# Integral Averages of Two Generalizations of the Poisson Kernel by Haruki and Rassias

Authors: Serap Bulut
Journal: Journal of Applied Mathematics and Stochastic Analysis
 (2008-02-14)

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## Abstract

In \tccheck{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 \tccheck{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.


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## Body

## 1. Introduction

It is well known that the Poisson kernel in two dimensions is
defined by(1.1)P(r,θ)=def1−r2(1−reiθ)(1−re−iθ),and the integral
formula(1.2)12π∫02πP(r,θ)dθ=1holds. 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(1.3)Q(θ;a,b)=def1−ab(1−aeiθ)(1−be−iθ),where a, b are complex
parameters satisfying |a|<1 and |b|<1.The second generalization is defined
by(1.4)R(θ;a,b,c,d)=L(a,b,c,d)(1−aeiθ)(1−be−iθ)(1−ceiθ)(1−de−iθ),where a, b, c, d are complex
parameters satisfying |a|<1, |b|<1, |c|<1, and |d|<1 as well
as(1.5)L(a,b,c,d)=def(1−ab)(1−ad)(1−bc)(1−cd)1−abcd.Then they proved the integral formulas(1.6)12π∫02πQ(θ;a,b)dθ=1,(1.7)12π∫02πR(θ;a,b,c,d)dθ=1.Remark 1.1.
If
we setc=a and d=b in (1.7), then we
obtain(1.8)12π∫02πQ(θ;a,b)2dθ=1+ab1−ab.
Afterwards, they set the following definition and open
problem.
Forn=0,1,2,…, let(1.9)In=def12π∫02πQ(θ;a,b)n+1dθ,where a, b are complex
parameters satisfying |a|<1 and |b|<1.Open Problem 1.2.
ComputeIn 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(2.1)In=∑k=0n(n+k)!(n−k)!(k!)2(ab1−ab)k,where In is defined by (1.9).Proof.
We
have(2.2)In=12π∫02π(1−ab(1−aeiθ)(1−be−iθ))n+1dθ=12π∫02π((1−ab)/(1−aeiθ))n+1(1−be−iθ)n+1dθ.By the change of variables z=eiθ and
setting(2.3)f(z)=def(1−ab1−az)n+1zn,we have(2.4)In=12πi∫|z|=1f(z)(z−b)n+1dz,where the complex integral of the
function f(z) along the unit
circle |z|=1 is in the
positive direction.
Sincef(z) is an analytic
function in |z|≤1, by Cauchy's integral formula for the derivative, we
obtain(2.5)In=f(n)(b)n!.So we must calculate f(n)(z). For this purpose, we will use the Leibniz rule
(generalized product rule).
Let(2.6)g(z)=defzn,h(z)=def(1−az)−(n+1).Thus by (2.3) and (2.6), we
have(2.7)f(z)=(1−ab)n+1g(z)h(z).Applying the Leibniz rule to
(2.7), we
get(2.8)f(n)(z)=(1−ab)n+1(gh)(n)(z)=(1−ab)n+1∑k=0n(nk)g(n−k)(z)h(k)(z)=n!(1−ab)n+1∑k=0n(n+k)!(n−k)!(k!)2(az)k(1−az)−(n+k+1), where(2.9)g(n−k)(z)=n!k!zk,h(k)(z)=ak(n+k)!n!(1−az)−(n+k+1).If we take z=b in (2.8), we
obtain(2.10)f(n)(b)n!=∑k=0n(n+k)!(n−k)!(k!)2(ab1−ab)k.Thus by (2.5) and (2.10), we get the
desired result.


## 3. New Generalizations of the Open Problem

In [3], the authors gave the values of the
integral(3.1)12π∫02πPn+1(r,θ)dθfor all real n>−1.In this section, we will generalizeIn, and hence above integral as follows.Theorem 3.1  Main theorem.
For any real numberu, it holds
that
(3.2)Ju:=12π∫02πQ(θ;a,b)udθ=(1−ab)uF12(u,u;1;ab),where  F12 is the usual hypergeometric
function.Proof.
Letu be any real
number. Define the shifted factorial (or the Pochhammer symbol)
by(3.3)(u)k:=Γ(u+k)Γ(u)(u⁢≠−n,⁢n=0,1,2,…),where Γ is the gamma
function. If u=−n is a
nonpositive integer, define (−n)k:=(−n)(−n+1)⋯(−n+k−1) so that (−n)k=0 for k=n+1,n+2,…. Then(3.4)1(1−w)u=∑k=0∞(u)kk!wk(|w|<1). For z=eiθ, one computes that(3.5)Ju=12π∫02πQ(θ;a,b)udθ=12π∫02π(1−ab)u(1−aeiθ)u(1−be−iθ)udθ=(1−ab)u2πi∫|z|=1dzz(1−az)u(1−b/z)u=(1−ab)u2πi∫|z|=11z(∑k=0∞(u)kk!akzk)(∑l=0∞(u)ll!blzl)dz.The integral of the terms with k≠l is 0 by residue
theorem, and thus(3.6)Ju=(1−ab)u∑k=0∞(u)k(u)k(1)kk!(ab)k=(1−ab)uF12(u,u;1;ab),where F12 is the usual hypergeometric
function.It is routine to check that(3.7)J1=1,J2=1+ab1−ab,as obtained in [1] because, then, the series
above is summable via elementary functions. Also for n=0,1,2,…, one has(3.8)Jn=(1−ab)n∑k=0∞(n−1+kk)2(ab)k,J−n=1(1−ab)n∑k=0n(nk)2(ab)k.Moreover, settinga=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(3.9)Ku:=12π∫02πR(θ;a,b,c,d)udθ=L(a,b,c,d)u∑j+l=k+m(u)j(u)k(u)l(u)mj!k!l!m!ajbkcldm.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(3.10)Du(xp)=dmdxm[1Γ(m−u)∫0x(x−t)m−u−1tpdt]=dmdxm[xm−u+pΓ(m−u)∫01(1−s)m−u−1spds]=dmdxm[xm−u+pΓ(m−u)B(m−u,p+1)]=dmdxm[Γ(p+1)Γ(m−u+p+1)xm−u+p]=xp−u(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,….Ifu≠0,−1,−2,…, then(3.11)1Γ(u)2Du−1(xu−1Du−1(xu−11−x))=1Γ(u)2Du−1(xu−1Du−1(∑k=0∞xk+u−1))=∑k=0∞(u)k2(k!)2xkby
successively applying the above fractional
differentiation formula. Thus(3.12)Ju=(1−x)uΓ(u)2Du−1(xu−1Du−1(xu−11−x))|x=ab.



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*Source: 760214-2008-03-16.xml*