Some new inequalities for Generalized Mathieu type series and Riemann zeta functions

0.14 MB·

by  Khaled Mehrez

SOME NEW INEQUALITIES FOR GENERALIZED MATHIEU TYPE SERIES
AND RIEMANN ZETA FUNCTIONS.
KHALEDMEHREZANDZˇIVORADTOMOVSKI
7
1
0
2

n Abstract. OuraiminthispaperistoshowsomenewinequalitiesforMathieu’stypeseriesand
a Riemannzetafunction. Inparticular,someTura´ntypeinequalities,somemonotonicityandlog-
J convexityresultsforthesespecialfunctionsaregiven. NewLaplacetypeintegralrepresentations

4 forMathietypeseriesandRiemannzetafunctionarealsopresented.
1
Keywords: Mathieu series, Generalized Mathieu series, Riemann zeta function, Bessel function,

] Mellin and Laplace integral representation, Tur´an type inequalities, Inequalities.
A
C Mathematics Subject Classification (2010): 3B15, 33E20, 60E10,11M35.
.
h
t 1. Introduction
a
m The infinite series
[ ∞ 2n
(1) S(r)= ,

1 (n2+r2)2
n=1
v X
1 is called a Mathieu series. It was introduced and studied by E´mile Leonard Mathieu in his book
7 [8] devoted to the elasticity of solid bodies. Bounds for this series are needed for the solution of
9
boundary value problems for the biharmonic equations in a two–dimensional rectangular domain,
3
see [[14], Eq. (54), p. 258]. A remarkable useful integral representation for S(r) is given by
0
. Emersleben [6] in the following form
1
0 1 ∞ xsin(rx)dx
7 (2) S(r)= .
r ex 1
1 Z0 −
: The so-called generalized Mathieu series with a fractional power reads [2]
v
i
X
∞ 2n
r (3) S (r)= , µ>0, r >0,
a µ (n2+r2)µ+1
n=1
X
such series has been widely considered in mathematical literature, see [2, 15, 17]. Cerone and
Lenard derived also the next integral expression [2]
∞ xµ+1/2
(4) S (r)=C (r) J (rx)dx, µ>0,
µ µ Z0 ex−1 µ−1/2
where
√π
C (r)= .
µ (2r)µ 1/2Γ(µ+1)
−
In the literature, the study of Mathieus series and its inequalities has a rich literature, many
interesting refinements and extensions of Mathieus inequality can be found in [16, 17].
1
2 K.MEHREZ,Zˇ.TOMOVSKI
In this paper is organized as follows. In section 2, we state some useful Lemmas, which are
usefulintheproofsofourresults. Insection3,weprovesomenewinequalitiesforMathieu’sseries.
In particular, we present the Tur´an type inequality for this function. Moreover, we present some
monotonicity and convexity results for the function µ S (r). As consequence we establish some
µ
7→
functional inequalities. At the end of this section, we derive the Laplace integral representation of
such series. In section 3, as applications of our main results in the section 2, we derive some new
inequalities for Riemann zeta function.
Before we present the main results of this paper we recall some denitions, which will be used in
the sequel. A function f :(0, ) R is said to be completely monotonic if f has derivatives of all
∞ →
orders and satises
( 1)nf(n)(x) 0,
− ≥
for all x 0 and n N. We say that a function g : [a,b] R R is said to be log-convex if its
≥ ∈ ⊆ →
natural logarithm logg is convex, that is, for all x,y [a,b] and α [0,1] we have
∈ ∈
g(αx+(1 α)y) [g(x)]α[g(y)]1 α.
−
− ≤
2. Some preliminary Lemmas
In this section, we state the following Lemmas, which are useful in the proofs of our results.
Lemma1. [4](Jenseninequality)Letµbeaprobability measureandletϕ 0beaconvexfunction.
≥
Then, for all f be a integrable function we have
(5) ϕ fdν ϕ fdν .
◦ ≥
Z (cid:18)Z (cid:19)
Our next Lemma is well-known and is stated only for easy reference, see for example [[7]. Eq.
10, p. 313]
Lemma 2. Let µ>2. Then the hollowing identity
∞ tµ−1
(6) dt=Γ(µ)(ζ(µ 1) ζ(µ)),
(et 1)2 − −
Z0 −
holds.
The following inequality for completely monotonic functions is due to Kimberling [9].
Lemma 3. Let f;(0, ) (0,1) be continuous. If f is completely monotonic, then
∞ −→
f(x)f(y) f(x+y), x,y 0.
≤ ≥
The next Lemma is given in [3].
Lemma 4. For all µ>0 and r >0. Then the integral of the Mathieu’s series on (0, ) holds true:
∞
∞ √πΓ(µ+1/2)ζ(2µ)
(7) S (r)dr = .
µ
Γ(µ+1)
Z0
3. Some new inequalities for Mathieu’s series
Our main results is the following Theorem.
Theorem 1. Let p>1. Then the following inequalities
Cp(r)Γp 1(µ+1/2)ζp 1(µ+1/2)Γ(p+µ+1/2)ζ(p+µ+1/2), ifµ 1
(8) Sµp(r)≤( C2µµpp2(r)Γp−−1(µ+1/2)ζp−−1(µ+1/2)Γ(p+µ+1/2)ζ(p+µ+1/2) ifµ≥≥ 32
NEW INEQUALITIES FOR FOR GENERALIZED MATHIEU SERIES AND RIEMANN ZETA FUNCTIONS 3
holds true for all r (0, ), where ζ(.) denotes the Riemann zeta function defined by
∈ ∞
∞ 1
ζ(p)= .
np
n=1
X
Moreover, the following inequalities holds true
15πζ(3/2)ζ(7/2) π3
(9) S(r) π, and S .
≤ 4√2r 3/2 ≤ 9√10r2
p
for all r (0, ).
∈ ∞
Proof. Let p>1, we define the function ϕ:(0, ) R by ϕ(x)=xp and we set
∞ →
xµ 1/2 J (rx)
− µ 1/2
f(x)=x, dν(x)= − dx,
K (r)(ex 1)
µ (cid:12) − (cid:12)
(cid:12) (cid:12)
where
xµ 1/2 J (rx)
∞ − µ 1/2
Kµ(r)= ex −1 dx.
Z0 (cid:12) − (cid:12)
(cid:12) (cid:12)
So, by means of Lemma 1 and the representation integral (4) we get
p
xµ+1/2 J (rx)
Sp(x) Cp(r) ∞ µ−1/2 dx
≤ µ Z0 e(cid:12)x−1 (cid:12) !
(cid:12) (cid:12)
p
(10) =Cp(r)Kp(r) ∞xdν(x)
µ µ
(cid:18)Z0 (cid:19)
xp+µ 1/2 J (rx)
≤Cµp(r)Kµp−1(r) ∞ − ex µ−11/2 dx.
Z0 (cid:12)− (cid:12)
(cid:12) (cid:12)
Combining the previous inequality and von Lommel’s uniform bounds [[12],[18], p. 406]
1
J (x) 1, ν 0 and J (x) , ν 1
ν ν
| |≤ ≥ | |≤ √2 ≥
and the representation,
∞ xµ
(11) dx=Γ(µ+1)ζ(µ+1),
ex 1
Z0 −
we obtain the desired inequality (8). Finally, let p= 2, µ = 1, and p= 2, µ = 3/2 respectively in
(8) we obtain the inequalities (9). (cid:4)
Theorem 2. Let µ>3/2. Then the following inequality
(2µ 1) (2µ 1)√πΓ(2µ)ζ(2µ 1)
(12) S (r) − S (r) − −
µ ≥ 2µr3 µ−1 − 22µ−2r3Γ(µ+1)Γ(µ+1/2)
is valid for all r >0.
Proof. Let us consider the function (x):R ( ,1] defined by
µ
J −→ −∞
2µΓ(µ+1)J (x)
µ
(x)= ,µ> 1.
Jµ xµ −
Thus, by (4), we can write S (r) in the following form
µ
∞ x2µ
(13) S (r)=c (rx)dx, µ 1,
µ µ,1Z0 ex−1Jµ−1/2 ≥
4 K.MEHREZ,Zˇ.TOMOVSKI
where
√π
c = .
µ,1 22µ 1Γ(µ+1/2)Γ(µ+1)
−
By using the differentiation formula [[18], p.18]
x
Jµ′(x)=−2(µ+1)Jµ+1(x),
and the integrating by parts in the right hand side of (13), we get
(14)
(2µ 1)cµ,1 ∞ x2µ−1
Sµ(r)=− −r2 Z0 ex−1Jµ′−3/2(rx)dx
(2µ 1)cµ,1 ∞ (2µ 1)x2µ−2 ∞ x2µ−1ex
= − − (rx)dx (rx)dx
r2 (cid:20)Z0 r(ex−1) Jµ−3/2 −Z0 r(ex−1)2Jµ−3/2 (cid:21)
(2µ 1)2cµ,1 (2µ 1)cµ,1 ∞ x2µ−1 ∞ x2µ−1
= − S (r) − (rx)dx+ (rx)dx
cµ−1,1r3 µ−1 − r3 (cid:20)Z0 ex−1Jµ−3/2 Z0 (ex−1)2Jµ−3/2 (cid:21)
(2µ 1)2cµ,1 (2µ 1)cµ,1 ∞ x2µ−1 ∞ x2µ−1
− S (r) − dx+ dx
≥ cµ−1,1r3 µ−1 − r3 (cid:20)Z0 ex−1 Z0 (ex−1)2 (cid:21)
(2µ 1)2c (2µ 1)c
µ,1 µ,1
= − S (r) − [Γ(2µ)ζ(2µ)+Γ(2µ)(ζ(2µ 1) ζ(2µ))]
cµ 1,1r3 µ−1 − r3 − −
−
(2µ 1)2c (2µ 1)c Γ(2µ)ζ(2µ 1)
µ,1 µ,1
= − S (r) − − .
cµ 1,1r3 µ−1 − r3
−
In the equation (14), we use the bound by Minakshisundaram and Sz´asz [13]
1
(x) 1, µ> − , x R,
µ
|J |≤ 2 ∈
and Lemma 2. The desired inequality (25) is established. (cid:4)
In the next Theorem we establish the Tur´an type inequalities for the Mathieu’s serie S (r).
µ
Theorem 3. Let µ>0. Then the Tur´an type inequality
(15) S (r)S (r) S2 (r) 0,
µ+2 µ − µ+1 ≥
holds true for all r (0, ).
∈ ∞
Proof. The Cauchy product reveals
S (r)S (r) S2 (r)=
µ+2 µ − µ+1
(16)
n n
∞ 4k(n k) ∞ 4k(n k)
= − −
(k2+r2)µ+1((n k)2+r2)µ+3 − (k2+r2)µ+2((n k)2+r2)µ+2
n=1k=1 − n=1k=1 −
XX XX
n
∞ 4nk(n k)(2k n)
= − −
(k2+r2)µ+2((n k)2+r2)µ+3
n=1k=1 −
XX
n
∞
= 4n(2k n)T ,
n,k
−
n=1k=0
XX
where
k(n k)
T = − .
n,k ((n k)2+r2)µ+3(k2+r2)µ+2
−
NEW INEQUALITIES FOR FOR GENERALIZED MATHIEU SERIES AND RIEMANN ZETA FUNCTIONS 5
If n is even, then
n n/2 1 n
−
T (2k n)= T (2k n)+ T (2k n)
n,k n,k n,k
− − −
(17) Xk=0 Xk=0 k=Xn/2+1
[(n 1)/2]
−
= (T T )(2k n),
n,k n,n k
− − −
k=0
X
where [.] denotes the greatest integer function. Similarly, if n is odd, then
n [(n 1)/2]
−
T (2k n)= (T T )(2k n).
n,k n,k n,n k
− − − −
k=0 k=0
X X
Thus,
[(n 1)/2]
∞ −
(18) S (r)S (r) S2 (r)= (T T )(2k n).
µ+2 µ − µ+1 n,k− n,n−k −
n=1 k=0
X X
Simplifying, we find that
n2k(n k)(2k n)
(19) T T = − − .
n,k− n,n−k ((n k)2+r2)µ+3(k2+r2)µ+3
−
In view of (18) and (19), we deduce that the Tur´an type inequality (15) holds. (cid:4)
Theorem 4. The following assertion are true:
1. The function µ S (r) is completely monotonic and log-convex on (0, ) for each r >0.
µ
7→ ∞
2. The function µ Sµ+1(r) is increasing on (0, ).
7→ Sµ(r) ∞
3. Furthermore, for all r>0, the following inequalities are valid
(20) S (r)S(r) S (r)S (r), µ,ν >0.
µ+ν µ ν
≥
1 1
Sν(r) ν+1 Sµ(r) µ+1
(21) , µ ν >0.
ζ(2ν+1) ≥ ζ(2µ+1) ≥
(cid:20) (cid:21) (cid:20) (cid:21)
1
Sµ(r) µ+1 ζ(2µ+3)Sµ(r)
(22) + 2,µ>0.
ζ(2µ+1) ζ(2µ+1)S (r) ≥
(cid:20) (cid:21) µ+1
for all µ,ν >0.
Proof. 1. For all m N, µ>0, we have
∈
∂mS (r) 2nlogm(n2+r2)
( 1)m µ = 0.
− ∂µm (n2+r2)µ+1 ≥
n 1
X≥
Thus, the function µ S (r) is completely monotonic and Log-convex on (0, ), since every
µ
7→ ∞
completely monotonic function is log-convex,see [[19], p. 167].
2. From part 1. of this Theorem, the function µ logS (r) is convex and hence, it follows that
µ
7→
the function µ logS (r) logS (r) is increasing.
µ+1 µ
7→ −
3. Since the function µ Sµ(r) is completely monotonic on (0, ) and maps (0, ) to (0,1),
7→ S(r) ∞ ∞
according to Lemma 3, we conclude the asserted inequality (20). Newt, we prove the inequality
(21). Suppose that µ ν >0 and define the function H :(0, ) R with relation
≥ ∞ −→
ν+1
H(r)= logS (r) logS (r).
µ ν
µ+1 −
6 K.MEHREZ,Zˇ.TOMOVSKI
On the other hand, by using the fact
S (r)= 2r(µ+1)S (r),
µ′ − µ+1
we have
S (r) S (r)
ν+1 µ+1
H (r)=2(ν+1)r 0.
′
S (r) − S (r) ≤
(cid:18) ν µ (cid:19)
So, by part 2. in this Theorem we conclude that the function H(r) is decreasing on (0, ).
∞
Consequently, H(r) H(0). Replacing µ by µ +1 and ν by µ in inequality (22), we use the
≤
inequality (21) and the arithmetic–geometric mean inequality
1 1
1 Sµ(r) µ+1 ζ(2µ+3)Sµ(r) Sµ(r) µ+1 ζ(2µ+3)Sµ(r)
+ . 1.
2 (cid:20)ζ(2µ+1)(cid:21) ζ(2µ+1)Sµ+1(r)!≥s(cid:20)ζ(2µ+1)(cid:21) ζ(2µ+1)Sµ+1(r) ≥
(cid:4)
Remark 1. We note that there are another proofs of the Tur´an type inequality (15). Indeed, since
thefunction ν S (r) is log-convex on (0, )for r >0, it follows that for all µ ,µ 1, α [0,1]
µ 1 2
7→ ∞ ≥ ∈
and r >0 we have
Sαµ1+(1−α)µ2(r)≤[Sµ1(r)]α[Sµ2(r)]1−α.
Choosing µ = µ, µ = µ+2 and α = 1, the above inequality reduces to the Tur´an inequality
1 2 2
(15). A third proof of this inequality can be obtained as follows. By using the fact that the function
µ Sµ+1(r) is increasing on (0, ), we have
7→ Sµ(r) ∞
S (r) S (r)
µ+2 µ+1
,
S (r) ≥ S (r)
µ+1 µ
and hence the required result follows.
Theorem 5. For µ 1. The Mathieu series admits the following integral representation
≥
(23) S (r)=c ∞e rtK (t)dt
µ µ − µ
Z0
and
cµ ∞
(24) ζ(2µ+1)= K (t)dt.
µ
2
Z0
and,
(25) S(r)= ∞e rtK(t)dt
−
Z0
where c = √π and K (t) and K(t) are defined by
µ 2µ−1/2Γ(µ+1) µ
Kµ(t)=tµ+1/2gµ(t)andK(t)=h(t) th′(t),
−
with g and h(t) are Kapteyn series, defined as
µ
∞ Jµ+1/2(nt) ∞ sin(nt)
g (t)= , andh(t)=
µ nµ 1/2 n2
n=1 − n=1
X X
where J (.) is the Bessel function.
µ
NEW INEQUALITIES FOR FOR GENERALIZED MATHIEU SERIES AND RIEMANN ZETA FUNCTIONS 7
Proof. By using the formula [[5], eq. 42, p. 397]
(26) 1 = √π ∞tµe stJ (at)dt, µ> 1/2
(a2+s2)µ+1/2 (2a)µ+1/2Γ(µ+1/2) − µ −
Z0
we get
(27) 2n = √π ∞tµ+1/2e rtJ (nt)dt.
(n2+r2)µ+1 (2n)µ 1/2Γ(µ+1) − µ+1/2
− Z0
The interchanging between integral and summation gives (23). Now, let r tends to 0 in (23) we
obtain that (24) holds true. Finally, let µ=1 in (23) and using the fact
2 sinx cosx
J (t)= ,
3/2 π x3/2 − x1/2
r (cid:18) (cid:19)
we obtain
S(r)= ∞e−rt ∞ sin(nt) t ∞ cos(nt) dt.
Z0 n=1 n2 − n=1 n !
X X
The proof of Theorem 5 is complete. (cid:4)
Example 1. Let r tends to 0 in (25) we get integral formula for the Apery constant
1 ∞ 1 ∞ ∞ sin(nt) ∞ cos(nt)
(28) ζ(3)= K(t)dt= t dt.
2Z0 2Z0 n=1 n2 − n=1 n !
X X
Corollary 1. Let µ>7/6. Then the following inequality
c √πΓ(µ+7/6)ζ(µ 1/6)
L
(29) S (r) − ,
µ ≤ 2µ 1/2Γ(µ+1)rµ+7/6
−
is valid for all r >0, where c =sup x1/3J (x) =0.78574687...
L x>0{ 0 }
Proof. By using the Landeau estimate (see [10])
J (x) c x 1/3,
µ L −
| |≤
where c = sup x1/3J (x) = 0.78574687... uniformly in µ, and using the integral representa-
tion (23L), we dedx>u0c{e that0the }inequality (29) holds true. (cid:4)
4. some new inequalities for Riemann Zeta Functions
Firstlyresultsinthissection,wepresentnewTur´antypeinequalityforRiemannZetafunctions.
Theorem 6. Let µ>1. Then the Tur´an type inequality
(30) ζ(µ)ζ(µ+2) ζ2(µ+1) 0.
− ≥
holds true.
Proof. Letting r tends to 0 in (15), we get
ζ(2µ+1)ζ(2µ+3) ζ2(2µ+2) 0.
− ≥
Replacing 2µ+1 by µ in the previous inequality we find that the inequality (30) is valid. (cid:4)
8 K.MEHREZ,Zˇ.TOMOVSKI
Remark 2. In [11], Laforgia and Natalini by using the generalization of the Schwarz inequality
proved the following Tur´an type inequality for Riemann zeta function
µ
(31) ζ(µ)ζ(µ+2) ζ2(µ+1), µ>1.
≥ µ+1
We note that the inequality (30) is better than the inequality (31).
In the next Theorem we establish a simple upper bounds of the Zeta function. Our main tool
will be the formula (7) in Lemma 4.
Theorem 7. Let µ 1. Then the following inequalities holds true,
≥
3π Γ(µ+1)
(32) ζ(2µ)
≤ 2 Γ(µ+1/2)
r
and
ζ32(2µ+1) (µ+1)Γ2(µ+1/2)
(33)
ζ2(2µ)ζ(2µ+3) ≤ Γ2(µ+1)
Proof. In [1], the authors proved that
1
(34) S(r)< , r >0,
r2+1/6
and since the function µ S (r) is decreasing on [1, ), we get
µ
7→ ∞
1
(35) S (r)< , r>0,
µ r2+1/6
for all µ 1. Integrating (35) on the interval (0, ), we obtain
≥ ∞
∞ 3
S (r)dr π.
µ
≤ 2
Z0 r
So, Lemma 4 completes the proof of inequality (32). Now, we proved the inequality (33). In [17],
the authors of this paper obtained that
ζ(2µ+3)
(36) 2ζ(2µ+1)exp (µ+1) r2 S (r), r>0.
µ
− ζ(2µ+1) ≤
(cid:26) (cid:27)
Therefore, integrating (36) and from the Lemma 1, we deduce that the inequality (33) holds.
(cid:4)
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KhaledMehrez. D´epartementde Math´ematiquesISSATKasserine, Universit´e de Kairouan,Tunisia.
E-mail address: [e-mail]
Zˇivorad Tomovski. University ”St. Cyril and Methodius”, Faculty of Natural Sciences and Mathe-
matics,Institute of Mathematics,Repubicof Macedonia.
E-mail address: [e-mail]