Some new extensions for fractional integral operator having exponential in the kernel and their applications in physical systems

Definition 1.1

[25] Let a real interval K ⊂ ℝ \ { 0 } and a function P : K → ℝ is said to be harmonically convex, if

for all x , y ∈ K and ζ ∈ [ 0 , 1 ] . If inequality (1.4) holds in the reversed direction, then P is called the harmonically concave function.

Definition 1.2

[25] A function P : K ⊂ ℝ \ { 0 } → ℝ is said to be harmonically symmetric with respect to 2 η 1 η 2 η 1 + η 2 , if

holds for all z ∈ ℐ .

Definition 1.3

[2] Let P ∈ L 1 ( [ η 1 , η 2 ] ) . The fractional integrals J η 1 γ and J η 2 γ of order γ > 0 are stated as:

and

Definition 1.4

Let P : ℐ → ℝ , ( 0 < η 1 < η 2 ) be a function such that P be a positive and integrable, also Ψ be a differentiable and strictly increasing on ( η 1 , η 2 ) . Then, the fractional integral operators J η 1 γ ,Ψ and J η 2 γ ,Ψ of order γ > 0 are stated as:

and

Definition 1.5

Let P : ℐ → ℝ , ( 0 < η 1 < η 2 ) be a function such that P be a positive and integrable, also Ψ be a differentiable and strictly increasing on ( η 1 , η 2 ) . Then, the one-sided fractional integral operator J η 1 γ ,Ψ is stated as:

Theorem 2.1

For γ > 0 and let there is a positive function P : ℐ ⊆ ℝ \ { 0 } → ℝ with η 2 > η 1 and P ∈ L 1 ( [ η 1 , η 2 ] ) . If P is a harmonically convex function on ℐ , then

where Q ( r ) = 1 r , r ∈ 1 η 2 , 1 η 1 .

Proof

By utilizing harmonically convexity of P on ℐ , we have for every z 1 , z 2 ∈ ℐ having ζ = 1 2 ,

choosing z 1 = η 1 η 2 ζ η 2 + ( 1 − ζ ) η 1 , z 2 = η 1 η 2 ζ η 1 + ( 1 − ζ ) η 2 , takes the form:

Conducting product on both sides of (2.7) by e − ϑ ζ , then integrating with respect to ζ from 0 to 1 , we get

and established the first inequality.

For the proof of the second inequality in (2.1), we first note that if P is a harmonically convex function, then, for ζ ∈ [ 0 , 1 ] , it yields

and

By adding the above inequalities, we have

Then, multiplying on both sides of (2.4) by e − ϑ ζ and integrating the inequality with respect to ζ from 0 to 1 , one obtains

As a result, we have

The proof is completed.□

Remark

In the limiting case, when γ → 1 , observe that

which is proposed by Iscan in [34].

Lemma 3.1

For γ > 0 and let there is a function Q : ℐ ⊆ ℝ \ { 0 } → ℝ integrable and harmonically symmetric with respect to 2 η 1 η 2 η 1 + η 2 , then

where γ > 0 and U ( z ) = 1 z , z ∈ 1 η 2 , 1 η 1 .

Proof

By the given assumption and substituting ζ = 1 η 1 + 1 η 2 − z in the following integral and performing some computation we get

the required result.□

Theorem 3.2

For γ > 0 and let there is a harmonically convex function P : ℐ ⊆ ℝ \ { 0 } → ℝ such that P ∈ L 1 ( [ η 1 , η 2 ] ) with η 2 > η 1 and η 1 , η 2 ∈ ℐ . Also, if there is non-negative, integrable and harmonically symmetric with respect to 2 η 1 η 2 η 1 + η 2 , then

where U ( z ) = 1 z , z ∈ 1 η 2 , 1 η 1 .

Proof

Since P is a harmonically convex function on ℐ , we have inequality (2.4) for all ζ ∈ [ 0 , 1 ] . Multiplying on both sides of (2.4) by e − ϑ ζ Q η 1 η 2 ζ η 2 + ( 1 − ζ ) η 1 , and then integrating the inequality with respect to ζ from 0 to 1 , we obtain

Setting z = ζ η 2 + ( 1 − ζ ) η 1 η 1 η 2 and utilizing that Q is harmonically symmetric, we have

It follows that

Applying Lemma 3.1 on the left hand side of (3.6), we have

For the proof of the second inequality in (3.3), multiplying on both sides of (2.4) by e − ϑ ζ Q η 1 η 2 ζ η 2 + ( 1 − ζ ) η 1 and integrating the inequality with respect to ζ from 0 to 1 , we obtain

Again, setting z = ζ η 2 + ( 1 − ζ ) η 1 η 1 η 2 and after simple calculations, we conclude the second inequality (3.3).□

Remark 3.1

In Theorem 3.2:

1. If one takes γ → 1 , then one has Theorem 4 in [35].

2. If one takes Q ( r ) = 1 and γ → 1 , then one has Theorem 2 in [35].

Theorem 4.1

For γ > 0 and let there are two harmonically convex functions P , W : ℐ ⊆ ℝ \ { 0 } → ℝ such that P , W ∈ L 1 ( [ η 1 , η 2 ] ) with η 2 > η 1 and η 1 , η 2 ∈ ℐ , then

and

where

and

Proof

Since P and W are harmonically convex functions on ℐ , then, for ζ ∈ [ 0 , 1 ] ,

and

Adding (4.5) and (4.6), we have

Multiplying on both sides of (4.7) by e − ϑ ζ and integrating the inequality with respect to ζ from 0 to 1 , we have

Consequently, we get