Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 4, Issue 5, Article 100, 2003
ON INTEGRAL FORMS OF GENERALISED MATHIEU SERIES
P. CERONE AND C.T. LENARD
S CHOOL OF C OMPUTER S CIENCES AND M ATHEMATICS
V ICTORIA U NIVERSITY OF T ECHNOLOGY
PO B OX 14428, MCMC 8001
VIC, AUSTRALIA .
pc@csm.vu.edu.au
URL: http://rgmia.vu.edu.au/cerone
D EPARTMENT OF M ATHEMATICS
L AT ROBE U NIVERSITY
PO B OX 199, B ENDIGO ,
V ICTORIA , 3552, AUSTRALIA .
C.Lenard@bendigo.latrobe.edu.au
URL: http://www.bendigo.latrobe.edu.au/mte/maths/staff/lenard/
Received 16 June, 2003; accepted 08 December, 2003
Communicated by F. Qi
A BSTRACT. Integral representations for generalised Mathieu series are obtained which recapture the Mathieu series as a special case. Bounds are obtained through the use of the integral
representations.
Key words and phrases: Mathieu Series, Bounds, Identities.
2000 Mathematics Subject Classification. Primary 26D15, 33E20; Secondary 26A42, 40A30.
1. I NTRODUCTION
The series
(1.1)
S (r) =
∞
X
n=1
(n2
2n
, r>0
+ r2 )2
is well known in the literature as Mathieu’s series. It has been extensively studied in the past
since its introduction by Mathieu [12] in 1890, where it arose in connection with work on
elasticity of solid bodies. The reader is directed to the references for further illustration.
One of the main questions addressed in relation (1.1) is to obtain sharp bounds. Alzer, Brenner and Ruehr [2] showed that the best constants a and b in
1
1
< S (x) < 2
, x 6= 0
2
x +a
x +b
ISSN (electronic): 1443-5756
c 2003 Victoria University. All rights reserved.
080-03
2
are a =
P. C ERONE AND C.T. L ENARD
1
2ζ(3)
and b =
1
6
where ζ (·) denotes the Riemann zeta function defined by
∞
X
1
.
ζ (p) =
p
n
n=1
(1.2)
An integral representation for S (r) as given in (1.1) was presented in [6] and [7] as
Z
1 ∞ x
(1.3)
S (r) =
sin (rx) dx.
r 0 ex − 1
Guo [10] utilised (1.3) and a lemma [3, pp. 89–90] to obtain bounds on S (r) , namely,
!
∞
∞
k
k
π X (−1) k + 12
1
π X (−1) k + 12
.
< S (r) < 2 1 +
(1.4)
1 π
1 π
r
r
r3
e(k+ 2 ) r − 1
e(k+ 2 ) r − 1
k=0
k=0
The following results were obtained by Qi and coworkers (see [4], [15] – [17])
π
π 4 (1 + r2 ) e− r + e− 2r − 4r2 − 1
(1.5)
≤ S (r)
π
e− r − 1 (1 + r2 ) (1 + 4r2 )
π
π (1 + 4r2 ) e− r − e− 2r − 4 (1 + r2 )
≤
π
e− r − 1 (1 + r2 ) (1 + 4r2 )
Z π
π
1 r x
1 + e− 2r
S (r) <
sin (rx) dx < 2 1 ,
r 0 ex − 1
r +4
and
h
πr i
3
1
2
3
2
2 πr
S (r) ≥
16r
r
−
3
+
π
r
+
1
sech
tanh
.
2
2
8r (1 + r2 )3
Guo in [10] poses the interesting problem as to whether there is an integral representation of the
generalised Mathieu series
∞
X
2n
(1.6)
Sµ (r) =
, r > 0, µ > 0.
2
(n + r2 )1+µ
n=1
This is resolved in Section 2.
Recently in [18] an integral representation was obtained for Sm (r) , where m ∈ N, namely
Z ∞ m
mπ
2
t
(1.7) Sm (r) =
cos
−
rt
dt
(2r)m m! 0 et − 1
2
"
m
X
(k − 1) (2r)k−2m−1 − (m + 1)
−2
k! (m − k + 1)
m−k
k=1
#
Z ∞ k
t cos π2 (2m − k + 1) − rt
×
dt .
et − 1
0
Bounds were obtained by Tomovski and Trenčevski [18] using (1.3).
It is the intention of the current paper to investigate further integral representations of the
generalised Mathieu series (1.6).
Bounds are obtained in Section 3 for Sµ (r) . In Section 4 the open problem of obtaining an
integral representation for
∞
X
2nγ
S (r; µ, γ) =
(n2γ + r2 )µ+1
n=1
J. Inequal. Pure and Appl. Math., 4(5) Art. 100, 2003
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M ATHIEU S ERIES
3
posed by Qi [15] is addressed.
We notice that
S (r; 1, 1) = S1 (r) = S (r) ,
the Mathieu series.
2. I NTEGRAL R EPRESENTATION
OF THE
G ENERALISED M ATHIEU S ERIES Sµ (r)
Before proceeding to obtain an integral representation for Sµ (r) as given by (1.6), it is instructive to present an alternative representation in terms of the zeta function ζ (p) presented in
(1.2). Namely, a straightforward series expansion gives
∞
X
µ+k
k
2k
(2.1)
Sµ (r) = 2
r (−1)
ζ (2µ + 2k + 1)
k
k=0
k k−α−1 α
with α = − (µ + 1) .
on using the result k = (−1)
k
Theorem 2.1. The generalised Mathieu series Sµ (r) defined by (1.6) may be represented in the
integral form
Z ∞ µ+ 1
x 2
(2.2)
Sµ (r) = Cµ (r)
J 1 (rx) dx, µ > 0,
ex − 1 µ− 2
0
where
√
π
(2.3)
Cµ (r) =
µ− 12
(2r)
Γ (µ + 1)
and Jν (z) is the ν th order Bessel function of the first kind.
Proof (A). Consider
Z
Tµ (r) =
(2.4)
0
∞
1
xµ+ 2
J 1 (rx) dx.
ex − 1 µ− 2
Then using the series definition for Jν (z) (Gradshtein and Ryzhik [9]),
ν+2k
∞
X
(−1)k z2
Jν (z) =
k!Γ (ν + k + 1)
k=0
in (2.4) produces after the permissible interchange of summation and integral,
µ+2k− 12 Z ∞
∞
X
(−1)k 2r
x2(µ+k)
dx.
(2.5)
Tµ (r) =
1
x−1
e
k!Γ
µ
+
k
+
0
2
k=0
Now, the well known representation [9]
Z ∞
xp
(2.6)
dx = Γ (p + 1) ζ (p + 1)
ex − 1
0
gives from (2.5) with p = 2 (µ + k)
(2.7)
Tµ (r) =
∞
X
(−1)k
k=0
1
r µ+2k− 2
2
Γ (2µ + 2k + 1) ζ (2µ + 2k + 1)
.
k!Γ µ + k + 12
An application of the duplication identity for the gamma function
√
1
2z−1
πΓ (2z) = 2
Γ (z) Γ z +
,
2
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4
P. C ERONE AND C.T. L ENARD
with z = µ + k +
1
2
simplifies the expression in (2.7) to
1
∞
(2r)µ− 2 2 X
Γ (µ + k + 1)
√
Tµ (r) =
(−1)k r2k
ζ (2µ + 2k + 1) .
k!
π
k=0
(2.8)
Repeated use of the identity Γ (z + 1) = zΓ (z) gives
µ+k
Γ (µ + k + 1)
=
Γ (µ + 1)
k!
k
and so from (2.8)
1
∞
µ+k
(2r)µ− 2 Γ (µ + 1) X
k 2k
√
2
(−1) r
ζ (2µ + 2k + 1)
Tµ (r) =
k
π
k=0
produces the result (2.2) on reference to (2.1), (2.3) and (2.4).
Proof (A) is now complete.
Proof (B). From (2.4) we have
∞
e−x
µ+ 12
x
Jµ− 1 (rx) dx
2
1 − e−x
0
Z
∞
X ∞
1
=
e−nx xµ+ 2 Jµ− 1 (rx) dx.
Z
Tµ (r) =
(2.9)
k=1
0
2
Now Gradshtein and Ryzhik [9] on page 712 has the result
Z ∞
2α (2β)ν Γ ν + 23
−αx ν+1
e x Jν (βx) dx = √
(2.10)
3 ,
0
π [α2 + β 2 ]ν+ 2
Re (ν) > −1, Re (α) > |Im β| ,
which is referred to in Watson [20] whom in turn attributes the result to an 1875 result of
Gegenbauer.
Taking α = n, ν = µ − 21 and β = r, all real, in (2.10) and substituting in (2.9) readily
produces
1
∞
2n
(2r)µ− 2 Γ (µ + 1) X
√
Tµ (r) =
,
2
π
[n + r2 ]µ+1
n=1
giving from (1.6), (2.4) and (2.3) the result (2.2).
We note that the more restrictive condition of µ > 0 needs to be imposed for the convergence
of the series although (2.10) requires Re (ν) = µ − 12 > −1.
Remark 2.2. If we take µ = 1 in (1.6) and (2.2) – (2.3) then S1 (r) ≡ S (r) , the Mathieu
series given
q by (1.1) and its integral representation (1.3). This is easily seen to be the case since
J 1 (z) =
2
2
πz
sin z and taking µ = 1 in (2.2) – (2.3) produces (1.3).
Remark 2.3. Gradshtein and Ryzhik [9] on page 712 also quote the result
Z ∞
ν
1
(2β)
Γ
ν
+
2
(2.11)
e−αx xν Jν (βx) dx = √
ν+ 12
2
2
0
π (α + β )
1
Re (ν) > − , Re (α) > |Im (β)| ,
2
which Watson [20] again attributes to an 1875 result by Gegenbauer.
We note that formal differentiation of (2.11) with respect to α produces the result (2.10).
J. Inequal. Pure and Appl. Math., 4(5) Art. 100, 2003
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M ATHIEU S ERIES
5
Following a similar process as in Proof (B) above, we may show that
1
Z ∞ µ− 1
∞
(2r)µ− 2 Γ (µ) X
x 2
1
√
(2.12)
Jµ− 1 (rx) dx =
.
x
2
2 )µ
2
e −1
(n
+
r
π
0
n=1
Gradshtein and Ryzhik [9] have an explicit expression which can be transformed by a simple
change of variables to (2.12). Namely,
∞
Z ∞ ν
(2b)ν Γ ν + 12 X
1
x Jν (bx)
√
(2.13)
dx =
1 ,
πx
2 2
2 ν+ 2
e −1
π
0
n=1 (n π + b )
Re (ν) > 0, |Im (b)| < π, which is attributed by Watson [20] to a 1906 result by Kapteyn.
An explicit integral expression for Sµ (r) of the current form does not seem to have been
available previously.
Finally, we note that (2.10) or (2.11) may be looked upon as an integral transform such as the
Laplace or Hankel transform and the results may be found in tables of such.
Remark 2.4. Sµ (r) as given in (2.2) – (2.3) may be written in the alternate form
√
Z ∞
i
π
x h
µ− 12
1 (rx) dx,
(2.14)
Sµ (r) = µ− 1
(rx)
J
µ− 2
2 2 r2µ−1 Γ (µ + 1) 0 ex − 1
which, for µ = m, a positive integer
(2.15)
r Z ∞
π
x
Sm (r) = m−1 2m−1
Rm (rx) dx,
x
2
r
m! 2 0 e − 1
1
where
r
(2.16)
π
Rm (z) =
2
r
π m− 1
z 2 Jm− 1 (z) .
2
2
For m = 1, 2, 3, 4 we have
r
π
Rm (z) = sin z, sin z − z cos z, 3 sin z − 3z cos z − z 2 sin z,
2
and
15 sin z − 15z cos z − 6z 2 sin z + z 3 cos z,
respectively.
Thus, for example,
Z
1 ∞ x
S1 (r) =
sin (rx) dx,
r 0 ex − 1
Z ∞
1
x
S2 (r) = 3
[sin (rx) − (rx) cos (rx)] dx,
x
4r 0 e − 1
Z ∞
1
x S3 (r) =
3 sin (rx) − 3 (rx) cos (rx) − (rx)2 sin (rx) dx,
5
x
24r 0 e − 1
and
1
S4 (r) =
192r7
Z
0
∞
ex
x
[15 sin (rx) − 15 (rx) cos (rx)
−1
−6 (rx)2 sin (rx) + (rx) cos (rx) dx.
J. Inequal. Pure and Appl. Math., 4(5) Art. 100, 2003
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6
P. C ERONE AND C.T. L ENARD
The above results for integer m can also be obtained using the relationship from (1.1) and
(1.3)
(2.17)
S1 (r) = S (r) =
∞
X
n=1
1
2n
=
2
r
(n2 + r2 )
Z
∞
x
sin (rx) dx.
ex − 1
0
Formal differentiation with respect to r of (2.17) gives
Z ∞
x cos rx sin rx
x
(−4r) S2 (r) =
−
dx
ex − 1
r
r2
0
Z
1 ∞ x
=− 2
(sin rx − rx cos rx) dx
r 0 ex − 1
producing the result above. Continuing in this manner would produce further representations
for Sm (r) .
The following theorem gives an explicit representation for Sm (r) , m ∈ N.
Theorem 2.5. For m a positive integer we have
(2.18)
Sm (r) =
1
1
·
2m−1 r2m−1
3k
m−1
1 X (−1)b 2 c k
·
r [δk even Ak (r) + δk odd Bk (r)] ,
m k=0
k!
where
∞
Z
(2.19)
Ak (r) =
0
xk+1
sin (rx) dx, Bk (r) =
ex − 1
Z
0
∞
xk+1
cos (rx) dx,
ex − 1
with δcondition = 1 if condition holds and zero otherwise and bxc is the greatest integer less than
or equal to x.
Proof. From (2.17) we may differentiate m − 1 times with respect to r to produce
(m−1)
S1
(2.20)
(r) = (−1)m−1 m! (2r)m−1 Sm (r)
Z ∞
x
dm−1 sin rx
=
·
dx.
ex − 1 drm−1
r
0
Now,
(2.21)
dm−1
drm−1
sin rx
r
=
m−1
X
k=0
m−1
k
dm−1−k −1 dk
r
· k (sin rx)
drm−1−k
dr
and
dl −1 r
= (−1)l l!r−(l+1) ,
l
dr
k
dk
(sin rx) = (−1)b 2 c xk [δk even sin (rx) + δk odd cos (rx)]
k
dr
where δcondition = 1 if condition is true and zero otherwise.
J. Inequal. Pure and Appl. Math., 4(5) Art. 100, 2003
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M ATHIEU S ERIES
7
Thus from (2.21)
dm−1 sin (rx)
(2.22)
drm−1
r
m−1 k
1 X m−1
= m
(−1)m−1−k+b 2 c
r k=0
k
× (m − 1 − k)!rk xk [δk even sin (rx) + δk odd cos (rx)]
m−1
X (−1)b 3k2 c
(−1)m−1
rk xk [δk even sin (rx) + δk odd cos (rx)] .
=
(m − 1)!
k!
rm
k=0
Substitution of (2.22) into (2.20) and simplifying produces the stated result (2.18).
Remark 2.6. The integral representation for Sm (r) given in Theorem 2.5 is simpler than that
obtained in [18] as given by (1.7). Further, the derivation here is much more straight forward.
3. B OUNDS FOR Sµ (r)
It was stated in the introduction that considerable effort has been expended in determining
bounds for the generalised Mathieu series. More recently, bounds for the generalised Mathieu series (1.6) has been investigated in particular by Qi and coworkers and by Tomovski and
Trenčevski [18].
In a recent article Landau [11] obtained the best possible uniform bounds for Bessel functions
using monotonicity arguments. Of particular interest to us here is that he showed that
bL
(3.1)
|Jν (x)| < 1
ν3
uniformly in the argument x and is best possible in the exponent 13 and constant
1
bL = 2 3 sup Ai (x) = 0.674885 · · · ,
(3.2)
x
where Ai (x) is the Airy function satisfying
w00 − xw = 0.
Landau also showed that
|Jν (x)| ≤
(3.3)
uniformly in the order ν > 0 and the exponent
1
3
cL
1
x3
is best possible with
1
cL = sup x 3 J0 (x)
(3.4)
x
= 0.78574687 . . . .
The following theorem is based on the Landau bounds (3.1) – (3.4).
Theorem 3.1. The generalised Mathieu series Sµ (r) satisfies the bounds for µ >
√
Γ µ + 32
1
3
π
ζ µ+
,
(3.5)
Sµ (r) ≤ bL
1 ·
1 ·
2
(2r)µ− 2
µ − 12 3 Γ (µ + 1)
and
(3.6)
√
Sµ (r) ≤ cL ·
1
π
1
2µ− 2 rµ− 6
J. Inequal. Pure and Appl. Math., 4(5) Art. 100, 2003
7
·Γ µ+
6
1
2
and r > 0
7
ζ µ+
,
6
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8
P. C ERONE AND C.T. L ENARD
where bL and cL are given by (4.2) and (4.4) respectively.
Proof. From (2.2) and (2.3) we have
Z
(3.7)
1
xµ+ 2 1
Jµ− 2 (rx) dx, r > 0
x
e −1
∞
Sµ (r) ≤ Cµ (r)
0
and so from (3.1) we obtain, on utilising (2.6)
3
µ+
Sµ (r) ≤ Cµ (r) ·
1 Γ
2
µ − 12 3
bL
3
ζ µ+
,
2
which simplifies down to (3.5).
Further, using (3.3) into (3.7) gives
1
∞
Z
xµ+ 2
1
·
dx
ex − 1 |rx| 13
Sµ (r) ≤ Cµ (r) · cL ·
0
= Cµ (r) ·
cL
∞
Z
1
r3
0
1
xµ+ 6
dx
ex − 1
which upon using (2.6) produces
(3.8)
cL
7
Sµ (r) ≤ Cµ (r) · 1 · Γ µ +
6
r3
7
ζ µ+
.
6
Simplifying (3.8) and using (2.3) gives the stated result (3.6).
Corollary 3.2. The Mathieu series S (r) satisfies the following bounds
3π
5
(3.9)
S (r) ≤ 11 bL ζ
2
2 12
and
(3.10)
7cL
S (r) ≤
·
36
r
π
·Γ
2
5
1
13
ζ
· r− 6 ,
6
6
where bL and cL are given by (3.2) and (3.4) respectively.
Proof. Taking µ = 1 in (3.5) and (3.6), noting that S (r) = S1 (r) gives the stated results after
some simplification.
The following corollary gives coarser bounds than Theorem 3.1 without the presence of the
zeta function.
Corollary 3.3. The generalised Mathieu series Sµ (r) satisfies the bounds for µ >
√
Γ µ + 12
bL
1
(3.11)
Sµ (r) ≤ 2 π ·
1 · µ− 1 ·
µ − 1 3 r 2 Γ (µ + 1)
1
2
and r > 0
2
and
(3.12)
Sµ (r) ≤ 2
2
3
√
π·
cL
1
rµ− 6
Γ µ + 61
·
Γ (µ + 1)
with bL and cL given by (3.2) and (3.4).
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M ATHIEU S ERIES
9
Proof. We use the well known inequality
e−x <
ex
x
x
< e− 2
−1
to produce from (3.7)
Z
(3.13)
Sµ (r) ≤ Cµ (r)
0
∞
1 x
e− 2 xµ− 2 Jµ− 1 (rx) dx.
2
We know from Laplace transforms or the definition of the gamma function that
Z ∞
Γ (s + 1)
.
(3.14)
e−αx xs dx =
αs+1
0
Hence, placing (3.1) into (3.13) and utilising (3.14) we obtain (3.11) after simplification. A
similar approach produces (3.12) starting from (3.3) rather than (3.1).
4. F URTHER I NTEGRAL E XPRESSIONS FOR G ENERALISED M ATHIEU S ERIES
In [18], Tomovski and Trenčevski gave the integral representation
Z ∞
2
2
xµ e−r x f (x) dx,
(4.1)
Sµ (r) =
Γ (µ + 1) 0
where
∞
X
2
(4.2)
f (x) =
ne−n x , convergent for finite x > 0,
n=1
by effectively utilising the result (3.14).
They leave the summation of the series in (4.2) as an open problem.
If we place ν = µ − 21 and β = r, all real in (2.9) then we obtain the identity
Z ∞
1
2α
(4.3)
Cµ (r)
e−αx xµ+ 2 Jµ− 1 (rx) dx =
,
2
2
[α + r2 ]µ+1
0
where Cµ (r) is as given by (2.3).
Proof B of Theorem 2.1 takes α = n and sums to produce the identity (2.1) – (2.2).
If we take α = nγ then we have from (4.3) on summing
∞
X
2nγ
(4.4)
S (r; µ, γ) =
(n2γ + r2 )µ+1
n=1
!
Z ∞ X
∞
1
γ
= Cµ (r)
e−n x xµ+ 2 Jµ− 1 (rx) dx,
0
2
n=1
giving an integral representation that was left as an open problem by Qi [15].
As a matter of fact, if we take α = an where a = (a1 , a2 , . . . , an , . . . ) is a positive sequence,
then
∞
X
2an
(4.5)
S (r; µ; a) =
(a2n + r2 )µ+1
n=1
!
Z ∞ X
∞
1
= Cµ (r)
e−an x xµ+ 2 Jµ− 1 (rx) dx.
0
2
n=1
We note that for a+ = (1γ , 2γ , . . . ) then
S r; µ; a+ = S (r; µ, γ) .
J. Inequal. Pure and Appl. Math., 4(5) Art. 100, 2003
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10
P. C ERONE AND C.T. L ENARD
The series
∞
X
n=1
2an
(a2n + r2 )2
has been investigated in [16].
A closed form expression for
F (a) =
∞
X
e−an x ,
x>0
n=1
where an is a positive sequence, remains an open problem.
If a∗ = (1, 2, 3, . . . , n, . . . ), then
1
F (a∗ ) = x
.
e −1
R EFERENCES
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