Page 166 - IJOCTA-15-1

J. Khan, M. Adil Khan, S. Sarwar / IJOCTA, Vol.15, No.1, pp.155-165 (2025)

Remark 1. If we substitute α = κ = 1, u = δ 1
and v = δ 2 in (11), we obtain α −1 2
(α + κ) κ Γ κ (α) α α,κ
α α J αu+κv + ω(v)
Z
1 δ 2 ω(δ 1 ) + ω(δ 2 ) (v − u) κ κ κ ( )
ω(s)ds − α+κ
δ 2 − δ 1 2 κ
δ 1 + J α,κ αω(u) + κω(v)
1 −1
Z α αu+κv − ω(u) −
δ 2 − δ 1 ′ t t α κ ( α+κ ) α + κ
= (t − 1) ω δ 1 + 1 − δ 2
4 0 2 2

t t ακ(v − u)
′
−ω 1 − δ 1 + δ 2 dt. ≤
2 2 (α + κ) 2
1
Z
α αt αt
′
(1 − t κ ) ω ( )u + (1 − )v
α + κ α + κ
0

′
Theorem 4. Suppose |ω | is a convex function + ω (1 − κt )u + ( κt )v dt.
′

defined on [δ 1 , δ 2 ], u, v ∈ [δ 1 , δ 2 ] such that u < v α + κ α + κ
′
and α, κ > 0, Then Since |ω | is convex, therefore
α
−1 2
(α + κ) κ Γ κ (α) α α,κ
α J
α αu+κv + ω(v)
κ κ ( )
(v − u) κ
α −1 2 α+κ

(α + κ) κ Γ κ (α) α α,κ
α α J αu+κv + ω(v) κ α,κ αω(u) + κω(v)
+ J
(v − u) κ κ κ ( ) α αu+κv − ω(u) −
α+κ −1
α κ ( α+κ ) α + κ

κ α,κ αω(u) + κω(v) Z 1
+ α J αu+κv − ω(u) − ≤ ακ(v − u) (1 − t κ ) αt |ω (u)|
α
′
α κ −1 ( α+κ ) α + κ (α + κ) 2 0 α + κ

αt ′
′
′

≤ (v − u) P 1 (α, κ) ω (u) + P 2 (α, κ) ω (v) , + (1 − α + κ )|ω (v)|

(14) Z 1
α κt ′
+ (1 − t κ ) (1 − )|ω (u)|
0 α + κ
where κt
′
+ |ω (v)| dt
α + κ
2
3α k
P 1 (α, κ) = , and ακ α − κ Z 1 α
2(α + 2κ)(α + κ) 2 = (v − u) t(1 − t κ )dt
(α + κ) 2 α + κ 0
Z 1
2
α
α κ (α + 5κ) + (1 − t κ )dt |ω (u)|
′
P 2 (α, κ) = .
3
2(α + κ) (α + 2κ) 0
Z 1
ακ α
+ (1 − t κ )dt
(α + κ) 2 0
Proof. Taking the absolute of (11), we have
α α − κ Z 1 α
−1 2 ′
(α + κ) κ Γ κ (α) α α,κ − t(1 − t κ )dt |ω (v)|
α 0
α J αu+κv + ω(v) α + κ
(v − u) κ α+κ
κ κ ( )

κ α,κ αω(u) + κω(v) = (v − u) P 1 (α, κ) ω (u) + P 2 (α, κ) ω (v) .
′
′

+ α J αu+κv − ω(u) −
α κ −1 ( α+κ ) α + κ
Z
ακ(v − u) 1 α
= t κ − 1
(α + κ) 2 0
Remark 2. If we put α = κ = 1, u = δ 1 and
αt αt v = δ 2 in (14), we obtain Theorem 2.2 of. 43
′
ω u + (1 − )v
α + κ α + κ
κt κt
− ω ′ (1 − )u + ( )v dt dt . (15)

α + κ α + κ ′ q
Theorem 5. Let |ω | be a convex function for
α α
As |t κ − 1| = (1 − t κ ) for t ∈ [0, 1], therefore q ≥ 1 and let u, v ∈ [δ 1 , δ 2 ] such that u < v. Then
160