(k, s)-Riemann-Liouville fractional integral and
applications
Mehmet Zeki SARIKAYA∗†, Zoubir DAHMANI‡, Mehmet Ey¨up KIRIS§and Farooq AHMAD¶
Abstract
In this paper, we introduce a new approach on fractional integration, which generalizes the Riemann-Liouville fractional integral. We prove some properties for this new approach. We also establish some new integral inequalities using this new fractional integration.
Keywords: Riemann-Liouville fractional integrals, synchronous function, Cheby-shev inequality, H¨older inequality.
2000 AMS Classification: 26A33;26A51;26D15.
1. Introduction
Fractional calculus and its widely application have recently been paid more and more attentions. For more recent development on fractional calculus, we refer the reader to [7, 12, 15, 16, 19]. There are several known forms of the fractional integrals of which two have been studied extensively for their applications [5, 10, 11, 14, 21]. The first is the Riemann-Liouville fractional integral of α ≥ 0 for a continuous function f on [a, b] which is defined by
Jα af (x) = 1 Γ(α) Z x a (x − t)α−1f (t)dt, α ≥ 0, a < x ≤ b. This integral is motivated by the well known Cauchy formula:
Z x a dt1 Z t1 a dt2... Z tn−1 a f (tn)dtn= 1 Γ(n) Z x a (x − t)n−1f (t)dt, n ∈ N∗. The second is the Hadamard fractional integral introduced by Hadamard [9]. It is given by: Jα af (x) = 1 Γ(α) Z x a ³ logx t ´α−1 f (t)dt t , α > 0, x > a.
∗Department of Mathematics, Faculty of Science and Arts, D¨uzce University, D¨uzce, Turkey,
Email: sarikayamz@gmail.com
†Corresponding Author.
‡Laboratory of Pure and Applied Mathematics, UMAB, University of Mostaganem, Algeria,
Email: zzdahmani@yahoo.fr
§Department of Mathematics, Faculty of Science and Arts, Afyon Kocatepe University,
Afyon-TURKEY, Email: mkiris@gmail.com, kiris@aku.edu.tr
¶Centre for Advanced Studies in Pure and Applied Mathematics, Bahauddin Zakariya
The Hadamard integral is based on the generalization of the integral Z x a dt1 t1 Z t1 a dt2 t2 ... Z tn−1 a f (tn) tn dtn= 1 Γ(α) 1 Γ(α) Z x a ³ logx t ´n−1 f (t)dt t for n ∈ N∗.
In [10], Katugampola gave a new fractional integration which generalizes both the Riemann-Liouville and Hadamard fractional integrals into a single form. This generalization is based on the observation that, for n ∈ N∗,
Z x a ts1dt1 Z t1 a ts2dt2... Z tn−1 a tsnf (tn)dtn= (s + 1) 1−n Γ(α) Z x a ¡ xs+1− ts+1¢n−1tsf (t)dt, which gives the following fractional version
sJα af (x) = (s + 1)1−n Γ(α) Z x a ¡ xs+1− ts+1¢α−1tsf (t)dt, where α and s 6= −1 are real numbers.
Recently, in [6], Diaz and Pariguan have defined new functions called k-gamma and k-beta functions and the Pochhammer k-symbol that is respectively generalization of the classical gamma and beta functions and the classical Pochhammer symbol:
Γk(x) = lim n→∞
n!kn(nk)x k−1
(x)n,k , k > 0,
where (x)n,k is the Pochhammer k-symbol for factorial function. It has been shown that the Mellin transform of the exponential function e−tk
k is the k-gamma function, explicitly given by
Γk(x) = Z ∞ 0 tx−1e−tk kdt, x > 0. Clearly, Γ(x) = lim k→1Γk(x), Γk(x) = k x k−1Γ(x k) and Γk(x + k) = xΓk(x). Further-more, k-beta function is defined as follows
Bk(x, y) = 1 k Z 1 0 txk−1(1 − t) y k−1dt,
so that Bk(x, y) = 1kB(xk,yk) and Bk(x, y) = ΓΓk(x)Γk(x+y)k(y).
Later, under the above definitions, in [13], Mubeen and Habibullah have introduced the k-fractional integral of the Riemann-Liouville type as follows:
kJαf (x) = 1
kΓk(α) Z x
0
(x − t)αk−1f (t)dt, α > 0, x > 0.
Note that when k → 1, then it reduces to the classical Riemann-liouville fractional integral.
2. (k, s)-Riemann-Liouville fractional integral
In this section, we present the (k, s) fractional integration which generalizes all of the above Riemann-Liouville fractional integrals as follows:
2.1. Definition. Let f be a continuous function on on a the finite real interval [a, b]. Then (k, s)-Riemann-Liouville fractional integral of f of order α > 0 is defined by: (2.1) s kJaαf (x) := (s + 1)1−α k kΓk(α) Z x a ¡ xs+1− ts+1¢αk−1tsf (t)dt, x ∈ [a, b], where k > 0, s ∈ R\{−1}.
In the following theorem, we prove that the (k, s) fractional integral is well defined:
2.2. Theorem. Let f ∈ L1[a, b], s ∈ R\{−1} and k > 0. ThenskJaαf (x) exists for
any x ∈ [a, b], α > 0.
Proof. Let ∆ := [a, b] × [a, b] and P : ∆ → R ; P (x, t) =h¡xs+1− ts+1¢αk−1tsi. It clear to see that P = P++ P−, where
P+(x, t) := ½ ¡ xs+1− ts+1¢αk−1ts , a ≤ t ≤ x ≤ b 0 , a ≤ x ≤ t ≤ b and P−(x, t) := ½ ¡ ts+1− xs+1¢αk−1xs , a ≤ t ≤ x ≤ b 0 , a ≤ x ≤ t ≤ b. Since P is measurable on ∆, then we can write
Z b a P (x, t)dt = Z x a P (x, t)dt = Z x a ¡ xs+1− ts+1¢αk−1tsdt = k α ¡ xs+1− as+1¢αk . By using the repeated integral, we obtain
Z b a ÃZ b a P (x, t) |f (x)| dt ! dx = Z b a |f (x)| ÃZ b a P (x, t)dt ! dx = k α Z b a ¡ xs+1− as+1¢αk|f (x)| dx ≤ k α ¡ bs+1− as+1¢αk Z b a |f (x)| dx. That is Z b a ÃZ b a P (x, t) |f (x)| dt ! dx = Z b a |f (x)| ÃZ b a P (x, t)dt ! dx ≤ k α ¡ bs+1− as+1¢αkkf (x)k L1[a,b]< ∞.
Therefore, the function Q : ∆ → R; Q(x, t) := P (x, t)f (x) is integrable over ∆ by Tonelli’s theorem. Hence, by Fubini’s theorem RabP (x, t)f (x)dx is an integrable function on [a, b], as a function of t ∈ [a, b]. That is, s
kJaαf (x) exists. ¤ Now, we prove the commutativity and the semigroup properties of the (k, s)-Riemann-Liouville fractional integral. We have:
2.3. Theorem. Let f be continuous on [a, b], k > 0 and s ∈ R\{−1}. Then, s kJaα £ s kJaβf (x) ¤ = skJaα+βf (x) = skJaβ[skJaαf (x)] ,
for all α > 0, β > 0, x ∈ [a, b].
Proof. Thanks to Definition 1 and by Dirichlet’s formula, we have
s kJaα £s kJaβf (x) ¤ =(s + 1)1− α k kΓk(α) Z x a ¡ xs+1− ts+1¢αk−1ts s kJaβf (t)dt = (s+1)1− αk kΓk(α) Rx a ¡ xs+1− ts+1¢αk−1ts · (s+1)1− βk kΓk(β) Rt a ¡ ts+1− τs+1¢βk−1τsf (τ )dτ ¸ dt. = (s+1)1− αk kΓk(α) Rx a ¡ xs+1− ts+1¢αk−1ts · (s+1)1− βk kΓk(β) Rt a ¡ ts+1− τs+1¢βk−1τsf (τ )dτ ¸ dt. That is (2.2) s kJaα £s kJaβf (x) ¤ = (s + 1) 2−α+β k k2Γ k(α)Γk(β) Z x a τsf (τ ) ·Z x τ ¡ xs+1− ts+1¢αk−1ts¡ts+1− τs+1¢βk−1dt ¸ dτ. Using the change of variable y =¡ts+1− τs+1¢/¡xs+1− τs+1¢, we can write
(2.3) Z x τ ¡ xs+1− ts+1¢αk−1¡ts+1− τs+1¢βk−1tsdt = (xs+1−τs+1) α+β k −1 s+1 R1 0(1 − y) α k−1y β k−1dy = (xs+1−τs+1) α+β k −1 s+1 R1 0(1 − y) α k−1yβk−1dy = (x s+1−τs+1)α+βk −1 s+1 kBk(α, β). According to the k-beta function and by (2.2) and (2.3), we obtain
s kJaα £s kJaβf (x) ¤ = (s + 1) 1−α+β k kΓk(α + β) Z x a ¡ xs+1− τs+1¢α+βk −1τsf (τ )dτ = skJaα+βf (x).
This completes the proof of the Theorem 2.3. ¤
2.4. Theorem. Let α, β > 0, k > 0 and s ∈ R\{−1}. Then, we have (2.4) skJaα h ¡ xs+1− as+1¢βk−1 i = Γk(β) (s + 1)αkΓk(α + β) ¡ xs+1− as+1¢α+βk −1, where Γk denotes the k-gamma function.
Proof. By Definition 1 and using the change of variable y =¡xs+1− ts+1¢/¡xs+1− as+1¢; x ∈ ]a, b], we get s kJaα h ¡ xs+1− as+1¢βk−1 i = (s + 1) 1−α k kΓk(α) Z x a ¡ xs+1− ts+1¢αk−1ts¡ts+1− as+1¢α+βk −1dt = (s + 1) −α k ¡ xs+1− as+1¢α+βk −1 kΓk(α) Z 1 0 (1 − y)αk−1yβk−1dy = ¡ xs+1− as+1¢α+βk −1 (s + 1)α kΓk(α) Bk(α, β).
The case a = x is trivial. The proof of the Theorem 2.4 is complete. ¤ 2.5. Remark. (i :) Taking s = 0, k > 0 in (2.4), we obtain
(2.5) kJaα h (x − a)βk−1 i = Γk(β) Γk(α + β) (x − a)α+βk −1. (ii :) The formula (2.4) for s = 0, k = 1 becomes
Jα a h (x − a)β−1 i = Γ(β) Γ(α + β)(x − a) α+β−1 .
2.6. Corollary. Let k > 0 and s ∈ R\{−1}. Then the formula (2.6) s kJaα(1) = 1 (s + 1)αkΓk(α + k) ¡ xs+1− as+1¢αk−2 is valid for any α > 0.
2.7. Remark. (a :) For s = 0, k > 0 in (2.6), we get (2.7) kJaα(1) = 1 Γk(α + k)(x − a) α k−2. (b :) For s = 0, k = 1 we have Jα a(1) = 1 Γ(α + 1)(x − a) α−2 .
3. Some new (k, s)-Riemann-Liouville fractional integral
inequal-ities
Chebyshev inequalities can be represented in (k, s)-fractional integral forms as follows:
3.1. Theorem. Let f, g be two synchronous on [0, ∞), then for all t > a ≥ 0, α > 0, β > 0, the following inequalities for (k, s)-fractional integrals hold:
(3.1) skJaαf g(t) ≥ 1 Jα a(1) s kJaαf (t) skJaαg(t) (3.2) skJaαf g(t)skJaβ(1)+skJaβf g(t)skJaα(1) ≥ skJaαf (t)skJaβg(t)+skJaαg(t)skJaβf (t).
Proof. Since the functions f and g are synchronous on [0, ∞), then for all x, y ≥ 0, we have
(f (x) − f (y)) (g (x) − g(y)) ≥ 0. Therefore
(3.3) f (x) g (x) + f (y) g (y) ≥ f (x) g (y) + f (y) g (x) Multiplying both sides of (3.3) by (s+1)1− αk
kΓk(α) ¡
ts+1− xs+1¢αk−1xs, then integrating the resulting inequality with respest to x over (a, t), we obtain
(s + 1)1−α k kΓk(α) Z t a ¡ ts+1− xs+1¢αk−1xsf (x) g (x) dx +(s + 1)1− α k kΓk(α) Z t a ¡ ts+1− xs+1¢αk−1xsf (y) g (y) dx ≥ (s + 1) 1−α k kΓk(α) Z t a ¡ ts+1− xs+1¢αk−1xsf (x) g (y) dx +(s + 1)1− α k kΓk(α) Z t a ¡ ts+1− xs+1¢αk−1xsf (y) g (x) dx, i.e. (3.4) s
kJaαf g(t) + f (y) g (y) skJaα(1) ≥ g (y) skJaαf (t) + f (y) skJaαg(t). Multiplying both sides of (3.3) by (s+1)1− αk
kΓk(α) ¡
ts+1− ys+1¢αk−1ys, then integrating the resulting inequality with respest to y over (a, t), we obtain
s kJaαf g(t) (s + 1)1−α k kΓk(α) Z t a ¡ ts+1− ys+1¢αk−1ysdy +skJaα(1) (s + 1)1−α k kΓk(α) Z t a ¡ ts+1− ys+1¢αk−1ysf (y) g (y) dy ≥ s kJaαf (t) (s + 1)1−α k kΓk(α) Z t a ¡ ts+1− ys+1¢αk−1ysg(y)dy +skJaαg(t) (s + 1)1−α k kΓk(α) Z t a ¡ ts+1− ys+1¢αk−1ysf (y) dy, that is s kJaαf g(t) ≥ 1 Jα a(1) s kJaαf (t)skJaαg(t) and the first inequality is proved.
Multiplying both sides of (3.3) by (s+1)1− βk kΓk(α)
¡
ts+1− ys+1¢βk−1ys, then integrat-ing the resultintegrat-ing inequality with respest to y over (a, t), we obtain
s kJaαf g(t) (s + 1)1−βk kΓk(α) Z t a ¡ ts+1− ys+1¢βk−1ysdy +s kJaα(1) (s + 1)1−β k kΓk(α) Z t a ¡ ts+1− ys+1¢βk−1ysf (y) g (y) dy ≥ s kJaαf (t) (s + 1)1−β k kΓk(α) Z t a ¡ ts+1− ys+1¢βk−1ysg(y)dy +s kJaαg(t) (s + 1)1−β k kΓk(α) Z t a ¡ ts+1− ys+1¢βk−1ysf (y) dy, that is s kJaαf g(t)skJaβ(1)+skJaβf g(t)skJaα(1) ≥ skJaαf (t)skJaβg(t)+skJaαg(t)skJaβf (t) and the second inequality is proved. The proof is completed. ¤ 3.2. Theorem. Let f, g be two synchronous on [0, ∞), h ≥ 0, then for all t > a ≥ 0, α > 0, β > 0, the following inequalities for (k, s)-fractional integrals hold:
1 (s + 1)βkΓk(β + k) ¡ ts+1− as+1¢βk−2 s kJaαf gh(t) + 1 (s + 1)αkΓk(α + k) ¡ ts+1− as+1¢αk−2 s kJaβf gh(t) ≥ s kJaαf h(t)skJaβg(t) + skJaαgh(t) skJaβf (t) − skJaαh(t)skJaβf g(t) − skJaαf g(t)skJaβh(t) +skJaαf (t)skJaβgh(t) + skJaαg(t)skJaβf h(t).
Proof. Since the functions f and g are synchronous on [0, ∞) and h ≥ 0, then for all x, y ≥ 0, we have
(f (x) − f (y)) (g (x) − g(y)) (h (x) + h (y)) ≥ 0. By opening the above, we get
f (x) g (x) h (x) + f (y) g (y) h (y) (3.5)
≥ f (x) g (y) h (x) + f (y) g (x) h (x) − f (y) g (y) h (x) −f (x) g (x) h (y) + f (x) g (y) h (y) + f (y) g (x) h (y) .
Multiplying both sides of (3.5) by (s+1)1− αk kΓk(α)
¡
ts+1− xs+1¢αk−1xs, then integrating the resulting inequality with respest to x over (a, t), we obtain
(s + 1)1−α k kΓk(α) Z t a ¡ ts+1− xs+1¢αk−1xsf (x) g (x) h (x) dx +f (y) g (y) h (y)(s + 1)1−
α k kΓk(α) Z t a ¡ ts+1− xs+1¢αk−1xsdx ≥ g (y)(s + 1) 1−α k kΓk(α) Z t a ¡ ts+1− xs+1¢αk−1xsf (x) h (x) dx +f (y)(s + 1)1− α k kΓk(α) Z t a ¡ ts+1− xs+1¢αk−1xsg (x) h (x) dx −f (y) g (y)(s + 1) 1−α k kΓk(α) Z t a ¡ ts+1− xs+1¢αk−1xsh (x) dx −h (y)(s + 1)1− α k kΓk(α) Z t a ¡ ts+1− xs+1¢αk−1xsf (x) g (x) dx +g (y) h (y)(s + 1) 1−α k kΓk(α) Z t a ¡ ts+1− xs+1¢αk−1xsf (x) dx +f (y) h (y)(s + 1)1− α k kΓk(α) Z t a ¡ ts+1− xs+1¢αk−1xsg (x) dx. i.e, s
kJaαf gh(t) + f (y) g (y) h(y) skJaα(1) (3.6)
≥ g (y) skJaαf h(t) + f (y) skJaαgh(t) − f (y) g (y) skJaαh(t) − h(y)skJaαf g(t) +g (y) h (y) s
Multiplying both sides of (3.6) by (s+1)1− βk kΓk(β)
¡
ts+1− ys+1¢βk−1ys, then integrating the resulting inequality with respest to y over (a, t), we obtain
s kJaαf gh(t) (s + 1)1−βk kΓk(β) Z t a ¡ ts+1− ys+1¢βk−1ysdy +s kJaα(1) (s + 1)1−βk kΓk(β) Z t a ¡
ts+1− ys+1¢βk−1ysf (y) g (y) h(y)dy
≥ s kJaαf h(t) (s + 1)1−βk kΓk(β) Z t a ¡ ts+1− ys+1¢βk−1ysg (y) dy +s kJaαgh(t) (s + 1)1−βk kΓk(β) Z t a ¡ ts+1− ys+1¢βk−1ysf (y) dy −s kJaαh(t) (s + 1)1−βk kΓk(β) Z t a ¡ ts+1− ys+1¢βk−1ysf (y) g (y) dy −s kJaαf g(t) (s + 1)1−β k kΓk(β) Z t a ¡ ts+1− ys+1¢βk−1ysh(y)dy +s kJaαf (t) (s + 1)1−β k kΓk(β) Z t a ¡ ts+1− ys+1¢βk−1ysg (y) h (y) dy +s kJaαg(t) (s + 1)1−β k kΓk(β) Z t a ¡
ts+1− ys+1¢βk−1ysf (y) h (y) dy, that is s kJaαf gh(t)skJaβ(1) +skJaα(1)skJaβf gh(t) ≥ skJaαf h(t)skJaβg(t) + skJaαgh(t)skJaβf (t) −s kJaαh(t)skJaβf g(t) − skJaαf g(t)skJaβh(t) +skJaαf (t)skJaβgh(t) + skJaαg(t)skJaβf h(t)
which this completes the proof. ¤
3.3. Corollary. Let f, g be two synchronous on [0, ∞), h ≥ 0, then for all t > a ≥ 0, α > 0, the following inequalities for (k, s)-fractional integrals hold:
1 (s + 1)αkΓk(α + k) ¡ ts+1− as+1¢αk−2 s kJaαf gh(t) ≥ skJaαf h(t)skJaαg(t) + skJaαgh(t)skJaαf (t) − skJaαh(t)skJaαf g(t). 3.4. Theorem. Let f, g and h be three monotonic functions defined on [0, ∞) satisfying the following
for all x, y ∈ [a, t] , then for all t > a ≥ 0, α > 0, β > 0, the following inequalities for (k, s)-fractional integrals hold:
1 (s + 1)βkΓk(β + k) ¡ ts+1− as+1¢βk−2 s kJaαf gh(t) − 1 (s + 1)αkΓk(α + k) ¡ ts+1− as+1¢αk−2 s kJaβf gh(t) ≥ s kJaαf h(t)skJaβg(t) + skJaαgh(t) skJaβf (t) − skJaαh(t)skJaβf g(t) + skJaαf g(t)skJaβh(t) −s kJaαf (t)skJaβgh(t) − skJaαg(t)skJaβf h(t).
Proof. The proof is similar to that given in Theorem 3.2. ¤ 3.5. Theorem. Let f and g be two functions on [0, ∞), then for all t > a ≥ 0, α > 0, β > 0, the following inequalities for (k, s)-fractional integrals hold:
1 (s + 1)βkΓk(β + k) ¡ ts+1− as+1¢βk−2 s kJaαf2(t) (3.7) + 1 (s + 1)α kΓk(α + k) ¡ ts+1− as+1¢αk−2 s kJaβg2(t) ≥ 2s kJaαf (t)skJaβg(t) (3.8) s kJaαf2(t)skJaβg2(t) + skJaβf2(t)skJaαg2(t) ≥ 2 ksJaαf g(t)skJaβf g(t). Proof. Since, (f (x) − g(y))2≥ 0 then we have (3.9) f2(x) + g2(y) ≥ 2f (x)g(y).
Multiplying both sides of (3.9) by(s+1)1− αk kΓk(α) ¡ ts+1− xs+1¢αk−1xsand(s+1)1− βk kΓk(β) ¡ ts+1− ys+1¢βk−1ys, then integrating the resulting inequality with respest to x and y over (a, t)
respec-tively, we obtain (3.7). On the other hand, since
(f (x)g(y) − f (y)g(x))2≥ 0
then under procedures similar to the above we obtain (3.8). ¤ 3.6. Corollary. Let f and g be two functions on [0, ∞), then for all t > a ≥ 0, α > 0, the following inequalities for (k, s)-fractional integrals hold:
1 (s + 1)αkΓk(α + k) ¡ ts+1− as+1¢αk−2 £s kJaαf2(t) + skJaβg2(t) ¤ ≥ 2skJaαf (t)skJaαg(t) s kJaαf2(t)skJaαg2(t) ≥ [skJaαf g(t)]2.
3.7. Theorem. Let f : R → R and defined by ¯ f (x) = Z x a tsf (t)dt, x > a ≥ 0, s ∈ R\{−1} then for α ≥ k > 0 s kJaαf (x) = 1 k s kJaα−kf (x)¯
Proof. By definition of the (k, s)-fractional integral and by using Dirichlet’s for-mula, we have s kJaαf (x) =¯ (s + 1)1−α k kΓk(α) Z x a ¡ xs+1− ts+1¢αk−1ts Z t a usf (u)dudt = (s + 1) 1−α k kΓk(α) Z x a usf (u) Z x u ¡ xs+1− ts+1¢αk−1tsdtdu = (s + 1)− α k Γk(α + k) Z x a ¡ xs+1− us+1¢αkusf (u)du = k s kJaα+kf (x).
This completes the proof of Theorem 3.7. ¤
We give the generalized Cauchy-Buniakovsky-Schwarz inequality as follows: 3.8. Lemma. Let f, g, h : [a, b] → (0, ∞) be three functions 0 ≤ a < b. Then (3.10) ÃZ b a gm(t)hx(t)f (t)dt ! ÃZ b a gn(t)hy(t)f (t)dt ! ≥ ÃZ b a gm+n2 (t)h x+y 2 (t)f (t)dt !2 . where m, n, x, y arbitrary real numbers.
Proof. Z b a pgm(t)hx(t)f (t) sZ b a gn(t)hy(t)f (t)dt −pgn(t)hy(t)f (t) sZ b a gm(t)hx(t)f (t)dt 2 dt ≥ 0 Z b a " gm(t)hx(t)f (t) Z b a gn(t)hy(t)f (t)dt + gn(t)hy(t)f (t) Z b a gm(t)hx(t)f (t)dt −2 gm+n2 (t)hx+y2 (t)f (t) sZ b a gm(t)hx(t)f (t)dt sZ b a gn(t)hy(t)f (t)dt dt ≥ 0
2 ÃZ b a gm(t)hx(t)f (t)dt ! ÃZ b a gn(t)hy(t)f (t)dt ! ≥ 2 ÃZ b a gm+n2 (t)h x+y 2 (t)f (t)dt ! sZ b a gm(t)hx(t)f (t)dt sZ b a gn(t)hy(t)f (t)dt
which this give the requair inequality. ¤
3.9. Theorem. Let f ∈ L1[a, b]. Then
(3.11) ³ s kJ m(α k−1)+1 a fr(x) ´ ³ s kJ n(α k−1)+1 a fp(x) ´ ≥ ³ s kJ m+n 2 (αk−1)+1 a f r+p 2 (x) ´2 for k, m, n, r, p > 0 and α > 1. Proof. By taking g(t) =¡xs+1− ts+1¢αk−1, f (t) = ts(s+1)1− αk kΓk(α) and h(t) = f (t) in (3.10), we obtain µ (s + 1)1−α k kΓk(α) Z x a ¡ xs+1− ts+1¢m(αk−1)tsfr(t)dt ¶ µ (s + 1)1−α k kΓk(α) Z x a ¡ xs+1− ts+1¢n(αk−1)tsfp(t)dt ¶ ≥ µ (s + 1)1−α k kΓk(α) Z x a ¡ xs+1− ts+1¢m+n2 (αk−1)tsfr+p2 (t)dt ¶2
which can be written as (3.11). ¤
3.10. Remark. For k = 1 in (3.11), we get the following inequalities ³ Jam(α−1)+1fr(x) ´ ³ kJan(α−1)+1fs(x) ´ ≥ ³ kJ m+n 2 (α−1)+1 a f r+s 2 (x) ´2 .
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