AND AN

(1)

Journal

of

Applied Mathematics and Stochastic Analysis, 12:3

(1999),

^261-263.

THE SOLUTION OF AN OPEN PROBLEM GIVEN BY H. HARUKI AND T.M. RASSIAS

BARA KIM

Korea

Advanced Institute

of

Science and Technology

(KAIST)

Department of

Mathematics and

Center for

^AppliedMathematics 39’3-1

Kusong-Dong, Yusong-Gu,

Taejon

305-?’01, Korea

e-mail: [email protected]

(Received

^July,

1998;

Revised

December, 1998)

Haruki and Rassias

[1]

generalized the Poisson kernel in two dimensions and discussed integral formulas for each case. They presented an open problem for an integralformula.

In

this paper, wegive asolution to that problem.

Key

words: Poisson

Kernel,

Integral Formula.

AMS

subject classifications:

31A05,

31A10.

1. Introduction

Haruki and Rassias

[1]

^introduced ^two

^types

of generalizations of the Poisson kernel.

One

of them is defined by

Q(O;

a,

b)

^A_ ¹ ^ab

(1-aei)(1 -be-i)

where

a,b

^are complex parameters satisfying

]a <

^{1 and}

]b] <

^1.

They proved the integralformulas:

27r

l/ ^)dO ^1,

2r

Q(O;a,

0

(1)

27t"

l_J_27r / ^Q(O;

^a,

^b)2dO

0

1

+ab

They set the open problem as follows:

"Let

(2)

262

BARA KIM

27I"

0 2-

2--7

₀

(1 )(1 o) ^eo, ⁽ⁿ ^O, ^1,...), ⁽³⁾

where

a,b

^are complex parameters

satisfying

a

<

1 and

bl ^< . ^{Compute I}

ⁿ^for

n

2, 3, 4, "

In

the next section, we will give the solution tothe problem.

2. Solution of the Problem

Theorem 1"

In, defined

^by

(3), satisfies

n

In-- E ^(2n--j)! (

^ab

)n-j

j o

j!((n- j)!)2

¹ ^ab

for

ⁿ

0, 1, 2,...,

and complex values a,b are such that a

<

¹ ^and

b[ <

^1.

Proof:

By

the

change

of

variables,

^with ^z-^e

iO, (3)

^becomes

-1./(1-ab)

^n+l

In

_2ri

(1 az)(1

^bz

1) ^z-ldz

Let

1

/ ^(1-ab)

^n+l

2ri 1

az/ zn(z b)

ⁿ

ldz"

f(z)

^A__.

(--ab)

⁺¹

Then

f(z)

is analytic on

{zEC’]zl <l,z-7 6b}

ând ^has â ^pole ât ^z-b.

Therefore,

by the residue

theorem, I

_n is the residue of

f(z)

^at ^z- ^b.

The

Laurent

series expansion of

f(z)

^at ^z ^b ^gives:

(

¹

)n+l

f(z)

1

1,,,ab(Z ^b) ^(b ⁺ ^(z ^b))n(z ^b)

n 1

E ^n+k

^a ^k

^)k

ⁿ

k 0 k ¹

ab (z

^b

E

^bⁿ

^{J(z b)J(z} ^b)

ⁿ ¹

3=0

J

k=O j=O k j 1-ab

Therefore,

the residue of

f(z)

^at

b,

which is

In,

^is ^{given by}

(3)

Solution

of

^an

^Open

Problem Given by Haruki and Rassias 263

I

_n

Z

^2n-j ⁿ

.i..ab

^n-j

3-0 n-j j -ab

n _j!((n ^(2n_ _j)v)2 ^j)! ⁽

^1-ab^ab

^)r_

^j

j--o

Note

that weobtain

(1)

^and

(2)

by substituting n 0 and n

1,

respectively.

References

[1]

Haruki,

H.

and

Rassias, T.M., ^New

generalizations of the Poisson

kernel, J.

Appl. Math. Stoch. Anal. 10:2

(1997),

^191-196.

AND AN

of

(1999),

THE SOLUTION OF AN OPEN PROBLEM GIVEN BY H. HARUKI AND T.M. RASSIAS

BARA KIM

Korea

of

(KAIST)

Department of

Center for

Kusong-Dong, Yusong-Gu,

305-?’01, Korea

(Received

1998;

December, 1998)

[1]

In

Key

Kernel,

AMS

31A05,

1. Introduction

[1]

types

One

Q(O;

b)

(1-aei)(1 -be-i)

a,b

]a <

]b] <

l/ )dO 1,

Q(O;a,

(1)

l_J_27r / Q(O;

b)2dO

+ab

"Let

BARA KIM

2--7

(1 )(1 o) eo, (n O, 1,...), (3)

a,b

satisfying

<

bl < . Compute I

2, 3, 4, "

In

2. Solution of the Problem

In, defined

(3), satisfies

In-- E (2n--j)! (

)n-j

j!((n- j)!)2

for

0, 1, 2,...,

<

b[ <

By

change

variables,

iO, (3)

-1./(1-ab)

In

(1 az)(1

1) z-ldz

Let

/ (1-ab)

az/ zn(z b)

ldz"

f(z)

(--ab)

f(z)

{zEC’]zl <l,z-7 6b}

Therefore,

theorem, I

f(z)

Laurent

f(z)

(

)n+l

^types

l/ ^)dO ^1,

l_J_27r / ^Q(O;

^b)2dO

(1 )(1 o) ^eo, ⁽ⁿ ^O, ^1,...), ⁽³⁾

bl ^< . ^{Compute I}

In-- E ^(2n--j)! (

1) ^z-ldz

/ ^(1-ab)

1,,,ab(Z ^b) ^(b ⁺ ^(z ^b))n(z ^b)

E ^n+k

^)k

^{J(z b)J(z} ^b)

^Open

n _j!((n ^(2n_ _j)v)2 ^j)! ⁽

^)r_

Rassias, T.M., ^New