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REMAINDERS OF PSI FUNCTION

FENG QI, DA-WEI NIU, AND BAI-NI GUO

Abstract. Let Λp,q(x) = λ(px)−qλ(x) and Φp,q(x) = φ(px)−qφ(x) in

x ∈ (0,∞) for p > 0 and q ∈ R, where λ(x) = R0∞

tdt

(t2+x2)(e2πt−1) and

φ(x) =R∞

0

tdt

(t2+4x2)(eπt+1) are related toψ(x) andψ x+ 1 2

. In this article, some sufficient conditions onp >0 andq∈Rsuch that Λp,q(x) and Φp,q(x) are monotonic inx∈(0,∞) are obtained. Moreover, as by-product, an inequality involving the exponential function is established.

1. Introduction

Recall [7, 11] that a functionf is said to be completely monotonic on an interval

Iiff has derivatives of all orders and 0≤(−1)kf(k)(x)<∞for allk≥0 onI. For

our own convenience, the class of completely monotonic functions onIis denoted by

C[I]. The well known Bernstein’s Theorem [11] states thatf ∈ C[(0,∞)] if and only

iff(x) = R∞

0 e

−xtdµ(t), whereµ(t) is a nonnegative measure on [0,) such that

the integral converges for allx >0. Note that a completely monotonic function in

(0,∞) which is non-identically zero cannot vanish at any point in (0,∞), see [7, 8]

and the references therein.

The noted Binet’s formula (see [9] and [10, p. 103]) states that forx >0,

ln Γ(x) =

x−1

2

lnx−x+ ln√2π+θ(x), (1)

where

θ(x) = Z ∞

0

t

et1 −1 + t

2 e−xt

t2 dt (2)

is called the remainder of Binet’s formula (1).

2000Mathematics Subject Classification. Primary 33B15; Secondary 26D07, 26D20. Key words and phrases. Monotonicity, difference, psi function, remainder, inequality. The first author was supported in part by the Science Foundation of Project for Fostering Innovation Talents at Universities of Henan Province, China.

This paper was typeset usingAMS-LATEX.

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Letp >0 andq∈Rbe real numbers and

hp,q(x) =θ(px)−qθ(x) (3)

in (0,∞).

It is clear thath1,q(x)∈ C[(0,∞)] forq≤1 and−h1,q(x)∈ C[(0,∞)] forq≥1.

Among other things, the following was proved in [1, 2].

Theorem A ([1, 2]). hp,p(x)∈ C[(0,∞)]if 0< p <1 and −hp,p(x)∈ C[(0,∞)] if p >1.

As a further generalization of Theorem A above, the following conclusion was

obtained recently.

Theorem B ([4]). hp,q(x) ∈ C[(0,∞)] if either q ≤ 1

p ≤ 1 or q ≤ 1 ≤

1

p and

−hp,q(x)∈ C[(0,∞)]if eitherq≥1

p ≥1 orq≥1≥

1

p.

In [3, p. 892] and [6, p. 17], it is given that forx >0,

ψ(x) = lnx− 1

2x−2

Z ∞

0

tdt

(t2+x2)(e2πt1) (4)

and

ψx+1 2

= lnx+ 2 Z ∞

0

tdt

(t2+ 4x2)(eπt+ 1). (5)

Let

Λp,q(x) =λ(px)−qλ(x) (6)

and

Φp,q(x) =φ(px)−qφ(x) (7)

inx∈(0,∞) forp >0 andq∈R, where

λ(x) = Z ∞

0

tdt

(t2+x2)(e2πt1) (8)

and

φ(x) = Z ∞

0

tdt

(t2+ 4x2)(eπt+ 1). (9)

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Theorem 1. The functionΛp,q(x)is positive and decreasing inx∈(0,∞)if either p≥1andq≤0or0< p <1andpq≤1; it is negative and increasing inx∈(0,∞)

if p≥1 andpq≥1.

The functionΦp,q(x)is positive and decreasing inx∈(0,∞)if eitherp≥1and

q≤0 or 0< p <1 and q≤1; it is negative and increasing inx∈(0,∞)if p >1

andq≥1.

Theorem 2. The function Λp,q(x) is positive and decreasing in x ∈ (0,∞) for

either q ≤ 0 or 0 < q = 1

p2 ≤1, it is negative and increasing in x∈ (0,∞) for

1

p2 =q≥1.

The functionΦp,q(x)is positive and decreasing in x∈(0,∞)for either p2q <1

and q(p21)[(1 + 3q)p24] 0 or p2q = 1 and 0 < q 1, it is negative and

increasing in x∈(0,∞) for either4≤p2(1 + 3q)1 + 3qor 1

p2 =q≥1.

As by-product, we obtain the following inequality.

Theorem 3. Let τ∈Rbe a nonzero constant. Then inequality

ea+b> b

τebaτea

aτ (10)

for alla >0 andb > 0 with a 6=b holds if and only ifτ ≥1 and reverses if and

only ifτ <0.

In particular, inequality

ea+b>be baea

b−a (11)

is valid for all a > 0 and b > 0 with a 6= b, which is equivalent to the following

integral inequality

ea+b > 1 a−b

Z a

b

(1 +u)eudu. (12)

2. Proofs of main results

Proof of Theorem 1. Direct calculation arrives at

Λp,q(x) =

Z ∞

0

t t2+x2

1

e2πpt1− q e2πt1

dt

= Z ∞

0 1

t2+x2

t e2πt1

e2πt1

e2πpt1−q

dt

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and

Φp,q(x) =

Z ∞

0

t t2+ 4x2

1

eπpt+ 1 − q eπt+ 1

dt

= Z ∞

0 1

t2+ 4x2

t eπt+ 1

eπt+ 1

eπpt+ 1 −q

dt.

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Let

ωr,s(t) =e

rt1

est1 and χr,s(t) = ert+ 1

est+ 1 (15)

int∈(0,∞) for positive real numbersr >0 ands >0. The L’Hˆospital rule yields

lim

t→0+ωr,s(t) =

r

s, t→lim0+χr,s(t) = 1, (16)

and

lim

t→∞ωr,s(t) = limt→∞χr,s(t) = 

 

 

0, r < s,

∞, r > s.

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Direct differentiation and standard argument gives that

dωr,s(t)

dt =

(r−s)e(r+s)t−(rert−sest) (est1)2

and

dχr,s(t)

dt =

(r−s)e(r+s)t(sestrert)

(est+ 1)2 .

The requirement dωr,s(t)

dt Q0 is equivalent with

(r−s)e(r+s)tQrert−sest,

(u−v)eu+vQueu−vev,

ueu(ev−1)Qvev(eu−1),

ueu eu1 Q

vev ev1,

whereu=rt >0 andv=st >0. Since

d dx

xex

ex1

=e

x(exx1)

(ex1)2 >0

forx >0, the function exexx1 is increasing inx∈(0,∞). This implies that

dωr,s(t)

dt Q

0 in (0,∞) if and only if rQs, and then ωr,s(t) is 

 

 

decreasing

increasing

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and only ifrQs. Therefore, whenp >1,

−q < e

2πt1

e2πpt1−q <

1

p−q;

when 0< p <1,

1

p−q <

e2πt1

e2πpt1 −q <∞.

Thus, ifp >1 andq≤0 or 0< p <1 andpq ≤1, the function Λp,q(x) is positive

and decreasing; ifp >1 andpq≥1, it is negative and increasing in (0,∞).

The requirement dχr,s(t)

dt Q0 is equivalent with

(r−s)e(r+s)tQsest−rert,

(u−v)eu+vQvev−ueu,

ueu(ev+ 1)Qvev(eu+ 1),

ueu eu+ 1 Q

vev ev+ 1,

whereu=rt >0 andv=st >0. Since

d dx

xex

ex+ 1

=e

x(ex+x+ 1)

(ex+ 1)2 >0

forx >0, the function exex+1x is increasing inx∈(0,∞). This implies that

dχr,s(t)

dt Q

0 in (0,∞) if and only if rQs, and then χr,s(t) is 

 

 

decreasing

increasing

int ∈(0,∞) if

and only ifrQs. Therefore, whenp >1,

−q < e πt+ 1

eπpt+ 1 −q <1−q;

when 0< p <1,

1−q < e πt+ 1

eπpt+ 1−q <∞.

Thus, if p >1 andq≤0 or 0< p <1 and q≤1, the function Φp,q(x) is positive

and decreasing; if p > 1 and q ≥1, it is negative and increasing in (0,∞). The

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Proof of Theorem 2. Straightforward computation yields

Λp,q(x) =

Z ∞

0

t e2πt1

1

t2+p2x2−

q t2+x2

dt = Z ∞ 0 t e2πt1

(1−q)t2+ (1−p2q)x2

(t2+p2x2)(t2+x2) dt

,

Z ∞

0

t

e2πt1ρp,q;t(x) dt,

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Φp,q(x) =

Z ∞

0

t eπt+ 1

1

t2+ 4p2x2 −

q t2+ 4x2

dt = Z ∞ 0 t eπt+ 1

(1−q)t2+ 4(1p2q)x2 (t2+p2x2)(t2+ 4x2) dt

,

Z ∞

0

t

eπt+ 1%p,q;t(x) dt.

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By standard argument, we have

dρp,q;t(x)

dx =

2xt4[p2(p2q1)u2+ 2p2(q1)u+ (qp2)] (t2+p2x2)2(t2+x2)2 ,

d%p,q;t(x)

dx =

2xt4[16p2(p2q−1)u2+ 8p2(q−1)u+ (4q−p2−3p2q)] (t2+p2x2)2(t2+ 4x2)2 ,

whereu= x

t

2

>0. Hence, if either

 

 

p2q1

≶0

q≤0

or           

p2q1 = 0

q−1Q0

q−p2Q0,

then the derivative dρp,q;t(x)

dx Q0; if either

 

 

p2q1

≶0

q(p2−1)[(1 + 3q)p2−4]≤0 or           

p2q1 = 0

q−1Q0

4q−p23p2q

Q0,

then the derivative d%p,q;t(x)

dx Q0.

Consequently, if eitherq ≤0 or 0 < q= p12 ≤1 and p4≥1 then

dρp,q;t(x)

dx ≤0

and the function Λp,q(x) is decreasing in x ∈ (0,∞); if q = p12 ≥ 1 and p

4 1

then dρp,q;t(x)

dx ≥0 and the function Λp,q(x) is increasing in x∈ (0,∞); if either

p2q <1 andq(p2−1)[(1 + 3q)p2−4]≤0 orp2q= 1 and 0< q ≤1 then d%p,q;t(x)

dx ≤

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(p21)[(1 + 3q)p24]0 orp2q= 1 andq1 thend%p,q;t(x)

dx ≥0 and the function

Φp,q(x) is increasing in x∈(0,∞). The proof of Theorem 2 is complete.

Proof of Theorem 3. Without loss of generality, assumeb > a >0 in (10). Then it

can be rearranged as

eb eb1 >

ea ea1.

Direct calculation gives

d dx

xτex

ex1

= x

τ−1ex τ x ex1

ex1 .

It is easy to see that the function exx1 is decreasing in (0,∞), with

lim

x→0+

x

ex1 = 1 and xlim→∞ x ex1 = 0.

Hence, the function exxτex1 is increasing (or decreasing, respectively) inx∈(0,∞) if

and only ifτ≥1 (orτ <0, respectively). Inequality (10) follows.

References

[1] Ch.-P. Chen and F. Qi, Completely monotonic functions related to the gamma functions, RGMIA Res. Rep. Coll.8 (2005), no. 2, Art. 3. Available online athttp://rgmia.vu.edu. au/v8n2.html.

[2] Ch.-P. Chen and F. Qi, Logarithmically completely monotonic functions related to gamma function and remainder of Binet’s formula with an application, submitted.

[3] I. S. Gradshteyn and I. M. Ryzhik,Table of Integrals, Series, and Products, Sixth Edition, Academic Press, New York, 2000.

[4] S. Guo,Complete monotonicity of difference of remainder of Binet’s formula, private com-munication.

[5] J.-Ch. Kuang,Ch´angy`ong B`udˇeng Sh`ı(Applied Inequalities), 3rd ed., Shandong Science and Technology Press, Ji’nan City, Shandong Province, China, 2004. (Chinese)

[6] W. Magnus, F. Oberhettinger and R. P. Soni,Formulas and Theorems for the Special Func-tions of Mathematical Physics, Springer, Berlin-New York, 1966.

[7] F. Qi Certain logarithmicallyN-alternating monotonic functions involving gamma andq -gamma functions, RGMIA Res. Rep. Coll.8(2005), no. 3, Art. 5. Available online athttp: //rgmia.vu.edu.au/v8n3.html.

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[9] Zh.-X. Wang and D.-R. Guo,Special Functions, Translated from the Chinese by D.-R. Guo and X.-J. Xia, World Scientific Publishing, Singapore, 1989.

[10] Zh.-X. Wang and D.-R. Guo, T`esh¯u H´ansh`u G`ail`un, The Series of Advanced Physics of Peking University, Peking University Press, Beijing, China, 2000. (Chinese)

[11] D. V. Widder,The Laplace Transform, Princeton Univ. Press, Princeton, New Jersey, 1941.

(F. Qi)Research Institute of Mathematical Inequality Theory, Henan Polytechnic University, Jiaozuo City, Henan Province, 454010, China

E-mail address: [email protected], [email protected], [email protected],

[email protected]

URL:http://rgmia.vu.edu.au/qi.html

(D.-W. Niu)School of Mathematics and Informatics, Henan Polytechnic University, Jiaozuo City, Henan Province, 454010, China

E-mail address:[email protected]

(B.-N. Guo)School of Mathematics and Informatics, Henan Polytechnic University, Jiaozuo City, Henan Province, 454010, China

References

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