The monotonicity rules for the ratio of two Laplace transforms with applications
aa r X i v : . [ m a t h . C A ] M a r THE MONOTONICITY RULES FOR THE RATIO OF TWOLAPLACE TRANSFORMS WITH APPLICATIONS
ZHEN-HANG YANG AND JING-FENG TIAN
Abstract.
Let f and g be both continuous functions on (0 , ∞ ) with g ( t ) > t ∈ (0 , ∞ ) and let F ( x ) = L ( f ), G ( x ) = L ( g ) be respectively the Laplacetransforms of f and g converging for x >
0. We prove that if there is a t ∗ ∈ (0 , ∞ ) such that f/g is strictly increasing on (0 , t ∗ ) and strictly decreasingon ( t ∗ , ∞ ), then the ratio F/G is decreasing on (0 , ∞ ) if and only if H F,G (cid:0) + (cid:1) = lim x → + (cid:18) F ′ ( x ) G ′ ( x ) G ( x ) − F ( x ) (cid:19) ≥ , with lim x → + F ( x ) G ( x ) = lim t →∞ f ( t ) g ( t ) and lim x →∞ F ( x ) G ( x ) = lim t → + f ( t ) g ( t )provide the indicated limits exist. While H F,G (cid:0) + (cid:1) <
0, there is at leas one x ∗ > F/G is increasing on (0 , x ∗ ) and decreasing on ( x ∗ , ∞ ). Asapplications of this monotonicity rule, a unified treatment for certain bounds ofpsi function is presented, and some properties of the modified Bessel functionsof the second are established. These show that the monotonicity rules inthis paper may contribute to study for certain special functions because manyspecial functions can be expressed as corresponding Laplace transforms. Introduction
The Laplace transform of a function f ( t ) defined on [0 , ∞ ) is the function F ( s ),which is a unilateral transform defined by F ( s ) = Z ∞ f ( t ) e − st dt, where s is a complex number frequency parameter. The Laplace transform of afunction f ( t ) is also denoted by L ( f ).It is known that some special functions can be represented as correspondingLaplace transforms, for example, Binet formula for gamma function:ln Γ ( z ) − (cid:18) z − (cid:19) ln z + z −
12 ln (2 π ) = Z ∞ (cid:18) e t − − t + 12 (cid:19) e − zt t dt R ( z ) > K v ( x ) = Z ∞ e − x cosh t cosh ( vt ) dt, Mathematics Subject Classification.
Primary 44A10, 26A48; Secondary 33B15, 33C10.
Key words and phrases.
Laplace transform, monotonicity rule, psi function, modified Besselfunctions of the second kind.This paper is in final form and no version of it will be submitted for publication elsewhere. which, by replacing cosh t − t , can be expressed as K v ( x ) = e − x Z ∞ e − xt cosh ( v arccosh ( t + 1)) p t ( t + 2) dt ;the Gaussian Q-function [3] defined by Q ( x ) = 1 √ π Z ∞ x e − t / dt for x > t = u + x , is represented as Q ( x ) = 1 √ π e − x / Z ∞ e − u / e − ux du. More examples can be found in [4], [5].An important notion related to the Laplace transform is the completely mono-tonic functions. A function F is said to be completely monotonic on an interval I ,if F has derivatives of all orders on I and satisfies(1.2) ( − n F ( n ) ( x ) ≥ x ∈ I and n = 0 , , , .... If the inequality (1.2) is strict, then F is said to be strictly completely monotonic on I . The classical Bernstein’s theorem [6], [7] states that a function F is completelymonotonic (for short, CM) on (0 , ∞ ) if and only if it is a Laplace transform of somenonnegative measure µ , that is, F ( x ) = Z ∞ e − xt dµ ( t ) , where µ ( t ) is non-decreasing and the integral converges for 0 < x < ∞ .Another important one is the Bernstein functions [8]. A non-negative function F is said to be a Bernstein function on an interval I , if F has derivatives of allorders on I and satisfies(1.3) ( − n F ( n ) ( x ) ≤ x ∈ I and n = 0 , , , .... Clearly, a function F is a Bernstein function on I if and only if F ′ is CM on I .Very recently, Yang and Tian [9] established a monotonicity rule for the ratio oftwo Laplace transforms as follows. Theorem 1.
Let the functions f, g be defined on (0 , ∞ ) such that their Laplacetransforms L ( f ) = R ∞ f ( t ) e − xt dt and L ( g ) = R ∞ g ( t ) e − xt dt exist with g ( t ) = 0 for all t > . Then the ratio L ( f ) / L ( g ) is decreasing (increasing) on (0 , ∞ ) if f /g is increasing (decreasing) on (0 , ∞ ) . By using this monotonicity rule, Yang and Tian proved that the function x x (cid:0) ln Γ ( x + 1 / − x ln x + x − ln √ π (cid:1) + 1 − x is strictly increasing from (0 , ∞ ) onto (1 , / x ψ ( n +1) ( x ) ψ ( n ) ( x ) ψ ( n +2) ( x )on (0 , ∞ ), where ψ ( n ) for n ∈ N is the polygamma functions, and obtained somenew properties of polygamma functions. These show that Theorem 1 is an efficienttool of studying special functions. ONOTONICITY RULES FOR THE RATIO OF TWO LAPLACE TRANSFORMS 3
Moreover, as shown in [9, Remark 4], if g ( t ) > t >
0, then by Theorem1 and Bernstein’s theorem, both the functions x
7→ L ( f ) − β L ( g ) and x α L ( g ) − L ( f )are CM on (0 , ∞ ), where β = inf x> ( L ( f ) / L ( g )) > −∞ and α = sup x> ( L ( f ) / L ( g )) < ∞ .Inspired by the above comments, the aim of this paper is to further establishthe monotonicity rule of the ratio of two Laplace transforms L ( f ) / L ( g ) under thecondition that there is a t ∗ > f /g is increasing (decreasing) on (0 , t ∗ )and decreasing (increasing) on ( t ∗ , ∞ ).The rest of this paper is organized as follows. In Section 2, some lemmas aregiven, which containing monotonicity rules for ratios of two power series (poly-nomials). In Section 3, our main results (Theorems 1–3) are proved by means ofdefinition of integral and lemmas. As applications, two monotonicity results involv-ing psi function and modified Bessel functions of the second kind are presented.2. Lemmas
To state needed lemmas, we recall a useful auxiliary function H f,g , which wasintroduced in [11]. For −∞ ≤ a < b ≤ ∞ , let f and g be differentiable functionson ( a, b ) with g ′ = 0 on ( a, b ). Then we define(2.1) H f,g := f ′ g ′ g − f. The auxiliary function H f,g has the following well properties [11, Property 1]:(i) H f,g is even with respect to g and odd with respect to f , that is,(2.2) H f,g ( x ) = H f, − g ( x ) = − H − f,g ( x ) = − H − f, − g ( x ) . (ii) If g = 0 on ( a, b ), then(2.3) (cid:18) fg (cid:19) ′ = g ′ g H f,g , and therefore,(2.4) sgn (cid:18) fg (cid:19) ′ = sgn ( g ′ ) sgn ( H f,g ) . (iii) If f and g are twice differentiable on ( a, b ), then(2.5) H ′ f,g = (cid:18) f ′ g ′ (cid:19) ′ g. The auxiliary function H f,g and its properties are very helpful to investigatethose monotonicity of ratios of two functions, see [11], [12], [13], [14], [15], [16],[17], [18], [19], [20]. Recently, in [21] they were successfully applied to establishmonotonicity rules for ratios of two power series and of two polynomials under thecondition that the ratio of coefficients of two power series is increasing (decreasing)then decreasing (increasing). The following monotonicity rule for ration of twopolynomials will be used in proof of our main results. Lemma 1 ([21, Theorem 2.5]) . Let A n ( t ) = P nk =0 a k t k and B n ( t ) = P nk =0 b k t k be two real polynomials defined on (0 , r ) ( r > ) with b k > for all ≤ k ≤ n .Suppose that for certain m ∈ N with m < n , the sequences { a k /b k } ≤ k ≤ m and ZHEN-HANG YANG AND JING-FENG TIAN { a k /b k } m ≤ k ≤ n are both non-constants, and are respectively increasing (decreasing)and decreasing (increasing). Then the function A n /B n is increasing (decreasing)on (0 , r ) if and only if H A n ,B n ( r − ) ≥ ( ≤ ) 0 . While H A n ,B n ( r − ) < ( > ) 0 , there is aunique t ∈ (0 , r ) such that the function A n /B n is increasing (decreasing) on (0 , t ) and decreasing (increasing) on ( t , r ) . The following monotonicity rule will be used in Proposition 1, which first ap-peared in [22, Lemma 6.4] without giving the details of the proof. Two strict proofswere given in [21] and [23].
Lemma 2 ([21, Corollary 2.6]) . Let A ( t ) = P ∞ k =0 a k t k and B ( t ) = P ∞ k =0 b k t k betwo real power series converging on R with b k > for all k . If for certain m ∈ N , thenon-constant sequences { a k /b k } ≤ k ≤ m and { a k /b k } k ≥ m are respectively increasing(decreasing) and decreasing (increasing), then there is a unique t ∈ (0 , ∞ ) such thatthe function A/B is increasing (decreasing) on (0 , t ) and decreasing (increasing)on ( t , ∞ ) . The following lemma [27, Lemma 2] offers a simple criterion to determine thesign of a class of special series, which will be used in proof of Proposition 2.
Lemma 3 ([27, Lemma 2]) . Let { a k } ∞ k =0 be a nonnegative real sequence with a m > and P ∞ k = m +1 a k > and let S ( t ) = − m X k =0 a k t k + ∞ X k = m +1 a k t k be a convergent power series on the interval (0 , r ) ( r > ). (i) If S ( r − ) ≤ , then S ( t ) < for all t ∈ (0 , r ) . (ii) If S ( r − ) > , then there is a unique t ∈ (0 , r ) suchthat S ( t ) < for t ∈ (0 , t ) and S ( t ) > for t ∈ ( t , r ) . Remark 1.
Clearly, when r = ∞ in Lemma 3, there is a unique t ∈ (0 , ∞ ) suchthat S ( t ) < for t ∈ (0 , t ) and S ( t ) > for t ∈ ( t , ∞ ) . This result appeared in [22, Lemma 6.3] without proof (see also [24] , [25] ). If a k = 0 for k ≥ n ≥ m + 1 ,then Lemma 3 is reduced to a polynomial version, which appeared in [26] (see also [20] ). Main results
We are in a position to state and prove our main results.
Theorem 2.
For < a < b < ∞ , let the functions F and G be defined on (0 , ∞ ) by F ( x ) = Z ba f ( t ) e − xt dt and G ( x ) = Z ba g ( t ) e − xt dt, where the functions f, g are both continuous on [ a, b ] with g ( t ) > for t ∈ [ a, b ] .If there is t ∗ ∈ ( a, b ) such that f /g is strictly increasing (decreasing) on [ a, t ∗ ] and strictly decreasing (increasing) on [ t ∗ , b ] , then the ratio x F ( x ) /G ( x ) isdecreasing (increasing) on (0 , ∞ ) if and only if H F,G (cid:0) + (cid:1) = lim x → + (cid:18) F ′ ( x ) G ′ ( x ) G ( x ) − F ( x ) (cid:19) ≥ ( ≤ ) 0 , ONOTONICITY RULES FOR THE RATIO OF TWO LAPLACE TRANSFORMS 5 with (3.1) lim x → + F ( x ) G ( x ) = R ba f ( t ) dt R ba g ( t ) dt and lim x →∞ F ( x ) G ( x ) = f ( a ) g ( a ) . While H F,G (0 + ) < ( > ) 0 , then there is at least one x ∗ > such that F/G isincreasing (decreasing) on (0 , x ∗ ) and decreasing (increasing) on ( x ∗ , ∞ ) .Proof of Theorem 2. We only prove this theorem under the condition that f /g isstrictly increasing on [ a, t ∗ ] and strictly decreasing on [ t ∗ , b ]. If f /g is strictly de-creasing on [ a, t ∗ ] and strictly increasing on [ t ∗ , b ], then ( − f ) /g is strictly increasingon [ a, t ∗ ] and strictly decreasing on [ t ∗ , b ], and then corresponding conclusion of thistheorem is also true, which suffices to note that H − F,G ( x ) = − H F,G ( x ) due to (2.2).For n ∈ N , given a partition of the interval [ a, b ]: a = t < t < t < · · · < t n = b, with ∆ t i = t i − t i − = ( b − a ) /n , and so t i = a + ( b − a ) i/n . Then we have n − X i =0 f ( t i ) e − xt i ∆ t i = n − X i =0 f ( t i ) e − ax h e − ( b − a ) x/n i i b − an = b − an e − ax n − X i =0 f ( t i ) y i =: b − an e − ax F n ( y ) , n − X i =0 g ( t i ) e − xt i ∆ t i = b − an e − ax n − X i =0 g ( t i ) y i := b − an e − ax G n ( y ) , where y ≡ y n ( x ) = e − ( b − a ) x/n ∈ (0 , P n − i =0 f ( t i ) e − xt i ∆ t i P n − i =0 g ( t i ) e − xt i ∆ t i = P n − i =0 f ( t i ) y i P n − i =0 g ( t i ) y i = F n ( y ) G n ( y ) , (3.2) F ( x ) G ( x ) = R ba f ( t ) e − xt dt R ba g ( t ) e − xt dt = lim n →∞ P n − i =0 f ( t i ) e − xt i ∆ t i lim n →∞ P n − i =0 g ( t i ) e − xt i ∆ t i = lim n →∞ F n ( y ) G n ( y ) . Also, we have n − X i =0 ( t i − a ) f ( t i ) e − xt i ∆ t i = n − X i =0 ( b − a ) in f ( t i ) e − ax h e − ( b − a ) x/n i i b − an = e − ax (cid:18) b − an (cid:19) y n − X i =0 if ( t i ) y i − = e − ax (cid:18) b − an (cid:19) yF ′ n ( y ) , n − X i =0 ( t i − a ) g ( t i ) e − xt i ∆ t i = e − ax (cid:18) b − an (cid:19) yG ′ n ( y ) . Therefore, we obtain n − P i =0 ( t i − a ) f ( t i ) e − xt i ∆ t in − P i =0 ( t i − a ) g ( t i ) e − xt i ∆ t i = F ′ n ( y ) G ′ n ( y ) , ZHEN-HANG YANG AND JING-FENG TIAN (3.3) R ba ( t − a ) f ( t ) e − xt dt R ba ( t − a ) g ( t ) e − xt dt = lim n →∞ n − P i =0 ( t i − a ) f ( t i ) e − xt i ∆ t i lim n →∞ n − P i =0 ( t i − a ) g ( t i ) e − xt i ∆ t i = lim n →∞ F ′ n ( y ) G ′ n ( y ) . Since Z ba ( t − a ) f ( t ) e − xt dt = Z ba tf ( t ) e − xt dt − a Z ba f ( t ) e − xt dt = − F ′ ( x ) − aF ( x ) , Z ba ( t − a ) g ( t ) e − xt dt = Z ba tg ( t ) e − xt dt − a Z ba g ( t ) e − xt dt = − G ′ ( x ) − aG ( x ) , equation (3.3) also can be written as(3.4) F ′ ( x ) + aF ( x ) G ′ ( x ) + aG ( x ) = lim n →∞ F ′ n ( y ) G ′ n ( y ) . It then follows that H F n ,G n ( y ) = F ′ n ( y ) G ′ n ( y ) G n ( y ) − F n ( y ) = n − P i =0 ( t i − a ) f ( t i ) e − xt i ∆ t in − P i =0 ( t i − a ) g ( t i ) e − xt i ∆ t i n − X i =0 g ( t i ) y i − n − X i =0 f ( t i ) y i = ne ax b − a P n − i =0 ( t i − a ) f ( t i ) e − xt i ∆ t i P n − i =0 ( t i − a ) g ( t i ) e − xt i ∆ t i n − X i =0 b − an e − ax g ( t i ) y i − n − X i =0 b − an e − ax f ( t i ) y i ! = ne ax b − a P n − i =0 ( t i − a ) f ( t i ) e − xt i ∆ t i P n − i =0 ( t i − a ) g ( t i ) e − xt i ∆ t i n − X i =0 g ( t i ) e − xt i ∆ t i − n − X i =0 f ( t i ) e − xt i ∆ t i ! : = ne ax b − a C [ n ] f,g ( x ) , and so(3.5) sgn ( H F n ,G n ( y )) = sgn (cid:16) C [ n ] f,g ( x ) (cid:17) . Clearly, we havelim n →∞ C [ n ] f,g ( x ) = R ba ( t − a ) f ( t ) e − xt dt R ba ( t − a ) g ( t ) e − xt dt Z ba g ( t ) e − xt dt − Z ba f ( t ) e − xt dt = F ′ ( x ) + aF ( x ) G ′ ( x ) + aG ( x ) G ( x ) − F ( x ) = G ′ ( x ) G ′ ( x ) + aG ( x ) H F,G ( x ) , which, in view of G ′ ( x ) < G ′ ( x ) + aG ( x ) = − Z ba ( t − a ) g ( t ) e − xt dt < x ∈ (0 , ∞ ), implies that(3.7) sgn (cid:16) lim n →∞ C [ n ] f,g ( x ) (cid:17) = sgn ( H F,G ( x ))for x ∈ (0 , ∞ ). ONOTONICITY RULES FOR THE RATIO OF TWO LAPLACE TRANSFORMS 7 (i) The necessity follows from (cid:18) F ( x ) G ( x ) (cid:19) ′ = G ′ ( x ) G ( x ) H F,G ( x ) ≤ x >
0, which, due to G ′ ( x ) < x ∈ (0 , ∞ ), implies that H F,G (0 + ) ≥ H F,G (0 + ) ≥
0, then by the relation (3.7), there is a large N ∈ N such that C [ n ] f,g (0 + ) ≥ n > N , which in combination with the relation (3.5)gives that H F n ,G n ( y ) ≥ y → − for n > N . On the other hand, since f /g isincreasing on [ a, t ∗ ] and decreasing on [ t ∗ , b ], we easily see that there is a i ≥ { f ( t i ) /g ( t i ) } is strictly increasing for 0 ≤ i ≤ i and decreasingfor i < i ≤ n −
1. By Lemma 1, the ratio F n ( y ) /G n ( y ) is strictly increasing withrespect to y on (0 , ddy F n ( y ) G n ( y ) = G ′ n ( y ) G n ( y ) (cid:18) F ′ n ( y ) G ′ n ( y ) − F n ( y ) G n ( y ) (cid:19) > n > N and y ∈ (0 , , which, due to G n ( y ) , G ′ n ( y ) >
0, yields F ′ n ( y ) G ′ n ( y ) − F n ( y ) G n ( y ) > n > N and y ∈ (0 , . This together with (3.2) and (3.4) gives F ′ ( x ) + aF ( x ) G ′ ( x ) + aG ( x ) − F ( x ) G ( x ) ≥ x ∈ (0 , ∞ ) , which indicates that (cid:18) F ( x ) G ( x ) (cid:19) ′ = G ′ ( x ) + aG ( x ) G ( x ) (cid:18) F ′ ( x ) + aF ( x ) G ′ ( x ) + aG ( x ) − F ( x ) G ( x ) (cid:19) ≤ x ∈ (0 , ∞ ) , where the inequality holds due to G ( x ) > G ′ ( x ) + aG ( x ) < x →∞ F ( x ) G ( x ) = lim x →∞ lim n →∞ F n ( y ) G n ( y ) = lim n →∞ lim x →∞ F n ( y ) G n ( y )= lim n →∞ lim y → F n ( y ) G n ( y ) = f ( t ) g ( t ) = f ( a ) g ( a ) . (ii) If H F,G (0 + ) <
0, by the relations (3.7) and (3.5), there is a large enough N ∈ N such that H F n ,G n ( y ) < y → − for n > N . By Lemma 1, thereis a unique y [ n ]0 ∈ (0 ,
1) for given n > N such that the function F n ( y ) /G n ( y ) isincreasing on (cid:16) , y [ n ]0 (cid:17) and decreasing on (cid:16) y [ n ]0 , (cid:17) , that is, ddy F n ( y ) G n ( y ) = G ′ n ( y ) G n ( y ) (cid:18) F ′ n ( y ) G ′ n ( y ) − F n ( y ) G n ( y ) (cid:19) > n > N and y ∈ (cid:16) , y [ n ]0 (cid:17) ,ddy F n ( y ) G n ( y ) = G ′ n ( y ) G n ( y ) (cid:18) F ′ n ( y ) G ′ n ( y ) − F n ( y ) G n ( y ) (cid:19) < n > N and y ∈ (cid:16) y [ n ]0 , (cid:17) , ZHEN-HANG YANG AND JING-FENG TIAN where y [ n ]0 is the unique solution of the equation [ F n ( y ) /G n ( y )] ′ = 0 on (0 , (cid:20) ddy F n ( y ) G n ( y ) (cid:21) y = y [ n ]0 = G ′ n (cid:16) y [ n ]0 (cid:17) G n (cid:16) y [ n ]0 (cid:17) F ′ n (cid:16) y [ n ]0 (cid:17) G ′ n (cid:16) y [ n ]0 (cid:17) − F n (cid:16) y [ n ]0 (cid:17) G n (cid:16) y [ n ]0 (cid:17) = 0 for n > N. In the same treatment as part (i) of the proof of this theorem, the above threerelations imply that (cid:18) F ( x ) G ( x ) (cid:19) ′ ≤ x ∈ ( x ∗ , ∞ ) , (cid:18) F ( x ) G ( x ) (cid:19) ′ ≥ x ∈ (0 , x ∗ ) , where x ∗ = lim n →∞ x [ n ]0 , x [ n ]0 = − n (cid:16) ln y [ n ]0 (cid:17) / ( b − a ), and satisfies "(cid:18) F ( x ) G ( x ) (cid:19) ′ x = x ∗ = 0 . Thus it remains to prove x ∗ = lim n →∞ x [ n ]0 = 0 , ∞ . First, we claim that x ∗ =lim n →∞ x [ n ]0 = 0. If not, that is, lim n →∞ x [ n ]0 = 0, then F ( x ) /G ( x ) is decreasingin x on (0 , ∞ ). This, by part (i) of this theorem, implies that H F,G (0 + ) ≥
0, whichyields a contraction with the assumption that H F,G (0 + ) < x ∗ = lim n →∞ x [ n ]0 = ∞ . If not, that is, lim n →∞ x [ n ]0 = ∞ , then F ( x ) /G ( x ) is increasing in x on (0 , ∞ ). It then follows that for all x > F ( x ) G ( x ) < lim x →∞ F ( x ) G ( x ) = f ( a ) g ( a ) . Since lim n →∞ ( F n ( y ) /G n ( y )) = F ( x ) /G ( x ), there exists a large enough N ∈ N such that for n > N the inequality(3.8) F n ( y ) G n ( y ) < f ( a ) g ( a )holds for all y ∈ (0 , F n ( y ) /G n ( y ) is increasing on (cid:16) , y [ n ]0 (cid:17) and decreasing on (cid:16) y [ n ]0 , (cid:17) , which suggeststhat there exists a small enough δ ∈ (cid:16) , y [ n ]0 (cid:17) such that F n ( y ) /G n ( y ) is increasingon (0 , δ ). Therefore, for y ∈ (0 , δ ) and n > N , F n ( y ) G n ( y ) ≥ F n (0) G n (0) = f ( t ) g ( t ) = f ( a ) g ( a ) . This is in contradiction with the inequality (3.8) for all y ∈ (0 ,
1) and n > N .Consequently, x ∗ = lim n →∞ x [ n ]0 = 0 , ∞ , which ends the proof. (cid:3) Letting a → + , b → ∞ in Theorem 2. Then F ( x ) and G ( x ) are Laplace trans-forms of the functions f and g , respectively. By properties of uniformly convergentimproper integral with a parameter, we have the following monotonicity rule, where ONOTONICITY RULES FOR THE RATIO OF TWO LAPLACE TRANSFORMS 9 the first limit of (3.9) follows from the first one of (3.1) and Cauchy mean valuetheorem, that is,lim x → + F ( x ) G ( x ) = lim b →∞ R ba f ( t ) dt R b g ( t ) dt = lim b →∞ (cid:2)R sa f ( t ) dt (cid:3) s = b − (cid:2)R sa f ( t ) dt (cid:3) s = a (cid:2)R sa g ( t ) dt (cid:3) s = b − (cid:2)R s g ( t ) dt (cid:3) s = a = lim b →∞ f ( a + θ ( b − a )) g ( a + θ ( b − a )) = lim t →∞ f ( t ) g ( t ) , here θ ∈ (0 , Theorem 3.
Let f and g be both continuous functions on (0 , ∞ ) with g ( t ) > for t ∈ (0 , ∞ ) and let F ( x ) = L ( f ) and G ( x ) = L ( g ) converge for x > . If there isa t ∗ ∈ (0 , ∞ ) such that f /g is strictly increasing (decreasing) on (0 , t ∗ ) and strictlydecreasing (increasing) on ( t ∗ , ∞ ) , then the function F/G is decreasing (increasing)on (0 , ∞ ) if and only if H F,G (cid:0) + (cid:1) = lim x → + (cid:18) F ′ ( x ) G ′ ( x ) G ( x ) − F ( x ) (cid:19) ≥ ( ≤ ) 0 , with (3.9) lim x → + F ( x ) G ( x ) = lim t →∞ f ( t ) g ( t ) and lim x →∞ F ( x ) G ( x ) = lim t → + f ( t ) g ( t ) provide the indicated limits exist. While H F,G (0 + ) < ( > ) 0 , there is at leas one x ∗ > such that F/G is increasing (decreasing) on (0 , x ∗ ) and decreasing (increasing)on ( x ∗ , ∞ ) . Theorem 4.
Suppose that (i) both the functions f and g are continuous on (0 , ∞ ) with g ( t ) > for t ∈ (0 , ∞ ) ; (ii) the function µ is positive, differentiable andincreasing from (0 , ∞ ) onto ( µ (0 + ) , ∞ ) ; (iii) both the functions F ( x ) = Z ∞ f ( t ) e − xµ ( t ) dt and G ( x ) = Z ∞ g ( t ) e − xµ ( t ) dt converge for all x > . Then the following statements are valid:(i) If the ratio f /g is increasing (decreasing) on (0 , ∞ ) , then F/G is decreasing(increasing) on (0 , ∞ ) with lim x → + F ( x ) G ( x ) = lim t →∞ f ( t ) g ( t ) and lim x →∞ F ( x ) G ( x ) = lim t → + f ( t ) g ( t ) . (ii) If there is a t ∗ ∈ (0 , ∞ ) such that f /g is strictly increasing (decreasing)on (0 , t ∗ ) and strictly decreasing (increasing) on ( t ∗ , ∞ ) , then the ratio F/G isdecreasing (increasing) on (0 , ∞ ) if and only if H F,G (cid:0) + (cid:1) = lim x → + (cid:18) F ′ ( x ) G ′ ( x ) G ( x ) − F ( x ) (cid:19) ≥ ( ≤ ) 0 . While H F,G (0 + ) < ( > ) 0 , there is at least one x ∗ > such that F/G is increasing(decreasing) on (0 , x ∗ ) and decreasing (increasing) on ( x ∗ , ∞ ) .Proof. Let µ ( t ) − a = s , where a = µ (0 + ). Then F ( x ) and G ( x ) are expressed as F ( x ) = e − xa Z ∞ f ( t ( s )) µ ′ ( t ( s )) e − xs ds and G ( x ) = e − xa Z ∞ g ( t ( s )) µ ′ ( t ( s )) e − xs ds, where t ( s ) = µ − ( s + a ), and F ( x ) /G ( x ) can be represented in the form of ratioof two Laplace transforms: F ( x ) G ( x ) = e xa F ( x ) e xa G ( x ) = R ∞ [ f ( t ( s )) /µ ′ ( t ( s ))] e − xs ds R ∞ [ g ( t ( s )) /µ ′ ( t ( s ))] e − xs ds := R ∞ f ∗ ( s ) e − xs ds R ∞ g ∗ ( s ) e − xs ds . It is easy to verify that(3.10) f ∗ ( s ) g ∗ ( s ) = f ( t ( s )) g ( t ( s )) , (cid:18) f ∗ ( s ) g ∗ ( s ) (cid:19) ′ = (cid:18) f ( t ) g ( t ) (cid:19) ′ × dtds = 1 µ ′ ( t ) (cid:18) f ( t ) g ( t ) (cid:19) ′ , where µ ′ ( t ) > t > f ( t ) /g ( t ) is increasing (decreasing) on (0 , ∞ ), then so is f ∗ ( s ) /g ∗ ( s ).By Theorem 1, we easily find that ( e xa F ) / ( e xa G ) = F/G is decreasing (increasing)on (0 , ∞ ).By the limit relations (3.9) we easily getlim x → + F ( x ) G ( x ) = lim s →∞ f ( t ( s )) /µ ′ ( t ( s )) g ( t ( s )) /µ ′ ( t ( s )) = lim t →∞ f ( t ) g ( t ) , lim x →∞ F ( x ) G ( x ) = lim s → + f ( t ( s )) /µ ′ ( t ( s )) g ( t ( s )) /µ ′ ( t ( s )) = lim t → + f ( t ) g ( t ) . (ii) If there is a t ∗ ∈ (0 , ∞ ) such that f ( t ) /g ( t ) is strictly increasing (decreasing)on (0 , t ∗ ) and strictly decreasing (increasing) on ( t ∗ , ∞ ), then by the second relationof (3.10), there is a s ∗ ∈ (0 , ∞ ) such that f ∗ ( s ) /g ∗ ( s ) is strictly increasing (decreas-ing) on (0 , s ∗ ) and strictly decreasing (increasing) on ( s ∗ , ∞ ), where s ∗ = µ ( t ∗ ) − a .By Theorem 3, the ratio e xa F/ ( e xa G ) = F/G is decreasing (increasing) on (0 , ∞ )if and only if(3.11) lim x → + H e xa F,e xa G ( x ) = lim x → + (cid:18) ( e xa F ( x )) ′ ( e xa G ( x )) ′ e xa G ( x ) − e xa F ( x ) (cid:19) ≥ ( ≤ ) 0 . We claim that the limit relation is equivalent to lim x → + H F,G ( x ) ≥ ( ≤ ) 0. In fact,we easily check that H e xa F,e xa G ( x ) = e xa ( F ′ ( x ) + aF ( x )) e xa ( G ′ ( x ) + aG ( x )) e xa G ( x ) − e xa F ( x )= e xa F ′ ( x ) G ( x ) + aF ( x ) G ( x ) − F ( x ) G ′ ( x ) − aF ( x ) G ( x ) G ′ ( x ) + aG ( x )= e xa G ′ ( x ) G ′ ( x ) + aG ( x ) (cid:18) F ′ ( x ) G ′ ( x ) G ( x ) − F ( x ) (cid:19) = e xa G ′ ( x ) G ′ ( x ) + aG ( x ) H F,G ( x ) . This together with G ′ ( x ) = − Z ∞ µ ( t ) g ( t ) e − xµ ( t ) dt < ,G ′ ( x ) + aG ( x ) = − Z ∞ (cid:2) µ ( t ) − µ (cid:0) + (cid:1)(cid:3) g ( t ) e − xµ ( t ) dt < x > H e xa F,e xa G ( x )) = sgn ( H F,G ( x )) , which proves the claim just now. ONOTONICITY RULES FOR THE RATIO OF TWO LAPLACE TRANSFORMS 11 (iii) If lim x → + H F,G ( x ) < ( > ) 0, then lim x → + H e xa F,e xa G ( x ) < ( > ) 0. By The-orem 3, there is at least one x ∗ > e xa F ( x ) / ( e xa G ( x )) = F ( x ) /G ( x )is increasing (decreasing) on (0 , x ∗ ) and decreasing (increasing) on ( x ∗ , ∞ ).Thus we complete the proof. (cid:3) A unified treatment for certain bounds of harmonic number
The Euler-Mascheroni constant is defined by γ = lim n →∞ ( H n − ln n ) = 0 . ..., where H n = P nk =1 k − is the n ’th harmonic number. There is a close connectionbetween H n and the psi (or digamma) function. Indeed, we have H n = ψ ( n + 1)+ γ .Several bounds for H n or ψ ( n + 1) can see [28], [29], [30], [31], [32], [33], [34], [35],[36], [37], [38], [39], [40] [41].In particular, Alzer [32] obtained the double inequality,12 ( n + a ) ≤ H n − ln n − γ <
12 ( n + b )holds for n ∈ N with the best constants a = 12 (1 − γ ) − b = 16by proving the sequence(4.1) A ( n ) = 12 1 ψ ( n + 1) − ln n − n is strictly decreasing for n ≥ H n = ln p n ( n + 1) + γ + 16 n ( n + 1) + L ( n ) , (4.2) = ln (cid:18) n + 12 (cid:19) + γ + 124 ( n + 1 / + D ( n ) , (4.3)where both the sequences L ( n ) and D ( n ) are increasing for n ∈ N . Qi [37] showedthat the sequence(4.4) Q ( n ) = 12 1ln n + 1 / (2 n ) − ψ ( n + 1) − n is strictly increasing for n ∈ N . These monotonicity of sequences L ( n ), D ( n ) and Q ( n ) similarly yield corresponding sharp bounds for H n or ψ ( n + 1).We remark that it is difficult to deal with the monotonicity of the function A ( x ), L ( x ), D ( x ) and Q ( x ) on (0 , ∞ ) by usual approach. However, if we write them asratios of two Laplace transforms, then we easily prove their monotonicity on (0 , ∞ )by Theorems 1 and 3. Here we chose Φ ( x ) = D ( x − /
2) defined by (4.3) andprove its monotonicity on (0 , ∞ ). As far as A ( x ), L ( x ) and Q ( x ), we only listtheir expressions in the form of ratios of Laplace transforms. In fact, By means of the formulas ψ ( x ) = Z ∞ (cid:18) e − t t − e − xt − e − t (cid:19) dt, ln x = Z ∞ e − t − e − xt t dt ,1 x n = 1( n − Z ∞ t n − e − xt dt, we have A ( x ) = 12 1 ψ ( x + 1) − ln x − x = R ∞ [ − p ′ ( t )] e − xt dt R ∞ p ( t ) e − xt dt , where p ( t ) = 2 (cid:18) t − e t − (cid:19) ; L ( x ) = 22 ψ ( x + 1) − ln ( x ( x + 1)) − x ( x + 1) = R ∞ [ − p ′′ ( t ) + p ′ ( t ))] e − xt dt R ∞ p ( t ) e − xt dt , where p ( t ) = e t − te t − t ( e t − e t ; Q ( x ) = 1ln x + 1 / (2 x ) − ψ ( x + 1) − x = R ∞ [ − p ′′ ( t )] e − xt dt R ∞ p ( t ) e − xt dt , where p ( t ) = 12 (cid:18) coth t − t (cid:19) . Next we prove the monotonicity of Φ ( x ) = D ( x − /
2) on (0 , ∞ ) by Theorem 3. Proposition 1.
The function
Φ ( x ) = 1 ψ ( x + 1 / − ln x − x is strictly increasing from (0 , ∞ ) onto (0 , / .Proof. We write Φ ( x ) = F ( x ) /G ( x ), where F ( x ) = 1 − x ( ψ ( x + 1 / − ln x ) ,G ( x ) = ψ ( x + 1 / − ln x. It has been shown in [ ? ] that G ( x ) = Z ∞ q ( t ) e − xt dt,x G ( x ) = 124 + Z ∞ e − xt q ′′ ( t ) dt, where(4.5) q ( t ) = 1 t −
12 sinh ( t/ . Then F ( x ) can be written as F ( x ) = 1 − x G ( x ) = Z ∞ ( − q ′′ ( t )) e − xt dt. ONOTONICITY RULES FOR THE RATIO OF TWO LAPLACE TRANSFORMS 13
Thus Φ ( x ) is expressed asΦ ( x ) = F ( x ) G ( x ) = R ∞ [ − q ′′ ( t )] e − xt dt R ∞ q ( t ) e − xt dt . We first show that there is a t ∗ > − q ′′ /q is decreasingon (0 , t ∗ ) and increasing on ( t ∗ , ∞ ). Direct computations give q ′′ ( t ) q ( t ) = 14 2 cosh s sinh s − s cosh s − s − s s ( s − sinh s ) sinh s = 12 sinh 3 s − s cosh 2 s − s − s s (sinh 3 s − s cosh 2 s − s + 2 s ) , where s = t/
2. Expanding in power series yields q ′′ ( t ) q ( t ) = 12 P ∞ n =3 3 n − − (2 n − n − n − n − − n − s n − P ∞ n =2 3 n − − (2 n − n − − n − s n − := 12 P ∞ n =4 a n (cid:0) s (cid:1) n − P ∞ n =4 b n ( s ) n − . Since (2 n − b n +1 − n − b n = (10 n −
23) 2 n − + 24 > n ≥ b = 0, we see that b n > n ≥
4. Thus, to prove the function − q ′′ /q is decreasing on (0 , t ∗ ) and increasing on ( t ∗ , ∞ ), by Lemma 2 it suffices to provethat there is an integer n > { a n /b n } n ≥ is increasingfor 4 ≤ n ≤ n and decreasing for n > n . For this end, we have to prove d n = [(2 n − ( a n b n +1 − a n +1 b n ) < ≤ n ≤ d n > n ≥ d n = 18 (4 n −
1) + 29 (4 n −
1) 3 n − n − (cid:0) n − n − n + 464 n − (cid:1) n + 49 (cid:0) n − n + 41 (cid:1) n −
34 (2 n −
1) (2 n − (cid:0) n + 5 n + 2 n − (cid:1) n . We find that d = −
66 802 176 , d = −
13 774 616 064 , d = − ,d = −
127 269 822 161 664 , d = − , d = −
240 861 038 835 686 400 . To prove d n > n ≥
10, we write d n as d n = 18 (4 n −
1) + (4 n −
1) 6 n × a ∗ n + (cid:0) n − n + 41 (cid:1) n × b ∗ n , where a ∗ n = 29 (cid:18) (cid:19) n − n − (cid:0) n − n − n + 464 n − (cid:1) n − ,b ∗ n = 49 (cid:18) (cid:19) n −
34 (2 n −
1) (2 n − (cid:0) n + 5 n + 2 n − (cid:1) n − n + 41 . It is easy to verify that a ∗ = 710 697 1416815 744 > b ∗ = 174 443 916 097149 159 936 > , and for n ≥ a ∗ n +1 − a ∗ n = 1432 (cid:0) n + n − (cid:1) (cid:0) n − n + 1448 n + 1789 n − (cid:1) (4 n −
1) (4 n + 3) > , b ∗ n +1 − b ∗ n = 316 2 n − n − n −
27) (64 n − n + 41) × (cid:0) n (cid:0) n − n − (cid:1) + 19 204 n + 16 517 n − n − (cid:1) > , which yield d n > n ≥ x → + F ( x ) = 1, lim x → + F ′ ( x ) = 0 , and lim x → + G ( x ) G ′ ( x ) = lim x → + ψ ( x + 1 / − ln xψ ′ ( x + 1 / − /x = 0 , which yield lim x → + H F,G ( x ) = lim x → + (cid:18) F ′ ( x ) G ′ ( x ) G ( x ) − F ( x ) (cid:19) = − < . It then follows by Theorem 3 that the function Φ =
F/G is strictly increasing on(0 , ∞ ).An easy computation giveslim x → + Φ ( x ) = lim t →∞ − q ′′ ( t ) q ( t ) = 0 and lim x →∞ Φ ( x ) = lim t → − q ′′ ( t ) q ( t ) = 215 , which completes the proof. (cid:3) An application to Bessel functions
The modified Bessel functions of the second kind K v is defined as [2, p. 78](5.1) K v ( x ) = π I − v ( x ) − I v ( x )sin ( vπ ) , where I v ( x ) is the modified Bessel functions of the first kind which can be repre-sented by the infinite series as(5.2) I v ( x ) = ∞ X n =0 ( x/ n + v n !Γ ( v + n + 1) , x ∈ R , v ∈ R \{− , − , ... } , and the right-hand side of (5.1) is replaced by its limiting value if v is an integer orzero.We easily see that K v ( x ) = K − v ( x ) for all v ∈ R and x > v ≥ − K v − ( x ) K v +1 ( x ) K v ( x ) = 1 x (cid:18) xK ′ v ( x ) K v ( x ) (cid:19) ′ holds for v ∈ R and x >
0, and the function x xK ′ v ( x ) K v ( x )is strictly decreasing on (0 , ∞ ) for all v ∈ R (see also [43]). As another appli-cation, in this section we will determine the monotonicity of the function x x + xK ′ v ( x ) /K v ( x ) on (0 , ∞ ) by Theorem 4. More precisely, we have ONOTONICITY RULES FOR THE RATIO OF TWO LAPLACE TRANSFORMS 15
Proposition 2.
For v ≥ , let K v ( x ) be the modified Bessel functions of the secondkind.(i) If v ∈ (1 / , ∞ ) , then the function x Λ ( x ) = x + xK ′ v ( x ) K v ( x ) is strictly increasing from (0 , ∞ ) onto ( − v, − / .(ii) If v ∈ [0 , / , then the function x Λ ( x ) is strictly decreasing from (0 , ∞ ) onto ( − / , − v ) .(iii) If v = 1 / , then the double inequality (5.4) − x − max (cid:18) v, (cid:19) < xK ′ v ( x ) K v ( x ) < − x − min (cid:18) v, (cid:19) holds for x > with the best constants min ( v, / and max ( v, / . Before proving Proposition 2, we give the following lemmas.
Lemma 4.
For v > with v = 1 / , let the function h v be defined on (0 , ∞ ) by (5.5) h v ( t ) = cosh ( tv ) + v sinh ( tv ) sinh t (cosh t + 1) cosh ( tv ) . (i) If v ∈ [1 , ∞ ) then h v is increasing from (0 , ∞ ) onto (1 / , v ) .(ii) If v ∈ (1 / , , then there is a t ∗ ∈ (0 , ∞ ) such that h v is increasing on (0 , t ∗ ) and decreasing on ( t ∗ , ∞ ) . Consequently, the inequalities
12 = min (cid:18) , v (cid:19) < h v ( t ) ≤ θ v hold for x > , where θ v = h v ( t ∗ ) , here t ∗ is the unique solution of the equation h ′ v ( t ) = 0 on (0 , ∞ ) .(iii) If v ∈ (0 , / , then there is a t ∗ ∈ (0 , ∞ ) such that h v is decreasing on (0 , t ∗ ) and increasing on ( t ∗ , ∞ ) . Therefore, it holds that for x > , θ v ≤ h v ( t ) < max (cid:18) , v (cid:19) = 12 , where θ v is as in (ii).Proof. Differentiating and simplifying yield h ′ v ( t ) = r v ( t )(cosh t + 1) cosh ( tv ) , where r v ( t ) = v (1 + cosh t ) sinh t + v (1 + cosh t ) cosh ( tv ) sinh ( tv ) − cosh ( tv ) sinh t. Using ”product into sum” formulas and expanding in power series gives4 r v ( t ) = ( v + 1) sinh (2 vt − t ) + ( v −
1) sinh (2 tv + t )+2 v sinh (2 tv ) + 2 v sinh (2 t ) + 2 (cid:0) v − (cid:1) sinh t : = ∞ X n =1 a n (2 n − t n − , where a n = ( v + 1) (2 v − n − + ( v −
1) (2 v + 1) n − + (2 v ) n + v n + 2 (cid:0) v − (cid:1) . To confirm the monotonicity of h v , we need to deal with the sign of a n . For thisend, we first give two recurrence relations. It is easy to check that(5.6) a n +1 − a n n = 2 v (cid:0) v − (cid:1) (cid:0) v − (cid:1) n − + 2 v (cid:0) v − (cid:1) (cid:0) v + (cid:1) n − + (cid:0) v − (cid:1) v n + 3 v := b n , (5.7) b n +1 − b n v ( v −
1) ( v − /
4) = (2 v − (cid:18) v − (cid:19) n − +(2 v + 3) (cid:18) v + 12 (cid:19) n − +4 v n − > v = 1 , /
2. We now distinguish three cases to prove the desired monotonicity.
Case 1 : v ≥
1. It is clear that a n > n ≥
1, which implies that h ′ v ( t ) > t ∈ (0 , ∞ ). Case 2 : v ∈ (1 / , { b n } n ≥ is decreasing, which together with b = 2 v (2 v −
1) (2 v + 1) > , lim n →∞ b n = sgn (cid:0) v − (cid:1) ∞ < n > b n ≥ ≤ n ≤ n and b n ≤ n ≥ n . This, by the recurrence relation (5.6), in turn implies that the sequence { a n } n ≥ is increasing for 1 ≤ n ≤ n and decreasing for n ≥ n . Therefore, weobtain a n ≥ a = 4 (2 v −
1) (2 v + 1) > ≤ n ≤ n . On the other hand, it is seen thatlim n →∞ a n n = sgn ( v − ∞ < . It then follows that there is an integer n > n such that a n ≥ ≤ n ≤ n and a n ≤ n ≥ n . By Lemma 3, there is a t ∗ > h ′ v ( t ) > t ∈ (0 , t ∗ ) and h ′ v ( t ) < t ∈ ( t ∗ , ∞ ). Case 3 : v ∈ (0 , / { b n } n ≥ is increasing,which in combination with b = 2 v (2 v −
1) (2 v + 1) < , lim n →∞ b n = 3 v > n > b n ≤ ≤ n ≤ n and b n ≥ n ≥ n . This, by the recurrence relation (5.6), means that the sequence { a n } n ≥ is decreasing for 1 ≤ n ≤ n and increasing for n ≥ n . Hence, we deducethat a n ≤ a = 4 (2 v −
1) (2 v + 1) < ≤ n ≤ n . Moreover, it is seen that lim n →∞ a n (2 v + 1) n = sgn (cid:0) v (cid:1) ∞ > . It then follows that there is an integer n > n such that a n ≤ ≤ n ≤ n and a n ≥ n ≥ n . By Lemma 3, there is a t ∗ > h ′ v ( t ) < t ∈ (0 , t ∗ ) and h ′ v ( t ) > t ∈ ( t ∗ , ∞ ).This completes the proof. (cid:3) ONOTONICITY RULES FOR THE RATIO OF TWO LAPLACE TRANSFORMS 17
Lemma 5.
Let K v ( x ) be the modified Bessel functions of the second kind and let F ( x ) = x [ K v ( x ) + K ′ v ( x )] and G ( x ) = K v ( x ) . Then for v ∈ (0 , , we have lim x → + H F,G ( x ) = lim x → + (cid:18) F ′ ( x ) G ′ ( x ) G ( x ) − F ( x ) (cid:19) = 0 . Proof.
By the asymptotic formulas [44, p. 375, (9.6.9)](5.8) K v ( x ) ∼
12 Γ ( v ) (cid:16) x (cid:17) − v for v > x → x → K ′ v ( x ) ∼ −
14 Γ ( v + 1) (cid:16) x (cid:17) − v − for v > ,K ′′ v ( x ) ∼
18 Γ ( v + 2) (cid:16) x (cid:17) − v − for v > . Then, for v >
0, as x → + , F ( x ) = x [ K v ( x ) + K ′ v ( x )] ∼ x (cid:20)
12 Γ ( v ) (cid:16) x (cid:17) − v −
14 Γ ( v + 1) (cid:16) x (cid:17) − v − (cid:21) = Γ ( v ) x − v (cid:16) x (cid:17) − v ,F ′ ( x ) = K v ( x ) + ( x + 1) K ′ v ( x ) + xK ′′ v ( x ) ∼
12 Γ ( v ) (cid:16) x (cid:17) − v − x + 14 Γ ( v + 1) (cid:16) x (cid:17) − v − + x
18 Γ ( v + 2) (cid:16) x (cid:17) − v − = 12 Γ ( v ) (1 − v ) x + v x (cid:16) x (cid:17) − v , which yield H F,G ( x ) = F ′ ( x ) G ′ ( x ) G ( x ) − F ( x ) ∼ Γ ( v ) (1 − v ) x + v x (cid:0) x (cid:1) − v − Γ ( v + 1) (cid:0) x (cid:1) − v −
12 Γ ( v ) (cid:16) x (cid:17) − v − Γ ( v ) x − v (cid:16) x (cid:17) − v = − Γ ( v ) v (cid:16) x (cid:17) − v → x → + if v ∈ (0 , . This completes the proof. (cid:3)
We now are in a position to prove Proposition 2.
Proof of Proposition 2.
We haveΛ ( x ) = x [ K v ( x ) + K ′ v ( x )] K v ( x ) = F ( x ) G ( x ) . By the integral representation (1.1) we get that(5.9) G ( x ) = K v ( x ) = Z ∞ cosh ( vt ) e − x cosh t dt ; K v ( x ) + K ′ v ( x ) = Z ∞ cosh ( vt ) e − x cosh t dt − Z ∞ cosh ( vt ) (cosh t ) e − x cosh t dt = − Z ∞ cosh ( vt ) (cosh t − e − x cosh t dt, then integration by parts yields F ( x ) = x ( K v ( x ) + K ′ v ( x )) = Z ∞ cosh ( vt ) cosh t − t de − x cosh t = − Z ∞ cosh ( vt ) + v sinh t sinh ( vt )cosh t + 1 e − x cosh t dt. (5.10)Thus Λ ( x ) can be written asΛ ( x ) = F ( x ) G ( x ) = R ∞ (cid:16) − cosh( vt )+ v sinh t sinh( vt )cosh t +1 (cid:17) e − x cosh t dt R ∞ cosh ( vt ) e − x cosh t dt := R ∞ f ( t ) e − xµ ( t ) dt R ∞ g ( t ) e − xµ ( t ) dt , where µ ( t ) = cosh t and f ( t ) = − cosh ( vt ) + v sinh t sinh ( vt )cosh t + 1 , g ( t ) = cosh ( vt ) . Clearly, µ ( t ) is positive, differentiable and increasing on (0 , ∞ ), while f ( t ) and g ( t ) are differentiable on (0 , ∞ ) with g ( t ) > t >
0. Also, we have f ( t ) g ( t ) = − [cosh ( vt ) + v sinh t sinh ( vt )] / (cosh t + 1)cosh ( vt ) = − h v ( t ) . (i) If v ≥
1, then by the first assertion of Lemma 4 we see that f /g is strictlydecreasing on (0 , ∞ ). It follows by part (i) of Theorem 4 that Λ = F/G is strictlyincreasing on (0 , ∞ ).(ii) If v ∈ (1 / , t ∗ ∈ (0 , ∞ ) such that f /g is decreasing on (0 , t ∗ ) and increasing on ( t ∗ , ∞ ). Also,lim x → + H F,G ( x ) = 0 due to Lemma 5. These, by part (ii) of Theorem 4, yieldthat Λ = F/G is strictly increasing on (0 , ∞ ).(iii) If v ∈ (0 , / t ∗ ∈ (0 , ∞ ) such that f /g is increasing on (0 , t ∗ ) and decreasing on ( t ∗ , ∞ ). And,lim x → + H F,G ( x ) = 0 due to Lemma 5. It then follows from part (ii) of Theorem4 that Λ = F/G is strictly decreasing on (0 , ∞ ).(iv) If v = 0, then f ( t ) /g ( t ) = − / (cosh t + 1) is clearly increasing on (0 , ∞ ).It follows from part (i) of Theorem 4 that Λ = F/G is strictly decreasing on (0 , ∞ ).The limit values areΛ (cid:0) + (cid:1) = lim x → + F ( x ) G ( x ) = − lim t →∞ cosh ( vt ) + v sinh t sinh ( vt )(cosh t + 1) cosh ( vt ) = − v, Λ ( ∞ ) = lim x →∞ F ( x ) G ( x ) = − lim t → + cosh ( vt ) + v sinh t sinh ( vt )(cosh t + 1) cosh ( vt ) = − . The double inequality (5.4) follows from the monotonicity of
F/G on (0 , ∞ ), whichends the proof. (cid:3) As a consequence of [45, Theorem 5], Miller and Samko showed that the function x
7→ √ xe x K v ( x ) is strictly decreasing on (0 , ∞ ) for | v | > /
2. Yang and Zhengin [43, Corollary 3.2] reproved this assertion and further proved this function is
ONOTONICITY RULES FOR THE RATIO OF TWO LAPLACE TRANSFORMS 19 strictly increasing on (0 , ∞ ) for | v | < /
2. Now we have a more general result byProposition 2.
Corollary 1.
For v ≥ , let K v ( x ) be the modified Bessel functions of the secondkind. Then the function x x r e x K v ( x ) is strictly increasing on (0 , ∞ ) if and only if r ≥ max ( v, / , and decreasing ifand only if r ≤ min ( v, / . While v > ( < ) 1 / and / < r < v ( v < r < / ),there is an x > such that this function is decreasing (increasing) on (0 , x ) andincreasing (decreasing) on ( x , ∞ ) .Proof. Differentiation yields[ x r e x K v ( x )] ′ = x r e x K v ( x ) + rx r − e x K v ( x ) + x r e x K ′ v ( x )= x r − e x K v ( x ) (cid:18) r + x + xK ′ v ( x ) K v ( x ) (cid:19) := x r − e x K v ( x ) × φ ( x ) . By Proposition 2, [ x r e x K v ( x )] ′ ≥ ( ≤ ) 0 for all x > r ≥ − inf x> (cid:18) x + xK ′ v ( x ) K v ( x ) (cid:19) = (cid:26) v if v ∈ (1 / , ∞ ) if v ∈ (0 , /
2) = max (cid:18) v, (cid:19) ,r ≤ − sup x> (cid:18) x + xK ′ v ( x ) K v ( x ) (cid:19) = (cid:26) if v ∈ (1 / , ∞ ) v if v ∈ (0 , /
2) = min (cid:18) v, (cid:19) . When v > / / < r < v , by part (i) of Proposition 2, we see that x r + x + xK ′ v ( x ) /K v ( x ) = φ ( x ) is increasing on (0 , ∞ ), which in combination with φ (0 + ) = r − v < φ ( ∞ ) = r − / > x > φ ( x ) < x ∈ (0 , x ) and φ ( x ) > x ∈ ( x , ∞ ). That is to say, thefunction x x r e x K v ( x ) is decreasing on (0 , x ) and increasing on ( x , ∞ ).Similarly, by part (ii) of Proposition 2 we can prove that for v ∈ [0 , /
2) and v < r < /
2, there is an x > x x r e x K v ( x ) is increasingon (0 , x ) and decreasing on ( x , ∞ ). This completes the proof. (cid:3) Remark 2.
Corollary 1 implies that for all < x < y , the double inequality (5.11) e y − x (cid:16) yx (cid:17) r < K v ( x ) K v ( y ) < e y − x (cid:16) yx (cid:17) r if and only if r ≤ min ( | v | , / and r ≥ max ( | v | , / . Obviously, our inequali-ties (5.11) are superior to those earlier results appeared in [46] , [47] , [48] , [50] , [49] .Detailed comments can see [51, Section 3] . By Lemma 4 and Bernstein’s theorem, we give a class of completely monotonicfunction.
Corollary 2.
For v ≥ , the function P λ ( x ) = ( x + λ ) e x K v ( x ) + xe x K ′ v ( x ) is CM on (0 , ∞ ) if and only if λ ≥ v if v ∈ [1 , ∞ ) ,θ v if v ∈ (cid:0) , (cid:1) , if v ∈ (cid:0) , (cid:1) , and so is − P λ ( x ) if and only if λ ≤ (cid:26) if v ∈ (cid:0) , ∞ (cid:1) ,θ v if v ∈ (cid:0) , (cid:1) , where θ v is as in Lemma 4.Proof. From integral representations (5.10) and (5.9), we have P λ ( x ) = e x F ( x ) + λe x G ( x )= Z ∞ (cid:18) − cosh ( vt ) + v sinh t sinh ( vt )cosh t + 1 (cid:19) e − x (cosh t − dt + λ Z ∞ cosh ( vt ) e − x (cosh t − dt = Z ∞ ( λ − h v ( t )) cosh ( vt ) e − x (cosh t − dt = Z ∞ ( λ − h v ( t ( s ))) cosh ( vt ( s ))sinh ( t ( s )) e − xs ds, where h v ( t ) is defined by (5.5) and t ( s ) = cosh − ( s + 1). By Bernstein’s theoremand Lemma 4 P λ ( x ) is CM on (0 , ∞ ) if and only if λ ≥ sup t> ( h v ( t )) = v if v ∈ [1 , ∞ ) ,θ v if v ∈ (cid:0) , (cid:1) , if v ∈ (cid:0) , (cid:1) , and so is − P λ ( x ) if and only if λ ≤ inf t> ( h v ( t )) = if v ∈ [1 , ∞ ) , if v ∈ (cid:0) , (cid:1) ,θ v if v ∈ (cid:0) , (cid:1) , which completes the proof. (cid:3) Remark 3.
It was proved in [45, Theorem 5] that the function x
7→ √ xe x K v ( x ) is CM on (0 , ∞ ) if | v | > / . Now we present a new proof by Corollary 2. Differ-entiation yields − (cid:2) √ xe x K v ( x ) (cid:3) ′ = − √ x (cid:20)(cid:18) x + 12 (cid:19) e x K v ( x ) + xe x K ′ v ( x ) (cid:21) = 1 √ x (cid:2) − P / ( x ) (cid:3) . Since x / √ x and − P / ( x ) are CM on (0 , ∞ ) by Corollary 2, we find that sois x − / (cid:2) − P / ( x ) (cid:3) , and so is √ xe x K v ( x ) on (0 , ∞ ) . Remark 4.
Baricz [51] conjectured that x
7→ √ xe x K v ( x ) for all | v | < / is aBernstein function, which was proved in [43, Remark 3.3] . By Corollary 2, we cangive a simple proof. Indeed, it suffices to prove [ √ xe x K v ( x )] ′ for v ∈ (0 , / isCM on (0 , ∞ ) . Differentiation yields (cid:2) √ xe x K v ( x ) (cid:3) ′ = 1 √ x (cid:20)(cid:18) x + 12 (cid:19) e x K v ( x ) + xe x K ′ v ( x ) (cid:21) = 1 √ x P / ( x ) . Since the function x / √ x is CM on (0 , ∞ ) , while P / ( x ) is CM on (0 , ∞ ) inview of Corollary 2, it then follows from [45, Theorem 1] that so is x − / P / ( x ) on (0 , ∞ ) . We note that (5.3) can be written as1 + 1 x − K v − ( x ) K v +1 ( x ) K v ( x ) = 1 x (cid:18) x + xK ′ v ( x ) K v ( x ) (cid:19) ′ = 1 x Λ ′ ( x ) , which, by Λ ′ ( x ) > ( <
0) if | v | > ( < ) 1 / ONOTONICITY RULES FOR THE RATIO OF TWO LAPLACE TRANSFORMS 21
Corollary 3.
Let v ∈ R with | v | 6 = 1 / . Then the following inequality K v − ( x ) K v +1 ( x ) K v ( x ) < ( > ) 1 + 1 x , or equivalently, (5.12) K v ( x ) − K v − ( x ) K v +1 ( x ) > ( < ) − x K v ( x ) holds for x > if | v | > ( < ) 1 / . Remark 5.
The inequality (5.12) is the Tur´an type inequality for modified Besselfunctions of the second kind, which first appeared in [52] . More such inequalitiescan be found in [53] , [54] , [55] , [42] , [56] , [57] , [58] [52] , [43] . Finally, we give an improvement of the double inequality (5.4).
Corollary 4.
Let v ∈ R with | v | 6 = 1 / . Then the following inequality (5.13) − s x + x + max (cid:18) | v | , (cid:19) < xK ′ v ( x ) K v ( x ) < − s x + x + min (cid:18) | v | , (cid:19) holds for x > .Proof. The desired inequalities are equivalent to − p x + x + v < xK ′ v ( x ) K v ( x ) < − x −
12 if | v | > , − x − < xK ′ v ( x ) K v ( x ) < − p x + x + v if | v | < . By the double inequality (5.4), it suffices to prove(5.14) xK ′ v ( x ) K v ( x ) > ( < ) − p x + x + v if | v | > ( < ) 1 / . To this end, we use the recurrence relations (see [2, p. 79]) xK ′ v ( x ) K v ( x ) + v = − xK v − ( x ) K v ( x ) , (5.15) xK ′ v ( x ) K v ( x ) − v = − xK v +1 ( x ) K v ( x )(5.16)to get the identity (cid:18) xK ′ v ( x ) K v ( x ) (cid:19) − v = x K v − ( x ) K v +1 ( x ) K v ( x ) . This in combination with the identity (5.3) yields x + x + v − (cid:18) xK ′ v ( x ) K v ( x ) (cid:19) = x (cid:18) x + xK ′ v ( x ) K v ( x ) (cid:19) ′ = x Λ ′ ( x ) . Using the inequality Λ ′ ( x ) > ( < ) 0 if | v | > ( < ) 1 / K v ( x ) > K ′ v ( x ) <
0, the inequality (5.14) follows. This completes theproof. (cid:3)
Remark 6.
The bound −√ x + x + v for xK ′ v ( x ) /K v ( x ) given in inequality 5.14is better than some known ones, which refer to [52, p. 242–243] . While the anotherone − x − / seems to be a new and simple one. Conclusions
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Forum Math. (2014),no. 1, 295–322. Zhen-Hang Yang, College of Science and Technology, North China Electric PowerUniversity, Baoding, Hebei Province, 071051, P. R. China and Department of Scienceand Technology, State Grid Zhejiang Electric Power Company Research Institute,Hangzhou, Zhejiang, 310014, China
E-mail address : [email protected] Jing-Feng Tian, College of Science and Technology, North China Electric PowerUniversity, Baoding, Hebei Province, 071051, P. R. China
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