Page 167 - IJOCTA-15-1

Improvements of Hermite-Hadamard-Mercer inequality using k-fractional integral
α
α
′ q
Since |ω | is convex and |t κ − 1| = (1 − t κ ) for

α t ∈ [0, 1], therefore
−1 Γ κ (α) α 2
(α + κ) κ α,κ
α α J ω(v) α −1 2
)
αu+κv + (α + κ) κ Γ κ (α) α α,κ
(v − u) κ κ κ (
α+κ α α J αu+κv + ω(v)
κ κ ( )
(v − u) κ α+κ

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

1 − t κ dt ω (u)
0 α + κ
1 Z 1 1
q
q
′
αt
′

× L 1 (α, κ) ω (u) + L 2 (α, κ) ω (v) q + 1 − t κ (1 − α + κ )dt ω (v) q q
α
′

q
q
′
′

+ L 3 (α, κ) ω (u) + L 4 (α, κ) ω (v) 1 q 0

(16) Z 1 α κt q
′
+ 1 − t κ (1 − )dt ω (u)
0 α + κ
where 1 !
Z
1 q
α κt q
′
+ 1 − t κ ( ) ω (v)
α 2 α(α + 4κ) 0 α + κ
L 1 (α, κ) = , L 2 (α, κ) = ,
2(α + κ)(α + 2κ) 2(α + κ)(α + 2κ)
α(2α + 3κ) ακ
L 3 (α, κ) = , L 4 (α, κ) = . 1− 1
2(α + κ)(α + 2κ) 2(α + κ)(α + 2κ) ακ(v − u) α q
≤ 2
(α + κ) α + κ
Proof. From (15), we have
q
q
′
′

× L 1 (α, κ) ω (u) + L 2 (α, κ) ω (v) 1 q

α −1 2
(α + κ) κ Γ κ (α) α α,κ
q
α J q 1
α αu+κv + ω(v) ′ ′ q

(v − u) κ α+κ
κ κ ( ) + L 3 (α, κ) ω (u) + L 4 (α, κ) ω (v) .

κ α,κ αω(u) + κω(v)

+ α J αu+κv − ω(u) −
α κ −1 ( α+κ ) α + κ Remark 3. Substituting α = κ = 1, u = δ 1 and
v = δ 2 in (16), we obtain

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

κt κt ω(s)ds − ω(δ 1 ) + ω(δ 2 )
′

+ ω (1 − )u + ( )v dt . (17) δ 2 − δ 1 2
α + κ α + κ δ 1
q
 ′ q ′  1 q
1− 1
δ 2 − δ 1 1 q ω (δ 1 ) + 5 ω (δ 2 )
≤  
As 1 ≤ q therefore by applying power mean in- 4 2 12
equality, we have
q
 q  1
′ ′
q !
α 5 ω (δ 1 ) + ω (δ 2 )
−1 2
(α + κ) κ Γ κ (α) α α,κ +   .
α J 12
α αu+κv + ω(v)
κ κ ( )
(v − u) κ α+κ

κ α,κ αω(u) + κω(v)
+ α J αu+κv − ω(u) − ′ q
α κ −1 ( α+κ ) α + κ Theorem 6. Let q ≥ p be such that |ω | is a con-
vex function and u, v ∈ [δ 1 , δ 2 ] for u < v. Then
1− 1
Z
ακ(v − u) 1 α q
≤ t κ − 1 the inequality
(α + κ) 2 0
q 1 α
Z
1 q −1 2
α ′ αt αt (α + κ) κ Γ κ (α) α α,κ

t κ − 1 ω u + (1 − )v dt α J
α + κ α + κ α αu+κv + ω(v)
0 κ κ ( )
(v − u) κ α+κ
q 1
Z
1 q
α κt κt
′
+ t κ − 1 ω (1 − )u + ( )v dt κ α,κ αω(u) + κω(v)

0 α + κ α + κ + α −1 J αu+κv − ω(u) −
α κ ( α+κ ) α + κ
161