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On The N-Fractional Calculus of Some Products of Some Power Functions(Sakaguchi Functions in Univalent Function Theory and Its Applications)

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(1)

On

The

N-Fractional Calculus of Some Products

of Some

Power

Functions

Katsuyuki

Nishimoto

Abstract

In

this

article,

N-

fractional calculus

of products

of

power

functions

$((z-c)^{\alpha}.(z-c)^{\beta})$

,

,

((z-

$c)^{\beta}\cdot$

$(z-c)^{a}$

),

,

and

$((z-c)^{a+\beta})_{\gamma}$

are

discussed

again.

\S

0.

Introduction

(

Definition

of

Fractional

Calculus

)

(

I

)

Definition.

(

by

K. Nishimoto

)(

[

1

]Vol.

1)

Let

$D=\{D_{-}, D_{+}\}$

,

$C=\{C_{-}, C_{+}\}$

,

$C_{-}$

be

a

curve

along

the cut joining two points

$z$

and

$-\infty+i{\rm Im}(z)$

,

$C_{+}$

be

a

curve

along

the

cut joining two points

$’.f$

and

$\infty$

$+i{\rm Im}(z)$

,

$D_{-}$

be

a

domain surrounded

by C-,

$D_{+}$

be

a

domain

surrounded

by

$C_{+}$

.

(

Here

$D$

contains

the

points

over

the

curve

$C$

).

Moreover,

let

$f=f(z)$

be

a

regular

function

$\mathrm{i}\mathrm{n}\mathrm{Z}$

)

$(\mathrm{z}\in D)$

,

$f_{\mathrm{v}}( \mathrm{z})=(\int)_{\nu}=_{c}(f)_{v}=\frac{\Gamma(\mathrm{v}+1)}{2\pi \mathrm{i}}\int_{c}\frac{f(\zeta)}{(\zeta-z)^{\mathrm{v}+1}}d\zeta$ $(\mathrm{v} \not\in T)$

,

(1)

$(f)_{-m}$

$-$

$\lim_{\mathrm{v}arrow-m}(f)_{v}$

$(m\in Z^{+})$

,

(

2

)

where

$-\pi\leq$

$\arg(\zeta-z)$

$\leq\pi$

for C-,

$0\leq\arg(\zeta-z)\leq 2\pi$

for

$C_{+}$

$\zeta\neq z$

,

$z$

$\in C$

,

$\mathrm{v}$

$\in R$

,

$\Gamma$

;Gamma

function,

then

$(f)_{v}$

is

the fractional

differintegration of arbitrary order

$\mathrm{v}$

(

derivatives of

order

$\mathrm{v}$

for

$\mathrm{v}$

$>0$

,

and

integrals

of

order

$-\mathrm{v}$

for

$\mathrm{v}$

$<0$

),

with

respect

to

$z$

,

of

the function

$f$

,

if

$|(f)_{\nu}|<\infty$

.

(II)

On

the

fractional

calculus

operator

$N^{\mathrm{Y}’}[3 ]$

Theorem A. Let

fractional

calculus

operator

(

Nishimoto’s

operator

\rangle

$N^{\mathrm{v}}$

be

$N^{\mathrm{v}}=( \frac{\Gamma(\mathrm{v}+1)}{2\pi \mathrm{i}}\int_{c}\frac{d\zeta}{(\zeta-z)^{\mathrm{v}+1}})$

(v

(2)

with

$N^{-m}=$

Jim

$N^{\mathrm{v}}$

$(m\in Z^{\star})$

,

(4)

$varrow-\}’\iota$

artd

define

the

binary operation

$\circ$

as

$N^{\beta}\circ N^{a}f=N^{\beta}N^{a}f\approx$

$N^{\beta}(N^{\alpha}f)$ $\langle$$\alpha$

,

$\beta\in E)$

,

(5)

then

the

set

$\{N^{\tau\prime}\}=\{N^{\mathrm{v}}|\mathrm{v}$

$\in R\}$

$\mathrm{t}$

$6)$

is

an

Abelian

product

group

(

having

continuous index

$\mathrm{v}$

)

which has the inverse

transform

operator

$(N^{\mathrm{V}})^{-?}=$$N^{-\mathrm{v}}$

to

the

fractional

calculus

operator

$N^{\mathrm{v}}$

,

for

the

function

$f$

such that

$f\in F=\{f$

;

$0\neq$$|f_{\mathrm{v}}|<\infty$

,

$\mathrm{v}$

$\in R\}$

,

where

$f\approx$

$f(z)$

and

$z$

$\in C$

.

(vis.

$-\infty<\mathrm{v}$ $<\infty$

).

(

For

our

convenience,

we

call

$N^{\beta}\circ N^{a}$

as

product of

$N^{\beta}$

and

$N^{\alpha}$

.

)

Theorem

B.

F.O.G.

$\{N^{\nu}\}$

is

art

.,

Action

product

group which has continuous

irtdcx

v

for

the set

of

F

.

(

F.O.G.

;

Fractional

calculus

operator

group

)

Theorem

C.

Let

$S:=\{\pm N^{v}\}\cup\{0\}=$

$\{N^{\mathrm{y}}\}\cup\{-N^{\mathrm{v}}\}\cup\{0\}$ $(\cdot \mathrm{v} \in R)$

.

$\langle$

7)

Then the

set

$S$

is

a

commutative

ring

for

the

function

$f\in F$

,

when the

identity

$N^{\alpha}+N^{\beta}=N^{\gamma}$

$(N^{a}, N^{\beta}, N^{\gamma}\in S)$

(8)

holds.

[

51

(

III

)

Lemma.

We have

[1]

$(\mathrm{i})$ $((z-c)^{\beta})_{\alpha}=e^{-i\pi a} \frac{\Gamma(\alpha-\beta)}{\Gamma(-\beta)}(z-c)^{\beta-\alpha}$ $\mathrm{f}$ $| \frac{\Gamma(\alpha-\beta)}{\Gamma(-\beta)}|<\infty)$

,

$(\mathrm{i}\mathrm{i})$

$(\log (z -c))_{c\mathrm{r}}=-e^{-i\pi a}\Gamma(\alpha)(z -c)$

$(|\Gamma(\alpha)1 <\infty)$

,

$(\mathrm{i}\mathrm{i}\mathrm{i})$

$((z-c)^{-\alpha})_{-a}=-e^{i\pi\alpha} \frac{1}{\Gamma(\alpha)}\log(z -c)$

$(1 \Gamma(\alpha)|<\infty)$

,

where

$z$

$-c\neq 0$

in

(i),

and

$z$

$-c\neq 0$

,

1

in

(

$1^{1}\mathrm{i}\}$

and

$(\mathrm{i}\mathrm{i}\mathrm{i})$

.

(

$\Gamma$

;Gamma

function

),

$(i\mathrm{v})$ $(u\cdot v)_{a}$ $:= \sum_{k-0}^{\infty}\frac{\Gamma(\alpha+1)}{k!\Gamma(\alpha+1-k)}u_{a-k}v_{k}$ $(\begin{array}{l}u=u(\mathrm{z})v=v(z)\end{array})$

.

\S 1.

N- fractional calculus of

products

of

some

power

functions

(3)

Theorem

1.

Let

$P=P( \alpha, \beta, \gamma):=\frac{\sin_{J\mathrm{K}\mathrm{X}}\cdot\sin\pi(\gamma-\alpha-\beta)}{\sin\pi(\alpha+\beta)\cdot\sin\pi(\gamma-\alpha)}$

(I

$P$

(

$\alpha$

,

$\beta$

,

$\gamma$

)

$|=M<\infty$

)

$\langle$

1)

and

$Q\approx$

$Q(\alpha, \beta, \gamma):=P(\beta, \alpha, \gamma)$

(I

$P$

(

$\beta$

,

$\alpha$

,

$\gamma$

)

$\mathrm{I}=M$ $<\infty$

)

(2)

Vllhert

$\alpha$

,

$\beta$

,

$\gamma\not\in Z_{0}^{+}$

,

we

have

;

(i)

$((z-c)^{a} \cdot(z-c)^{\beta})_{\gamma}=e^{-i\pi\gamma}P(\alpha,\beta,\gamma)\frac{\Gamma(\gamma-\alpha-\beta)}{\Gamma\langle-\alpha-\beta)}(z-c)^{a+\beta-\gamma}$

(3)

$({\rm Re}(\alpha+\beta+1)>0, (1+\alpha-\gamma)\not\in T_{0})$

,

$(\mathrm{i}\mathrm{i})$ $((\mathrm{Z}-C)^{\beta}\cdot(\mathrm{Z} -c)^{\alpha})_{\gamma}=$$e^{-i\pi\gamma}Q( \alpha, \beta,\gamma)\frac{\Gamma(\gamma-\alpha-\beta)}{\Gamma(-\alpha-\beta)}(z-c)^{\alpha+\beta-\gamma}$

,

(4)

$({\rm Re}(\alpha+\beta+1)>0, (1+\beta-\gamma)\not\in Z_{0}^{-}$

$)$

$(\mathrm{i}\mathrm{i}\mathrm{i})$ $((z-c)^{a+\beta})_{\gamma}=e^{-i\pi \mathrm{y}} \frac{\Gamma(\gamma-\alpha-\beta)}{\Gamma(-\alpha-\beta)}(z-c)^{o+\beta-\gamma}$

,

$\mathrm{t}5$ $)$

where

$z$

$-c\neq 0$

,

$| \frac{\Gamma(\gamma-\alpha-\beta)}{\Gamma(-\alpha-\beta)}|<\infty$

Proof

of

(i).

We have

$((z-c)^{\alpha} \cdot(z-c)^{\beta})_{\gamma}=.\sum_{\mathrm{A}arrow 0}^{\infty}\frac{\Gamma(\gamma+1)}{k!\Gamma(\gamma+1-k)}((z -c)^{\alpha})_{\gamma-\mathrm{A}}((z -c)^{\beta})‘$

.

(6)

by

Lemma

$(\mathrm{i}\mathrm{v})$

.

Next

we

have

$((Z -c)^{a})_{\gamma-k}=e^{-\mathrm{i}\pi(\gamma-k)} \frac{\Gamma(\gamma-k-\alpha)}{\Gamma(-\alpha)}(z -c)^{a-\gamma+k}$ $\mathrm{f}$

(7)

and

$((z-c)^{\beta})_{\mathrm{A}}=e^{-\mathrm{i}\pi k_{\frac{\Gamma(k-\beta)}{\Gamma(-\beta)}}}(z -c)^{\beta- \mathrm{A}}$

(8)

by

Lemma

(i),

respectively.

(4)

$((z-c)^{a} \cdot(z-c)^{\beta})_{\gamma}=e^{-I\pi\gamma}\sum_{\Rightarrow 0}\frac{\Gamma(\gamma+1)\Gamma(\gamma-\alpha-k)\Gamma(k-\beta)}{k!\Gamma(\gamma+1-k)\Gamma(-\alpha)\Gamma(-\beta)}(z-c)^{a+\beta- \mathrm{y}}\infty$

(9)

$=e^{-in\gamma} \frac{\Gamma(\gamma-\alpha)}{\Gamma(-\alpha)}$

(z

$-c)^{\alpha+\beta-\gamma}.\sum_{\mathrm{A}-0}^{\infty}\frac{[-\beta]_{\mathrm{A}}[-\gamma]_{k}}{k![1+\alpha-\gamma]_{k}}$

(10)

using

the

relationship

$\Gamma(\lambda+1-k)=(-1)^{-k}\frac{\Gamma(\lambda+1)\Gamma(-\lambda)}{\Gamma(k-\lambda)}$

,

(11)

where

$[\lambda]_{k}=\lambda(\lambda+1)\cdots(\lambda+k-1)=\Gamma(\lambda+k)/\Gamma(\lambda)$

,

with

$[\lambda]_{0}\subset$$1$

.

(

notation

of

Pochhammer

).

Therefore,

applying

the following relationships

$\sum_{k=0}^{\infty}\frac{[a]_{k}[b]_{\mathrm{f}\mathrm{f}}}{\overline{k![c]_{R}}}=\mathrm{z}^{F_{1}(a,b;c}|$

,

$1)= \frac{\Gamma(c\rangle\Gamma(c-a-b)}{\Gamma(c-a)\Gamma(c-b)}$ $($

${\rm Re}(c-a-b)>0c\not\in Z_{0}^{-)}$

(12)

and

$\Gamma(\lambda)\Gamma(1-\ovalbox{\tt\small REJECT}\%)=\frac{\pi}{\sin\pi\lambda}$ $(\lambda\not\in Z)$

(13)

to

(

10

),

we

obtain

$((z -c)^{\alpha} \cdot(z-c)^{\beta})_{\gamma}=e^{-l\pi\gamma}\frac{\Gamma(\gamma-\alpha)}{\Gamma(-\alpha)}(z -c)^{\alpha+\beta-\gamma}\mathrm{z}^{F_{1}(-\beta,-\gamma}$

;

$1+\alpha-\gamma$

;

1)

{14)

$=e^{-i\pi\gamma} \frac{\Gamma(\gamma-\alpha)\Gamma(1+\alpha+\beta)\Gamma(1+\alpha-\gamma)}{\Gamma(-\alpha)\Gamma(1+\alpha)\Gamma(1+\alpha+\beta-\gamma)}(_{\mathrm{t}}z-c)^{\alpha+\beta-\gamma}$

( IS)

$\{|(1+\alpha-\gamma)\not\in Z_{0}^{-)}{\rm Re}(\alpha+\beta+1)>0$

,

$=e^{-i\pi\gamma} \frac{\sin m\cdot\sin\pi(\gamma-\alpha-\beta)}{\sin\pi(\alpha+\beta)\cdot\sin\pi(\gamma-\alpha)}$

.

$\frac{\Gamma(\gamma-\alpha-\beta)}{\Gamma(-\alpha-\beta)}$

$(z -c)^{a+\beta-\gamma}$

(16)

We

have then

(3)

from

{16)

using the notation

(

1

),

under the conditions.

Proof of

$(\mathrm{i}\mathrm{i})$

.

In

the

same

way

as

the

proof

of

(i)we

obtain

(4)

using

(2)

instead of {1), ubder the conditions.

(

Simply, changing

$\alpha$

and

$\beta$

in

(3)we

(5)

Proof

of

We have

$((z-c)^{\alpha+\beta}),’=e^{-\{\pi\gamma} \frac{\Gamma(\gamma-\alpha-\beta)}{\Gamma(-\alpha-\beta)}(z-c)^{a+\beta-\gamma}$ $(17\rangle$

$[$

$<\infty)$

directly by

Lemma

(i).

Corollary

1.

When

$\alpha_{:}/3$

,

$\gamma\not\in Z_{0l}^{+}$

we

have ;

(i)

(

$\mathrm{Z}^{a}\cdot \mathrm{Z}^{\beta}\grave{)}_{\gamma}=e^{-i\pi\gamma}P(\alpha,\beta,\gamma)\frac{\Gamma(\gamma-\alpha-\beta)}{\Gamma(-\alpha-\beta)}z^{\alpha+\beta-\gamma}$

,

$\langle$

18}

$({\rm Re}(\alpha+\beta+1)>0, (1+\alpha-\gamma)\not\in T_{0})$

,

$(\mathrm{i}\mathrm{i})$ $(z^{\beta}\cdot z^{a})_{\gamma}=e^{-i\pi\gamma}\mathfrak{g}\alpha$

,

$\beta,\gamma)\frac{\Gamma(\gamma-\alpha-\beta)}{\Gamma(-\alpha-\beta)}z^{\alpha+\beta-\gamma}$

(19)

$({\rm Re}(\alpha+\beta+1)>0, (1+\beta-\gamma)\not\in \mathrm{Z}_{0}^{-})$

(

$\mathrm{i}i\mathrm{i}\}$ $(z^{a+\beta})_{\gamma}=e^{-i\pi\gamma} \frac{\Gamma(\gamma-\alpha-\beta)}{\Gamma(-\alpha-\beta)}z^{\alpha+\beta-\gamma}$

(20)

where

$<\infty$

,

and P

and

Q

are

the

on

es

show

n

by

{1) and

(2

)respectively.

Proof. Set

c

$=0$

in

Theorem

1.

Theorem

2.

When

$\alpha$

,

$\beta$

,

$\gamma\not\in Z_{0}^{+}$

,

we

have ;

(i)

$((z-c)^{\alpha}\cdot(z-c)^{\beta})_{\gamma}=P(\alpha,\beta,\gamma)((z-c)^{\mathrm{o}+\beta})_{\gamma}$

,

(21

$l$

$({\rm Re}(\alpha+\beta+ 1)>0, (1+\alpha-\gamma)\not\in T_{0})$

,

$(\mathrm{i}\mathrm{i})$ $((z-c)^{\beta}\cdot(z -c)^{\alpha})_{\gamma}=\mathfrak{g}\alpha,\beta,\gamma)((z -c)^{a+\beta})_{\gamma}$

,

(22)

$({\rm Re}(\alpha+\beta+1)>0, (1+\beta-\gamma)\not\in Z_{0}^{-}$

$)$

$(\mathrm{i}\mathrm{i}\mathrm{i})$ $\frac{1}{P}((\mathrm{z}-c)^{\alpha}\cdot(z-c)^{\beta})_{\gamma}=\frac{1}{Q}((z-c)^{\beta}.(z-c)^{a})_{\gamma}=((z-c)^{\alpha+\beta})_{\gamma}$

,

(23)

$({\rm Re}(\alpha+\beta+1)>0, (1+\alpha-\gamma)\not\in Z_{0}^{-}$

,

$(1+\beta-\gamma)\not\in Z_{0}^{-}$

,

$PQ\neq 0)$

(6)

$z-c\neq 0$

,

$| \frac{\Gamma(\gamma-\alpha-\beta)}{\Gamma(-\alpha-\beta)}|<\infty$

and P and Q

are

the

ones

shown by

(

1

)and

(

2

)

respectively.

Proof.

It

is clear

by

Theorem

1.

Corollary

2.

When

$\alpha$

,

$\beta$

,

$\gamma\not\in Z_{0}^{+}$

,

we

have

$j$

(i)

$(z^{\alpha}\cdot d)_{\gamma}=P(\alpha,\beta,\gamma)(z^{a\dashv\cdot\beta})_{\gamma}$

,

(24)

$({\rm Re}(\alpha+\beta+1)>0,$

(1

$+\alpha-\gamma)\not\in T_{0}$

),

$(\mathrm{i}\mathrm{i})$ $(z^{\beta}\cdot z^{a})_{\gamma}=Q(\alpha,\beta,\gamma)(z^{a+\beta})_{)’}$

,

$l$

$25)$

$({\rm Re}(\alpha+\beta+1)>0, (1+\beta-\gamma)\not\in Z_{0}^{-}$

$)$

$(\mathrm{i}\mathrm{i}\mathrm{i})$ $\frac{1}{P}(Z’ .z^{\beta})_{\gamma}=\frac{1}{Q}(\mathrm{Z}^{\beta}.\mathrm{z}^{\mathrm{Q}})_{\gamma}=(\mathrm{Z}^{\alpha+\beta})_{\gamma\prime}$

(26)

$({\rm Re}(\alpha+\beta+1)>0, (1+\alpha-\gamma)\not\in Z_{0}^{-}$

,

$(1+\beta-\gamma)\not\in Z_{0}^{-}$

,

$PQ\neq 0)$

where

$<\infty$

,

artd

P

and

Q

are

the

ones

shown

by

(1

)

artd

(2)

respectively.

Proof. S

et

c

$=0$

in

Theorem 2.

Corollary

3.

(i)

When

$\alpha$

,

$\beta$

,

$\gamma\not\in Z_{0}^{+}$

and

$P(\alpha, \beta,\gamma)=QCcx$

,

$\beta,\gamma)=1$

,

(27)

we

have

$((z-c)^{a}\cdot(z -c)^{\beta})_{\gamma}=((z-c)^{\beta}\cdot(z -c)^{\alpha})_{\gamma}=((z$

–$c)^{a+\beta})_{\gamma}$

.

(28)

$({\rm Re}(\alpha+\beta+1)>0, (1+\alpha-\gamma)\not\in Z_{0}^{-}$

,

$(1+\beta-\gamma)\not\in Z_{0}^{-}$

$)$

$(\mathrm{i}\mathrm{i})$

When

$\gamma=m\in Z_{0}^{+}$

we

have

$((z-c)^{a}\cdot(z-c)^{\beta})_{m}=((z-c)^{\beta}\cdot(z -c)^{\alpha})_{m}=((z-c)^{a+\beta})_{m’}$

(29)

Proof

of

(i).

We

have

(28 )from

Theorem

2(23

),

clearly.

Proof of

$(\mathrm{i}\mathrm{i})$

.

When

$\gamma=m\in Z_{0}^{+}$

we

have

(29)from

Theorem

2

(23 ),

since

$P(\alpha, \beta, m)=Q(\alpha, \beta, m)-1$

.

(30)

(7)

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\langle

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