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

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.

1. Introduction

It is well known that the Poisson kernel in two dimensions is defined by𝑃(𝑟,𝜃)=^def1−𝑟²1−𝑟𝑒^𝑖𝜃1−𝑟𝑒^−𝑖𝜃,(1.1)and the integral formula12𝜋^02𝜋𝑃(𝑟,𝜃)𝑑𝜃=1(1.2)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 by𝑄(𝜃;𝑎,𝑏)=^def1−𝑎𝑏1−𝑎𝑒^𝑖𝜃1−𝑏𝑒^−𝑖𝜃,(1.3)where 𝑎, 𝑏 are complex parameters satisfying |𝑎|<1 and |𝑏|<1.

The second generalization is defined by𝑅(𝜃;𝑎,𝑏,𝑐,𝑑)=𝐿(𝑎,𝑏,𝑐,𝑑)1−𝑎𝑒^𝑖𝜃1−𝑏𝑒^−𝑖𝜃1−𝑐𝑒^𝑖𝜃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 as𝐿(𝑎,𝑏,𝑐,𝑑)=^def(1−𝑎𝑏)(1−𝑎𝑑)(1−𝑏𝑐)(1−𝑐𝑑)1−𝑎𝑏𝑐𝑑.(1.5)

Then they proved the integral formulas12𝜋^02𝜋1𝑄(𝜃;𝑎,𝑏)𝑑𝜃=1,(1.6)2𝜋^02𝜋𝑅(𝜃;𝑎,𝑏,𝑐,𝑑)𝑑𝜃=1.(1.7)𝑐=𝑎

Remark 1.1. If we set 𝑑=𝑏 and 12𝜋^02𝜋𝑄(𝜃;𝑎,𝑏)²𝑑𝜃=1+𝑎𝑏1−𝑎𝑏.(1.8) in (1.7), then we obtain𝑛=0,1,2,…
Afterwards, they set the following definition and open problem.
For 𝐼=_𝑛^def12𝜋^02𝜋𝑄(𝜃;𝑎,𝑏)^𝑛+1𝑑𝜃,(1.9), let𝑎where 𝑏, |𝑎|<1 are complex parameters satisfying |𝑏|<1 and 𝐼_𝑛.

Open Problem 1.2. Compute 𝑛=2,3,4,…. for 𝐼_𝑛=^𝑛_𝑘=0(𝑛+𝑘)!(𝑛−𝑘)!(𝑘!)²1−𝑎𝑏𝑎𝑏^𝑘,(2.1)
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 𝐼_𝑛where 𝐼_𝑛=12𝜋∫^02𝜋(1−𝑎𝑏1−𝑎𝑒^𝑖𝜃1−𝑏𝑒^−𝑖𝜃)^𝑛+1𝑑𝜃=12𝜋∫^02𝜋(1−𝑎𝑏)/1−𝑎𝑒^𝑖𝜃^𝑛+11−𝑏𝑒^−𝑖𝜃^𝑛+1𝑑𝜃.(2.2) is defined by (1.9). Proof. We have𝑧=𝑒^𝑖𝜃By the change of variables 𝑓(𝑧)=^def1−𝑎𝑏1−𝑎𝑧^𝑛+1𝑧^𝑛,(2.3) and setting𝐼_𝑛=12𝜋𝑖||_𝑧||₌₁𝑓(𝑧)(𝑧−𝑏)^𝑛+1𝑑𝑧,(2.4)we have𝑓(𝑧)where the complex integral of the function |𝑧|=1 along the unit circle 𝑓(𝑧) is in the positive direction.
Since |𝑧|≤1 is an analytic function in 𝐼_𝑛=𝑓^(𝑛)(𝑏)𝑛!.(2.5), by Cauchy's integral formula for the derivative, we obtain𝑓^(𝑛)(𝑧)So we must calculate ℎ(𝑧)𝑔_def(𝑧)=^def𝑧^𝑛,=(1−𝑎𝑧)^−(𝑛+1).(2.6). For this purpose, we will use the Leibniz rule (generalized product rule).
Let𝑓(𝑧)=(1−𝑎𝑏)^𝑛+1𝑔(𝑧)ℎ(𝑧).(2.7)Thus by (2.3) and (2.6), we have𝑓^(𝑛)(𝑧)==((11−−𝑎𝑎𝑏𝑏))^𝑛+1_𝑛𝑛+1(𝑔ℎ)^(𝑛)(𝑧)_𝑘=0𝑘𝑛𝑔^{(𝑛−𝑘)}=𝑛!(1−𝑎𝑏)_𝑛𝑛+1(𝑧)ℎ^(𝑘)(𝑧)_𝑘=0(𝑛+𝑘)!(𝑛−𝑘)!(𝑘!)²(𝑎𝑧)^𝑘(1−𝑎𝑧)^{−(𝑛+𝑘+1)},(2.8)Applying the Leibniz rule to (2.7), we get𝑔^{(𝑛−𝑘)}(𝑧)=𝑛!𝑘!𝑧^𝑘ℎ_(𝑘),(𝑧)=𝑎^𝑘(𝑛+𝑘)!𝑛!(1−𝑎𝑧)^{−(𝑛+𝑘+1)}.(2.9) where𝑧=𝑏If we take 𝑓^(𝑛)(𝑏)𝑛!=^𝑛_𝑘=0(𝑛+𝑘)!(𝑛−𝑘)!(𝑘!)²1−𝑎𝑏𝑎𝑏^𝑘.(2.10) in (2.8), we obtain12𝜋^02𝜋𝑃^𝑛+1(𝑟,𝜃)𝑑𝜃(3.1)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 integral𝑛>−1for all real 𝐼_𝑛.

In this section, we will generalize 𝑢, and hence above integral as follows.

Theorem 3.1 Main theorem. For any real number  𝐽_𝑢∶=12𝜋^02𝜋𝑄𝜃;𝑎,𝑏^𝑢𝑑𝜃=(1−𝑎𝑏)^𝑢2𝐹₁𝑢,𝑢;1;𝑎𝑏,(3.2), it holds that ₂𝐹₁where  𝑢 is the usual hypergeometric function.Proof. Let 𝑢_𝑘∶=Γ(Γ𝑢(+𝑢)𝑘)𝑢≠−𝑛,𝑛=0,1,2,…,(3.3) be any real number. Define the shifted factorial (or the Pochhammer symbol) byΓwhere 𝑢=−𝑛 is the gamma function. If (−𝑛)_𝑘∶=(−𝑛)(−𝑛+1)⋯(−𝑛+𝑘−1) is a nonpositive integer, define (−𝑛)_𝑘=0 so that 𝑘=𝑛+1,𝑛+2,…. for 11−𝑤^𝑢=^∞_𝑘=0𝑢_𝑘𝑘!𝑤^𝑘||𝑤||<1.(3.4) Then𝑧=𝑒^𝑖𝜃 For 𝐽_𝑢=12𝜋^02𝜋𝑄𝜃;𝑎,𝑏^𝑢𝑑𝜃=12𝜋^02𝜋1−𝑎𝑏^𝑢1−𝑎𝑒^𝑖𝜃^𝑢1−𝑏𝑒^−𝑖𝜃^𝑢=1−𝑎𝑏_𝑢𝑑𝜃2𝜋𝑖||_𝑧||₌₁𝑑𝑧𝑧1−𝑎𝑧^𝑢1−𝑏/𝑧^𝑢=1−𝑎𝑏^𝑢2𝜋𝑖||_𝑧||₌₁1𝑧(^∞_𝑘=0𝑢_𝑘𝑘!𝑎^𝑘𝑧^𝑘)(^∞_𝑙=0𝑢_𝑙𝑙!𝑧𝑏^𝑙𝑙)𝑑𝑧.(3.5), one computes that𝑘≠𝑙The integral of the terms with 0 is 𝐽_𝑢=1−𝑎𝑏^𝑢∞_𝑘=0𝑢_𝑘𝑢_𝑘1_𝑘𝑘!𝑎𝑏^𝑘=(1−𝑎𝑏)^𝑢2𝐹₁𝑢,𝑢;1;𝑎𝑏,(3.6) by residue theorem, and thus₂𝐹₁where 𝐽₁=1,𝐽₂=1+𝑎𝑏1−𝑎𝑏,(3.7) is the usual hypergeometric function.

It is routine to check that𝑛=0,1,2,…as obtained in [1] because, then, the series above is summable via elementary functions. Also for 𝐽_𝑛=1−𝑎𝑏^𝑛∞_𝑘=0𝑛−1𝑘+𝑘²𝑎𝑏^𝑘𝐽_−𝑛,=11−𝑎𝑏^𝑛𝑛_𝑘=0𝑘𝑛²𝑎𝑏^𝑘.(3.8), one has𝑎=𝑏=𝑟

Moreover, setting 𝑢 generalizes the results of [3] to all real powers 𝐾_𝑢∶=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!𝑎^𝑗𝑏^𝑘𝑐^𝑙𝑑^𝑚.(3.9) of the Poisson kernel.

The same method applied to the integral averages of the second generalization of the Poisson kernel yields𝐷^𝑢

There is a further connection with the fractional-order derivative in [3] which is called 𝑢 here for any real number 𝑝. If 𝑚=⌈𝑝⌉ is also any real number, let 𝑝 be the least integer greater than or equal to 𝑠=𝑡/𝑥. Then one can compute with 𝐷^𝑢𝑥^𝑝=𝑑^𝑚𝑑𝑥^𝑚Γ𝑚1−𝑢^𝑥0𝑥−𝑡^{𝑚−𝑢−1}=𝑑𝑡^𝑝𝑑𝑡_𝑚𝑑𝑥^𝑚Γ𝑥𝑚−𝑢^{𝑚−𝑢+𝑝}¹⁰1−𝑠^{𝑚−𝑢−1}=𝑑𝑠^𝑝𝑑𝑠_𝑚=𝑑𝑑𝑥^𝑚Γ𝑥𝑚−𝑢^{𝑚−𝑢+𝑝}B𝑚−𝑢,𝑝+1_𝑚𝑑𝑥^𝑚Γ𝑚Γ−𝑝𝑢++1𝑝+1𝑥^{𝑚−𝑢+𝑝}=𝑝𝑥+1^𝑝−𝑢_−𝑢,(3.10) that𝑢which agrees with the usual derivative when 𝑢≠𝑝+1,𝑝+2,… is a positive integer, where B is the beta function, 𝑝≠0,−1,−2,…., and 𝑢≠0,−1,−2,…

If 1Γ𝑢²𝐷^𝑢−1𝑥^𝑢−1𝐷^𝑢−1𝑥^𝑢−11−𝑥=Γ1𝑢²𝐷^𝑢−1(𝑥^𝑢−1𝐷^𝑢−1(^∞_𝑘=0𝑥^{𝑘+𝑢−1}))=^∞_𝑘=0𝑢^2𝑘𝑘!²𝑥^𝑘(3.11), then𝐽_𝑢=1−𝑥^𝑢Γ𝑢²𝐷^𝑢−1𝑥^𝑢−1𝐷^𝑢−1𝑥^𝑢−11−𝑥|||_{𝑥=𝑎𝑏}.(3.12)by successively applying the above fractional differentiation formula. Thus

Acknowledgment

The author is grateful to the referee for useful comments and suggestions.