) = 0. In the previous part, Equation (10)
gm(x) = 2 fm(2x) − fm(x) is derived. As a direct consequence, we deduce
x Z hm(x) = 2 fm(2t) − fm(t)dt. 0
Let 0 < x < π. Due to result ([7] p. 113)
∞ Z −t m fm(x) = e t s(xt)dt f or | x |< 2π, 0 we obtain
x ∞ ∞ Z Z Z −t m −t m hm(x) = 2 e t s(2ut)dt − e t s(ut)dt du 0 0 0 ∞ 2x x ∞ 2x Z Z Z Z Z − − = e ttm s(vt)dv − s(vt)dv dt = e ttm s(vt)dv dt. 0 0 0 0 x
To prove our theorem, it suffices to show that
2x Z s(vt)dv ≥ 0 x Mathematics 2020, 8, 537 12 of 14 for 0 < x < π and t > 0. Simple computation yields
xt 2x 2x π Z Z 1 1 vt x 2 Z s(vt)dv = + H dv = + H(u)du. 2 π 2π 2 t x x xt 2π
The proof would be done if we show
2y y Z Z H(v)dv − H(v)dv ≥ 0. 0 0
However, it follows from Segal [13] that
y Z ∞ y H(v)dv = 2 ∑ sin2 = 2n 0 n 1 and Lemma2. Let x ≥ 2 log 2. One can easily determine
x x 2x Z Z Z hm(x) = gm(t)dt = 2 fm(2t) − fm(t)dt = fm(t)dt. 0 0 x
H. Alzer et al. [7] (p. 112) derived that fm(x) ≥ 0 for x ≥ 2 log 2. Thus, hm(x) ≥ 0 for x ≥ 2 log 2. This implies the proof of our theorem.
Remark 2. We note that, in [8,31], it was proved that the function
xm|β(m)(x)| is strictly completely monotonic on (0, ∞) for m = 2, 3, respectively, for m = 1.
Remark 3. We note that it is easy to show that functions
xm|β(m)(x)| are strictly completely monotonic on (0, ∞) for m = 4, 5, 6 by using Theorem4.
Remark 4. We note that our results and results obtained in [7] imply that
t (m) m − 1 ≤ degcm|ψ (t)| ≤ m and t (m) m − 1 ≤ degcm|β (t)| ≤ m + 1 for m ∈ N, m ≥ 3.
3. Open Problem It is natural to pose the following problem. Mathematics 2020, 8, 537 13 of 14
m (m) 1. Find the value mβ such that x |β (x)| is not completely monotonic on (0, +∞) for m ≥ mβ, m (m) and mβ = min{m0; x |β (x)| is not completely monotonic on (0, +∞) f or m ≥ m0}. m (m) 2. Find the value mψ such that x |ψ (x)| is not completely monotonic on (0, +∞) for m ≥ mψ, m (m) and mψ = min{m1; x |ψ (x)| is not completely monotonic on (0, +∞) f or m ≥ m1}. 3. What is the relation between mβ and mψ?
4. Materials and Methods In this paper, MATLAB software and methods of mathematical analysis were used.
5. Conclusions
m (m) The main result of this paper is proving that the functions Υm(x) = x | β (x) | are not − ( ) − ( ) completely monotonic on (0, ∞) for all m ∈ N and that the functions xm 1|β m (x)| and xm 1|ψ m (x)| are completely monotonic on (0, ∞) for all m ∈ N, m ≥ 3.
Author Contributions: Author has read and agreed to the published version of the manuscript. Funding: The work was supported by VEGA grants Nos. 1/0589/17, 1/0649/17, and 1/0185/19 and by Kega grant No. 007 TnUAD-4/2017. Acknowledgments: The author thanks to Ondrušová, dean of FPT TnUAD, and Vavro, deputy dean of FPT TnUAD, Slovakia, for their kind grant support. Conflicts of Interest: The author declares that he has no competing interests.
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