1. Introduction
In 1938, A. M. Ostrowski proved the following important inequality.
Theorem 1 (see [1]).
Let
f
:
I
⊆
R
→
R
be a differentiable mapping on
I
∘
(interior of
I
), and let
a
,
b
∈
I
∘
with
a
<
b
. If
f
′
≤
M
for all
x
∈
a
,
b
, then
(1)
f
x
-
1
b
-
a
∫
a
b
f
t
d
t
≤
M
b
-
a
1
4
+
x
-
a
+
b
/
2
2
b
-
a
2
,
∀
x
∈
a
,
b
.
This is well known in the literature as Ostrowski’s inequality. Due to its wide range of applications in numerical analysis and in probability, many researchers have established generalizations, extensions, and variants of inequality (1); we refer readers to [2–10] and the references cited therein.
In recent years, a lot of efforts have been made by many mathematicians to generalize classical convexity. Hanson [11] introduced a new class of generalized convexity, called invexity. In [12], the authors gave the concept of preinvex function which is a special case of invexity. Pini [13], Noor [14, 15], Yang and Li [16], and Weir and Mond [17] have studied the basic properties of the preinvex functions and their role in optimization, variational inequalities, and equilibrium problems.
In [5], Işcan established some Ostrowski type inequalities for functions whose derivatives in absolute value are preinvex, by using the following identity.
Lemma 2 (see [5]).
Let
A
⊂
R
be an open invex subset with respect to
η
:
A
×
A
→
R
and
a
,
b
∈
A
with
a
<
a
+
η
(
b
,
a
)
. Suppose that
f
:
A
→
R
is a differentiable function. If
f
′
is integrable on
[
a
,
a
+
η
(
b
,
a
)
]
, then the following equality holds:
(2)
f
x
-
1
η
b
,
a
∫
a
a
+
η
b
,
a
f
u
d
u
=
η
b
,
a
∫
0
x
-
a
/
η
b
,
a
t
f
′
a
+
t
η
b
,
a
d
t
+
∫
x
-
a
/
η
b
,
a
1
t
-
1
f
′
a
+
t
η
b
,
a
d
t
,
for all
x
∈
[
a
,
a
+
η
(
b
,
a
)
]
.
Motivated by the results given in [5], in the present paper, we establish some new Ostrowski type inequalities for functions whose first derivatives in absolute value are logarithmically preinvex.
3. Main Results
Theorem 7.
Let
K
⊆
0
,
∞
be an invex subset with respect to
η
:
K
×
K
→
R
and
a
,
b
∈
K
∘
(
K
∘
interior of
K
) with
η
(
b
,
a
)
>
0
and
a
,
a
+
η
b
,
a
⊂
K
. Let
f
:
a
,
a
+
η
b
,
a
→
0
,
∞
be a differentiable function such that
f
′
∈
L
a
,
a
+
η
b
,
a
and
f
′
(
a
)
≠
0
. If
f
′
is logarithmically preinvex function, then the following inequality holds:
(7)
f
x
-
1
η
b
,
a
∫
a
a
+
η
b
,
a
f
u
d
u
≤
η
b
,
a
f
′
a
2
x
-
a
η
b
,
a
2
+
1
-
x
-
a
η
b
,
a
2
i
f
A
=
1
,
2
2
x
-
a
η
b
,
a
-
1
A
x
-
a
/
η
b
,
a
ln
A
+
1
-
2
A
x
-
a
/
η
b
,
a
+
A
ln
2
A
i
f
A
≠
1
,
for all
x
∈
a
,
a
+
η
(
b
,
a
)
, where
A
=
f
′
(
b
)
/
f
′
(
a
)
.
Proof.
From Lemma 2 and properties of modulus, we have
(8)
f
x
-
1
η
b
,
a
∫
a
a
+
η
b
,
a
f
u
d
u
≤
η
b
,
a
∫
0
x
-
a
/
η
b
,
a
t
f
′
a
+
t
η
b
,
a
d
t
+
∫
x
-
a
/
η
b
,
a
1
1
-
t
f
′
a
+
t
η
b
,
a
d
t
.
Since
f
′
is a logarithmically preinvex function, we deduce
(9)
f
x
-
1
η
b
,
a
∫
a
a
+
η
b
,
a
f
u
d
u
≤
η
b
,
a
∫
0
x
-
a
/
η
b
,
a
t
f
′
a
1
-
t
f
′
b
t
d
t
+
∫
x
-
a
/
η
b
,
a
1
1
-
t
f
′
a
1
-
t
f
′
b
t
d
t
=
η
b
,
a
f
′
a
∫
0
x
-
a
/
η
b
,
a
t
A
t
d
t
+
∫
x
-
a
/
η
b
,
a
1
1
-
t
A
t
d
t
.
If
A
=
1
, then (9) gives
(10)
f
x
-
1
η
b
,
a
∫
a
a
+
η
b
,
a
f
u
d
u
≤
η
b
,
a
f
′
a
2
x
-
a
η
b
,
a
2
+
1
-
x
-
a
η
b
,
a
2
.
In the case where
A
≠
1
, (9) gives
(11)
f
x
-
1
η
b
,
a
∫
a
a
+
η
b
,
a
f
u
d
u
≤
η
b
,
a
f
′
a
2
x
-
a
η
b
,
a
-
1
A
x
-
a
/
η
b
,
a
ln
A
+
1
-
2
A
x
-
a
/
η
b
,
a
+
A
ln
A
2
,
where we use the fact that
(12)
∫
0
x
-
a
/
η
b
,
a
t
A
t
=
x
-
a
η
b
,
a
A
x
-
a
/
η
b
,
a
ln
A
-
A
x
-
a
/
η
b
,
a
-
1
ln
A
2
,
∫
x
-
a
/
η
b
,
a
1
1
-
t
A
t
d
t
=
-
1
-
x
-
a
η
b
,
a
A
x
-
a
/
η
b
,
a
ln
A
+
A
-
A
x
-
a
/
η
b
,
a
ln
A
2
.
The desired result follows from (10) and (11).
Corollary 8.
In Theorem 7, if we choose
x
=
2
a
+
η
b
,
a
/
2
, we obtain the following midpoint inequality:
(13)
f
2
a
+
η
b
,
a
2
-
1
η
b
,
a
∫
a
a
+
η
b
,
a
f
u
d
u
≤
η
b
,
a
f
′
a
4
i
f
A
=
1
,
η
b
,
a
f
′
a
A
-
1
ln
A
2
i
f
A
≠
1
.
Corollary 9.
Let
f
:
a
,
b
→
0
,
∞
be a differentiable function such that
f
′
∈
L
a
,
b
and
f
′
(
a
)
≠
0
. If
f
′
is a logarithmically convex function, then the following inequality holds:
(14)
f
x
-
1
b
-
a
∫
a
b
f
u
d
u
≤
b
-
a
f
′
a
2
x
-
a
b
-
a
2
+
b
-
x
b
-
a
2
i
f
A
=
1
2
2
x
-
a
b
-
a
-
1
A
x
-
a
/
b
-
a
ln
A
+
1
-
2
A
x
-
a
/
b
-
a
+
A
l
n
2
A
i
f
A
≠
1
,
for all
x
∈
a
,
b
, where
A
=
f
′
(
b
)
/
f
′
(
a
)
.
Example 10.
In Theorem 7, if we choose
η
(
b
,
a
)
=
a
b
, the geometric mean, we obtain the following inequality:
(15)
f
x
-
1
η
b
,
a
∫
a
a
+
a
b
f
u
d
u
≤
f
′
a
a
b
2
x
-
a
a
b
2
+
1
-
x
-
a
a
b
2
if
A
=
1
,
2
2
x
-
a
a
b
-
1
A
x
-
a
/
a
b
ln
A
+
1
-
2
A
x
-
a
/
a
b
+
A
l
n
2
A
if
A
≠
1
.
Theorem 11.
Let
K
⊆
0
,
∞
be an invex subset with respect to
η
:
K
×
K
→
R
and
a
,
b
∈
K
∘
(
K
∘
interior of
K
) with
η
(
b
,
a
)
>
0
and
a
,
a
+
η
b
,
a
⊂
K
. Let
f
:
a
,
a
+
η
b
,
a
→
0
,
∞
be a differentiable function such that
f
′
∈
L
a
,
a
+
η
b
,
a
and
f
′
(
a
)
≠
0
; let
q
>
1
with
1
/
p
+
1
/
q
=
1
. If
f
′
q
is a logarithmically preinvex function, then the following inequality holds:
(16)
f
x
-
1
η
b
,
a
∫
a
a
+
η
b
,
a
f
u
d
u
≤
η
b
,
a
f
′
a
p
+
1
1
/
p
x
-
a
η
b
,
a
2
+
1
-
x
-
a
η
b
,
a
2
i
f
A
=
1
,
x
-
a
η
b
,
a
1
+
1
/
p
A
q
x
-
a
/
η
b
,
a
-
1
q
ln
A
1
/
q
+
1
-
x
-
a
η
b
,
a
1
+
1
/
p
A
q
-
A
q
x
-
a
/
η
b
,
a
q
ln
A
1
/
q
i
f
A
≠
1
,
for all
x
∈
a
,
a
+
η
(
b
,
a
)
, where
A
=
f
′
(
b
)
/
f
′
(
a
)
.
Proof.
From Lemma 2, properties of modulus, and Hölder’s inequality, we have
(17)
f
x
-
1
η
b
,
a
∫
a
a
+
η
b
,
a
f
u
d
u
≤
η
b
,
a
∫
0
x
-
a
/
η
b
,
a
t
p
d
t
1
/
p
∫
0
x
-
a
/
η
b
,
a
f
′
a
+
t
η
b
,
a
q
d
t
1
/
q
+
∫
x
-
a
/
η
b
,
a
1
1
-
t
p
d
t
1
/
p
∫
x
-
a
/
η
b
,
a
1
f
′
a
+
t
η
b
,
a
q
d
t
1
/
q
=
η
b
,
a
p
+
1
1
/
p
x
-
a
η
b
,
a
1
+
1
/
p
∫
0
x
-
a
/
η
b
,
a
f
′
a
+
t
η
b
,
a
q
d
t
1
/
q
+
1
-
x
-
a
η
b
,
a
1
+
1
/
p
∫
x
-
a
/
η
b
,
a
1
f
′
a
+
t
η
b
,
a
q
d
t
1
/
q
.
Since
f
′
q
is a logarithmically preinvex function, we deduce
(18)
f
x
-
1
η
b
,
a
∫
a
a
+
η
b
,
a
f
u
d
u
≤
η
b
,
a
p
+
1
1
/
p
x
-
a
η
b
,
a
1
+
1
/
p
∫
0
x
-
a
/
η
b
,
a
f
′
a
q
1
-
t
f
′
b
q
t
d
t
1
/
q
+
1
-
x
-
a
η
b
,
a
1
+
1
/
p
∫
x
-
a
/
η
b
,
a
1
f
′
a
q
1
-
t
f
′
b
q
t
d
t
1
/
q
=
η
b
,
a
f
′
a
p
+
1
1
/
p
x
-
a
η
b
,
a
1
+
1
/
p
∫
0
x
-
a
/
η
b
,
a
A
q
t
d
t
1
/
q
+
1
-
x
-
a
η
b
,
a
1
+
1
/
p
∫
x
-
a
/
η
b
,
a
1
A
q
t
d
t
1
/
q
.
If
A
=
1
, then (18) gives
(19)
f
x
-
1
η
b
,
a
∫
a
a
+
η
b
,
a
f
u
d
u
≤
η
b
,
a
f
′
a
p
+
1
1
/
p
x
-
a
η
b
,
a
2
+
1
-
x
-
a
η
b
,
a
2
.
In the case where
A
≠
1
, (18) becomes
(20)
f
x
-
1
η
b
,
a
∫
a
a
+
η
b
,
a
f
u
d
u
≤
η
b
,
a
f
′
a
p
+
1
1
/
p
x
-
a
η
b
,
a
1
+
1
/
p
A
q
x
-
a
/
η
b
,
a
-
1
q
ln
A
1
/
q
+
1
-
x
-
a
η
b
,
a
1
+
1
/
p
A
q
-
A
q
x
-
a
/
η
b
,
a
q
ln
A
1
/
q
,
where we use the fact that
(21)
∫
0
x
-
a
/
η
b
,
a
A
q
t
d
t
=
A
q
x
-
a
/
η
b
,
a
-
1
q
ln
A
,
∫
x
-
a
/
η
b
,
a
1
A
q
t
d
t
=
A
q
-
A
q
x
-
a
/
η
b
,
a
q
ln
A
.
From (19) and (20), we get the desired result.
Corollary 12.
In Theorem 11, if we choose
x
=
2
a
+
η
b
,
a
/
2
, we obtain the following midpoint inequality:
(22)
f
2
a
+
η
b
,
a
2
-
1
η
b
,
a
∫
a
a
+
η
b
,
a
f
u
d
u
≤
η
b
,
a
f
′
a
2
1
+
1
/
p
p
+
1
1
/
p
2
1
/
p
i
f
A
=
1
,
1
+
A
A
q
-
1
q
ln
A
1
/
q
i
f
A
≠
1
.
Corollary 13.
Let
f
:
a
,
b
→
0
,
∞
be a differentiable function such that
f
′
∈
L
a
,
b
and
f
′
(
a
)
≠
0
; let
q
>
1
with
1
/
p
+
1
/
q
=
1
. If
f
′
q
is a logarithmically convex function, then the following inequality holds:
(23)
f
x
-
1
b
-
a
∫
a
b
f
u
d
u
≤
b
-
a
f
′
a
p
+
1
1
/
p
x
-
a
b
-
a
2
+
b
-
x
b
-
a
2
i
f
A
=
1
,
x
-
a
b
-
a
1
+
1
/
p
A
q
x
-
a
/
b
-
a
-
1
q
ln
A
1
/
q
+
b
-
x
b
-
a
1
+
1
/
p
A
q
-
A
q
x
-
a
/
b
-
a
q
ln
A
1
/
q
i
f
A
≠
1
,
for all
x
∈
a
,
b
, where
A
=
f
′
(
b
)
/
f
′
(
a
)
.
Example 14.
In Theorem 11, if we choose
η
(
b
,
a
)
=
a
+
b
/
2
, the arithmetic mean, we obtain the following inequality:
(24)
f
x
-
2
a
+
b
∫
a
a
+
a
+
b
/
2
f
u
d
u
≤
a
+
b
f
′
a
2
p
+
1
1
/
p
4
a
+
b
2
x
-
a
2
+
b
+
3
a
-
2
x
2
2
if
A
=
1
,
2
x
-
a
a
+
b
1
+
1
/
p
A
2
q
x
-
a
/
a
+
b
-
1
q
ln
A
1
/
q
+
b
+
3
a
-
2
x
2
1
+
1
/
p
A
q
-
A
2
q
x
-
a
/
a
+
b
q
ln
A
1
/
q
if
A
≠
1
.
Theorem 15.
Let
K
⊆
0
,
∞
be an invex subset with respect to
η
:
K
×
K
→
R
and
a
,
b
∈
K
∘
(
K
∘
interior of
K
) with
η
(
b
,
a
)
>
0
and
a
,
a
+
η
b
,
a
⊂
K
. Let
f
:
a
,
a
+
η
b
,
a
→
0
,
∞
be a differentiable function such that
f
′
∈
L
a
,
a
+
η
b
,
a
and
f
′
(
a
)
≠
0
; let
q
>
1
.
If
f
′
q
is a logarithmically preinvex function, then the following inequality holds:
(25)
f
x
-
1
η
b
,
a
∫
a
a
+
η
b
,
a
f
u
d
u
≤
η
b
,
a
2
1
-
1
/
q
f
′
a
1
2
1
/
q
x
-
a
η
b
,
a
2
+
1
-
x
-
a
η
b
,
a
2
i
f
A
=
1
,
x
-
a
η
b
,
a
2
-
2
/
q
x
-
a
η
b
,
a
A
q
x
-
a
/
η
b
,
a
ln
A
+
1
-
A
q
x
-
a
/
η
b
,
a
l
n
2
A
1
/
q
+
1
-
x
-
a
η
b
,
a
2
-
2
/
q
A
q
-
A
q
x
-
a
/
η
b
,
a
l
n
2
A
-
1
-
x
-
a
η
b
,
a
A
q
x
-
a
/
η
b
,
a
ln
A
1
/
q
i
f
A
≠
1
,
for all
x
∈
a
,
a
+
η
(
b
,
a
)
, where
A
=
f
′
(
b
)
/
f
′
(
a
)
.
Proof.
From Lemma 2, properties of modulus, and power mean inequality, we have
(26)
f
x
-
1
η
b
,
a
∫
a
a
+
η
b
,
a
f
u
d
u
≤
η
b
,
a
∫
0
x
-
a
/
η
b
,
a
t
d
t
1
-
1
/
q
∫
0
x
-
a
/
η
b
,
a
t
f
′
a
+
t
η
b
,
a
q
d
t
1
/
q
+
∫
x
-
a
/
η
b
,
a
1
1
-
t
d
t
1
-
1
/
q
∫
x
-
a
/
η
b
,
a
1
1
-
t
f
′
a
+
t
η
b
,
a
q
d
t
1
/
q
=
η
b
,
a
2
1
-
1
/
q
x
-
a
η
b
,
a
2
1
-
1
/
q
∫
0
x
-
a
/
η
b
,
a
t
f
′
a
+
t
η
b
,
a
q
d
t
1
/
q
+
1
-
x
-
a
η
b
,
a
2
1
-
1
/
q
∫
x
-
a
/
η
b
,
a
1
1
-
t
f
′
a
+
t
η
b
,
a
q
d
t
1
/
q
.
Since
f
′
q
is a logarithmically preinvex function, we deduce
(27)
f
x
-
1
η
b
,
a
∫
a
a
+
η
b
,
a
f
u
d
u
≤
η
b
,
a
2
1
-
1
/
q
f
′
a
x
-
a
η
b
,
a
2
-
2
/
q
∫
0
x
-
a
/
η
b
,
a
t
A
q
t
d
t
1
/
q
+
1
-
x
-
a
η
b
,
a
2
-
2
/
q
∫
x
-
a
/
η
b
,
a
1
1
-
t
A
q
t
d
t
1
/
q
.
In the case where
A
=
1
, (27) gives
(28)
f
x
-
1
η
b
,
a
∫
a
a
+
η
b
,
a
f
u
d
u
≤
η
b
,
a
2
f
′
a
x
-
a
η
b
,
a
2
+
1
-
x
-
a
η
b
,
a
2
.
For
A
≠
1
, (27) gives
(29)
f
x
-
1
η
b
,
a
∫
a
a
+
η
b
,
a
f
u
d
u
≤
η
b
,
a
2
1
-
1
/
q
f
′
a
x
-
a
η
b
,
a
2
-
2
/
q
x
-
a
η
b
,
a
A
q
x
-
a
/
η
b
,
a
ln
A
+
1
-
A
q
x
-
a
/
η
b
,
a
l
n
2
A
1
/
q
+
1
-
x
-
a
η
b
,
a
2
-
2
/
q
A
q
-
A
q
x
-
a
/
η
b
,
a
l
n
2
A
-
1
-
x
-
a
η
b
,
a
A
q
x
-
a
/
η
b
,
a
ln
A
1
/
q
,
where we use the fact that
(30)
∫
0
x
-
a
/
η
b
,
a
t
A
q
t
d
t
=
x
-
a
η
b
,
a
A
q
x
-
a
/
η
b
,
a
ln
A
+
1
-
A
q
x
-
a
/
η
b
,
a
l
n
2
A
,
∫
x
-
a
/
η
b
,
a
1
1
-
t
A
q
t
d
t
=
A
q
-
A
q
x
-
a
/
η
b
,
a
l
n
2
A
-
1
-
x
-
a
η
b
,
a
A
q
x
-
a
/
η
b
,
a
ln
A
.
From (28) and (29), we obtain the desired result.
Corollary 16.
In Theorem 15, if we choose
x
=
2
a
+
η
b
,
a
/
2
, we obtain the following midpoint inequality:
(31)
f
2
a
+
η
b
,
a
2
-
1
η
b
,
a
∫
a
a
+
η
b
,
a
f
u
d
u
≤
η
b
,
a
4
f
′
a
1
i
f
A
=
1
,
1
2
1
-
3
/
q
1
-
A
q
/
2
l
n
2
A
+
A
q
/
2
2
ln
A
1
/
q
+
A
1
/
2
A
q
/
2
-
1
l
n
2
A
-
1
2
ln
A
1
/
q
i
f
A
≠
1
.
Corollary 17.
Let
f
:
a
,
a
+
η
b
,
a
→
0
,
∞
be a differentiable function such that
f
′
∈
L
a
,
a
+
η
b
,
a
and
f
′
(
a
)
≠
0
; let
q
>
1
.
If
f
′
q
is a logarithmically convex function, then the following inequality holds:
(32)
f
x
-
1
b
-
a
∫
a
b
f
u
d
u
≤
b
-
a
2
1
-
1
/
q
f
′
a
1
2
1
/
q
x
-
a
b
-
a
2
+
b
-
x
b
-
a
2
i
f
A
=
1
,
x
-
a
b
-
a
2
-
2
/
q
x
-
a
b
-
a
A
q
x
-
a
/
b
-
a
ln
A
+
1
-
A
q
x
-
a
/
b
-
a
l
n
2
A
1
/
q
+
b
-
x
b
-
a
2
-
2
/
q
A
q
-
A
q
x
-
a
/
b
-
a
l
n
2
A
-
b
-
x
b
-
a
A
q
x
-
a
/
b
-
a
ln
A
1
/
q
i
f
A
≠
1
,
for all
x
∈
a
,
b
, where
A
=
f
′
(
b
)
/
f
′
(
a
)
.
Example 18.
In Theorem 15, if we choose
η
(
b
,
a
)
=
b
-
a
/
ln
b
-
ln
a
with
a
≠
b
, the logarithmic mean, we obtain the following inequality:
(33)
f
x
-
ln
b
-
ln
a
b
-
a
∫
a
a
+
b
-
a
/
ln
b
-
ln
a
f
u
d
u
≤
b
-
a
2
1
-
1
/
q
ln
b
-
ln
a
f
′
a
1
2
1
/
q
θ
2
+
1
-
θ
2
i
f
A
=
1
,
θ
2
-
2
/
q
θ
A
q
θ
ln
A
+
1
-
A
q
θ
l
n
2
A
1
/
q
+
1
-
θ
2
-
2
/
q
A
q
-
A
q
θ
l
n
2
A
-
1
-
θ
A
q
θ
ln
A
1
/
q
i
f
A
≠
1
,
where
θ
=
x
-
a
ln
b
-
ln
a
/
b
-
a
.
Theorem 19.
Suppose that all the assumptions of Theorem 15 are satisfied, then the following inequality holds:
(34)
f
x
-
1
η
b
,
a
∫
a
a
+
η
b
,
a
f
u
d
u
≤
η
b
,
a
f
′
a
1
q
+
1
1
/
q
x
-
a
η
b
,
a
2
+
1
-
x
-
a
η
b
,
a
2
i
f
A
=
1
,
A
x
-
a
/
η
b
,
a
1
-
x
-
a
η
b
,
a
x
-
a
η
b
,
a
∑
i
=
1
∞
ln
A
q
1
-
x
-
a
/
η
b
,
a
i
-
1
α
i
1
/
q
+
x
-
a
η
b
,
a
2
∑
i
=
1
∞
-
ln
A
q
x
-
a
/
η
b
,
a
i
-
1
α
i
1
/
q
i
f
A
≠
1
,
for all
x
∈
a
,
a
+
η
(
b
,
a
)
, where
A
=
f
′
(
b
)
/
f
′
(
a
)
and
q
+
1
i
=
∏
j
=
0
i
-
1
q
+
1
+
j
.
Proof.
From Lemma 2, properties of modulus, and power mean inequality, we have
(35)
f
x
-
1
η
b
,
a
∫
a
a
+
η
b
,
a
f
u
d
u
≤
η
b
,
a
∫
0
x
-
a
/
η
b
,
a
d
t
1
-
1
/
q
∫
0
x
-
a
/
η
b
,
a
t
q
f
′
a
+
t
η
b
,
a
q
d
t
1
/
q
+
∫
x
-
a
/
η
b
,
a
1
d
t
1
-
1
/
q
∫
x
-
a
/
η
b
,
a
1
1
-
t
q
f
′
a
+
t
η
b
,
a
q
d
t
1
/
q
=
η
b
,
a
x
-
a
η
b
,
a
1
-
1
/
q
∫
0
x
-
a
/
η
b
,
a
t
q
f
′
a
+
t
η
b
,
a
q
d
t
1
/
q
+
1
-
x
-
a
η
b
,
a
1
-
1
/
q
∫
x
-
a
/
η
b
,
a
1
1
-
t
q
f
′
a
+
t
η
b
,
a
q
d
t
1
/
q
.
Since
f
′
q
is a logarithmically preinvex function, we deduce
(36)
f
x
-
1
η
b
,
a
∫
a
a
+
η
b
,
a
f
u
d
u
≤
η
b
,
a
f
′
a
x
-
a
η
b
,
a
1
-
1
/
q
∫
0
x
-
a
/
η
b
,
a
t
q
A
q
t
1
/
q
+
1
-
x
-
a
η
b
,
a
1
-
1
/
q
∫
x
-
a
/
η
b
,
a
1
1
-
t
q
A
q
t
d
t
1
/
q
.
If
A
=
1
, (36) gives
(37)
f
x
-
1
η
b
,
a
∫
a
a
+
η
b
,
a
f
u
d
u
≤
η
b
,
a
f
′
a
q
+
1
1
/
q
x
-
a
η
b
,
a
2
+
1
-
x
-
a
η
b
,
a
2
.
In the case where
A
≠
1
, we can restate (36) as
(38)
f
x
-
1
η
b
,
a
∫
a
a
+
η
b
,
a
f
u
d
u
≤
η
b
,
a
f
′
a
x
-
a
η
b
,
a
2
∫
0
1
t
q
A
q
x
-
a
/
η
b
,
a
t
d
t
1
/
q
+
1
-
x
-
a
η
b
,
a
x
-
a
η
b
,
a
A
x
-
a
/
η
b
,
a
∫
0
1
t
-
1
q
A
q
1
-
x
-
a
/
η
b
,
a
t
d
t
1
/
q
.
Applying Lemma 6 with
z
=
1
, we get
(39)
∫
0
1
t
q
A
q
x
-
a
/
η
b
,
a
t
d
t
=
A
q
x
-
a
/
η
b
,
a
∑
i
=
1
∞
-
ln
A
q
x
-
a
/
η
b
,
a
i
-
1
α
i
∫
0
1
t
-
1
q
A
q
1
-
x
-
a
/
η
b
,
a
t
d
t
=
∑
i
=
1
∞
ln
A
q
1
-
x
-
a
/
η
b
,
a
i
-
1
α
i
.
Substituting (39) into (38), we obtain
(40)
f
x
-
1
η
b
,
a
∫
a
a
+
η
b
,
a
f
u
d
u
≤
η
b
,
a
f
′
a
x
-
a
η
b
,
a
2
A
x
-
a
/
η
b
,
a
∑
i
=
1
∞
-
ln
A
q
x
-
a
/
η
b
,
a
i
-
1
α
i
1
/
q
+
1
-
x
-
a
η
b
,
a
x
-
a
η
b
,
a
A
x
-
a
/
η
b
,
a
∑
i
=
1
∞
ln
A
q
1
-
x
-
a
/
η
b
,
a
i
-
1
α
i
1
/
q
.
The desired result follows from (37) and (40).
Corollary 20.
In Theorem 19, if we choose
x
=
2
a
+
η
b
,
a
/
2
, we obtain the following midpoint inequality:
(41)
f
2
a
+
η
b
,
a
2
-
1
η
b
,
a
∫
a
a
+
η
b
,
a
f
u
d
u
≤
η
b
,
a
f
′
a
2
1
q
+
1
1
/
q
i
f
A
=
1
,
A
2
∑
i
=
1
∞
ln
A
q
/
2
i
-
1
α
i
1
/
q
+
∑
i
=
1
∞
-
ln
A
q
/
2
i
-
1
α
i
1
/
q
i
f
A
≠
1
.
Corollary 21.
In Theorem 19, if we choose
η
b
,
a
=
b
-
a
, we obtain the following inequality:
(42)
f
x
-
1
b
-
a
∫
a
b
f
u
d
u
≤
b
-
a
f
′
a
1
q
+
1
1
/
q
x
-
a
b
-
a
2
+
b
-
x
b
-
a
2
i
f
A
=
1
,
A
x
-
a
/
b
-
a
b
-
x
x
-
a
b
-
a
2
∑
i
=
1
∞
ln
A
q
b
-
x
/
b
-
a
i
-
1
α
i
1
/
q
+
x
-
a
b
-
a
2
∑
i
=
1
∞
-
ln
A
q
x
-
a
/
b
-
a
i
-
1
α
i
1
/
q
i
f
A
≠
1
.
Remark 22.
In all the above theorems, inequalities for nonconvex functions could be drawn by just replacing
η
b
,
a
by other means than those in the previously mentioned examples.