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Appl. Math. Inf. Sci. 9, No. 4, 2161-2167 (2015) 2161 Applied Mathematics & Information Sciences An International Journal

http://dx.doi.org/10.12785/amis/090456

Certain Integral Transforms for the Incomplete Functions

1 2, Junesang Choi and Praveen Agarwal ∗ 1 Department of Mathematics, Dongguk University, Gyeongju 780-714, Republic of Korea 2 Department of Mathematics, Anand International College of Engineering, Jaipur-303012, India

Received: 25 Nov. 2014, Revised: 25 Feb. 2015, Accepted: 27 Feb. 2015 Published online: 1 Jul. 2015

Abstract: Srivastava et al. [14] introduced the incomplete Pochhammer symbols that lead to a natural generalization and decomposition of a class of hypergeometric and other related functions to mainly investigate certain potentially useful closed-form representations of definite and semi-definite integrals of various . Here, in this paper, we use the integral transforms like Beta transform, , Mellin transform, Whittaker transforms, K-transform and Hankel transform to investigate certain interesting and (potentially) useful integral transforms the incomplete hypergeometric type functions pγq[z] and pΓq[z]. Relevant connections of the various results presented here with those involving simpler and earlier ones are also pointed out.

Keywords: ; Beta function; Incomplete gamma functions; Incomplete Pochhammer symbols; incomplete hypergeometric functions; Beta transform; Laplace transform; Mellin transform; Whittaker transforms; K-transform; Hankel transform

1 Introduction and Preliminaries We also recall the Pochhammer symbol (λ)n defined (for λ C) by The incomplete Gamma type functions like γ(s,x) and ∈ Γ (s,x), both of which are certain generalizations of the classical Gamma function Γ (z), given in (1) and (2), 1 (n = 0) respectively, have been investigated by many authors. The (λ)n : = incomplete Gamma functions have proved to be important λ(λ + 1) (λ + n 1) (n N := 1,2,... ) ··· − ∈ { }(4) for physicists and engineers as well as mathematicians. Γ (λ + n) = (λ C Z−), For more details, one may refer to the books [1,2,5,20, Γ (λ) ∈ \ 0 22] and the recent papers [3,13,14,17,18]and [19] on the subject. The familiar incomplete Gamma functions γ(s,x) and Z Γ (s,x) defined by where 0− denotes the set of nonpositive integers (see, e.g., [15, p. 2 and p. 5]). x γ s 1 t ℜ (s,x) := t − e− dt ( (s) > 0; x 0), (1) The theory of the incomplete Gamma functions, as a Z0 ≥ part of the theory of confluent hypergeometric functions, and ∞ has received its first systematic exposition by Tricomi s 1 t [21] in the early 1950s. Musallam and Kalla [8,9] Γ (s,x) := t − e− dt (2) Zx considered a more general incomplete gamma function (x 0; ℜ(s) > 0 when x = 0), involving the Gauss hypergeometric function and ≥ respectively, satisfy the following decomposition formula established a number of analytic properties including recurrence relations, asymptotic expansions and γ Γ Γ ℜ (s,x)+ (s,x)= (s) ( (s) > 0), computation for special values of the parameters. Very where Γ (s) is the well-known Euler’s Gamma function recently, Srivastava et al.[14] introduced and studied defined by some fundamental properties and characteristics of a ∞ family of two potentially useful and generalized Γ s 1 t ℜ (s) := t − e− dt ( (s) > 0). (3) incomplete hypergeometric functions defined as follows: Z0

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2 of the Incomplete

(a1,x),a2, ,ap; Hypergeometric functions pγq ··· z :=  b1, ,bq ;  In this section, we shall prove six theorems, which exhibit ∞ ··· (a ;x) (a ) (a ) zn certain connections between the integral transforms such ∑ 1 n 2 n p n (5) ··· as Euler transform, Laplace transform, Mellin transform, n 0 (b1)n (bq)n · n! = ··· Whittaker transforms, K-transform and Hankel transform and and the generalized incomplete hypergeometric type γ Γ (a1,x),a2, ,ap; functions p q[z] and p q[z] given by (5) and (6), pΓq ··· z := respectively.  b1, ,bq ;  ∞ ··· n [a1;x]n(a2)n (ap)n z Theorem 1. Suppose that x 0, α, β C and p, q ∑ ··· , (6) N : N 0 . Then we have ≥ ∈ ∈ (b1)n (bq)n · n! 0 = n=0 ··· ∪{ } Γ (α)Γ (β) where (a1;x)n and [a1;x]n are interesting generalization of B pγq[yz] : α,β = λ Γ (α + β) × the Pochhammer symbol ( )n, in terms of the incomplete  gamma type functions γ s x and Γ s x defined as α a x a a ; ( , ) ( , ) γ ,( 1, ), 2, , p follows p+1 q+1 ··· y (11)  α + β,b1, ,bq ;  ··· and γ(λ + ν,x) Γ α Γ β (λ;x)ν := (λ, ν C; x 0) (7) Γ α β ( ) ( ) Γ (λ) B p q[yz] : , = ∈ ≥ Γ (α + β) ×  and α a x a a ; Γ ,( 1, ), 2, , p Γ λ ν x p+1 q+1 α β ··· y , (12) λ ( + , ) λ ν C  + ,b1, ,bq ;  [ ;x]ν := Γ λ ( , ; x 0). (8) ··· ( ) ∈ ≥ where B f (z) : α,β denotes the Euler (Beta) transform { } These incomplete Pochhammer symbols (λ;x)ν and of f (z) defined by (see, e.g.,[12]): [λ;x]ν satisfy the following decomposition relation: 1 α β α 1 β 1 λ λ λ λ ν C B f (z) : , = z − (1 z) − f (z)dz. (13) ( ;x)ν + [ ;x]ν = ( )ν ( , ;x 0). { } Z0 − ∈ ≥ (9) Proof.Applying (5) to the Euler (Beta) transform in (13), Remark 1. We repeat the remark given by Srivastava we get et al.[14, Remark 7] for completeness and an easier 1 α 1 β 1 reference. In (1),(2), (5),(6),(7) and (8), the argument z − (1 z) − pγq[yz]dz Z0 − x 0 is independent of the argument z C which occurs 1 in≥ the (5) and (6) and also in the result∈ presented in this α 1 β 1 = z − (1 z) − paper. Since (see, e.g.,[14, p. 675]) Z0 − × ∞ n λ λ λ λ (a1;x)n(a2)n (ap)n (yz) ( ;x)n ( )n and [ ;x]n ( )n ∑ ··· dz. (14) | | ≤ | | | | ≤ | | (b1)n (bq)n · n! (n N;λ C;x 0), (10) n=0 ··· ∈ ∈ ≥ By changing the order of integration and summation in and the precise (sufficient) conditions under which the infinite α β series in the definitions (5) and (6) would converge using the Beta function B( , ) (see, e.g., [15, p. 8]): absolutely can be derived from those that are 1 well-documented in the case of the generalized α 1 β 1 t − (1 t) − dt (ℜ(α) > 0;ℜ(β) > 0)  Z hypergeometric function pFq (p,q N) (see, for details, 0 − α β  [10, pp.73-74]and [16, p.20];seealso∈ [5]). B( , )=   Γ α Γ β ( ) ( ) α β C Z Γ α β ( , 0−),  ( + ) ∈ \ Integral transforms are widely used to solve  differential equations and integral equations. So a lot of  (15) work has been done on the theory and applications of we get integral transforms. Most popular integral transforms are 1 due to Euler, Laplace, Fourier, Mellin, Hankel and so on. α 1 β 1 z − (1 z) − pγq[yz]dz Here, in this paper, we use the integral transforms like Z0 − Beta transform, Laplace transform, Mellin transform, ∞ (a1;x)n(a2)n (ap)n Whittaker transforms, K-transform and Hankel transform = ∑ ··· (b1)n (bq)n × to investigate certain interesting and (potentially) useful n=0 ··· integral transforms for incomplete hypergeometric type Γ (α + n)Γ (β) (y)n γ Γ , (16) functions p q[z] and p q[z]. Γ (α + β + n) n!

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which, upon using (5), yields our desired result (11). and terminating at the end points σ + i∞ (σ R) with the It is easy to see that a similar argument as in the proof usual indentations in order to separate one∈ set of poles of (11) will establish the result (12). This completes the from the other set of poles of the integrand. Then we have proof of Theorem 1. (Ap,x); (a1,x),a2, ,ap; pγq z = pγq ··· z  Bq;   b1, ,bq ;  ℜ α C ··· Theorem 2. Suppose that x 0, (s) > 0, , Γ Γ N y ≥ ∈ 1 (b1) (bq) p, q 0 and s < 1. Then we obtain = ··· ∈ | | 2πi Γ (a ) Γ (a ) × α 1 1 p L z − pγq[yz] ··· γ(a1 + s,x)Γ (a2 + s) Γ (ap + s)Γ ( s)  Γ (α) α,(a1,x),a2, ,ap; y ··· − γ ZL Γ (b1 + s) Γ (bq + s) = n p+1 q ··· (17) ··· s  b1, ,bq ; s  Γ ( s)( z)sds, (23) ··· − − and and α 1 L z − pΓq[yz]  Γ α α,(a ,x),a , ,a ; y A x ; a x a a ; ( ) Γ 1 2 p Γ ( p, ) Γ ( 1, ), 2, , p = n p+1 q ··· , (18) p q z =p q ··· z s  b1, ,bq ; s   Bq;   b1, ,bq ;  ··· ··· where L f (z) denotes the Laplace transform of f (z) 1 Γ (b1) Γ (bq) { } = ··· defined by (see,[12]): 2πi Γ (a1) Γ (ap) × ∞ ··· sz Γ (a1 + s,x)Γ (a2 + s) Γ (ap + s)Γ ( s) L f (z) = e− f (z)dz. (19) ··· − { } Z0 ZL Γ (b1 + s) Γ (bq + s) ··· Here we assume both sides of above results exist. Γ ( s)( z)sds, (24) − − Proof.Applying (5) to the Laplace transform in (19), we where arg( z) < π. get | − | ∞ ∞ Remark 2 (see [14, Remark 7]). In their special case α 1 sz α 1 sz z − e− pγq[yz]dz = z − e− when p 1 = q = 1, the assertions (23) and (24) of Z0 Z0 Lemma− 1 would immediately yield the corresponding ∞ n (a1;x)n(a2)n (ap)n (yz) Mellin-Barnes type contour integral representations for ∑ ··· . (20) γ × n=0 (b1)n (bq)n · n! the incomplete Gauss hypergeometric functions 2 2 and ··· Γ , respectively. By changing the order of integration and summation and 2 2 using the known Laplace transform (see, e.g., [11, Theorem 3. Suppose x 0, ℜ(s) > 0, w C and ≥ ∈ Eq.(2.2)]): p, q N0. Then we have ∈ Γ (ν + 1) Γ Γ ν ℜ ν ℜ γ (b1) (bq) L z = ν+1 ( ( ) > 1; (s) > 0), M p q[ wt];s = ··· { } s − − Γ (a1) Γ (ap) (21)  ··· γ(a1 + s,x)Γ (a2 + s) Γ (ap + s)Γ ( s) ··· − ws (25) we get × Γ (b1 + s) Γ (bq + s) · ∞ ··· α 1 sz and z − e− pγq[yz]dz Z0 Γ (b1) Γ (bq) ∞ Γ α n M Γ wt ;s (a1;x)n(a2)n (ap)n ( + n) (y) p q[ ] = Γ ···Γ = ∑ ··· , (22) − (a1) (ap) × α+n  ··· n 0 (b1)n (bq)n · s n! Γ Γ Γ Γ = ··· (a1 + s,x) (a2 + s) (ap + s) ( s) ··· − ws, (26) which, upon using (5), yields our desired result (17). Γ (b1 + s) Γ (bq + s) · By following the same procedure as in the proof of ··· where M f (z);s denotes the Mellin transform of f (z) the result (17), one can easily establish the result (18). { } Therefore, we omit the details of the proof of the result defined by (see, e.g.,[7, p.46, Eqs. (2.1) and (2.2)]: ∞ (18). This completes the proof of Theorem 2. s 1 M f (z);s = z − f (z)dz = f (s),ℜ(s) > 0 (27) { } Z0 Mellin transform is based on the preliminary 1 assertions giving Mellin-Barnes contour integral and its inverse Mellin transform M− representations of the incomplete generalized 1 1 s hypergeometric functions γ [z] and Γ [z]. f (s)= M− [ f (s);t]= f (s)t− ds. (28) p q p q 2πi ZL L L Lemma 1 (see [14, Theorem 18]). Let = (σ; i∞) be Here we assume that both members of the Mellin a Mellin-Barnes-type contour starting at the point ∓σ i∞ transforms exist. −

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w Proof.Setting z = t in (23), we get also [6, p. 79]): − ∞ ν 1 Γ (b ) Γ (bq) 1 t/2 γ 1 t − e− Wλ ,µ (t)dt p q[ wt]= ··· Z0 − 2πi Γ (a1) Γ (ap) × 1 1 γ Γ ··· Γ Γ Γ ( + µ + ν)Γ ( µ + ν) (a1 + s,x) (a2 + s) (ap + s) ( s) = 2 2 − (33) Γ ···Γ − Γ (1 λ + ν) ZL (b1 + s) (bq + s) × − ··· w s 1 Γ s ds (29) ℜ(ν µ) > , ( ) .  2 −  t  ± − Using (28) in the resulting identity, Equation (29) we obtain immediately leads to (25). ∞ ρ (a1;x)n(a2)n (ap)n Following the same procedure, one can easily establish L = δ − ∑ ··· (b1)n (bq)n × the result (26). n=0 ··· (w)n Γ ( 1 + µ + ρ + n)Γ ( 1 µ + ρ + n) 2 2 − , (34) Theorem 4. Suppose that x 0, ρ, δ, λ, µ C and δ nn! Γ (1 λ + ρ + n) ≥ ∈ − p, q N0. Then the following integral relations hold: ∈ which, in view of (5), yields our desired result (30). ∞ It is easy to see that a similar argument as in the proof ρ 1 δt/2 δ γ t − e− Wλ ,µ ( t) p q[wt]dt of (30) will establish the result (31). This completes the Z0 Γ 1 µ ρ Γ 1 µ ρ proof of Theorem 4. ρ ( 2 + + ) ( 2 + ) = δ − − Γ (1 λ + ρ) − Theorem 5. Suppose that x 0, ℜ(u) > 0, ρ, ν, w C 1 1 ≥ ∈ + µ + ρ, µ + ρ,(a ,x),a , ,a ; w and p, q N0. Then the following integral formulas hold: γ 2 2 1 2 p ∈ p+2 q+1  − ··· δ  ∞ ρ × ρ 1 2 ρ ν (1 λ + ρ),b1, ,bq ; 1 γ 2 Γ  − ···  t − Kν (ut) p q[wt ]dt = ( ± ) (30) Z0 4 u 2 ρ + ν ρ ν (35) and , − ,(a1,x),a2, ,ap; 4w γ 2 2 ∞ p+2 q  ··· 2  ρ 1 δt/2 δ Γ × u t − e− Wλ,µ ( t) p q[wt]dt b1, ,bq ; Z0  ···  Γ ( 1 + µ + ρ)Γ ( 1 µ + ρ) and δ ρ 2 2 ρ = − − ∞ ρ ν Γ (1 λ + ρ) ρ 1 2 1 2 − t − Kν (ut) pΓq[wt ]dt = Γ ( ± ) 1 1 Z0 4 u 2 + µ + ρ, µ + ρ,(a ,x),a , ,a ; w Γ 2 2 1 2 p ρ ν ρ ν p+2 q+1  − ··· δ . + (36) × , − ,(a1,x),a2, ,ap; 4w (1 λ + ρ),b , ,bq ; 1 p+2Γq  2 2 ··· .  − ···  × u2 (31) b1, ,bq ;  ···  Here we assume that both members of the Whittaker Here we assume that both members of the K-transform transform transform exist. exist. Proof.For simplicity and convenience, let L be the left- Proof.By applying (5) to the integrand of (35) and then hand side of (30). Setting δ t = ν in L , we have interchanging the order of integral sign and summation, which is verified by uniform convergence of the involved ∞ ν ρ 1 series under the given conditions, we get − ν/2 L = e− Wλ µ (ν) ∞ δ , ρ 1 2 Z0   × t − Kν (ut) pγq[wt ]dt ∞ n Z0 (a1;x)n(a2)n (ap)n (wν) ∑ dν. ∞ n ··· δ nδ (a1;x)n(a2)n (ap)n (w) (b1)n (bq)n · n! ∑ n=0 ··· = ··· n=0 (b1)n (bq)n · n! × Changing the order of integration and summation, we get ···∞ ρ+2n 1 t − Kν (ut)dt. (37) ∞ n ρ (a1;x)n(a2)n (ap)n (w) Z0 L = δ ∑ ··· − δ n Now, if we use the integral formula involving the K- n=0 (b1)n (bq)n · n! × ∞ ··· function (see, e.g.,[7, p. 54, Eq.(2.37)];see also [6, p. 78]): νρ+n 1 ν/2 ν ν − e− Wλ ,µ ( )d . (32) ∞ ρ ν Z0 ρ 1 ρ 2 ρ t − Kν (at)dt = 2 − a− Γ ± (38) Now, if we use the integral formula involving the Z0  2  Whittaker function (see, e.g.,[7, p. 56, Eq.(2.41)]; see (ℜ(a) > 0; ℜ(ρ) > ℜ(ν) ), | |

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we find which, on using the known identity (see, e.g.,[15,p.5,Eq. ∞ ρ 1 2 ρ 2 ρ (23)]) t − Kν (ut) pγq[wt ]dt = 2 − (u)− n Z0 ( 1) (λ) n = − , ∞ n λ (a1;x)n(a2)n (ap)n (4w) − (1 )n ∑ ··· − 2n in view of (5), yields our desired result (40). × n=0 (b1)n (bq)n · u n! ··· It is easy to see that a similar argument as in the proof ρ + ν ρ ν Γ + n Γ − + n , (39) of (40) will establish the result (41). This completes the ×  2   2  proof of Theorem 6. which, on using (5), yields our desired result (35). It is easy to see that a similar argument as in the proof of (35) will establish the result (36). This completes the 3 Special Cases and Concluding Remarks proof of Theorem 5. We consider some further consequences of main results Theorem 6. Suppose x 0, ℜ(u) > 0, ρ, ν, w C and derived in the preceding sections. To do this, we recall the ≥ ∈ following two standard formulae [4, p. 79, Eqs. (14–15)]: p, q N0. Then the following integral formulas hold: ∈ ∞ ρ Γ ρ+ν 2 2 ρ 1 2 1 2 ( 2 ) t Jν ut γ wt dt J 1/2(z)= cosz and J1/2(z)= sinz. (45) − ( ) p q[ ] = ν ρ − rπ z rπ z Z0 2 u Γ (1 + − ) 2 ν 1 ρ + ν ν ρ Setting = 2 for the parameter values of the Bessel , − ,(a1,x),a2, ,ap; 4w function of first± kind in (40) and (41), we obtain two p+2γq  2 2 ···  (40) × − u2 further pairs of integral formulae as follows: b1, ,bq ; 1  ···  Case 1.ν = ; x 0, ℜ(u) > 0, ρ, u, w C and p, q N0. and 2 ≥ ∈ ∈ ∞ ρ Γ ρ+ν ∞ ρ 3 Γ 2ρ+1 ρ 1 2 1 2 ( ) ρ 1 2 (2) − 2 ( 4 ) ν Γ 2 γ √π t − J (ut) p q[wt ]dt = ν ρ t − sin(ut) p q[wt ]dt = 1 1 2ρ Z0 2 u Γ Z0 ρ Γ (1 + −2 ) u − 2 (1 + −4 ) ρ + ν ν ρ 2ρ + 1 1 2ρ a x a a ; Γ , − ,( 1, ), 2, , p 4w , − ,(a1,x),a2, ,ap; 4w p+2 q  2 2 ··· , (41) p+2γq  4 4 ···  × − u2 × − u2 b1, ,bq ; b1, ,bq ;  ···   ···  where ( ℜ(ν) < ℜ(ρ) < 3 ). Here we assume that both (46) − 2 members of the Hankel transform exist. and ∞ ρ 3 2ρ+1 Proof.By applying (5) to the integrand of (40) and then (2) 2 Γ ( ) ρ 1 Γ 2 √π − 4 interchanging the order of integral sign and summation, t − sin(ut) p q[wt ]dt = ρ 1 1 2ρ Z0 2 Γ 1 which is verified by uniform convergence of the involved u − ( + −4 ) series under the given conditions, we get 2ρ + 1 1 2ρ , − ,(a1,x),a2, ,ap; 4w p+2Γq  4 4 ··· . ∞ × − u2 ρ 1 2 b1, ,bq ; t − Jν (ut) pγq[wt ]dt  ···  Z0 (47) ∞ n (a1;x)n(a2)n (ap)n (w) ν 1 ℜ ρ C N = ∑ ··· Case 2. = 2 ; x 0, (u) > 0, , u, w and p, q 0. b b n! − ≥ ∈ ∈ n=0 ( 1)n ( q)n · × ∞ ρ 3 2ρ+1 ··· 2 Γ ∞ ρ 1 2 (2) − ( 4 ) ρ+2n 1 t − cos(ut) pγq[wt ]dt = √π ν ρ 1 1 2ρ t − J (ut)dt. (42) Z0 2 Γ (1 + − ) Z0 u − 4 ρ ρ Now, if we use the integral formula involving the K- 2 + 1 1 2 , − ,(a1,x),a2, ,ap; 4w function (see, e.g., [7, p. 57, Eq. (2.46)]): p+2γq  4 4 ···  × − u2 ∞ Γ ρ+ν b1, ,bq ; ρ 1 ρ 1 ρ ( 2 )  ···  t − Jν (at)dt = 2 − a− ν ρ (43) (48) Z0 Γ (1 + − ) 2 and 3 ℜ ℜ ν ℜ ρ ∞ ρ 3 2ρ+1 (a) > 0, ( ) < ( ) < , (2) 2 Γ ( )  − 2 ρ 1 Γ 2 √π − 4 t − cos(ut) p q[wt ]dt = 1 1 2ρ Z0 ρ Γ we obtain u − 2 (1 + −4 ) ∞ ρ 1 2 ρ 1 ρ 2ρ + 1 1 2ρ t − Jν (ut) pγq[wt ]dt = 2 − (u)− , − ,(a ,x),a , ,a ; 4w Z Γ 1 2 p 0 × p+2 q  4 4 ··· 2 . ∞ n ρ+ν × − u (a1;x)n(a2)n (ap)n (4w) Γ ( + n) b1, ,bq ; ∑ ··· 2 , (44)  ···  2n 2+ρ+ν (49) n 0 (b1)n (bq)n · u n! Γ ( n) = ··· 2 −

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Junesang Choi received Praveen Agarwal is his B.A. from Gyeongsang currently Associate Professor, National University (Republic Department of Mathematics, of Korea) in 1981 and Anand International College his Ph.D. from the Florida of Engineering, Jaipur, State University in 1991. He India . He completed Ph.D. currently teaches at Dongguk from MNIT, Jaipur in 2006. University (Gyeongju) He has more than 14 years in the Republic of Korea. For of academic and research his academic activities, see experience. His research field http://wwwk.dongguk.ac.kr/∼junesang/. is Special Functions, Integral Transforms and . He has published more than 72 research papers in the national and international journal of repute. He associated with more than 50 National/International journals Editorial board. In 2014 he received the Summer Research Fellowship Award from the Indian Academy of Science, INDIA. He also received the International Travel Grants from DST (India), NBHM (India), IMA (USA), CIMPA (Germany) and ICM (Korea) to attend International Conferences and Workshops. He is the Guest Editors of Artificial Intelligence and Its Applications (2013) and Swarm Intelligence and Its Applications respectively (2013-2014) and Recent Trends in Fractional Integral Inequalities and their Applications (2014) (Hindawi Publishing Corporation). He is the Editor-in Chief of the International Journal of Mathematical Research, Pakistan. He has presented several papers, delivered several expert lectures and performed task of judge in National/International conferences, Seminars, STTP and SDP. He is presently guiding 3 research scholars for Ph.D. degree. He published 5 books on Engineering Mathematics, one research level book entitled Study of New Trends and Analysis of Special Function, LAP-Lambert Academic Publishing, Germany, 2013, ISBN: 978-3-659-312861. He is reviewer for Mathematical Reviews, USA (American Mathematical Society), Zentralblatt MATH, Berlin and number of SCI & SCIE Journals since last ten years. For his research activities, see https://www.researchgate.net/profile/Praveen Agarwal2

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