Journal of Inequalities in Pure and
Applied Mathematics
ON AN OPEN PROBLEM OF BAI-NI GUO AND FENG QI
ŽIVORAD TOMOVSKI AND KOSTADIN TRENČEVSKI
Institute of Mathematics,
St. Cyril and Methodius University,
P. O. Box 162, 1000 Skopje,
MACEDONIA
E-Mail: tomovski@iunona.pmf.ukim.edu.mk
volume 4, issue 2, article 29,
2003.
Received 17 September, 2002;
accepted 28 March, 2003.
Communicated by: F. Qi
E-Mail: kostatre@iunona.pmf.ukim.edu.mk
Abstract
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c
2000
Victoria University
ISSN (electronic): 1443-5756
101-02
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Abstract
In this paper, an open problem posed respectively by B.-N. Guo and F. Qi in
[4, 6, 7] is partially solved: an integralPexpression and a new double inequality
2n
of the generalized Mathieu’s series ∞
n=1 (n2 +a2 )p+1 are established by using
some properties of gamma function and Fourier transform inequalities, where
a > 0, p ∈ N.
2000 Mathematics Subject Classification: Primary 26D15, 33E20; Secondary 40A30
Key words: Integral expression, Inequality, Mathieu’s series, Gamma function,
Fourier transform inequality.
On an Open Problem of Bai-Ni
Guo and Feng Qi
Živorad Tomovski and
Kostadin Trenčevski
Contents
Title Page
1
2
3
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
The Integral Expressions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
The Inequality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
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1.
Introduction
It is well-known that the following
S(a, 1) ,
(1.1)
∞
X
n=1
2n
,
+ a2 )2
(n2
a>0
is called the Mathieu’s series. The integral expression of Mathieu’s series (1.1)
was given in [3] as follows
Z
1 ∞ x sin ax
(1.2)
S(a, 1) =
dx.
a 0 ex − 1
The Mathieu’ series (1.1) and related inequalities have been studied by many
mathematicians for more than a century and there has been a vast amount of
literature. Please refer to [4, 6, 7] and the references therein.
The following Fourier transform inequalities can be found in [2, pp. 89–90]:
If f ∈ L([0, ∞)) with limt→∞ f (t) = 0, then
Z ∞
∞
∞
X
X
k
(1.3)
(−1) f (kπ) <
f (t) cos tdt <
(−1)k f (kπ),
0
k=1
(1.4)
∞
X
k=0
k
(−1) f
1
k+
2
k=0
Z
π <
On an Open Problem of Bai-Ni
Guo and Feng Qi
Živorad Tomovski and
Kostadin Trenčevski
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∞
f (t) sin tdt
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0
< f (0) +
∞
X
k=0
k
(−1) f
1
k+
2
π .
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By using the integral expression (1.2) and Fourier transform inequality (1.4),
Bai-Ni Guo established in [4] the following inequalities for Mathieu’s series
(1.1).
Theorem A ([4]). for a > 0, then
∞
π X (−1)k k + 12
(1.5) 3
< S(a, 1)
a k=0 exp k + 21 πa − 1
1
< 2
a
!
∞
π X (−1)k k + 12
.
1+
a k=0 exp k + 21 πa − 1
At the end of the short note [4], B.-N. Guo proposed an open problem: Let
(1.6)
S(a, p) =
∞
X
n=1
(n2
2n
,
+ a2 )p+1
(1.7)
S(r, t, α) =
n=1
Živorad Tomovski and
Kostadin Trenčevski
Title Page
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where p > 0 and a > 0. Can one establish an integral expression of S(a, p)?
Soon after, Feng Qi further proposed in [6, 7] a similar open problem: Let
∞
X
On an Open Problem of Bai-Ni
Guo and Feng Qi
2nα/2
(nα + r2 )t+1
for t > 0, r > 0 and α > 0. Can one obtain an integral expression of S(r, t, α)?
Give some sharp inequalities for the series S(r, t, α).
In this paper, using the well-known formula
Z ∞
1
1
=
xa e−xt dx,
(1.8)
a+1
t
Γ(a + 1) 0
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which can be deduced from the definition of a gamma function, and Fourier
transform inequalities (1.3) and (1.4), we will establish an integral expression
and a new double inequality of the generalized Mathieu’s series (1.6) for p ∈ N,
the set of all positive integers. Our results partially solve the open problems by
B.-N. Guo and F. Qi in [4] and [6, 7] mentioned above.
On an Open Problem of Bai-Ni
Guo and Feng Qi
Živorad Tomovski and
Kostadin Trenčevski
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2.
The Integral Expressions
One of our main results is to establish an integral expression of S(a, p) for a > 0
and p ∈ N, which can be stated as the following.
Theorem 2.1. Let a > 0 and p ∈ N. Then we have
∞
X
2n
(2.1)
S(a, p) =
2
(n + a2 )p+1
n=1
Z ∞ p
t cos pπ
− at
2
2
dt
=
et − 1
(2a)p p! 0
p
X
(k − 1)(2a)k−2p−1 −(p + 1)
−2
k!(p − k + 1)
p−k
k=2
Z ∞ k
t cos π2 (2p − k + 1) − at
dt.
×
et − 1
0
Proof. Let an =
2n
,
(n2 +a2 )p+1
On an Open Problem of Bai-Ni
Guo and Feng Qi
Živorad Tomovski and
Kostadin Trenčevski
Title Page
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where a > 0 and p ∈ N. Then
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J
n + ai + n − ai
an =
= bn + c n ,
(n + ai)p+1 (n − ai)p+1
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where
bn =
1
ai)p (n
ai)p+1
(n +
−
By putting n + ai = x, we obtain
bn =
xp (x
,
1
=
− 2ai)p+1
cn =
p
X
Ak
k=1
xk
1
(n +
+
ai)p+1 (n
p+1
X
k=1
−
ai)p
.
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Bk
,
(x − 2ai)k
J. Ineq. Pure and Appl. Math. 4(2) Art. 29, 2003
http://jipam.vu.edu.au
where Ak and Bk are constants.
Applying the binomial expansion, we get
x −(p+1)
−(p+1)
−(p+1)
(x − 2ai)
= (−2ai)
1−
2ai
∞ X
−(p
+ 1) x k
−(p+1)
= (−2ai)
−
k
2ai
k=0
∞
X
1
−(p + 1) k
−(p+1)
x
= (−2ai)
k
(−2ai)
k
k=0
for |x| < 2a, i.e.
Živorad Tomovski and
Kostadin Trenčevski
p−1
X
−(p + 1) k−p
1
x
bn ∼ (−2ai)
k
(−2ai)
k
k=0
p
X
1
−(p + 1) 1
−(p+1)
= (−2ai)
.
p−k
k
(−2ai)
p
−
k
x
k=1
−(p+1)
Hence,
−(p + 1)
Ak = (−2ai)
,
p−k
Further, by putting n − ai = y in bn , we obtain
−2p+k−1
1
+ 2ai)p
p (2ai)−p X −p y k
∼
y p+1 k=0 k
2ai
bn =
y p+1 (y
On an Open Problem of Bai-Ni
Guo and Feng Qi
k = 1, 2, . . . , p.
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−p
= (2ai)
p+1 X
k=1
−p
1
1
.
p−k+1
p − k + 1 (2ai)
yk
Hence,
−2p+k−1
Bk = (2ai)
−p
,
p−k+1
k = 1, 2, . . . , p + 1.
Analogously,
p+1
p
X Ck X
Dk
1
cn = p+1
=
+
.
p
k
k
x (x − 2ai)
x
(x
−
2ai)
k=1
k=1
Applying the same technique, for coefficients Ck , Dk we obtain
−p
−2p+k−1
Ck = (−2ai)
, k = 1, 2, . . . , p + 1,
p−k+1
−2p+k−1 −(p + 1)
Dk = (2ai)
, k = 1, 2, . . . , p.
p−k
Thus
an =
(2ai)−p
(−2ai)−p
+
(n − ai)p+1 (n + ai)p+1
p
X
−p
−(p + 1)
1
−2p+k−1
+
(2ai)
+
p−k+1
p−k
(n − ai)k
k=1
p
X
−p
−(p + 1)
1
−2p+k−1
+
(−2ai)
+
p−k+1
p−k
(n + ai)k
k=1
On an Open Problem of Bai-Ni
Guo and Feng Qi
Živorad Tomovski and
Kostadin Trenčevski
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Z
Z
(2ai)−p ∞ p −(n−ai)t
(−2ai)−p ∞ p −(n+ai)t
=
te
dt +
te
dt
p!
p!
0
0
Z ∞
p
X
−p
−(p + 1)
1
−2p+k−1
+
(2ai)
+
tk e−(n−ia)t dt
p−k+1
p−k
k! 0
k=1
Z
p
X
−p
−(p + 1)
1 ∞ k −(n+ia)t
−2p+k−1
(−2ai)
+
+
t e
dt.
p
−
k
+
1
p
−
k
k!
0
k=1
Since
∞
X
n=1
and
e−nt =
et
1
−1
−p
−(p + 1)
−(p + 1)
1−k
+
=
,
p−k+1
p−k
p−k
p−k+1
Živorad Tomovski and
Kostadin Trenčevski
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we obtain
∞
X
On an Open Problem of Bai-Ni
Guo and Feng Qi
Z
tp iat
(−2ai)−p ∞ tp −iat
e dt +
e dt
et − 1
p!
et − 1
0
0
Z ∞
p
X
1−k
tk iat
−2p+k−1 −(p + 1)
+
(2ai)
e dt
p−k
k!(p − k + 1) 0 et − 1
k=1
Z ∞
p
X
1−k
tk −iat
−2p+k−1 −(p + 1)
+
(−2ai)
e dt.
t−1
p
−
k
k!(p
−
k
+
1)
e
0
k=1
(2ai)−p
an =
p!
n=1
Z
∞
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Let z = (2ai)−p eiat and u = (2ai)−2p+k−1 eiat . Then
i
h
2
pπ
pπ z + z̄ =
Re cos
− i sin
(cos at + i sin at)
(2a)p
2
2
pπ
2
=
cos
−
at
(2a)p
2
and
2 Re
u + ū =
=
i
o
nh
(2p−k+1)π
cos (2p−k+1)π
−
i
sin
(cos
at
+
i
sin
at)
2
2
(2a)2p+1−k
(2p − k + 1)π
cos
− at .
2
2
(2a)2p−k+1
On an Open Problem of Bai-Ni
Guo and Feng Qi
Živorad Tomovski and
Kostadin Trenčevski
Title Page
Finally, we get
S(a, p) =
∞
X
Contents
an
n=1
Z
pπ
2(2a)−p ∞ tp
=
cos
− at dt
p!
et − 1
2
0
p
X
2
−(p + 1)
1−k
+
2p−k+1
(2a)
p−k
k!(p − k + 1)
k=1
Z ∞
h
i
tk
π
×
cos (2p − k + 1) − at dt.
et − 1
2
0
The proof is complete.
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Remark 2.1. Using the well-known formula for the polygamma function (see
[1])
Z ∞ n −zt
t e
(n)
n+1
ψ (z) = (−1)
dt
(n = 1, 2, 3, . . . , Re z > 0),
1 − e−t
0
where ψ(z) =
Z ∞
0
d ln Γ(z)
, we
dz
p
obtain
pπ
t
cos
−
at
dt
et − 1
2
pπ Z
pπ Z
∞ p −(1−ia)t
∞ p −t(1+ia)
te
ei 2
te
e−i 2
dt
=
dt
+
2
1 − e−t
2 0
1 − e−t
0
pπ
pπ
ei 2 (p)
e−i 2 (p)
ψ (1 − ia)
=
ψ (1 + ia) +
2
2
= Re[eipπ/2 ψ (p) (1 + ia)].
Analogously,
Z ∞
h
i
i(2p−k+1)π/2 (p)
tp
π
cos
(2p
−
k
+
1)
−
at
dt
=
Re
e
ψ
(1
+
ia)
.
et − 1
2
0
So for S(a, p) we have the following expression
(2.2) S(a, p) =
2
p!(2a)p
p
X
−(p + 1)
2(1 − k)
+
(2a)2p−k+1 k!(p − k + 1)
p−k
k=1
i(2p−k+1)π/2 (p)
× Re e
ψ (1 + ia) .
Re eipπ/2 ψ (p) (1 + ia)
On an Open Problem of Bai-Ni
Guo and Feng Qi
Živorad Tomovski and
Kostadin Trenčevski
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Remark 2.2. If p > 0, p ∈ R, then we have
Z ∞
2n
2
2
2
=
tp ne−(n +a )t dt.
2
2
p+1
(n + a )
Γ(p + 1) 0
Using the Cauchy integration test, we obtain that
P
−n2 t
. Thus
for all t > 0, i.e. f (t) = ∞
n=1 ne
(2.3)
2
S(a, p) =
Γ(p + 1)
Z
2
Γ(p + 1)
Z
=
∞
−a2 t
tp e
0
P∞
∞
X
n=1
2
ne−n t is convergent
!
−n2 t
ne
dt
On an Open Problem of Bai-Ni
Guo and Feng Qi
n=1
∞
2
tp e−a t f (t)dt.
Živorad Tomovski and
Kostadin Trenčevski
0
Remark 2.3. In addition we set an open problem for summing up the functional
series
P∞
−n2 t
for all t > 0.
n=1 ne
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3.
The Inequality
Another one of our main results is to obtain a double inequality of S(a, p) for
a > 0 and p ∈ N by using Fourier transform inequalities (1.3) and (1.4).
Theorem 3.1. For a > 0 and p ∈ N, we have
(3.1)
|S(a, p)|
"∞
#
∞
X (−1)k (kπ)p X
((k + 12 )π)p
2
k
≤ p+1
(−1)
+
a (2a)p p! k=0 exp kπ
exp((k + 12 ) πa ) − 1
−1
a
k=0
p
X
2(k − 1)(2a)−2p+k−1 −(p + 1) +
k!(p − k + 1)ak+1 p − k
k=1
"∞
k #
∞
X (−1)j (jπ)k X
(−1)j j + 12 π
×
+
.
jπ
1 π
−1
exp
j
+
exp
−
1
2 a
a
j=0
j=0
Proof. For all k = 1, 2, . . . , p, let
Z ∞ k
Z ∞ k
t cos at
t sin at
I(a, k) =
dt and J(a, k) =
dt.
t
e −1
et − 1
0
0
Then
S(a, p) =
On an Open Problem of Bai-Ni
Guo and Feng Qi
Živorad Tomovski and
Kostadin Trenčevski
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2
pπ
I(a, p) cos
+ J(a, p) sin
p
(2a) p!
2
2
p
k−2p−1
X 2(1 − k)(2a)
−(p + 1)
+
k!(p − k + 1)
p−k
k=1
h
pπ i
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(2p − k + 1)π
(2p − k + 1)π
× I(a, k) cos
+ J(a, k) sin
.
2
2
Since
I(a, k) =
1
Z
∞
ak+1 0
Z ∞
1
J(a, k) = k+1
a
0
tk cos t
dt,
et/a − 1
tk sin t
dt
et/a − 1
On an Open Problem of Bai-Ni
Guo and Feng Qi
for fixed a > 0 and k = 1, 2, . . . , p, and
Živorad Tomovski and
Kostadin Trenčevski
k
fk ∈ L([0, ∞)),
lim fk (t) = lim
t→∞
t
t→∞ et/a
−1
= 0,
lim fk (t) = 0,
t→0
Title Page
where fk (t) =
tk
,
et/a −1
then, using inequalities (1.3) and (1.4), we have
2
[I(a, p) + J(a, p)]
p!(2a)p
p
X
−(p + 1) 2(k − 1)
[I(a, k) + J(a, k)]
+
2p−k+1 k!(p
−
k
+
1)(2a)
p
−
k
k=1
"∞
p #
∞
2(2a)−p X (−1)k (kπ)p X (−1)k k + 12 π
+
≤
1 π
p!ap+1 k=0 exp kπ
exp
k
+
−1
−1
2
a
a
k=0
p
X
2(k − 1)(2a)−2p+k−1 −(p + 1) +
k!ak+1 (p − k + 1) p − k
k=1
|S(a, p)| ≤
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"
×
k #
∞
X
(−1)j j + 12 π
.
+
exp j + 12 πa − 1
−1
j=0
∞
X
(−1)j (jπ)k
j=0
exp jπ
a
The proof is complete.
Acknowledgements
The authors would like to thank Professor Feng Qi and the anonymous refereee
for some valuable suggestions which have improved the final version of this
paper.
On an Open Problem of Bai-Ni
Guo and Feng Qi
Živorad Tomovski and
Kostadin Trenčevski
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References
[1] M. ABRAMOWITZ AND I.A. STEGUN (Eds), Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables, National
Bureau of Standards, Applied Mathematics Series 55, 4th printing, Washington, 1965.
[2] P.S. BULLEN, A Dictionary of Inequalities, Pitman Monographs and Surveys in Pure and Applied Mathematics 97, Addison Wesley Longman Limited, 1998.
P
2
2 −2
[3] O.E. EMERSLEBEN, Über die reihe ∞
k=1 k(k + c ) , Math. Ann., 125
(1952), 165–171.
On an Open Problem of Bai-Ni
Guo and Feng Qi
[4] B.-N. GUO, Note on Mathieu’s inequality, RGMIA Res. Rep. Coll., 3(3)
(2000), Art. 5, 389–392. Available online at http://rgmia.vu.edu.
au/v3n3.html.
Title Page
[5] E. MATHIEU, Traité de physique mathématique, VI–VII: Théorie de
l’élasticité des corps solides, Gauthier-Villars, Paris, 1890.
[6] F. QI, Inequalities for Mathieu’s series, RGMIA Res. Rep. Coll., 4(2)
(2001), Art. 3, 187–193. Available online at http://rgmia.vu.edu.
au/v4n2.html.
[7] F. QI AND Ch.-P. CHEN, Notes on double inequalities of Mathieu’s series,
Internat. J. Pure Appl. Math., (2003), in press.
Živorad Tomovski and
Kostadin Trenčevski
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