Monotonicity rules for the ratio of two Laplace transforms with applications

J. Math. Anal. Appl. 470 (2019) 821–845

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Monotonicity rules for the ratio of two Laplace transforms with applications Zhenhang Yang, Jing-Feng Tian ∗ College of Science and Technology, North China Electric Power University, Baoding, Hebei Province, 071051, PR China

a r t i c l e

i n f o

Article history: Received 27 March 2018 Available online 16 October 2018 Submitted by J.A. Ball Keywords: Laplace transform Monotonicity rule Psi function Modified Bessel functions of the second kind

a b s t r a c t Let f and g be both continuous functions on (0, ∞) with g (t) > 0 for t ∈ (0, ∞) and let F (x) = L (f ), G (x) = L (g) be respectively the Laplace transforms 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 decreasing on (t∗ , ∞), then the ratio F/G is decreasing on (0, ∞) if and only if   HF,G 0+ = lim+ x→0



F  (x) G (x) − F (x) G (x)

 ≥ 0,

with lim

x→0+

F (x) f (t) = lim and t→∞ g (t) G (x)

lim

x→∞

F (x) f (t) = lim+ t→0 g (t) G (x)

provided the indicated limits exist. While HF,G (0+ ) < 0, there is at least one x∗ > 0 such that F/G is increasing on (0, x∗ ) and decreasing on (x∗ , ∞). As applications, a unified treatment for certain bounds of psi function is presented, and some properties of the modified Bessel functions of the second are established. These show that the monotonicity rules in this paper may contribute to study for certain special functions because many special functions can be expressed as corresponding Laplace transforms. © 2018 Elsevier Inc. All rights reserved.

1. Introduction The Laplace transform of a function f (t) defined on [0, ∞) is the function F (s), which is a unilateral transform defined by * Corresponding author. E-mail addresses: [email protected] (Z. Yang), [email protected] (J.-F. Tian). https://doi.org/10.1016/j.jmaa.2018.10.034 0022-247X/© 2018 Elsevier Inc. All rights reserved.

Z. Yang, J.-F. Tian / J. Math. Anal. Appl. 470 (2019) 821–845

822

∞ F (s) =

f (t) e−st dt,

0

where s is a complex number frequency parameter. The Laplace transform of a function f (t) is also denoted by L (f ). It is known that some special functions can be represented as corresponding Laplace transforms, for example, Binet formula for gamma function: 

1 ln Γ (z) − z − 2



1 ln z + z − ln (2π) = 2

∞  0

1 1 1 − + et − 1 t 2



e−zt dt R (z) > 0 t

(see [14, p. 21, Eq. (5)]); the integral representation of the modified Bessel functions of the second (see [36, p. 181]) ∞ Kv (x) =

e−x cosh t cosh (vt) dt,

(1.1)

0

which, by replacing cosh t − 1 with t, can be expressed as Kv (x) = e−x

∞

e−xt

0

cosh (varccosh (t + 1))  dt; t (t + 2)

the Gaussian Q-function [31] defined by 1 Q (x) = √ 2π

∞

e−t

2

/2

dt for x > 0,

x

which, by a change of variable t = u + x, is represented as 2 1 Q (x) = √ e−x /2 2π

∞

e−u

2

/2 −ux

e

du.

0

More examples can be found in [22], [23]. An important notion related to the Laplace transform is the completely monotonic 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)n F (n) (x) ≥ 0 for all x ∈ I and n = 0, 1, 2, ....

(1.2)

If the inequality (1.2) is strict, then F is said to be strictly completely monotonic on I. The classical Bernstein’s theorem [9], [37] states that a function F is completely monotonic (for short, CM) on (0, ∞) if and only if it is a Laplace transform of some nonnegative measure μ, that is, ∞ F (x) =

e−xt dμ (t) ,

0

where μ (t) is non-decreasing and the integral converges for 0 < x < ∞.

Z. Yang, J.-F. Tian / J. Math. Anal. Appl. 470 (2019) 821–845

823

Another important one is the Bernstein functions [29]. A non-negative function F is said to be a Bernstein function on an interval I, if F has derivatives of all orders on I and satisfies (−1)n F (n) (x) ≤ 0 for all x ∈ I and n = 0, 1, 2, ....

(1.3)

Clearly, a function F is a Bernstein function on I if and only if F  is CM on I. Very recently, Yang and Tian [49] established a monotonicity rule for the ratio of two Laplace transforms as follows. Theorem A. Let the functions f, g be defined on (0, ∞) such that their Laplace transforms L (f ) = ∞ ∞ f (t) e−xt dt and L (g) = 0 g (t) e−xt dt exist with g (t) = 0 for all t > 0. Then the ratio L (f ) /L (g) is 0 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 →

1 120 2 √   x − 7 24x ln Γ (x + 1/2) − x ln x + x − ln 2π + 1

is strictly increasing from (0, ∞) onto (1, 1860/343). In another paper [39], this monotonicity rule was applied to investigate the monotonicity of the function 2

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 some new properties of polygamma functions. These show that Theorem A is an efficient tool of studying special functions. Moreover, if g (t) > 0 for all t > 0, then by Bernstein’s theorem, both of the functions x → L (f ) − βL (g) and x → αL (g) − L (f ) are CM on (0, ∞), where β = inf x>0 (f /g) > −∞ and α = supx>0 (f /g) < ∞. Inspired by the above comments, the aim of this paper is to further establish the monotonicity rule of the ratio of two Laplace transforms L (f ) /L (g) under the condition that there is a t∗ > 0 such that 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 are given, which containing monotonicity rules for ratios of two power series (polynomials). In Section 3, our main results (Theorems 1–3) are proved by means of definition of integral and lemmas. As applications, two monotonicity results involving psi function and modified Bessel functions of the second kind are presented. 2. Lemmas To state needed lemmas, we recall a useful auxiliary function Hf,g , which was introduced in [40]. For −∞ ≤ a < b ≤ ∞, let f and g be differentiable functions on (a, b) with g  = 0 on (a, b). Then we define Hf,g :=

f g − f. g

(2.1)

The auxiliary function Hf,g has the following well properties [40, Property 1]: (i) Hf,g is even with respect to g and odd with respect to f , that is, Hf,g (x) = Hf,−g (x) = −H−f,g (x) = −H−f,−g (x) .

(2.2)

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(ii) If g = 0 on (a, b), then   f g = 2 Hf,g , g g

(2.3)

  f sgn = sgn (g  ) sgn (Hf,g ) . g

(2.4)

and therefore,

(iii) If f and g are twice differentiable on (a, b), then  Hf,g

 =

f g

 g.

(2.5)

The auxiliary function Hf,g and its properties are very helpful to investigate those monotonicities of ratios of two functions, see [20], [21], [40], [42], [43], [50], [51], [55], [56], [57]. Recently, in [45] they were successfully applied to establish monotonicity rules for ratios of two power series and of two polynomials under the condition that the ratio of coefficients of two power series is increasing (decreasing) then decreasing (increasing). The following monotonicity rule for the ratio of two polynomials will be used in the proof of our main results. n n Lemma 1 ([45, Theorem 2.5]). Let An (t) = k=0 ak tk and Bn (t) = k=0 bk tk be two real polynomials defined on (0, r) (r > 0) with bk > 0 for all 0 ≤ k ≤ n. Suppose that for certain m ∈ N with m < n, the sequences {ak /bk }0≤k≤m and {ak /bk }m≤k≤n are both non-constants, and are respectively increasing (decreasing) and decreasing (increasing). Then the function An /Bn is increasing (decreasing) on (0, r) if and only if HAn ,Bn (r− ) ≥ (≤) 0. While HAn ,Bn (r− ) < (>) 0, there is a unique t0 ∈ (0, r) such that the function An /Bn is increasing (decreasing) on (0, t0 ) and decreasing (increasing) on (t0 , r). The following monotonicity rule will be used in Proposition 1, which first appeared in [8, Lemma 6.4] without giving the details of the proof. Two strict proofs were given in [45] and [38]. ∞ ∞ k k Lemma 2 ([45, Corollary 2.6]). Let A (t) = k=0 ak t and B (t) = k=0 bk t be two real power series converging on R with bk > 0 for all k. If for certain m ∈ N, the non-constant sequences {ak /bk }0≤k≤m and {ak /bk }k≥m are respectively increasing (decreasing) and decreasing (increasing), then there is a unique t0 ∈ (0, ∞) such that the function A/B is increasing (decreasing) on (0, t0 ) and decreasing (increasing) on (t0 , ∞). Remark 1. As commented in [52, Remark 2], Theorem 2.1 in [45, Theorem 2.5] cannot be right in the case: one of two sequences {ak /bk }0≤k≤m and {ak /bk }k≥m is a constant. So it should be assumed that both of the sequences {ak /bk }0≤k≤m and {ak /bk }k≥m are non-constant. In the same way, it is needed to make similar assumptions in Corollary 2.3, Theorem 2.5 and Corollary 2.6 in [45]. By the way, the modification pointed out in [58, Remark 2.4] still cannot avoid the case mentioned just now. The following lemma [54, Lemma 2] offers a simple criterion to determine the sign of a class of special series, which will be used in proof of Proposition 2. Lemma 3 ([54, Lemma 2], [48, Lemma 2.1]). Let {ak }∞ k=0 be a nonnegative real sequence with am > 0 and ∞ k=m+1 ak > 0 and let

Z. Yang, J.-F. Tian / J. Math. Anal. Appl. 470 (2019) 821–845

S (t) = −

m

ak t k +

k=0

∞

825

ak t k

k=m+1

be a convergent power series on the interval (0, r) (r > 0). (i) If S (r− ) ≤ 0, then S (t) < 0 for all t ∈ (0, r). (ii) If S (r− ) > 0, then there is a unique t0 ∈ (0, r) such that S (t) < 0 for t ∈ (0, t0 ) and S (t) > 0 for t ∈ (t0 , r). Remark 2. Clearly, when r = ∞ in Lemma 3, there is a unique t0 ∈ (0, ∞) such that S (t) < 0 for t ∈ (0, t0 ) and S (t) > 0 for t ∈ (t0 , ∞). This result appeared in [8, Lemma 6.3] without proof (see also [41], [46]). If ak = 0 for k ≥ n ≥ m + 1, then Lemma 3 is reduced to a polynomial version, which appeared in [44] (see also [51], [53]). 3. Main results We are in a position to state and prove our main results. Theorem 1. For 0 < a < b < ∞, let the functions F and G be defined on (0, ∞) by b F (x) =

f (t) e−xt dt and G (x) =

a

b

g (t) e−xt dt,

a

where the functions f, g are both continuous on [a, b] with g (t) > 0 for t ∈ [a, b]. If there is a 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) is decreasing (increasing) on (0, ∞) if and only if     + F (x) G (x) − F (x) ≥ (≤) 0, HF,G 0 = lim+ x→0 G (x) with b f (t) dt F (x) = ab lim+ and x→0 G (x) g (t) dt a

lim

x→∞

f (a) F (x) = . G (x) g (a)

(3.1)

While HF,G (0+ ) < (>) 0, there is at least one x∗ > 0 such that F/G is increasing (decreasing) on (0, x∗ ) and decreasing (increasing) on (x∗ , ∞). Proof. We only prove this theorem under the condition that f /g is strictly increasing on [a, t∗ ] and strictly decreasing on [t∗ , b]. If f /g is strictly decreasing on [a, t∗ ] and strictly increasing on [t∗ , b], then (−f ) /g is strictly increasing on [a, t∗ ] and strictly decreasing on [t∗ , b], and then corresponding conclusion of this theorem is also true, which suffices to note that H−F,G (x) = −HF,G (x) due to (2.2). For n ∈ N, given a partition of the interval [a, b]: a = t0 < t1 < t2 < · · · < tn = b, with Δti = ti − ti−1 = (b − a) /n, and so ti = a + (b − a) i/n. Then we have n−1

f (ti ) e−xti Δti =

i=0

 i b − a f (ti ) e−ax e−(b−a)x/n n i=0

n−1

b − a −ax

b − a −ax e e f (ti ) y i =: Fn (y) , n n i=0 n−1

=

Z. Yang, J.-F. Tian / J. Math. Anal. Appl. 470 (2019) 821–845

826 n−1

b − a −ax

b − a −ax e e g (ti ) y i := Gn (y) , n n i=0 n−1

g (ti ) e−xti Δti =

i=0

where y ≡ yn (x) = e−(b−a)x/n ∈ (0, 1). These imply that n−1 b

n−1

f (ti ) e−xti Δti i=0 n−1 −xti Δt i i=0 g (ti ) e

= i=0 n−1 i=0

f (ti ) y i g (ti

) yi

=

Fn (y) , Gn (y)

n−1

f (t) e−xt dt limn→∞ i=0 f (ti ) e−xti Δti F (x) Fn (y) = ab . = lim = n−1 −xt n→∞ Gn (y) G (x) limn→∞ i=0 g (ti ) e i Δti g (t) e−xt dt a

(3.2)

Also, we have n−1

(ti − a) f (ti ) e−xti Δti =

i=0

n−1

i=0

= e−ax



n−1

 i b − a (b − a) i f (ti ) e−ax e−(b−a)x/n n n

 2 n−1 2

b−a b−a y if (ti ) y i−1 = e−ax yFn (y) , n n i=0

(ti − a) g (ti ) e−xti Δti = e−ax



i=0

b−a n

2

yGn (y) .

Therefore, we obtain n−1 i=0 n−1

(ti − a) f (ti ) e−xti Δti = (ti − a) g (ti ) e−xti Δti

Fn (y) , Gn (y)

i=0

b

−xt

a

(t − a) f (t) e

a

(t − a) g (t) e−xt dt

b

dt

limn→∞ = limn→∞

n−1 i=0 n−1

(ti − a) f (ti ) e−xti Δti

Fn (y) . n→∞ Gn (y)

= lim (ti − a) g (ti ) e−xti Δti

(3.3)

i=0

Since b

−xt

(t − a) f (t) e

b dt =

a

b

tf (t) e

−xt

b dt − a

a

(t − a) g (t) e−xt dt =

a

b a

f (t) e−xt dt = −F  (x) − aF (x) ,

a

tg (t) e−xt dt − a

b

g (t) e−xt dt = −G (x) − aG (x) ,

a

equation (3.3) can also be written as F  (x) + aF (x) Fn (y) = lim . G (x) + aG (x) n→∞ Gn (y) It then follows that

(3.4)

Z. Yang, J.-F. Tian / J. Math. Anal. Appl. 470 (2019) 821–845 n−1

HFn ,Gn (y) =

Fn (y) Gn Gn (y)

(y) − Fn (y) =

i=0 n−1

(ti − a) f (ti ) e−xti Δti n−1

(ti − a) g (ti ) e−xti Δti

g (ti ) y i −

i=0

827

n−1

f (ti ) y i

i=0

i=0



n−1 n−1

b−a − a) f (ti ) e−xti Δti b − a −ax i e e−ax f (ti ) y i g (ti ) y − n − a) g (ti ) e−xti Δti i=0 n i=0

 n−1 n−1 n−1 −xti ax

Δti

ne −xti −xti i=0 (ti − a) f (ti ) e = g (ti ) e Δti − f (ti ) e Δti n−1 −xti Δt b−a i i=0 i=0 (ti − a) g (ti ) e i=0

neax = b−a

: =

n−1 (ti i=0 n−1 i=0 (ti



neax [n] C (x) , b − a f,g

and so   [n] sgn (HFn ,Gn (y)) = sgn Cf,g (x) .

(3.5)

Clearly, we have b [n] lim C n→∞ f,g

(x) =

a b a

(t − a) f (t) e−xt dt (t − a) g (t) e−xt dt

b g (t) e

−xt

b dt −

a

f (t) e−xt dt

a



=

F (x) + aF (x) G (x) G (x) − F (x) = HF,G (x) , G (x) + aG (x) G (x) + aG (x)

which, in view of G (x) < 0 and b



G (x) + aG (x) = −

(t − a) g (t) e−xt dt < 0

(3.6)

a

for x ∈ (0, ∞), implies that  sgn

 [n] lim Cf,g (x) = sgn (HF,G (x))

n→∞

(3.7)

for x ∈ (0, ∞). (i) The necessity follows from 

F (x) G (x)

 =

G (x)

2 HF,G

G (x)

(x) ≤ 0

for x > 0, which, due to G (x) < 0 for all x ∈ (0, ∞), implies that HF,G (0+ ) ≥ 0. [n] Conversely, if HF,G (0+ ) ≥ 0, then by the relation (3.7), there is a large N ∈ N such that Cf,g (0+ ) ≥ 0 for n > N , which in combination with the relation (3.5) gives that HFn ,Gn (y) ≥ 0 as y → 1− for n > N . On the other hand, since f /g is increasing on [a, t∗ ] and decreasing on [t∗ , b], we easily see that there is an i0 ≥ 1 such that the sequence {f (ti ) /g (ti )} is strictly increasing for 0 ≤ i ≤ i0 and decreasing for i0 < i ≤ n − 1. By Lemma 1, the ratio Fn (y) /Gn (y) is strictly increasing with respect to y on (0, 1), that is, G (y) d Fn (y) = n dy Gn (y) Gn (y)



Fn (y) Fn (y) −  Gn (y) Gn (y)

 > 0 for n > N and y ∈ (0, 1) ,

Z. Yang, J.-F. Tian / J. Math. Anal. Appl. 470 (2019) 821–845

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which, due to Gn (y) , Gn (y) > 0, yields Fn (y) Fn (y) − > 0 for n > N and y ∈ (0, 1) . Gn (y) Gn (y) This together with (3.2) and (3.4) gives F  (x) + aF (x) F (x) − ≥ 0 for x ∈ (0, ∞) , G (x) + aG (x) G (x) which indicates that 

F (x) G (x)

 =

G (x) + aG (x) G (x)



F  (x) + aF (x) F (x) − G (x) + aG (x) G (x)

 ≤ 0 for x ∈ (0, ∞) ,

where the inequality holds due to G (x) > 0 and G (x) + aG (x) < 0 by (3.6). This proves the sufficiency. The first limit of (3.1) is clear. While the second follows from (3.2), which implies that F (x) Fn (y) Fn (y) = lim lim = lim lim x→∞ G (x) x→∞ n→∞ Gn (y) n→∞ x→∞ Gn (y) lim

= lim lim

n→∞ y→0

f (t0 ) f (a) Fn (y) = = . Gn (y) g (t0 ) g (a)

(ii) If HF,G (0+ ) < 0, by the relations (3.7) and (3.5), there is a large enough N ∈ N such that [n] HFn ,Gn (y) < 0 as y → 1− for n > N . By Lemma 1, there  is a unique y0 ∈(0, 1) for given n > N [n]

such that the function Fn (y) /Gn (y) is increasing on 0, y0 G (y) d Fn (y) = n dy Gn (y) Gn (y) G (y) d Fn (y) = n dy Gn (y) Gn (y)

 

Fn (y) Fn (y) −  Gn (y) Gn (y) Fn (y) Fn (y) − Gn (y) Gn (y)

[n]

and decreasing on y0 , 1 , that is,



  [n] > 0 for n > N and y ∈ 0, y0 ,



  [n] < 0 for n > N and y ∈ y0 , 1 , 

[n]

where y0 is the unique solution of the equation [Fn (y) /Gn (y)] = 0 on (0, 1), namely, 

d Fn (y) dy Gn (y)

 [n]

y=y0

 ⎛    ⎞ [n] [n] [n] Fn y0 Fn y0 Gn y0  ⎝  −   ⎠ = 0 for n > N. = [n] [n] [n] Gn y0 Gn y0 Gn y0

In the same treatment as part (i) of the proof of this theorem, the above three relations imply that  

F (x) G (x) F (x) G (x)

 

≤ 0 for x ∈ (x∗ , ∞) , ≥ 0 for x ∈ (0, x∗ ) ,

  [n] [n] [n] where x∗ = limn→∞ x0 , x0 = −n ln y0 / (b − a), satisfies 

F (x) G (x)

  = 0. x=x∗

Z. Yang, J.-F. Tian / J. Math. Anal. Appl. 470 (2019) 821–845

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Thus it remains to prove x∗ = limn→∞ x0 = 0, ∞. First, we claim that x∗ = limn→∞ x0 = 0. If not, that [n] is, limn→∞ x0 = 0, then F (x) /G (x) is decreasing in x on (0, ∞). This, by part (i) of this theorem, implies + that HF,G (0 ) ≥ 0, which yields a contraction with the assumption that HF,G (0+ ) < 0. [n] [n] Second, we also claim that x∗ = limn→∞ x0 = ∞. If not, that is, limn→∞ x0 = ∞, then F (x) /G (x) is increasing in x on (0, ∞). It then follows that for all x > 0, [n]

[n]

f (a) F (x) F (x) < lim = . G (x) x→∞ G (x) g (a) Since limn→∞ (Fn (y) /Gn (y)) = F (x) /G (x), there exists a large enough N1 ∈ N such that for n > N1 the inequality f (a) Fn (y) < Gn (y) g (a)

(3.8)

holds for on  hand, as shown just now, the function Fn (y) /Gn (y)  is increasing   all y ∈ (0, 1). On the  other  [n] [n] [n] 0, y0 and decreasing on y0 , 1 , which suggests that there exists a small enough δ ∈ 0, y0 such that Fn (y) /Gn (y) is increasing on (0, δ). Therefore, for y ∈ (0, δ) and n > N , Fn (0) f (t0 ) f (a) Fn (y) ≥ = = . Gn (y) Gn (0) g (t0 ) g (a) This is in contradiction with the inequality (3.8) for all y ∈ (0, 1) and n > N1 . [n] Consequently, x∗ = limn→∞ x0 = 0, ∞, which ends the proof. 2 Letting a → 0+ , b → ∞ in Theorem 1. Then F (x) and G (x) are Laplace transforms of the functions f and g, respectively. By properties of uniformly convergent improper integral with a parameter, we have the following monotonicity rule, where the first limit of (3.9) follows from the first one of (3.1) and Cauchy mean value theorem, that is,  s   s  b f (t) dt s=b − a f (t) dt s=a f (t) dt F (x) a a     = lim  b lim = lim s s b→∞ x→0+ G (x) g (t) dt b→∞ a g (t) dt s=b − a g (t) dt s=a a = lim

b→∞

f (a + θ (b − a)) f (t) = lim , t→∞ g (a + θ (b − a)) g (t)

here θ ∈ (0, 1). Theorem 2. Let f and g be both continuous functions on (0, ∞) with g (t) > 0 for t ∈ (0, ∞) and let F (x) = L (f ) and G (x) = L (g) converge for x > 0. 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 function F/G is decreasing (increasing) on (0, ∞) if and only if   HF,G 0+ = lim+ x→0





F  (x) G (x) − F (x) G (x)

≥ (≤) 0,

with lim+

x→0

F (x) f (t) = lim and G (x) t→∞ g (t)

F (x) f (t) = lim+ x→∞ G (x) t→0 g (t) lim

(3.9)

provided the indicated limits exist. While HF,G (0+ ) < (>) 0, there is at least one x∗ > 0 such that F/G is increasing (decreasing) on (0, x∗ ) and decreasing (increasing) on (x∗ , ∞).

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830

Theorem 3. Suppose that (i) both of the functions f and g are continuous on (0, ∞) with g (t) > 0 for t ∈ (0, ∞); (ii) the function μ is positive, differentiable and increasing from (0, ∞) onto (μ (0+ ) , ∞); (iii) both the functions ∞ F (x) =

f (t) e

−xμ(t)

∞ dt and G (x) =

0

g (t) e−xμ(t) dt

0

converge for all x > 0. 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→0+

F (x) f (t) = lim and G (x) t→∞ g (t)

lim

x→∞

F (x) f (t) = lim . G (x) t→0+ 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 is decreasing (increasing) on (0, ∞) if and only if   HF,G 0+ = lim+ x→0



 F  (x) G (x) − F (x) ≥ (≤) 0. G (x)

While HF,G (0+ ) < (>) 0, there is at least one x∗ > 0 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 −xa

∞

F (x) = e

f (t (s)) −xs e ds and G (x) = e−xa μ (t (s))

0

∞

g (t (s)) −xs e ds, μ (t (s))

0

where t (s) = μ−1 (s + a), and F (x) /G (x) can be represented in the form of the ratio of two Laplace transforms: ∞ ∞ ∗ [f (t (s)) /μ (t (s))] e−xs ds f (s) e−xs ds exa F (x) F (x) 0 0∞ = xa = ∞ := . G (x) e G (x) [g (t (s)) /μ (t (s))] e−xs ds g ∗ (s) e−xs ds 0 0 It is easy to verify that f (t (s)) f ∗ (s) = , g ∗ (s) g (t (s))



f ∗ (s) g ∗ (s)



 =

f (t) g (t)

 ×

1 dt =  ds μ (t)



f (t) g (t)

 ,

(3.10)

where μ (t) > 0 for all t > 0. (i) If the ratio f (t) /g (t) is increasing (decreasing) on (0, ∞), then so is f ∗ (s) /g ∗ (s). By Theorem A, we easily find that (exa F ) / (exa G) = F/G is decreasing (increasing) on (0, ∞). By the limit relations (3.9) we easily get lim+

x→0

F (x) f (t (s)) /μ (t (s)) f (t) = lim = lim , G (x) s→∞ g (t (s)) /μ (t (s)) t→∞ g (t)

F (x) f (t (s)) /μ (t (s)) f (t) = lim+ = lim+ .  x→∞ G (x) s→0 g (t (s)) /μ (t (s)) t→0 g (t) lim

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(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 relation of (3.10), there is an s∗ ∈ (0, ∞) such that f ∗ (s) /g ∗ (s) is strictly increasing (decreasing) on (0, s∗ ) and strictly decreasing (increasing) on (s∗ , ∞), where s∗ = μ (t∗ ) − a. By Theorem 2, the ratio exa F/ (exa G) = F/G is decreasing (increasing) on (0, ∞) if and only if  lim Hexa F,exa G (x) = lim+

x→0+

x→0





(exa F (x)) xa xa  e G (x) − e F (x) (exa G (x))

≥ (≤) 0.

(3.11)

We claim that the limit relation is equivalent to limx→0+ HF,G (x) ≥ (≤) 0. In fact, we easily check that Hexa F,exa G (x) =

exa (F  (x) + aF (x)) xa e G (x) − exa F (x) exa (G (x) + aG (x))

F  (x) G (x) + aF (x) G (x) − F (x) G (x) − aF (x) G (x) G (x) + aG (x)   exa G (x) exa G (x) F  (x) G (x) − F (x) =  HF,G (x) . =   G (x) + aG (x) G (x) G (x) + aG (x) = exa

This together with ∞



G (x) = −

μ (t) g (t) e−xμ(t) dt < 0,

0

G (x) + aG (x) = −

∞



  μ (t) − μ 0+ g (t) e−xμ(t) dt < 0

0

for x > 0 indicates that sgn (Hexa F,exa G (x)) = sgn (HF,G (x)) , which proves the claim just now. (iii) If limx→0+ HF,G (x) < (>) 0, then limx→0+ Hexa F,exa G (x) < (>) 0. By Theorem 2, there is at least one x∗ > 0 such that exa F (x) / (exa G (x)) = F (x) /G (x) is increasing (decreasing) on (0, x∗ ) and decreasing (increasing) on (x∗ , ∞). Thus we complete the proof. 2 4. A unified treatment for certain bounds of harmonic number The Euler–Mascheroni constant is defined by γ = lim (Hn − ln n) = 0.577215664..., n→∞

n where Hn = k=1 k−1 is the n’th harmonic number. There is a close connection between Hn and the psi (or digamma) function. Indeed, we have Hn = ψ (n + 1) + γ. Several bounds for Hn or ψ (n + 1) can see [2], [11], [12], [13], [19], [24], [25], [27], [32], [34], [35], [47], [60], [61]. In particular, Alzer [2] obtained the double inequality, 1 1 ≤ Hn − ln n − γ < 2 (n + a) 2 (n + b)

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holds for n ∈ N with the best constants a=

1 1 − 1 and b = 2 (1 − γ) 6

by proving the sequence A (n) =

1 1 −n 2 ψ (n + 1) − ln n

(4.1)

is strictly decreasing for n ∈ N. Villarino [35] showed that Hn = ln



n (n + 1) + γ +

1 , 6n (n + 1) + L (n)

  1 1 = ln n + +γ+ , 2 2 24 (n + 1/2) + D (n)

(4.2) (4.3)

where both of the sequences L (n) and D (n) are increasing for n ∈ N. Qi [27] showed that the sequence Q (n) =

1 1 − 12n2 2 ln n + 1/ (2n) − ψ (n + 1)

(4.4)

is strictly increasing for n ∈ N. These monotonicities of the sequences L(n), D(n) and Q(n) yield similarly corresponding sharp bounds for Hn 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 as ratios of two Laplace transforms, then we easily prove their monotonicity on (0, ∞) by Theorems A and 2. Here we choose Φ (x) = D (x − 1/2) defined by (4.3) and prove its monotonicity on (0, ∞). As far as A (x), L (x) and Q (x), we only list their expressions in the form of ratios of Laplace transforms. In fact, by means of the formulas   ∞  e−t  ∞ e−t −e−xt  ∞ n−1 −xt e−xt 1 ψ(x) = 0 dt, x1n = (n−1)! t e dt, t − 1−e−t dt, ln x = 0 t 0 we have 1 1 −x= A (x) = 2 ψ (x + 1) − ln x

∞ [−p1 (t)] e−xt dt 0 , ∞ p1 (t) e−xt dt 0

where  p1 (t) = 2

1 1 − t t e −1

2 L (x) = − 6x (x + 1) = 2ψ (x + 1) − ln (x (x + 1))

 ; ∞ 0

[−6 (p2 (t) + p2 (t))] e−xt dt ∞ , p2 (t) e−xt dt 0

where e2t − 2tet − 1 ; t (et − 1) et ∞ [−12p3 (t)] e−xt dt 1 2 , − 12x = 0  ∞ Q (x) = ln x + 1/ (2x) − ψ (x + 1) p3 (t) e−xt dt 0 p2 (t) =

where

Z. Yang, J.-F. Tian / J. Math. Anal. Appl. 470 (2019) 821–845

1 p3 (t) = 2

833

  t 2 coth − . 2 t

Next we prove the monotonicity of Φ (x) = D (x − 1/2) on (0, ∞) by Theorem 2. Proposition 1. The function Φ (x) =

1 − 24x2 ψ (x + 1/2) − ln x

is strictly increasing from (0, ∞) onto (0, 21/5). Proof. We write Φ (x) = F (x) /G (x), where     1 − ln x , F (x) = 1 − 24x2 ψ x + 2

  1 G (x) = ψ x + − ln x. 2

It has been shown in [46] that ∞ G (x) =

−xt

q (t) e

1 + x G (x) = 24

∞

2

dt,

0

e−xt q  (t) dt,

0

where q (t) =

1 1 − . t 2 sinh (t/2)

(4.5)

Then F (x) can be written as ∞ F (x) = 1 − 24x G (x) = 2

(−24q  (t)) e−xt dt.

0

Thus Φ (x) is expressed as F (x) = Φ (x) = G (x)

∞ [−24q  (t)] e−xt dt 0  . ∞ q (t) e−xt dt 0

We first show that there is a t∗ > 0 such that the function −24q  /q is decreasing on (0, t∗ ) and increasing on (t∗ , ∞). Direct computations give 1 2 cosh2 s sinh s − s3 cosh2 s − 2 sinh s − s3 q  (t) = q (t) 4 s2 (s − sinh s) sinh2 s =

1 sinh 3s − s3 cosh 2s − 3 sinh s − 3s3 , 2 s2 (sinh 3s − 2s cosh 2s − 3 sinh s + 2s)

where s = t/2. Expanding in power series yields  n−4 ∞ 1 n=4 an s2 q  (t) = ∞ , q (t) 2 n=4 bn (s2 )n−4 where

Z. Yang, J.-F. Tian / J. Math. Anal. Appl. 470 (2019) 821–845

834

an =

32n−1 − (2n − 1) (2n − 2) (2n − 3) 22n−4 − 3 , (2n − 1)!

bn =

32n−3 − (2n − 3) 22n−3 − 3 . (2n − 3)!

Since (2n − 1)!bn+1 − 9 (2n − 3)!bn = (10n − 23) 22n−3 + 24 > 0 for n ≥ 3 and b3 = 0, we see that bn > 0 for n ≥ 4. Thus, to prove the function −24q  /q is decreasing on (0, t∗ ) and increasing on (t∗ , ∞), by Lemma 2 it suffices to prove that there is an integer n0 > 4 such that the sequence {an /bn }n≥4 is increasing for 4 ≤ n ≤ n0 and decreasing for n > n0 . For this end, we have to prove 2 dn = [(2n − 1)!] (an bn+1 − an+1 bn ) < 0 for 4 ≤ n ≤ 9 and dn > 0 for n ≥ 10. Some elementary computations give 2 (4n − 1) 34n 9   1 (2n − 1) 20n4 − 56n3 − 77n2 + 464n − 243 62n − 108    4 3 + 64n2 − 132n + 41 32n − (2n − 1) (2n − 3) 6n3 + 5n2 + 2n − 1 22n . 9 4

dn = 18 (4n − 1) +

We find that d4 = −66 802 176, d6 = −1570 251 361 536, d8 = −7526 731 991 528 448, d5 = −13 774 616 064, d7 = −127 269 822 161 664, d9 = −240 861 038 835 686 400. To prove dn > 0 for n ≥ 10, we write dn as   dn = 18 (4n − 1) + (4n − 1) 62n × a∗n + 64n2 − 132n + 41 22n × b∗n , where  2n 3 − 2  2n 4 3 b∗n = − 9 2

a∗n

2 = 9

  1 (2n − 1) 20n4 − 56n3 − 77n2 + 464n − 243 , 108 4n − 1   3 (2n − 1) (2n − 3) 6n3 + 5n2 + 2n − 1 . 4 64n2 − 132n + 41

It is easy to verify that a∗10 =

710 697 141 > 0, 6815 744

b∗10 =

174 443 916 097 > 0, 149 159 936

and for n ≥ 10,    9 1 2n2 + n − 9 400n4 − 2500n3 + 1448n2 + 1789n − 777 > 0, a∗n+1 − a∗n = 4 432 (4n − 1) (4n + 3) 2n − 1 9 3 b∗n+1 − b∗n = 4 16 (64n2 − 4n − 27) (64n2 − 132n + 41)     × 4n4 960n2 − 3004n − 1181 + 19 204n3 + 16 517n2 − 684n − 2697 > 0, which yields dn > 0 for n ≥ 10.

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Second, it is easy to see that lim F  (x) = 0,

lim F (x) = 1,

x→0+

x→0+

and lim

x→0+

G (x) ψ (x + 1/2) − ln x = lim = 0, G (x) x→0+ ψ  (x + 1/2) − 1/x

which yields  lim+ HF,G (x) = lim+

x→0

x→0



F  (x) G (x) − F (x) G (x)

= −1 < 0.

It then follows by Theorem 2 that the function Φ = F/G is strictly increasing on (0, ∞). An easy computation gives −24q  (t) = 0 and t→∞ q (t)

lim+ Φ (x) = lim

x→0

21 −24q  (t) = , t→0 q (t) 5

lim Φ (x) = lim

x→∞

which completes the proof. 2 Remark 3. Using the piecewise monotonicity of −24q  /q on (0, t∗ ) and (t∗ , ∞), we obtain β0 :=

  −24q  (t) 21 −24q  (t∗ ) −24q  (t) −24q  (t) ≤ < max lim , lim = := α0 . t→∞ t→0 q (t∗ ) q (t) q (t) q (t) 5 

Solving the equation [−24q  (t) /q (t)] = 0 on (0, ∞) yields t∗ ≈ 13.53015819, and β0 ≈ −0.171 242 27. Taking into account the proof of Proposition 1, we immediately find that both of the functions     1 ψ x+ − ln x − 1, 2       1 2 F (x) − β0 G (x) = 1 − 24x + β0 ψ x + − ln x 2 

α0 G (x) − F (x) =

24x2 +

21 5

are CM on (0, ∞) (see [46, Theorem 2]). 5. An application to Bessel functions The modified Bessel functions of the second kind Kv is defined as [36, p. 78] Kv (x) =

π I−v (x) − Iv (x) , 2 sin (vπ)

(5.1)

where Iv (x) is the modified Bessel functions of the first kind which can be represented by the infinite series as Iv (x) =

∞

2n+v

(x/2) , x ∈ R, v ∈ R\{−1, −2, ...}, n!Γ (v + n + 1) n=0

and the right-hand side of (5.1) is replaced by its limiting value if v is an integer or zero.

(5.2)

836

Z. Yang, J.-F. Tian / J. Math. Anal. Appl. 470 (2019) 821–845

We easily see that Kv (x) = K−v (x) for all v ∈ R and x > 0 by (5.1), so we assume that v ≥ 0 in this section, unless otherwise specified. As shown in the proof of Theorem 3.1 in [3], the identity 1−

Kv−1 (x) Kv+1 (x) 2

Kv (x)

=

1 x



xKv (x) Kv (x)

 (5.3)

holds for v ∈ R and x > 0, and the function x →

xKv (x) Kv (x)

is strictly decreasing on (0, ∞) for all v ∈ R (see also [59]). As another application, in this section we will determine the monotonicity of the function x → x + xKv (x) /Kv (x) on (0, ∞) by Theorem 3. More precisely, we have Proposition 2. For v ≥ 0, let Kv (x) be the modified Bessel functions of the second kind. (i) If v ∈ (1/2, ∞), then the function x → Λ (x) = x +

xKv (x) Kv (x)

is strictly increasing from (0, ∞) onto (−v, −1/2). (ii) If v ∈ [0, 1/2), then the function x → Λ (x) is strictly decreasing from (0, ∞) onto (−1/2, −v). (iii) If v = 1/2, then the double inequality 

1 −x − max v, 2



  xKv (x) 1 < < −x − min v, Kv (x) 2

(5.4)

holds for x > 0 with the best constants min (v, 1/2) and max (v, 1/2). Before proving Proposition 2, we give the following lemmas. Lemma 4. For v > 0 with v = 1/2, let the function hv be defined on (0, ∞) by hv (t) =

cosh (vt) + v sinh (vt) sinh t . (cosh t + 1) cosh (vt)

(5.5)

(i) If v ∈ [1, ∞) then hv is increasing from (0, ∞) onto (1/2, v). (ii) If v ∈ (1/2, 1), then there is a t∗ ∈ (0, ∞) such that hv is increasing on (0, t∗ ) and decreasing on (t∗ , ∞). Consequently, the inequalities 1 = min 2



1 ,v 2

 < hv (t) ≤ θv

hold for x > 0, where θv = hv (t∗ ), here t∗ is the unique solution of the equation hv (t) = 0 on (0, ∞). (iii) If v ∈ (0, 1/2), then there is a t∗ ∈ (0, ∞) such that hv is decreasing on (0, t∗ ) and increasing on ∗ (t , ∞). Therefore, it holds that for x > 0,  θv ≤ hv (t) < max where θv is as in (ii).

1 ,v 2

 =

1 , 2

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Proof. Differentiating and simplifying yield hv (t) =

rv (t) 2

(cosh t + 1) cosh2 (vt)

,

where rv (t) = v 2 (1 + cosh t) sinh t + v (1 + cosh t) cosh (vt) sinh (vt) − cosh2 (vt) sinh t. Using “product into sum” formulas and expanding in power series gives 4rv (t) = (v + 1) sinh (2vt − t) + (v − 1) sinh (2vt + t)   +2v sinh (2vt) + 2v 2 sinh (2t) + 2 2v 2 − 1 sinh t : =

∞

an t2n−1 , (2n − 1)! n=1

where 2n−1

an = (v + 1) (2v − 1)

+ (v − 1) (2v + 1)

2n−1

+ (2v)

2n

  + v 2 22n + 2 2v 2 − 1 .

To confirm the monotonicity of hv , we need to deal with the sign of an . For this end, we first give two recurrence relations. It is easy to check that 2n−1 2n−1     an+1 − an = 2v v 2 − 1 v − 12 + 2v v 2 − 1 v + 12 22n   + 4v 2 − 1 v 2n + 3v 2 := bn ,  2n−2 2n−2  1 1 bn+1 − bn = (2v − 3) v − + (2v + 3) v + + 4v 2n−1 > 0 v (v 2 − 1) (v 2 − 1/4) 2 2

(5.6)

(5.7)

for v = 1, 1/2. We now distinguish three cases to prove the desired monotonicity. Case 1: v ≥ 1. It is clear that an > 0 for n ≥ 1, which implies that hv (t) > 0 for t ∈ (0, ∞). Case 2: v ∈ (1/2, 1). From the recurrence relation (5.7) we see that the sequence {bn }n≥1 is decreasing, which together with b1 = 2v 2 (2v − 1) (2v + 1) > 0,   lim bn = sgn v 2 − 1 ∞ < 0

n→∞

yields that there is an integer n1 > 1 such that bn ≥ 0 for 1 ≤ n ≤ n1 and bn ≤ 0 for n ≥ n1 . This, by the recurrence relation (5.6), in turn implies that the sequence {an }n≥1 is increasing for 1 ≤ n ≤ n1 and decreasing for n ≥ n1 . Therefore, we obtain an ≥ a1 = 4 (2v − 1) (2v + 1) > 0 for 1 ≤ n ≤ n1 . On the other hand, it is seen that lim

n→∞

an = sgn (v − 1) ∞ < 0. 22n

It then follows that there is an integer n0 > n1 such that an ≥ 0 for 1 ≤ n ≤ n0 and an ≤ 0 for n ≥ n0 . By Lemma 3, there is a t∗ > 0 such that hv (t) > 0 for t ∈ (0, t∗ ) and hv (t) < 0 for t ∈ (t∗ , ∞).

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838

Case 3: v ∈ (0, 1/2). In this case, we see that the sequence {bn }n≥1 is increasing, which in combination with b1 = 2v 2 (2v − 1) (2v + 1) < 0, lim bn = 3v 2 > 0

n→∞

indicates that there is an integer n1 > 1 such that bn ≤ 0 for 1 ≤ n ≤ n1 and bn ≥ 0 for n ≥ n1 . This, by the recurrence relation (5.6), means that the sequence {an }n≥1 is decreasing for 1 ≤ n ≤ n1 and increasing for n ≥ n1 . Hence, we deduce that an ≤ a1 = 4 (2v − 1) (2v + 1) < 0 for 1 ≤ n ≤ n1 . Moreover, it is seen that lim

n→∞

an 2n

(2v + 1)

  = sgn v 2 ∞ > 0.

It then follows that there is an integer n0 > n1 such that an ≤ 0 for 1 ≤ n ≤ n0 and an ≥ 0 for n ≥ n0 . By Lemma 3, there is a t∗ > 0 such that hv (t) < 0 for t ∈ (0, t∗ ) and hv (t) > 0 for t ∈ (t∗ , ∞). This completes the proof. 2 Lemma 5. Let Kv (x) be the modified Bessel functions of the second kind and let F (x) = x [Kv (x) + Kv (x)] and G (x) = Kv (x) . Then for v ∈ (0, 1), we have  lim HF,G (x) = lim+

x→0+

x→0



F  (x) G (x) − F (x) G (x)

= 0.

Proof. By the asymptotic formulas [1, p. 375, (9.6.9)] Kv (x) ∼

 x −v 1 Γ (v) for v > 0 as x → 0, 2 2

we have that for v > 0, as x → 0,  x −v−1 1 Kv (x) ∼ − Γ (v + 1) , 4 2

Kv (x) ∼

 x −v−2 1 Γ (v + 2) . 8 2

Then, for v > 0, as x → 0+ ,  x −v 1  x −v−1  1 F (x) = x [Kv (x) + (x)] ∼ x Γ (v) − Γ (v + 1) 2 2 4 2   x − v x −v = Γ (v) , 2 2 F  (x) = Kv (x) + (x + 1) Kv (x) + xKv (x)  x −v x + 1  x −v−1  x −v−2 1 1 Γ (v + 1) ∼ Γ (v) − + x Γ (v + 2) 2 2 4 2 8 2 2  −v (1 − v) x + v x 1 = Γ (v) , 2 x 2 Kv



(5.8)

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which yields F  (x) G (x) − F (x) G (x) 2  −v 1 x  x −v Γ (v) (1−v)x+v x − v  x −v 1 x 2 Γ (v) ∼ 2 − Γ (v)   −v−1 2 2 2 2 − 14 Γ (v + 1) x2 Γ (v)  x 1−v =− → 0 as x → 0+ if v ∈ (0, 1) . v 2

HF,G (x) =

This completes the proof. 2 We now are in a position to prove Proposition 2. Proof of Proposition 2. We have Λ (x) =

F (x) x [Kv (x) + Kv (x)] = . Kv (x) G (x)

By the integral representation (1.1) we get that ∞ G (x) = Kv (x) =

cosh (vt) e−x cosh t dt;

(5.9)

0

Kv (x) +

Kv

∞ (x) =

cosh (vt) e

−x cosh t

∞ dt −

0

cosh (vt) (cosh t) e−x cosh t dt

0

∞ =−

cosh (vt) (cosh t − 1) e−x cosh t dt,

0

then integration by parts yields

F (x) = x (Kv (x) +

Kv

∞ (x)) =

cosh (vt)

cosh t − 1 −x cosh t de sinh t

0

∞ =−

cosh (vt) + v sinh t sinh (vt) −x cosh t e dt. cosh t + 1

(5.10)

0

Thus Λ (x) can be written as

Λ (x) =

F (x) = G (x)

∞ 0

 ∞ sinh t sinh(vt) − cosh(vt)+v e−x cosh t dt f (t) e−xμ(t) dt cosh t+1 ∞ 0∞ := , cosh (vt) e−x cosh t dt g (t) e−xμ(t) dt 0 0

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) > 0 for t > 0. Also, we have

840

Z. Yang, J.-F. Tian / J. Math. Anal. Appl. 470 (2019) 821–845

− [cosh (vt) + v sinh t sinh (vt)] / (cosh t + 1) f (t) = = −hv (t) . g (t) cosh (vt) (i) If v ≥ 1, then by the first assertion of Lemma 4 we see that f /g is strictly decreasing on (0, ∞). It follows by part (i) of Theorem 3 that Λ = F/G is strictly increasing on (0, ∞). (ii) If v ∈ (1/2, 1), then by the second assertion of Lemma 4 we see that there is a t∗ ∈ (0, ∞) such that f /g is decreasing on (0, t∗ ) and increasing on (t∗ , ∞). Also, limx→0+ HF,G (x) = 0 due to Lemma 5. These, by part (ii) of Theorem 3, yield that Λ = F/G is strictly increasing on (0, ∞). (iii) If v ∈ (0, 1/2), then by the third assertion of Lemma 4 it is seen that there is a t∗ ∈ (0, ∞) such that f /g is increasing on (0, t∗ ) and decreasing on (t∗ , ∞). And, limx→0+ HF,G (x) = 0 due to Lemma 5. It then follows from part (ii) of Theorem 3 that Λ = F/G is strictly decreasing on (0, ∞). (iv) If v = 0, then f (t) /g (t) = −1/ (cosh t + 1) is clearly increasing on (0, ∞). It follows from part (i) of Theorem 3 that Λ = F/G is strictly decreasing on (0, ∞). The limit values are   F (x) cosh (vt) + v sinh t sinh (vt) = − lim = −v, Λ 0+ = lim+ t→∞ x→0 G (x) (cosh t + 1) cosh (vt) Λ (∞) = lim

x→∞

1 F (x) cosh (vt) + v sinh t sinh (vt) = − lim+ =− . t→0 G (x) (cosh t + 1) cosh (vt) 2

The double inequality (5.4) follows from the monotonicity of F/G on (0, ∞), which ends the proof. 2 √ As a consequence of [23, Theorem 5], Miller and Samko showed that the function x → xex Kv (x) is strictly decreasing on (0, ∞) for |v| > 1/2. Yang and Zheng in [59, Corollary 3.2] reproved this assertion and further proved this function is strictly increasing on (0, ∞) for |v| < 1/2. Now we have a more general result by Proposition 2. Corollary 1. For v ≥ 0, let Kv (x) be the modified Bessel functions of the second kind. Then the function x → xr ex Kv (x) is strictly increasing on (0, ∞) if and only if r ≥ max (v, 1/2), and decreasing if and only if r ≤ min (v, 1/2). While v > (<) 1/2 and 1/2 < r < v (v < r < 1/2), there is an x0 > 0 such that this function is decreasing (increasing) on (0, x0 ) and increasing (decreasing) on (x0 , ∞). Proof. Differentiation yields 

[xr ex Kv (x)] = xr ex Kv (x) + rxr−1 ex Kv (x) + xr ex Kv (x)   xKv (x) r−1 x := xr−1 ex Kv (x) × φ (x) . = x e Kv (x) r + x + Kv (x) 

By Proposition 2, [xr ex Kv (x)] ≥ (≤) 0 for all x > 0 if and only if  r ≥ − inf

x>0

xKv (x) x+ Kv (x)



xKv (x) r ≤ − sup x + Kv (x) x>0





v

= 

 =

1 2 1 2

v

  1 if v ∈ (1/2, ∞) , = max v, 2 if v ∈ (0, 1/2)   1 if v ∈ (1/2, ∞) = min v, . 2 if v ∈ (0, 1/2)

When v > 1/2 and 1/2 < r < v, by part (i) of Proposition 2, we see that x → r +x +xKv (x) /Kv (x) = φ (x) is increasing on (0, ∞), which in combination with φ (0+ ) = r − v < 0 and φ (∞) = r − 1/2 > 0 yields that

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there is an x0 > 0 such that φ (x) < 0 for x ∈ (0, x0 ) and φ (x) > 0 for x ∈ (x0 , ∞). That is to say, the function x → xr ex Kv (x) is decreasing on (0, x0 ) and increasing on (x0 , ∞). Similarly, by part (ii) of Proposition 2 we can prove that for v ∈ [0, 1/2) and v < r < 1/2, there is an x0 > 0 such that the function x → xr ex Kv (x) is increasing on (0, x0 ) and decreasing on (x0 , ∞). This completes the proof. 2 Remark 4. Corollary 1 implies that for all 0 < x < y, the double inequality ey−x

 y r1 x

<

 y r2 Kv (x) < ey−x Kv (y) x

(5.11)

if and only if r1 ≤ min (|v| , 1/2) and r2 ≥ max (|v| , 1/2). Obviously, our inequalities (5.11) are superior to those earlier results appeared in [10], [16], [17], [26], [28]. Detailed comments can see [4, Section 3]. By Lemma 4 and Bernstein’s theorem, we give a class of completely monotonic function. Corollary 2. For v ≥ 0, the function Pλ (x) = (x + λ) ex Kv (x) + xex Kv (x) is CM on (0, ∞) if and only if ⎧ ⎪ ⎨ v λ ≥ θv ⎪ ⎩ 1 2

if v ∈ [1, ∞),   if v ∈ 12 , 1 ,   if v ∈ 0, 12 ,

and so is −Pλ (x) if and only if  λ≤

1 2

θv

  if v ∈ 12 , ∞ ,  1 if v ∈ 0, 2 ,

where θv is as in Lemma 4. Proof. From the integral representations (5.10) and (5.9), we have Pλ (x) = ex F (x) + λex G (x)  ∞  ∞ cosh (vt) + v sinh t sinh (vt) −x(cosh t−1) = − e dt + λ cosh (vt) e−x(cosh t−1) dt cosh t + 1 0

0

∞ = 0

(λ − hv (t)) cosh (vt) e−x(cosh t−1) dt =

∞ (λ − hv (t (s)))

cosh (vt (s)) −xs e ds, sinh (t (s))

0

where hv (t) is defined by (5.5) and t (s) = cosh−1 (s + 1). By Bernstein’s theorem and Lemma 4 Pλ (x) is CM on (0, ∞) if and only if ⎧ ⎪ ⎨ v λ ≥ sup (hv (t)) = θv ⎪ t>0 ⎩ 1 2

if v ∈ [1, ∞),   if v ∈ 12 , 1 ,  1 if v ∈ 0, 2 ,

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and so is −Pλ (x) if and only if

λ ≤ inf (hv (t)) = t>0

⎧ ⎪ ⎨

if v ∈ [1, ∞),   if v ∈ 12 , 1 ,  1 if v ∈ 0, 2 ,

1 2 1 2

⎪ ⎩θ

v

which completes the proof. 2 √ Remark 5. It was proven in [23, Theorem 5] that the function x → xex Kv (x) is CM on (0, ∞) if |v| > 1/2. Now we present a new proof by Corollary 2. Differentiation yields −

√



1 xe Kv (x) = − √ x x



1 x+ 2

 x

x

e Kv (x) + xe

Kv



 1  (x) = √ −P1/2 (x) . x

  √ Since x → 1/ x and −P1/2 (x) are CM on (0, ∞) by Corollary 2, we find that so is x−1/2 −P1/2 (x) , and √ x so is xe Kv (x) on (0, ∞). √ Remark 6. Baricz [4] conjectured that x → xex Kv (x) for all |v| < 1/2 is a Bernstein function, which was proved in [59, Remark 3.3]. By Corollary 2, we can give a simple proof. Indeed, it suffices to prove √  [ xex Kv (x)] for v ∈ (0, 1/2) is CM on (0, ∞). Differentiation yields √



1 xe Kv (x) = √ x x

   1 x 1 x  x+ e Kv (x) + xe Kv (x) = √ P1/2 (x) . 2 x

√ Since the function x → 1/ x is CM on (0, ∞), while P1/2 (x) is CM on (0, ∞) in view of Corollary 2, it then follows from [23, Theorem 1] that so is x−1/2 P1/2 (x) on (0, ∞). We note that (5.3) can be written as 1+

1 Kv−1 (x) Kv+1 (x) 1 − = 2 x x Kv (x)

 x+

xKv (x) Kv (x)

 =

1  Λ (x) . x

Since Λ (x) > (< 0) if |v| > (<) 1/2 given in Proposition 2, we derive the following corollary. Corollary 3. Let v ∈ R with |v| = 1/2. Then the following inequality Kv−1 (x) Kv+1 (x) 2

Kv (x)

< (>) 1 +

1 , x

or equivalently, 2

Kv (x) − Kv−1 (x) Kv+1 (x) > (<) −

1 2 Kv (x) x

(5.12)

holds for x > 0 if |v| > (<) 1/2. Remark 7. The inequality (5.12) is the Turán type inequality for the modified Bessel functions of the second kind, which first appeared in [5]. More such inequalities can be found in [3], [5], [6], [7], [15], [18], [30], [33], [59]. Finally, we give an improvement of the double inequality (5.4).

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Corollary 4. Let v ∈ R with |v| = 1/2. Then the following double inequality −

2  1 xKv (x) <− x2 + x + max |v| , < 2 Kv (x)

2  1 x2 + x + min |v| , 2

(5.13)

holds for x > 0. Proof. The desired inequalities are equivalent to  1 1 xKv (x) < −x − if |v| > , − x2 + x + v 2 < Kv (x) 2 2   xKv (x) 1 1 < − x2 + x + v 2 if |v| < . −x − < 2 Kv (x) 2 By the double inequality (5.4), it suffices to prove  xKv (x) > (<) − x2 + x + v 2 if |v| > (<) 1/2. Kv (x)

(5.14)

To this end, we use the recurrence relations (see [36, p. 79]) xKv−1 (x) xKv (x) +v =− , Kv (x) Kv (x)

xKv (x) xKv+1 (x) −v =− Kv (x) Kv (x)

to get the identity 

xKv (x) Kv (x)

2 − v 2 = x2

Kv−1 (x) Kv+1 (x) 2

Kv (x)

.

This in combination with the identity (5.3) yields  x2 + x + v 2 −

xKv (x) Kv (x)

2

  xKv (x) =x x+ = xΛ (x) . Kv (x)

Using the inequality Λ (x) > (<) 0 if |v| > (<) 1/2 due to Proposition 2 and noting Kv (x) > 0 and Kv (x) < 0, the inequality (5.14) follows. This completes the proof. 2 √ Remark 8. The bound − x2 + x + v 2 for xKv (x) /Kv (x) given in inequality (5.14) is better than some known bounds, see [5, p. 242–243]. While the another bound −x − 1/2 seems to be a new and simple one. 6. Conclusions In this paper, by the monotonicity rule for the ratio of two polynomials with the same highest degree (Lemma 1) and definition of integral, we find that the decreasing (increasing) property of three ratios b f (t) e−xt dt F (x) := ab , G (x) g (t) e−xt dt a

∞ f (t) e−xt dt 0∞ , g (t) e−xt dt 0

∞ f (t) e−xμ(t) dt 0∞ , f (t) e−xμ(t) dt 0

under the condition that f /g has the monotonicity pattern that ( ), according as HF,G (0+ ) ≥ (≤) 0. Otherwise, the three ratios mentioned above are unimodal. Since many special functions have integral b representations in the form of a (·) e−xt dt, our three theorems in this paper are efficient tools to research for certain special functions.

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References [1] M. Abramowitz, I.A. Stegun (Eds.), Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables, National Bureau of Standards, Appl. Math. Ser., vol. 55, Dover Publications, New York and Washington, 1972. [2] H. Alzer, Inequalities for the gamma and polygamma functions, Abh. Math. Semin. Univ. Hambg. 68 (1998) 363–372. [3] Á. Baricz, Turán type inequalities for modified Bessel functions, Bull. Aust. Math. Soc. 82 (2) (2010) 254–264. [4] Á. Baricz, Bounds for modified Bessel functions of the first and second kinds, Proc. Edinb. Math. Soc. 53 (2010) 575–599. [5] Á. Baricz, Bounds for Turánians of modified Bessel functions, Expo. Math. 2015 (2) (2015) 223–251. [6] Á. Baricz, T.K. Pogány, Turán determinants of Bessel functions, Forum Math. 26 (1) (2014) 295–322. [7] Á. Baricz, S. Ponnusamy, On Turán type inequalities for modified Bessel functions, Proc. Amer. Math. Soc. 141 (2) (2013) 523–532. [8] F. Belzunce, E. Ortega, J.M. Ruiz, On non-monotonic ageing properties from the Laplace transform, with actuarial applications, Insurance Math. Econom. 40 (2007) 1–14. [9] S. Bernstein, Sur les fonctions absolument monotones, Acta Math. 52 (1929) 1–66. [10] D.J. Bordelon, Problem 72-15, inequalities for special functions, SIAM Rev. 15 (1973) 665–668. [11] C.-P. Chen, Inequalities for the Euler–Mascheroni constant, Appl. Math. Lett. 23 (2010) 161–164. [12] C.P. Chen, C. Mortici, New sequence converging towards the Euler–Mascheroni constant, Comput. Math. Appl. 64 (2012) 391–398. [13] D.W. DeTemple, A quicker convergence to Euler’s constant, Amer. Math. Monthly 100 (1993) 468–470. [14] A. Erdélyi, W. Magnus, F. Oberhettinger, F.G. Tricomi, Higher Transcendental Functions, Krieger, New York, 1981. [15] M.E.H. Ismail, M.E. Muldoon, Monotonicity of the zeros of a cross-product of Bessel functions, SIAM J. Math. Anal. 9 (4) (1978) 759–767. [16] C.M. Joshi, S.K. Bissu, Some inequalities of Bessel and modified Bessel functions, J. Aust. Math. Soc. A 50 (2) (1991) 333–342. [17] A. Laforgia, Bounds for modified Bessel functions, J. Comput. Appl. Math. 34 (3) (1991) 263–267. [18] A. Laforgia, P. Natalini, Some inequalities for modified Bessel functions, J. Inequal. Appl. 2010 (2010) 253035, 10 pages. [19] D. Lu, Some new convergent sequences and inequalities of Euler’s constant, J. Math. Anal. Appl. 419 (2014) 541–552. [20] T.-Q. Luo, H.-L. Lv, Z.-H. Yang, Sh.-Zh. Zheng, New sharp approximations involving incomplete gamma functions, Results Math. 72 (2017) 1007–1020. [21] H.-L. Lv, Z.-H. Yang, T.-Q. Luo, Sh.-Zh. Zheng, Sharp inequalities for tangent function with applications, J. Inequal. Appl. 2017 (2017) 94. [22] W. Magnus, F. Oberhettinger, R.P. Soni, Formulas and Theorems for the Special Functions of Mathematical Physics, Springer-Verlag, 1966. [23] K.S. Miller, S.G. Samko, Completely monotonic functions, Integral Transforms Spec. Funct. 12 (4) (2001) 389–402. [24] C. Mortici, On new sequences converging towards the Euler–Mascheroni constant, Comput. Math. Appl. 59 (2010) 2610–2614. [25] T. Negoi, A faster convergence to the constant of Euler, Gaz. Mat., Ser. A 15 (1997) 111–113 (in Romanian). [26] R.B. Paris, An inequality for the Bessel function Jv (vx), SIAM J. Math. Anal. 15 (1) (1984) 203–205. [27] F. Qi, B.-N. Guo, Sharp bounds for harmonic numbers, Appl. Math. Comput. 218 (2011) 991–995. [28] D.K. Ross, Problem 72-15, inequalities for special functions, SIAM Rev. 15 (1973) 668–670. [29] R. Schilling, R. Song, Z. Vondraček, Bernstein Functions: Theory and Applications, Stud. Math., vol. 37, de Gruyter, Berlin, 2010. [30] J. Segura, Bounds for ratios of modified Bessel functions and associated Turán-type inequalities, J. Math. Anal. Appl. 374 (2011) 516–528. [31] M.K. Simon, Probability Distributions Involving Gaussian Random Variables: A Handbook for Engineers and Scientists, Springer, New York, 2006. [32] S.R. Tims, J.A. Tyrrell, Approximate evaluation of Euler’s constant, Gaz. Math. 55 (1971) 65–67. [33] H. van Haeringen, Bound states for r-2-like potentials in one and three dimensions, J. Math. Phys. 19 (1978) 2171–2179. [34] M.B. Villarino, Ramanujan’s harmonic number expansion into negative powers of triangular number, J. Inequal. Pure Appl. Math. 9 (3) (2008) 89. [35] M.B. Villarino, Sharp bounds for the harmonic numbers, arXiv:math/0510585 [math.CA]. [36] G.N. Watson, A Treatise on the Theory of Bessel Functions, Cambridge University Press, Cambridge, 1922. [37] D.V. Widder, Necessary and sufficient conditions for the representation of a function as a Laplace integral, Trans. Amer. Math. Soc. 33 (1931) 851–892. [38] F.-L. Xia, Z.-H. Yang, Y.-M. Chu, A new proof for the monotonicity criterion of the quotient of two power series on the infinite interval, Pac. J. Appl. Math. 7 (2) (2016) 97–101. [39] Z.-H. Yang, Some properties of the divided difference of psi and polygamma functions, J. Math. Anal. Appl. 455 (2017) 761–777. [40] Z.-H. Yang, A new way to prove L’Hospital Monotone Rules with applications, arXiv:1409.6408 [math.CA]. [41] Z.-H. Yang, Y.-M. Chu, Inequalities for certain means in two arguments, J. Inequal. Appl. 2015 (2015) 299. [42] Z.-H. Yang, Yu.-M. Chu, On approximating the error function, J. Inequal. Appl. 2016 (2016) 311. [43] Z.-H. Yang, Y.-M. Chu, A monotonicity property involving the generalized elliptic integral of the first kind, Math. Inequal. Appl. 20 (3) (2017) 729–735. [44] Z.-H. Yang, Y.-M. Chu, X.-J. Tao, A double inequality for the trigamma function and its applications, Abstr. Appl. Anal. 2014 (2014) 702718, 9 pages. [45] Z.-H. Yang, Y.-M. Chu, M.-K. Wang, Monotonicity criterion for the quotient of power series with applications, J. Math. Anal. Appl. 428 (2015) 587–604.

Z. Yang, J.-F. Tian / J. Math. Anal. Appl. 470 (2019) 821–845

845

[46] Z.-H. Yang, Y.-M. Chu, X.-H. Zhang, Necessary and sufficient conditions for functions involving the psi function to be completely monotonic, J. Inequal. Appl. 2015 (2015) 157. [47] Z.-H. Yang, Y.-M. Chu, X.-H. Zhang, Sharp bounds for psi function, Appl. Math. Comput. 268 (2015) 1055–1063. [48] Z.-H. Yang, W.-M. Qian, Y.-M. Chu, W. Zhang, On approximating the arithmetic-geometric mean and complete elliptic integral of the first kind, J. Math. Anal. Appl. 462 (2018) 1714–1726, https://doi.org/10.1016/j.jmaa.2018.03.005. [49] Z.-H. Yang, J. Tian, Monotonicity and inequalities for the gamma function, J. Inequal. Appl. 2017 (2017) 317. [50] Z.-H. Yang, J. Tian, Optimal inequalities involving power-exponential mean, arithmetic mean and geometric mean, J. Math. Inequal. 11 (4) (2017) 1169–1183. [51] Z.-H. Yang, J. Tian, Monotonicity and sharp inequalities related to gamma function, J. Math. Inequal. 12 (1) (2018) 1–22. [52] Z.-H. Yang, J. Tian, Sharp inequalities for the generalized elliptic integrals of the first kind, Ramanujan J. (2018), https:// doi.org/10.1007/s11139-018-0061-4. [53] Z.-H. Yang, J.-F. Tian, An accurate approximation formula for gamma function, J. Inequal. Appl. 2018 (2018) 56. [54] Z.-H. Yang, J.-F. Tian, Convexity and monotonicity for the elliptic integrals of the first kind and applications, arXiv: 1705.05703 [math.CA]. [55] Z.-H. Yang, W. Zhang, Yu.-M. Chu, Monotonicity and inequalities involving the incomplete gamma function, J. Inequal. Appl. 2016 (2016) 221. [56] Z.-H. Yang, W. Zhang, Yu.-M. Chu, Monotonicity of the incomplete gamma function with applications, J. Inequal. Appl. 2016 (2016) 251. [57] Z.-H. Yang, W. Zhang, Yu.-M. Chu, Sharp Gautschi inequality for parameter 0 < p < 1 with applications, Math. Inequal. Appl. 20 (4) (2017) 1107–1120. [58] Z.-H. Yang, Sh.-Zh. Zheng, Sharp bounds for the ratio of modified Bessel functions, Mediterr. J. Math. 14 (2017) 169, https://doi.org/10.1007/s00009-017-0971-1. [59] Z.-H. Yang, Sh.-Zh. Zheng, The monotonicity and convexity for the ratios of modified Bessel functions of the second kind and applications, Proc. Amer. Math. Soc. 145 (2017) 2943–2958. [60] R.M. Young, Euler’s constant, Gaz. Math. 75 (1991) 187–190. [61] T.-H. Zhao, Z.-H. Yang, Y.-M. Chu, Monotonicity properties of a function involving the psi function with applications, J. Inequal. Appl. 2015 (2015) 193.