Research article

Completely monotonic integer degrees for a class of special functions

  • Received: 14 January 2020 Accepted: 26 March 2020 Published: 07 April 2020
  • MSC : 26A48, 33B15, 44A10

  • Let fn(x) (n=0,1,) be the remainders for the asymptotic formula of lnΓ(x) and Rn(x)=(1)nfn(x). This paper introduced the concept of completely monotonic integer degree and discussed the ones for the functions (1)mR(m)n(x), then demonstrated the correctness of the existing conjectures by using a elementary simple method. Finally, we propose some operational conjectures which involve the completely monotonic integer degrees for the functions (1)mR(m)n(x) for m=0,1,2,.

    Citation: Ling Zhu. Completely monotonic integer degrees for a class of special functions[J]. AIMS Mathematics, 2020, 5(4): 3456-3471. doi: 10.3934/math.2020224

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  • Let fn(x) (n=0,1,) be the remainders for the asymptotic formula of lnΓ(x) and Rn(x)=(1)nfn(x). This paper introduced the concept of completely monotonic integer degree and discussed the ones for the functions (1)mR(m)n(x), then demonstrated the correctness of the existing conjectures by using a elementary simple method. Finally, we propose some operational conjectures which involve the completely monotonic integer degrees for the functions (1)mR(m)n(x) for m=0,1,2,.


    It is well known that the convexity [1,2,3,4,5,6,8,9,11,15,16,40,55,63,64], monotonicity [7,12,13,14,41,42,43,44,45,46,47,48,49,50,51,52,53] and complete monotonicity [58,59,61,62] have widely applications in many branches of pure and applied mathematics [19,24,28,32,35,38,65]. In particular, many important inequalities [20,25,30,33,37,39,69] can be discovered by use of the convexity, monotonicity and complete monotonicity. The concept of complete monotonicity can be traced back to 1920s [18]. Recently, the complete monotonicity has attracted the attention of many researchers [23,34,56,67] due to it has become an important tool to study geometric function theory [26,31,36], its definition can be simply stated as follows.

    Definition 1.1. Let IR be an interval. Then a real-valued function f:IR is said to be completely monotonic on I if f has derivatives of all orders on I and satisfies

    (1)nf(n)(x)0 (1.1)

    for all xI and n=0,1,2,.

    If I=(0,), then a necessary and sufficient condition for the complete monotonicity can be found in the literature [54]: the real-valued function f:(0,)R is completely monotonic on (0,) if and only if

    f(x)=0extdα(t) (1.2)

    is a Laplace transform, where α(t) is non-decreasing and such that the integral of (1.2) converges for 0<x<.

    In 1997, Alzer [10] studied a class of completely monotonic functions involving the classical Euler gamma function [21,22,60,66,68] and obtained the following result.

    Theorem 1.1. Let n0 be an integer, κ(x) and fn(x) be defined on (0,) by

    κ(x)=lnΓ(x)(x12)lnx+x12ln(2π) (1.3)

    and

    fn(x)={κ(x)nk=1B2k2k(2k1)x2k1,n1,κ(x),n=0, (1.4)

    where Bn denotes the Bernoulli number. Then both the functions xf2n(x) and xf2n+1(x) are strictly completely monotonic on (0,).

    In 2009, Koumandos and Pedersen [27] first introduced the concept of completely monotonic functions of order r. In 2012, Guo and Qi [17] proposed the concept of completely monotonic degree of nonnegative functions on (0,). Since the completely monotonic degrees of many functions are integers, in this paper we introduce the concept of the completely monotonic integer degree as follows.

    Definition 1.2. Let f(x) be a completely monotonic function on (0,) and denote f()=limxf(x). If there is a most non-negative integer k () such that the function xk[f(x)f()] is completely monotonic on (0,), then k is called the completely monotonic integer degree of f(x) and denoted as degxcmi[f(x)]=k.

    Recently, Qi and Liu [29] gave a number of conjectures about the completely monotonic degrees of these fairly broad classes of functions. Based on thirty six figures of the completely monotonic degrees, the following conjectures for the functions (1)mR(m)n(x)=(1)m[(1)nfn(x)](m)=(1)m+nf(m)n(x) are shown in [29]:

    (ⅰ) If m=0, then

    degxcmi[Rn(x)]={0,if n=01,if n=12(n1),if n2; (1.5)

    (ⅱ) If m=1, then

    degxcmi[Rn(x)]={1,if n=02,if n=12n1,if n2; (1.6)

    (ⅲ) If m1, then

    degxcmi[(1)mR(m)n(x)]={m1,if n=0m,if n=1m+2(n1),if n2. (1.7)

    In this paper, we get the complete monotonicity of lower-order derivative and lower-scalar functions (1)mR(m)n(x) and their completely monotonic integer degrees using the Definition 1.2 and a common sense in Laplace transform that the original function has the one-to-one correspondence with the image function, and demonstrated the correctness of the existing conjectures by using a elementary simple method. The negative conclusion to the second clause of (1.7) is given. Finally, we propose some operational conjectures which involve the completely monotonic integer degrees for the functions (1)mR(m)n(x) for m=0,1,2,.

    In order to prove our main results, we need several lemmas and a corollary which we present in this section.

    Lemma 2.1. If the function xnf(x) (n1) is completely monotonic on (0,), so is the function xn1f(x).

    Proof. Since the function 1/x is completely monotonic on (0,), we have xn1f(x)=(1/x)[xnf(x)] is completely monotonic on (0,) too.

    Corollary 2.1. Let α(t)0 be given in (1.2). Then the functions xi1f(x) for i=n,n1,,2,1 are completely monotonic on (0,) if the function xnf(x) (nN) is completely monotonic on (0,).

    The above Corollary 2.1 is a theoretical cornerstone to find the completely monotonic integer degree of a function f(x). According to this theory and Definition 1.2, we only need to find a nonnegative integer k such that xkf(x) is completely monotonic on (0,) and xk+1f(x) is not, then degxcmi[f(x)]=k.

    The following lemma comes from Yang [57]:

    Lemma 2.2. Let fn(x) be defined as (1.4). Then fn(x) can be written as

    fn(x)=140pn(t2)extdt, (2.1)

    where

    pn(t)=cothttnk=022kB2k(2k)!t2k2. (2.2)

    Lemma 2.3. Let m,r0, n1, fn(x) and pn(t) be defined as (2.1) and (2.2). Then

    xr(1)mR(m)n(x)=xr(1)m+nf(m)n(x)=140[(1)ntmpn(t2)](r)extdt. (2.3)

    Proof. It follows from (2.1) that

    x(1)mR(m)n(x)=x(1)m+nf(m)n(x)=x(1)m+n140(t)mpn(t2)extdt=x(1)n140tmpn(t2)extdt=(1)n1140tmpn(t2)dext=(1)n114{[tmpn(t2)ext]t=00[tmpn(t2)]extdt}=(1)n140[tmpn(t2)]extdt.

    Repeat above process. Then we come to the conclusion that

    xr(1)mR(m)n(x)=(1)n140[tmpn(t2)](r)extdt,

    which completes the proof of Lemma 2.3.

    In recent paper [70] the reslut degxcmi[R1(x)]=degxcmi[f1(x)]=1 was proved. In this section, we mainly discuss degxcmi[R2(x)] and degxcmi[R3(x)]. Then discuss whether the most general conclusion exists about degxcmi[Rn(x)].

    Theorem 3.1. The function x3R2(x) is not completely monotonic on (0,), and

    degxcmi[R2(x)]=degxcmi[f2(x)]=2.

    Proof. Note that the function x2R2(x) is completely monotonic on (0,) due to

    x2R2(x)=140p(2)2(t2)extdt,p2(t)=cothtt+145t21t213,p2(t)=245t1tsinh2t+2t31t2coshtsinht,p2(t)=45A(t)+180t2B(t)+t4C(t)90t4sinh3t>0,

    where

    A(t)=tcosh3t3sinh3t+9sinhttcosht=n=52(n4)(32n1)(2n+1)!t2n+1>0,B(t)=tcosht+sinht>0,C(t)=sinh3t3sinht>0.

    So degxcmi[R2(x)]2.

    On the other hand, we can prove that the function x3f2(x)=x3R2(x) is not completely monotonic on (0,). By (2.3) we have

    x3f2(x)=140p(3)2(t2)extdt,

    then by (1.2), we can complete the staged argument since we can verify

    p(3)2(t2)>0p(3)2(t)>0

    is not true for all t>0 due to

    p2(t)=2tsinh2t2t3sinh2t+4t3+24t56tcosh2tsinh4t4t3cosh2tsinh2t2t2cosh3tsinh3t+2t2coshtsinht4t2coshtsinh3t6t4coshtsinht

    with p2(10)=0.00036.

    Theorem 3.2. The function x4R3(x) is completely monotonic on (0,), and

    degxcmi[R3(x)]=degxcmi[f3(x)]=4.

    Proof. By (2.3) we obtain that

    x4R3(x)=0[p(4)3(t2)]extdt.

    From (2.2) we clearly see that

    p3(t)=cothtt1t2+145t22945t413,
    p(4)3(t)=:1630H(t)t6sinh5t,

    or

    p(4)3(t)=1630H(t)t6sinh5t,

    where

    H(t)=(2t6+4725)sinh5t945tcosh5t(1260t5+3780t32835t)cosh3t(10t6+2520t4+3780t2+23625)sinh3t(13860t53780t3+1890t)cosht+(20t67560t4+11340t2+47250)sinht:=n=5hn(2n+3)!t2n+3

    with

    hn=2125[64n6+96n580n4120n3+16n22953101n+32484375]52n1027[64n6+96n5+5968n4+105720n3+393136n2+400515n+1760535]32n+20[64n6+96n511168n49696n3+9592n2+13191n+7749]>0

    for all n5. So x4R3(x) is completely monotonic on (0,), which implies degxcmi[R3(x)]4.

    Then we shall prove x5R3(x)=x5f3(x) is not completely monotonic on (0,). Since

    x5R3(x)=0[p(5)3(t2)]extdt,

    and

    p(5)3(t)=14K(t)t7sinh6t,

    where

    K(t)=540cosh4t1350cosh2t90cosh6t240t2cosh2t+60t2cosh4t+80t4cosh2t+40t4cosh4t+208t6cosh2t+8t6cosh4t120t3sinh2t+60t3sinh4t+200t5sinh2t+20t5sinh4t+75tsinh2t60tsinh4t+15tsinh6t+180t2120t4+264t6+900.

    We find K(5)2.6315×1013<0, which means p(5)3(5)<0. So the function x5R3(x)=x5f3(x) is not completely monotonic on (0,).

    In a word, degxcmi[R3(x)]=degxcmi[f3(x)]=4.

    Remark 3.1. So far, we have the results about the completely monotonic integer degrees of such functions, that is, degxcmi[R1(x)]=1 and degxcmi[Rn(x)]=2(n1) for n=2,3, and find that the existing conclusions support the conjecture (1.5).

    In this section, we shall calculate the completely monotonic degrees of the functions (1)mR(m)n(x), where m=1 and 1n3.

    Theorem 4.1 The function x2R1(x)=x2f1(x) is completely monotonic on (0,), and

    degxcmi[(1)1R1(x)]=2.

    Proof. By the integral representation (2.3) we obtain

    x2f1(x)=140[tp1(t2)]extdt.

    So we complete the proof of result that x2f1(x) is completely monotonic on (0,) when proving

    [tp1(t2)]>0[tp1(t2)]<0[tp1(t)]<0.

    In fact,

    tp1(t)=t(cothtt1t213)=coshtsinht13t1t,[tp1(t)]=1t2cosh2tsinh2t+23,[tp1(t)]=2sinh3t[cosht(sinhtt)3]<0.

    Then we have degxcmi[(1)1R1(x)]2.

    Here x3R1(x)=x3f1(x) is not completely monotonic on (0,). By (2.2) and (2.3) we have

    x3f1(x)=140[tp1(t2)]extdt,

    and

    [tp1(t)]=23t4cosh2tt4sinh2t3sinh4tt4sinh4t

    with [tp1(t)]|t=23.6237×102<0.

    Theorem 4.2. The function x3R2(x) is completely monotonic on (0,), and

    degxcmi[R2(x)]=degxcmi[f2(x)]=3.

    Proof. First, we can prove that the function x3R2(x) is completely monotonic on (0,). Using the integral representation (2.3) we obtain

    x3f2(x)=140[tp2(t2)](3)extdt,

    and complete the proof of the staged argument when proving

    [tp2(t2)](3)>0[tp2(t)](3)>0.

    In fact,

    p2(t)=cothtt+145t21t213,L(t):=tp2(t)=coshtsinht13t1t+145t3,
    L(t)=180cosh2t+45cosh4t124t4cosh2t+t4cosh4t237t4+13560t4sinh4t=160t4sinh4t[n=322n+2bn(2n+4)!t2n+4]>0,

    where

    bn=22n(4n4+20n3+35n2+25n+2886)4(775n+1085n2+620n3+124n4+366)>0

    for all n3.

    On the other hand, by (2.3) we obtain

    x4(1)1R2(x)=x4f2(x)=140[tp2(t2)](4)extdt,

    and

    L(4)(t)=[tp2(t)](4)=16coshtsinht40cosh3tsinh3t+24cosh5tsinh5t24t5

    is not positive on (0,) due to L(4)(10)2.3993×104<0, we have that x4R2(x) is not completely monotonic on (0,).

    Theorem 4.3. The function x5R3(x) is completely monotonic on (0,), and

    degxcmi[R3(x)]=degxcmi[f3(x)]=5.

    Proof. We shall prove that x5R3(x)=x5f3(x) is completely monotonic on (0,) and x6R3(x)=x6f3(x) is not. By (2.2) and (2.3) we obtain

    xrf3(x)=140[tp3(t2)](r)extdt, r0.

    and

    p3(t)=cothtt1t2+145t22945t413,M(t):=tp3(t)=coshtsinht13t1t+145t32945t5,M(5)(t)=1252p(t)t6sinh6t,

    we have

    [M(t)](5)=1252p(t)t6sinh6t,[M(t)](6)=14q(t)t7sinh7t,

    where

    p(t)=14175cosh2t+5670cosh4t945cosh6t+13134t6cosh2t+492t6cosh4t+2t6cosh6t+16612t6+9450,q(t)=945sinh3t315sinh5t+45sinh7t1575sinht456t7cosh3t8t7cosh5t2416t7cosht.

    Since

    p(t)=n=4262n+49242n+1313422n(2n)!t2n+6n=494562n+6567042n+6+1417522n+6(2n+6)!t2n+6>0,q(0.1)2.9625×105<0,

    we obtain the expected conclusions.

    Remark 4.1. The experimental results show that the conjecture (1.6) may be true.

    Theorem 5.1. The function x3R1(x) is completely monotonic on (0,), and

    degxcmi[R1(x)]=degxcmi[f1(x)]=3. (5.1)

    Proof. By (2.2) and (2.3) we obtain

    x3R1(x)=x3f1(x)=140[t2p1(t2)]extdt,

    and

    t2p1(t)=tcoshtsinht13t21,
    [t2p1(t)]=23sinh4t[n=23(n1)22n+1(2n+1)!t2n+1]>0.

    So x3R1(x) is completely monotonic on (0,).

    But x4R1(x) is not completely monotonic on (0,) due to

    x4R1(x)=140[t2p1(t2)](4)extdt,

    and

    [t2p1(t)](4)=1sinh5t(4sinh3t+12sinht22tcosht2tcosh3t)

    with [t2p1(t)](4)|t=105.2766×107<0.

    So

    degxcmi[R1(x)]=degxcmi[f1(x)]=3.

    Remark 5.1. Here, we actually give a negative answer to the second paragraph of conjecture (1.7).

    Theorem 5.2. The function x4R2(x) is completely monotonic on (0,), and

    degxcmi[R2(x)]=degxcmi[f2(x)]=4.

    Proof. By (2.2) and (2.3) we

    x4f2(x)=140[t2p2(t2)](4)extdt,

    and

    t2p2(t)=145t413t2+tcoshtsinht1,[t2p2(t)](4)=130(125sinh3t+sinh5t350sinht+660tcosht+60tcosh3t)sinh5t=130sinh5t[n=35(52n+(24n63)32n+264n+62)(2n+1)!t2n+1]>0.

    Since

    x5f2(x)=140[t2p2(t2)](5)extdt,

    and

    [t2p2(t)](5)=50sinh2t66t+5sinh4t52tcosh2t2tcosh4tsinh6t

    with [t2p2(t)](5)|t=109.8935×107<0, we have that x5f2(x) is not completely monotonic on (0,). So

    degxcmi[R2(x)]=4.

    Theorem 5.3. The function x6R3(x) is completely monotonic on (0,), and

    degxcmi[R3(x)]=degxcmi[f3(x)]=6.

    Proof. By the integral representation (2.3) we obtain

    x6f3(x)=140[t2p3(t2)](6)extdt,x7f3(x)=140[t2p3(t2)](7)extdt.

    It follows from (2.2) that

    p3(t)=cothtt1t2+145t22945t413,N(t):=t2p3(t)=145t413t22945t6+tcoshtsinht1,N(6)(t)=142r(t)sinh7t,N(7)(t)=(2416t1715sinh2t392sinh4t7sinh6t+2382tcosh2t+240tcosh4t+2tcosh6t)sinh8t,

    where

    r(t)=6321sinh3t+245sinh5t+sinh7t+10045sinht25368tcosht4788tcosh3t84tcosh5t=n=4cn(2n+1)!t2n+1

    with

    cn=772n(168n1141)52n(9576n14175)32n(50736n+15323).

    Since ci>0 for i=4,5,6,7, and

    cn+149cn=(4032n31584)52n+(383040n653184)32n+2435328n+684768>0

    for all n8. So cn>0 for all n4. Then r(t)>0 and N(6)(t)>0 for all t>0. So x6R3(x) is completely monotonic on (0,).

    In view of N(7)(1.5)0.57982<0, we get x7R3(x) is not completely monotonic on (0,). The proof of this theorem is complete.

    Remark 5.2. The experimental results show that the conjecture (1.7) may be true for n,m2.

    In this way, the first two paragraphs for conjectures (1.5) and (1.6) have been confirmed, leaving the following conjectures to be confirmed:

    degxcmi[Rn(x)]=2(n1), n4; (6.1)
    degxcmi[Rn(x)]=2n1, n4; (6.2)

    and for m1,

    degxcmi[(1)mR(m)n(x)]={m,if n=0m+1,if n=1m+2(n1),if n2, (6.3)

    where the first formula and second formula in (6.3) are two new conjectures which are different from the original ones.

    By the relationship (2.3) we propose the following operational conjectures.

    Conjecture 6.1. Let n4, and pn(t) be defined as (2.2). Then

    [(1)npn(t)](2n2)>0 (6.4)

    holds for all t(0,) and

    [(1)npn(t)](2n1)>0 (6.5)

    is not true for all t(0,).

    Conjecture 6.2. Let n4, and pn(t) be defined as (2.2). Then

    (1)n[tpn(t)](2n1)>0 (6.6)

    holds for all t(0,) and

    (1)n[tpn(t)](2n)>0 (6.7)

    is not true for all t(0,).

    Conjecture 6.3. Let m1, and pn(t) be defined as (2.2). Then

    [tmp0(t)](m)>0, (6.8)
    [tmp1(t)](m+1)>0 (6.9)

    hold for all t(0,), and

    [tmp0(t)](m+1)>0, (6.10)
    [tmp1(t)](m+2)>0 (6.11)

    are not true for all t(0,).

    Conjecture 6.4. Let m1, n2, and pn(t) be defined as (2.2). Then

    (1)n[tmpn(t)](m+2n2)>0 (6.12)

    holds for all t(0,) and

    (1)n[tmpn(t)](m+2n1)>0 (6.13)

    is not true for all t(0,).

    The author would like to thank the anonymous referees for their valuable comments and suggestions, which led to considerable improvement of the article.

    The research is supported by the Natural Science Foundation of China (Grant No. 61772025).

    The author declares no conflict of interest in this paper.



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