Hindawi Publishing Corporation
Journal of Applied Mathematics and Stochastic Analysis
Volume 2008, Article ID 760214, 6 pages
doi:10.1155/2008/760214
Research Article
Integral Averages of Two Generalizations of
the Poisson Kernel by Haruki and Rassias
Serap Bulut
Kocaeli University, Civil Aviation College, Arslanbey Campus, 41285 Izmit, Turkey
Correspondence should be addressed to Serap Bulut, bulutserap@yahoo.com
Received 15 November 2007; Revised 13 January 2008; Accepted 17 January 2008
Recommended by John Rassias
In 1997, Haruki and Rassias introduced two generalizations of the Poisson kernel in two dimensions
and discussed integral formulas for them. Furthermore, they presented an open problem. In 1999,
Kim gave a solution to that problem. Here, we give a solution to this open problem by means of a
different method. The purpose of this paper is to give integral averages of two generalizations of
the Poisson kernel, that is, we generalize the open problem.
Copyright q 2008 Serap Bulut. This is an open access article distributed under the Creative
Commons Attribution License, which permits unrestricted use, distribution, and reproduction in
any medium, provided the original work is properly cited.
1. Introduction
It is well known that the Poisson kernel in two dimensions is defined by
1 − r2
def
P r, θ ,
iθ
1 − re
1 − re−iθ
1.1
and the integral formula
1
2π
2π
P r, θdθ 1
1.2
0
holds. Here r is a real parameter satisfying |r| < 1.
In 1, Haruki and Rassias introduced two generalizations of the Poisson kernel.
The first generalization is defined by
def
Qθ; a, b 1−
1 − ab
,
1 − be−iθ
aeiθ
where a, b are complex parameters satisfying |a| < 1 and |b| < 1.
1.3
2
Journal of Applied Mathematics and Stochastic Analysis
The second generalization is defined by
Rθ; a, b, c, d 1−
aeiθ
La, b, c, d
,
1 − be−iθ 1 − ceiθ 1 − de−iθ
1.4
where a, b, c, d are complex parameters satisfying |a| < 1, |b| < 1, |c| < 1, and |d| < 1 as well as
1 − ab1 − ad1 − bc1 − cd
.
1 − abcd
def
La, b, c, d 1.5
Then they proved the integral formulas
2π
1
2π
1
2π
Qθ; a, bdθ 1,
1.6
Rθ; a, b, c, ddθ 1.
1.7
0
2π
0
Remark 1.1. If we set c a and d b in 1.7, then we obtain
1
2π
2π
Qθ; a, b2 dθ 0
1 ab
.
1 − ab
1.8
Afterwards, they set the following definition and open problem.
For n 0, 1, 2, . . ., let
1
In 2π
def
2π
Qθ; a, bn1 dθ,
1.9
0
where a, b are complex parameters satisfying |a| < 1 and |b| < 1.
Open Problem 1.2. Compute In for n 2, 3, 4, . . . .
In 2, Kim gave a solution to this open problem using the Laurent series expansion.
In the next section, we give a solution to the open problem by means of the Leibniz rule.
2. A different solution of the open problem
Theorem 2.1. It holds that
In n
n k!
k0
n − k!k!2
ab
1 − ab
n1
2π k
,
2.1
where In is defined by 1.9.
Proof. We have
1
In 2π
2π 0
1 − ab
iθ
1 − ae
1 − be−iθ
1
dθ 2π
0
n1
1 − ab/ 1 − aeiθ
dθ.
n1
1 − be−iθ
2.2
Serap Bulut
3
By the change of variables z eiθ and setting
1 − ab n1 n
def
fz z ,
1 − az
we have
1
In 2πi
fz
|z|1
z − bn1
dz,
2.3
2.4
where the complex integral of the function fz along the unit circle |z| 1 is in the positive
direction.
Since fz is an analytic function in |z| ≤ 1, by Cauchy’s integral formula for the derivative, we obtain
In f n b
.
n!
2.5
So we must calculate f n z. For this purpose, we will use the Leibniz rule generalized product rule.
Let
def
gz zn ,
−n1
def
hz 1 − az
2.6
.
Thus by 2.3 and 2.6, we have
fz 1 − abn1 gzhz.
2.7
Applying the Leibniz rule to 2.7, we get
f n z 1 − abn1 ghn z
1 − ab
n1
n
k0
n!1 − abn1
n
g n−k zhk z
k
n
n k!
k0
n − k!k!2
2.8
azk 1 − az−nk1 ,
where
g n−k z hk z ak
n! k
z ,
k!
n k!
1 − az−nk1 .
n!
2.9
If we take z b in 2.8, we obtain
k
n
f n b ab
n k!
.
2 1 − ab
n!
k0 n − k!k!
Thus by 2.5 and 2.10, we get the desired result.
2.10
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Journal of Applied Mathematics and Stochastic Analysis
3. New generalizations of the open problem
In 3, the authors gave the values of the integral
1
2π
2π
P n1 r, θdθ
3.1
0
for all real n > −1.
In this section, we will generalize In , and hence above integral as follows.
Theorem 3.1 Main theorem. For any real number u, it holds that
Ju :
1
2π
2π
Qθ; a, bu dθ 1 − abu 2 F1 u, u; 1; ab,
3.2
0
where 2 F1 is the usual hypergeometric function.
Proof. Let u be any real number. Define the shifted factorial or the Pochhammer symbol by
uk :
Γu k
Γu
u / − n, n 0, 1, 2, . . .,
3.3
where Γ is the gamma function. If u −n is a nonpositive integer, define −nk : −n−n 1 · · · −n k − 1 so that −nk 0 for k n 1, n 2, . . . . Then
∞
uk k
1
w
u 1 − w k0 k!
|w| < 1 .
3.4
For z eiθ , one computes that
1
Ju 2π
2π
0
1
Qθ; a, b dθ 2π
1 − abu
2πi
1 − abu
2πi
u
|z|1
|z|1
2π
0
1 − abu
u u dθ
1 − aeiθ 1 − be−iθ
dz
z1 − azu 1 − b/zu
1
z
∞
uk k k
a z
k!
k0
3.5
∞
ul bl
l! zl
l0
dz.
The integral of the terms with k /
l is 0 by residue theorem, and thus
Ju 1 − abu
∞
uk uk
abk 1 − abu 2 F1 u, u; 1; ab,
1k k!
k0
where 2 F1 is the usual hypergeometric function.
3.6
Serap Bulut
5
It is routine to check that
J1 1,
J2 1 ab
,
1 − ab
3.7
as obtained in 1 because, then, the series above is summable via elementary functions. Also
for n 0, 1, 2, . . ., one has
2
∞
n−1k
n
Jn 1 − ab
abk ,
k
k0
3.8
2
n
1
n
J−n abk .
1 − abn k0 k
Moreover, setting a b r generalizes the results of 3 to all real powers u of the
Poisson kernel.
The same method applied to the integral averages of the second generalization of the
Poisson kernel yields
Ku :
1
2π
2π
Rθ; a, b, c, du dθ La, b, c, du
0
uj uk ul um
jlkm
j!k!l!m!
aj bk cl dm .
3.9
There is a further connection with the fractional-order derivative in 3 which is called
Du here for any real number u. If p is also any real number, let m p be the least integer
greater than or equal to p. Then one can compute with s t/x that
x
p
1
dm
m−u−1 p
u
D x x − t
t dt
dxm Γm − u 0
m−up 1
x
dm
m−u−1 p
1
−
s
s
ds
dxm Γm − u 0
m−up
x
dm
Bm
−
u,
p
1
dxm Γm − u
3.10
Γp 1
xp−u
dm
m−up
,
x
m
dx Γm − u p 1
p 1−u
which agrees with the usual derivative when u is a positive integer, where B is the beta function, u /
p 1, p 2, . . ., and p /
0, −1, −2, . . . .
If u /
0, −1, −2, . . ., then
u−1 ∞
∞
u2k k
1
1
x
u−1
u−1 u−1
u−1
u−1 u−1
ku−1
x
D
D
D
D
x
x 3.11
x
2
1−x
Γu2
Γu2
k0
k0 k!
by successively applying the above fractional differentiation formula. Thus
1 − xu u−1 u−1 u−1 xu−1
x
Ju D
D
.
1−x
Γu2
xab
3.12
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Journal of Applied Mathematics and Stochastic Analysis
Acknowledgment
The author is grateful to the referee for useful comments and suggestions.
References
1 H. Haruki and T. M. Rassias, “New generalizations of the Poisson kernel,” Journal of Applied Mathematics
and Stochastic Analysis, vol. 10, no. 2, pp. 191–196, 1997.
2 B. Kim, “The solution of an open problem given by H. Haruki and T. M. Rassias,” Journal of Applied
Mathematics and Stochastic Analysis, vol. 12, no. 3, pp. 261–263, 1999.
3 A. S. Galbraith and J. W. Green, “A note on the mean value of the Poisson kernel,” Bulletin of the American Mathematical Society, vol. 53, pp. 314–320, 1947.