Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 4, Issue 2, Article 29, 2003
ON AN OPEN PROBLEM OF BAI-NI GUO AND FENG QI
ŽIVORAD TOMOVSKI AND KOSTADIN TRENČEVSKI
I NSTITUTE OF M ATHEMATICS ,
S T. C YRIL AND M ETHODIUS U NIVERSITY,
P. O. B OX 162, 1000 S KOPJE ,
MACEDONIA
tomovski@iunona.pmf.ukim.edu.mk
kostatre@iunona.pmf.ukim.edu.mk
Received 17 September, 2002; accepted 28 March, 2003
Communicated by F. Qi
A BSTRACT. In this paper, an open problem posed respectively by B.-N. Guo and F. Qi in [4,
6, 7] is partially P
solved: an integral expression and a new double inequality of the generalized
∞
2n
Mathieu’s series n=1 (n2 +a
2 )p+1 are established by using some properties of gamma function
and Fourier transform inequalities, where a > 0, p ∈ N.
Key words and phrases: Integral expression, Inequality, Mathieu’s series, gamma function, Fourier transform inequality.
2000 Mathematics Subject Classification. Primary 26D15, 33E20; Secondary 40A30.
1. I NTRODUCTION
It is well-known that the following
(1.1)
S(a, 1) ,
∞
X
n=1
(n2
2n
,
+ a2 )2
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.
ISSN (electronic): 1443-5756
c 2003 Victoria University. All rights reserved.
101-02
2
Ž IVORAD T OMOVSKI AND KOSTADIN T REN ČEVSKI
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
(−1)k f
k=0
Z ∞
∞
X
1
1
k
π <
f (t) sin tdt < f (0) +
(−1) f
k+
π .
k+
2
2
0
k=0
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
(1.5)
< S(a, 1) < 2
1 π
3
a
a k=0 exp k + 2 a − 1
!
∞
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
∞
X
2n
(1.6)
S(a, p) =
,
2
2 )p+1
(n
+
a
n=1
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
2nα/2
(1.7)
S(r, t, α) =
(nα + r2 )t+1
n=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
(1.8)
=
xa e−xt dx,
ta+1
Γ(a + 1) 0
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.
2. T HE I NTEGRAL E XPRESSIONS
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) =
(n2 + a2 )p+1
n=1
Z ∞ p
t cos pπ
−
at
2
2
=
dt
(2a)p p! 0
et − 1
Z ∞ k
p
X
t cos π2 (2p − k + 1) − at
(k − 1)(2a)k−2p−1 −(p + 1)
−2
dt.
k!(p − k + 1)
p−k
et − 1
0
k=2
J. Inequal. Pure and Appl. Math., 4(2) Art. 29, 2003
http://jipam.vu.edu.au/
O N AN O PEN P ROBLEM OF BAI -N I G UO AND F ENG Q I
Proof. Let an =
2n
,
(n2 +a2 )p+1
where a > 0 and p ∈ N. Then
an =
n + ai + n − ai
= bn + c n ,
(n + ai)p+1 (n − ai)p+1
where
bn =
3
1
ai)p (n
(n +
By putting n + ai = x, we obtain
−
ai)p+1
,
cn =
p
1
(n +
ai)p+1 (n
− ai)p
.
p+1
X Ak X
Bk
1
bn = p
=
+
,
p+1
k
x (x − 2ai)
x
(x − 2ai)k
k=1
k=1
where Ak and Bk are constants.
Applying the binomial expansion, we get
x −(p+1)
(x − 2ai)−(p+1) = (−2ai)−(p+1) 1 −
2ai
∞ X
−(p + 1) x k
−(p+1)
= (−2ai)
−
k
2ai
k=0
∞
X
1
−(p + 1) k
−(p+1)
= (−2ai)
x
k
(−2ai)
k
k=0
for |x| < 2a, i.e.
p−1
X
1
−(p + 1) k−p
bn ∼ (−2ai)
x
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)
, k = 1, 2, . . . , p.
p−k
Further, by putting n − ai = y in bn , we obtain
1
bn = p+1
y (y + 2ai)p
p (2ai)−p X −p y k
∼
y p+1 k=0 k
2ai
p+1 X
−p
1
1
−p
= (2ai)
.
p−k+1
k
(2ai)
y
p
−
k
+
1
k=1
−2p+k−1
Hence,
−2p+k−1
Bk = (2ai)
−p
,
p−k+1
k = 1, 2, . . . , p + 1.
Analogously,
p+1
p
X Ck X
1
Dk
cn = p+1
=
+
.
p
k
x (x − 2ai)
x
(x − 2ai)k
k=1
k=1
J. Inequal. Pure and Appl. Math., 4(2) Art. 29, 2003
http://jipam.vu.edu.au/
4
Ž IVORAD T OMOVSKI AND KOSTADIN T REN ČEVSKI
Applying the same technique, for coefficients Ck , Dk we obtain
−p
−2p+k−1
, k = 1, 2, . . . , p + 1,
Ck = (−2ai)
p−k+1
−2p+k−1 −(p + 1)
Dk = (2ai)
, k = 1, 2, . . . , p.
p−k
Thus
(2ai)−p
(−2ai)−p
+
(n − ai)p+1 (n + ai)p+1
p
X
−p
−(p + 1)
1
−2p+k−1
+
(2ai)
+
p−k
p−k+1
(n − ai)k
k=1
p
X
−p
−(p + 1)
1
−2p+k−1
+
(−2ai)
+
(n + ai)k
p−k+1
p−k
k=1
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
t e
dt.
+
(−2ai)
+
p
−
k
+
1
p
−
k
k!
0
k=1
an =
Since
∞
X
n=1
e−nt =
et
1
−1
and
−p
−(p + 1)
−(p + 1)
1−k
+
=
,
p−k+1
p−k
p−k
p−k+1
we obtain
∞
X
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.
p−k
k!(p − k + 1) 0 et − 1
k=1
(2ai)−p
an =
p!
n=1
Z
∞
Let z = (2ai)−p eiat and u = (2ai)−2p+k−1 eiat . Then
z + z̄ =
h
i
pπ
2
pπ
pπ 2
Re
cos
−
i
sin
(cos
at
+
i
sin
at)
=
cos
−
at
(2a)p
2
2
(2a)p
2
J. Inequal. Pure and Appl. Math., 4(2) Art. 29, 2003
http://jipam.vu.edu.au/
O N AN O PEN P ROBLEM OF BAI -N I G UO AND F ENG Q I
5
and
2 Re
u + ū =
nh
i
o
(2p−k+1)π
cos (2p−k+1)π
−
i
sin
(cos
at
+
i
sin
at)
2
2
2
=
(2a)2p−k+1
(2a)2p+1−k
(2p − k + 1)π
cos
− at .
2
Finally, we get
S(a, p) =
∞
X
an
n=1
p
pπ
X
2
tp
−(p + 1)
1−k
cos
− at dt +
t
2p−k+1
e −1
2
(2a)
p−k
k!(p − k + 1)
0
k=1
Z ∞
h
i
π
tk
cos
(2p
−
k
+
1)
−
at
dt.
×
et − 1
2
0
2(2a)−p
=
p!
Z
∞
The proof is complete.
Remark 2.2. 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) =
d ln Γ(z)
,
dz
Z
0
Analogously,
Z ∞
0
∞
we obtain
pπ
tp
cos
−
at
dt
et − 1
2
pπ Z
pπ Z
∞ p −t(1+ia)
∞ p −(1−ia)t
ei 2
te
e−i 2
te
=
dt +
dt
−t
2 0
1−e
2
1 − e−t
0
pπ
pπ
e−i 2 (p)
ei 2 (p)
ψ (1 + ia) +
ψ (1 − ia)
=
2
2
= Re[eipπ/2 ψ (p) (1 + ia)].
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
So for S(a, p) we have the following expression
2
Re eipπ/2 ψ (p) (1 + ia)
p
p!(2a)
p
X
2(1 − k)
−(p + 1)
Re ei(2p−k+1)π/2 ψ (p) (1 + ia) .
+
2p−k+1
(2a)
k!(p − k + 1)
p−k
k=1
(2.2) S(a, p) =
Remark 2.3. If p > 0, p ∈ R, then we have
2n
2
=
2
2
p+1
(n + a )
Γ(p + 1)
J. Inequal. Pure and Appl. Math., 4(2) Art. 29, 2003
Z
∞
2 +a2 )t
tp ne−(n
dt.
0
http://jipam.vu.edu.au/
6
Ž IVORAD T OMOVSKI AND KOSTADIN T REN ČEVSKI
P∞
Using the Cauchy integration test, we obtain that
P
−n2 t
. Thus
f (t) = ∞
n=1 ne
(2.3)
2
S(a, p) =
Γ(p + 1)
∞
Z
−a2 t
p
te
0
∞
X
n=1
!
ne
−n2 t
n=1
2
ne−n t is convergent for all t > 0, i.e.
2
dt =
Γ(p + 1)
Z
∞
2
tp e−a t f (t)dt.
0
Remark
2.4. In addition we set an open problem for summing up the functional series
P∞
−n2 t
for all t > 0.
n=1 ne
3. T HE I NEQUALITY
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
"∞
#
∞
1
p
X (−1)k (kπ)p X
((k
+
)π)
2
2
+
(−1)k
(3.1) |S(a, p)| ≤ p+1
1 π
a (2a)p p! 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!(p − k + 1)ak+1
p−k
k=1
"∞
k #
∞
X (−1)j (jπ)k X
(−1)j j + 12 π
×
+
.
jπ
1 π
exp
j
+
−1
exp
−
1
2 a
a
j=0
j=0
Proof. For all k = 1, 2, . . . , p, let
Z ∞ k
t cos at
dt and
I(a, k) =
et − 1
0
Z
J(a, k) =
0
∞ k
t sin at
dt.
et − 1
Then
S(a, p) =
2 h
pπ
pπ i
I(a,
p)
cos
+
J(a,
p)
sin
(2a)p p!
2
2
p
X 2(1 − k)(2a)k−2p−1 −(p + 1)
+
k!(p − k + 1)
p−k
k=1
(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
for fixed a > 0 and k = 1, 2, . . . , p, and
fk ∈ L([0, ∞)),
tk
= 0,
t→∞ et/a − 1
lim fk (t) = lim
t→∞
J. Inequal. Pure and Appl. Math., 4(2) Art. 29, 2003
lim fk (t) = 0,
t→0
http://jipam.vu.edu.au/
O N AN O PEN P ROBLEM OF BAI -N I G UO AND F ENG Q I
where fk (t) =
tk
,
et/a −1
7
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
"∞
k #
∞
X (−1)j (jπ)k X
(−1)j j + 12 π
×
+
.
jπ
1 π
exp
j
+
−1
exp
−
1
2 a
a
j=0
j=0
|S(a, p)| ≤
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.
R EFERENCES
[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.
[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.
[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.
J. Inequal. Pure and Appl. Math., 4(2) Art. 29, 2003
http://jipam.vu.edu.au/