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Research article

The switching and learning behavior of an octopus cell implemented on FPGA


  • Received: 16 January 2024 Revised: 08 March 2024 Accepted: 22 March 2024 Published: 25 April 2024
  • A dendrocentric backpropagation spike timing-dependent plasticity learning rule has been derived based on temporal logic for a single octopus neuron. It receives parallel spike trains and collectively adjusts its synaptic weights in the range [0, 1] during training. After the training phase, it spikes in reaction to event signaling input patterns in sensory streams. The learning and switching behavior of the octopus cell has been implemented in field-programmable gate array (FPGA) hardware. The application in an FPGA is described and the proof of concept for its application in hardware that was obtained by feeding it with spike cochleagrams is given; also, it is verified by performing a comparison with the pre-computed standard software simulation results.

    Citation: Alexej Tschumak, Frank Feldhoff, Frank Klefenz. The switching and learning behavior of an octopus cell implemented on FPGA[J]. Mathematical Biosciences and Engineering, 2024, 21(4): 5762-5781. doi: 10.3934/mbe.2024254

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  • A dendrocentric backpropagation spike timing-dependent plasticity learning rule has been derived based on temporal logic for a single octopus neuron. It receives parallel spike trains and collectively adjusts its synaptic weights in the range [0, 1] during training. After the training phase, it spikes in reaction to event signaling input patterns in sensory streams. The learning and switching behavior of the octopus cell has been implemented in field-programmable gate array (FPGA) hardware. The application in an FPGA is described and the proof of concept for its application in hardware that was obtained by feeding it with spike cochleagrams is given; also, it is verified by performing a comparison with the pre-computed standard software simulation results.



    Historically, hypergeometric functions emerged during the 18th century in the work of mathematicians such as Euler and Gauss, who developed theories around hypergeometric series and associated integrals. These functions were first considered as generalizations of geometric series and solutions of higher-order linear differential equations. Today, hypergeometric functions are at the center of an entire field of research, that of special functions [1]. Moreover, it is now well known that most special functions can be expressed in terms of hypergeometric functions, and these functions can be represented using either series or integrals. This dual representation makes it an excellent tool for evaluating series and integrals or solving differential equations. Hypergeometric functions have a wide range of applications. From our literature review, we found that these functions are utilized to express solutions in various fields, including probability and statistics [2,3,4], combinatorics and number theory [5,6,7], random walks [8,9,10], random graphs [11], quantum mechanics (see [12], p. 89, 96, 127 and [13], p. 235, 290, 333), conformal mapping [14], and fractional hypergeometric differential equations [15,16], among many other problems.

    Hypergeometric functions usually have no explicit expression and are only represented by power series or integrals, which makes their evaluation time-consuming. Research on hypergeometric functions is divided into two main branches: continuous and discrete. The continuous branch focuses on the analytical study of these functions, treating their arguments as continuous variables (see [17,18,19,20,21,22,23,24]). In contrast, the discrete branch examines these functions by substituting their arguments with integers or specific values. Research in the discrete branch has surged in the past decade, driven by advancements in powerful computer algebra software and the wide range of problems that can be solved using hypergeometric functions. This study belongs to the discrete branch and aims to provide new results for different families of the generalized hypergeometric function 3F2(1). The literature includes several explicit forms of 3F2(a1,a2,a3;b1;b2;1), for particular choices of the parameters (a1,a2,a3,b1,b2). Thus, in [25], the authors used the Gamma function to derive explicit forms for 3F2(a,b,c;1+ab,1+ac;1) and 3F2(a,b,c;1+ab,a+2bc1;1). Moreover, in [26], the authors used specialized software and managed to generate, without any mathematical proof, about thirty explicit formulas of 3F2(a1,a2,a3;b1;b2;1), for particular choices of the parameters (a1,a2,a3,b1,b2). Furthermore, in [27], the authors exhibited the explicit expressions of 3F2(2n,a,1+d;2a+1,d;2) and 3F2(2n1,a,1+d;2a+1,d;2) for nN. Besides, in [28], the authors succeeded in determining the explicit expressions of the following sequences:

    3F2(16,56,12+n;1,32+n;1),3F2(16,56,12n;1,12n;1),3F2(16,56,13+n;1,43+n;1),3F2(16,56,13n;1,23n;1),3F2(16,56,23+n;1,53+n;1),3F2(16,56,23n;1,13n;1),3F2(16,56,14+n;1,54+n;1),3F2(16,56,14n;1,34+n;1),3F2(16,56,34+n;1,74+n;1),3F2(16,56,14n;1,14n;1),

    for all nN. Finally, in [29], the authors provided the explicit forms of the sets 3F2(2x,2x+12,x;12,1+x;1) and 3F2(2x,2x12,x;32,1+x;1), for all x(,14).

    The main objective of our study is to find explicit forms for the sets:

    K(α)=3F2(1α,1,α+1;α+1,α+2;1),α(12,+),G(α)=3F2(1α,1,2+α;1+α,3+α;1),α(12,+),H(p)=3F2(12p,1,1+2p;32+p,2p+2;1),pN. (1.1)

    These sets emerge naturally from an exact evaluation of certain classes of fractional integrals, as we show in Section 4.

    In order to understand the notation used above and throughout, we present in this section some basic notations, definitions, and intermediate results, which will be useful to justify certain passages in the proofs of this manuscript. Let us now present a set of symbols and notations.

    N, Z, R, and C denote the sets of non-negative integers, integers, real numbers, and complex numbers, respectively,

    X denotes any set X{0},

    Z0 denotes set N={,n,,1,0},

    B(0,1)={zC||z|<1},

    ¯B(0,1)={zC||z|1},

    R(z) denotes the real part of z.

    Now we present some basic notations, definitions, and intermediate results related to the so-called the Gauss hypergeometric function.

    Definition 2.1. [30] The Euler gamma function Γ(z) is defined by

    Γ(z)=0tz1etdt,  zC|R(z)>0. (2.1)

    Using integration by parts, one sees that

    Γ(z+1)=zΓ(z),  R(z)>0. (2.2)

    The extension of the Euler gamma function to the half-plane R(z)0 is given by

    Γ(z)=Γ(z+k)(z)k,   (R(z)>k;  kN;zZ0),

    where (z)k is the Pochhammer symbol defined for all zC and kN by

    (z)0=1  and  (z)k=z(z+1)(z+k1),  kN. (2.3)

    Relations (2.2) and (2.3) give

    Γ(k+1)=(1)k=k!,  kN. (2.4)

    Definition 2.2. [30] The Gauss hypergeometric function 2F1(a,b;c;z) is defined in the unit disk as the sum of the hypergeometric series as follows:

    2F1(a,b;c;z)=k=0(a)k(b)k(c)kzkk!,(a,bC;cCZ0;z¯B;R(cba)>0). (2.5)

    Furthermore, if 0<R(b)<R(c) and |arg(1z)|<π, then 2F1(a,b;c;z) is given by the following Euler integral representation:

    2F1(a,b;c;z)=Γ(c)Γ(b)Γ(cb)10xb1(1x)cb1(1zx)adx. (2.6)

    If z=1 with R(cba)>0, the Gauss hypergeometric function has the following property:

    2F1(a,b;c;1)=Γ(c)Γ(cab)Γ(ca)Γ(cb). (2.7)

    A natural extension of 2F1 to 3F2 is defined by

    3F2(a,b,c;d,e;z)=+k=0(a)k(b)k(c)k(d)k(e)kzkk!,
    (z¯BandR(d+eabc)>0).

    In [31] Theorem 38, Rainville proves a general integral representation for p+kFq+k, but here we state the following three special cases,

    Case 1: Let p=2, q=k=1, and choose a=1α, b=1, c=α+1, d=α+1, e=α+2, and z=1. Then, 3F2 is given by the following integral representation:

    3F2(1α,1,α+1;α+1,α+2;1)=(α+1)10xα2F1(1α,1;α+1;x)dx. (2.8)

    Case 2: Let p=2, q=k=1, and choose a=1α, b=1, c=α+2, d=α+1, e=α+3, and z=1. Then, 3F2 is given by the following integral representation:

    3F2(1α,1,α+2;α+1,α+3;1)=(α+2)10xα+12F1(1α,1;α+1;x)dx. (2.9)

    Case 3: Let p=2, q=k=1, and choose a=1α, b=1, c=2α, d=α+1, e=2α+1, and z=1. Then, 3F2 is given by the following integral representation:

    3F2(1α,1,2α;α+1,2α+1;1)=2α10x2α12F1(1α,1;α+1;x)dx. (2.10)

    The following is the definition of the Riemann-Liouville fractional integral Iαf of order α.

    Definition 2.3. [30] Let Ω=[τ,η]. The Riemann-Liouville fractional integral Iαf of order αC (R(α)>0) is defined by

    Iαf(t)=1Γ(α)tτ(ts)α1f(s)ds,t>τandR(α)>0. (2.11)

    In this section, we establish some results which will play an important role herein. We believe that some of these results may be new.

    To justify the interchangeability between the integral and the sum, or to rewrite certain integrals, we give the following two lemmas that we will refer to several times in our work.

    Lemma 2.1. Let a,b,cC, cZ0, and R(cab)>0. Then, the series

    k=0(a)k(b)k(c)kzk,

    is normally convergent on the interval [1,1].

    Proof of Lemma 2.1. Since a,b,cC, cZ0, and R(cab)>0, the Gauss hypergeometric function

    2F1(a,b;c;z)=k=0(a)k(b)k(c)kzkk!, (2.12)

    is defined for any complex number z¯B(0,1). Moreover, the series (2.12) is absolutely convergent for all z=1. Therefore, the series uk is convergent, where uk is defined by

    uk=|(a)k(b)k(c)k|,kN.

    Furthermore, it is clear that for all z[1,1],kN, we have

    |(a)k(b)k(c)kzk|uk. (2.13)

    The convergence of the series uk together with the relation (2.13) leads to the normal (therefore uniform) convergence of the series (a)k(b)k(c)kzk on the interval [1,1]. The proof is complete.

    Lemma 2.2. Let Ω=[τ,η] (<τ<η<) be a finite interval on the real axis R, and α>1/2. Then,

    ητ+k=0(1α)k(α+1)k(tτητ)kdt=+k=0(1α)k(α+1)kητ(tτητ)kdt. (2.14)

    Proof of Lemma 2.2. If we take a=1α, b=1, and c=α+1, then cZ0 and R(cab)=2α1>0, since α>1/2. Then by Lemma 2.1, the series (a)k(b)k(c)kzk converges normally on the interval [0,1][1,1]. Consequently, for all α>1/2, we have

    10+k=0(1α)k(α+1)kzkdz=+k=0(1α)k(α+1)k10zkdz. (2.15)

    By using the change of variable with z=tτητ, for relation (2.15) we have

    1(ητ)ητ+k=0(1α)k(α+1)k(tτητ)kdt=+k=0(1α)k(α+1)k1(ητ)ητ(tτητ)kdt. (2.16)

    Thus, we have obtained (2.14). The proof is complete.

    We now establish some results, which we believe are new. These results give the limit of some series.

    The following lemma gives the sum of the series Sα,0 defined by the left-hand side of relation (2.17).

    Lemma 2.3. For all α>1/2, we have

    Sα,0=+k=0(1α)k(α+1)k(k+α+1)=12α. (2.17)

    Proof of Lemma 2.3. For all kN, let uk be the general term of the series Sα,0 given by the relation (2.17). If we take a=1α, b=1, and c=α+1, then cZ0 and R(cab)=2α1>0, since α>1/2. Then, by Lemma 2.1 the series (1α)k(α+1)k is absolutely convergent, and since 0<uk(1α)k(α+1)k for all kN, we then deduce that the series Sα,0 is absolutely convergent. Thus, multiplying and dividing the term uk by (0α), we obtain

    Sα,0=+k=0(0α)(1α)k(0α)(α+1)k(k+α+1)=+k=0(α)k+1α(α+1)k+1=1α[+k=1(α)k(α+1)k]=1α[+k=0(α)k(α+1)k1]=1α[12F1(α,1;α+1;1)]. (2.18)

    From the property of 2F1, given by (2.7) we have

    2F1(α,1;α+1;1)=12, (2.19)

    and so substituting (2.19) into (2.18), we obtain (2.17). This completes the proof.

    The following lemma gives the sum of the series Sα,1 defined by the left-hand side of relation (2.20).

    Lemma 2.4. For all α>1/2, we have

    Sα,1=+k=0(1α)k(α+1)k(α+2+k)=1α+1+12α22α+1. (2.20)

    Proof of Lemma 2.4. The proof is similar to the proof of Lemma 2.3. If we take a=1α, b=1, and c=α+1, then cZ0 and R(cab)=2α1>0, since α>1/2. We deduce that the series Sα,1 is absolutely convergent. Thus, multiplying and dividing the term uk by the same quantity (k+α+1), we obtain

    uk=(1α)k(k+α+1)(α+1)k(α+1+k)(α+1+k+1)=(1α)k(1α+k+2α)(α+1)k+2=(1α)k+1(α+1)k+2+2α(1α)k(α+1)k+2. (2.21)

    Applying the identity (a)k+1=a(1+a)k to α+1, we obtain the following two relations:

    (α+1)k+2=(α+1)(α+2)k+1,(α+1)k+2=(α+1)(α+2)(α+3)k, (2.22)

    and using the above relations and the fact that (a)0=1, the series Sα,1 can be rewritten as follows:

    Sα,1=1α+1+k=0(1α)k+1(α+2)k+1+2α(α+1)(α+2)+k=0(1α)k(α+3)k=1α+1+k=1(1α)k(α+2)k+2α(α+1)(α+2)+k=0(1α)k(α+3)k=1α+1[+k=0(1α)k(α+2)k1]+2α(α+1)(α+2)+k=0(1α)k(α+3)k=1α+1[2F1(1α,1;α+2;1)1]+2α(α+1)(α+2)2F1(1α,1;α+3;1).

    From the property of 2F1 given by (2.7), we have

    2F1(1α,1;α+2;1)=α+12α, (2.23)

    and

    2F1(1α,1;α+3;1)=α+22α+1. (2.24)

    Thus, we obtain

    Sα,1=1α+1[α+12α1]+2α(α+1)(α+2)α+22α+1=1α+1+12α2(2α+1). (2.25)

    The proof is complete.

    The following lemma gives the limit of the series Sα,2 defined by the left-hand side of relation (2.26).

    Lemma 2.5. For all α>1/2, we have

    Sα,2=+k=0(1α)k(α+1)k(k+2α)=12α3F2(1α,1,2α;α+1,2α+1;1). (2.26)

    Proof of Lemma 2.5. If a=1α, b=1, and c=α+1, then cZ0, and R(cab)=2α1>0, since α>1/2. Then, expressing the rightside of (2.10) as a series and changing the order of integration and summation which is justified by Lemma 2.1 (due to the uniform convergence of the series) gives

    2α10x2α12F1(1α,1;α+1;x)dx=2α+k=0(1α)k(1)k(α+1)k10x2α1xkk!dx=2α+k=0(1α)k(α+1)k1(k+2α).

    Thus, we obtained (2.26). The proof is complete.

    The following lemma gives the sum of the series Sα,3 defined by the left-hand side of relation (2.27).

    Corollary 2.1. For all α>1, we have

    Sα,3=+k=0(2α)k(α+1)k(2α+k)=α(1α)(2α1)+[13α2α(1α)]3F2(1α,1,2α;α+1,2α+1;1). (2.27)

    Proof of Corollary 2.1. The proof is similar to the proof of Lemma 2.4, and so we just sketch the basic idea. If we take a=2α, b=1, and c=α+1, then cZ0 and R(cab)=2α2>0, since α>1. Then, by Lemma 2.1 we deduce that the series Sα,3 is absolutely convergent. Thus, multiplying and dividing the term uk by the same quantity (1α), we obtain

    Sα,3=+k=0(2α)k(α+1)k(2α+k)=+k=0(1α)(2α)k(α+1)k(1α)(2α+k)=+k=0(1α)k+1(1α)(α+1)k(2α+k)=11α[+k=0(1α)k(k+1α)(α+1)k(2α+k)]=11α[+k=0(1α)k(k+2α+13α)(α+1)k(2α+k)]=11α[+k=0(1α)k(α+1)k+(13α)+k=0(1α)k(α+1)k(2α+k)]=11α[2F1(1α,1;α+1;1)+(13α)+k=0(1α)k(α+1)k(2α+k)].

    Above we use the property of 2F1 (2.7) and the result of Lemma 2.5 to obtain (2.27). The proof is complete.

    In the following corollary, we give the calculation of 3F2(1,1,4;3,5;1), which we believe may be new.

    Corollary 2.1.

    3F2(1,1,4;3,5;1)=1115. (2.28)

    Proof of Corollary 2.1. Note that when α=2, the numerical series introduced in Lemmas 2.4 and 2.5 coincide. Therefore, the right-hand terms of relations (2.20) and (2.26) are equal when α=2. Thus,

    3F2(1,1,4;3,5;1)=4(13+1425)=7385=1115. (2.29)

    The proof is complete.

    This section explores specific subfamilies of 3F2(1) and is structured into two subsections. The first subsection presents the explicit forms of the subfamilies {3F2(1α,1,α+1;α+1,α+2;1)} and {3F2(1α,1,α+2;α+1,α+3;1)} for all α>12, while the second subsection provides the explicit form of the subfamily {3F2(12p,1,1+2p;32+p,2p+2;1)} for all pN.

    This part aims to find the explicit form of the function K(α)=:3F2(1α,1,α+1;α+1,α+2;1) (Theorem 3.1). To do this, we write K in the form of a series of functions (Lemma 3.1), and then we use the result of Lemma 2.3 to prove our main result of this section, Theorem 3.1.

    In the following lemma, we express the function K(α) as a series.

    Lemma 3.1. For all α>1/2, we have

    3F2(1α,1,α+1;α+1,α+2;1)=(α+1)+k=0(1α)k(α+1)k(k+α+1). (3.1)

    Proof of Lemma 3.1. If a=1α, b=1, and c=α+1, then cZ0 and R(cab)=2α1>0, since α>1/2. Then, expressing the rightside of (2.8) as a series, changing the order of integration and summation, which is justified by Lemma 2.1, and applying the steps of the proof of Lemma 2.5, we can readily derive the proof of Lemma 3.1.

    Now we state and prove our main result of this section, Theorem 3.1.

    Theorem 3.1. For all α>1/2, we have

    3F2(1α,1,α+1;α+1,α+2;1)=α+12α. (3.2)

    Proof of Theorem 3.1. By identifying relations (2.17) and (3.1), we obtain

    3F2(1α,1,α+2;α+1,α+3;1)=(α+1)(12α)=α+12α.

    The proof is complete.

    This part aims to find the explicit form of the function G(α)=:3F2(1α,1,α+2;α+1,α+3;1) (Theorem 3.2). To do this, we write G in the form of a series of functions (Lemma 3.2), then we use the result of Lemma 2.4 to prove our main result of this section, Theorem 3.2. Our argument in this section is similar to the previous section.

    In the following lemma, we express the function G(α) as a series.

    Lemma 3.2. For all α>1/2, we have

    3F2(1α,1,α+2;α+1,α+3;1)=(α+2)+k=0(1α)k(α+1)k(k+α+2). (3.3)

    Proof of Lemma 3.2. If a=1α, b=1, and c=α+1, then cZ0 and R(cab)=2α1>0, since α>1/2. Then, expressing the rightside of (2.9) as a series, changing the order of integration and summation, which is justified by Lemma 2.1, and applying the steps of the proof of Lemma 2.5, we can readily derive the proof of Lemma 3.2.

    Now we state and prove our main result of this section, Theorem 3.2.

    Theorem 3.2. For all α>1/2, we have

    3F2(1α,1,α+2;α+1,α+3;1)=12+1α+1α+132α+1. (3.4)

    Proof of Theorem 3.2. By identifying relations (2.20) and (3.3), we obtain

    3F2(1α,1,α+2;α+1,α+3;1)=(α+2)(1α+1+12α22α+1)=12+1α+1α+132α+1.

    The proof is complete.

    This section deals with the family of functions F(α)=3F2(1α,1,2α;α+1,2α+1;1). It is easy to find a representation for it by a series of functions +k=0fk(α) (see Lemma 2.5). However, it is not obvious to find an explicit one unless αN or a rational number of the form α=2p+12 and pN. Note that when α=pN, then +k=0fk(α) is equal to p1k=0fk(α). Therefore, this case will be excluded from this study. In summary, this part aims to determine the explicit form of F(α) when α=2p+12, which coincides with 3F2(12p,1,1+2p;32+p,2p+2;1) and which we will designate by H(p). To do this, we first calculate H(1) in Section 3.3.1 to make the calculation of H(p) easy to follow in Section 3.3.2.

    The main result of this section is summarized in the following theorem. The proof of this theorem is postponed to the end of this section.

    Theorem 3.3. For all pN, we have

    3F2(12p,1,1+2p;32+p,2p+2;1)=(2p+1)Kp(2a2p+1ln(2)+2a2p+1(Ep+Bp)+2pj=1ajBj),

    where

    Kp=(12p)(32p)(52p)(1+2p)(1+2p)a2p+1=22p(2p)!(3p)!(6p+1)!(p)!Ep=2pk=112kBj=j1k=012k2p+1,j=1,,2paj=2(1)j(2)2p(j)!(2pj)!(6p2j+1),j=1,,2p. (3.5)

    This part aims to find the explicit form of the function H(1). To do this, we write H(1) in the form of a numerical series. Then, we determine the limit of this series.

    In Lemma 3.3, we give the exact value of 3F2(12,1,3;52,4;1).

    Lemma 3.3.

    3F2(12,1,3;52,4;1)=3335635ln(2). (3.6)

    Proof of Lemma 3.3. For p=1 (i.e., α=32), after simplifications, relation (2.26) gives

    3+k=01(2k1)(2k+1)(2k+3)(k+3)=133F2(12,1,3;52,4;1), (3.7)

    that is, 3F2(12,1,3;52,4;1)=9S, where

    S=+k=01(2k1)(2k+1)(2k+3)(k+3). (3.8)

    The decomposition into partial fractions of the general term uk, of the series S=uk, gives

    uk=1(2k1)(2k+1)(2k+3)(k+3)=128(2k1)110(2k+1)+112(2k+3)1105(k+3). (3.9)

    Let n be a fixed positive integer. Then, (Sn)n0 and (Tn)n1 are sequences defined by

    Sn=nk=01(2k1)(2k+1)(2k+3)(k+3)Tn=nk=112k1. (3.10)

    We will simplify the partial sum Sn in order to find its limit when n tends to infinity. To do this, we will express the partial sums,

    Un=nk=012k1;Vn=nk=012k+1;Wn=nk=012k+3,

    as a function of Tn. It is easy to check the following equalities:

    Un=1+TnVn=Tn+12n+1Wn=1+Tn+12n+1+12n+3. (3.11)

    We then deduce Sn as a function of Tn:

    Sn=Un28Vn10+Wn121105nk=01k+3=542+An+Bn, (3.12)

    where

    An=2105Tn1105nk=01k+3Bn=160(2n+1)+112(2n+3). (3.13)

    Furthermore, we have

    nk=01k+3=2nk=012(k+3)=2n+3k=312k=2(1214+nk=112k+12(n+1)+12(n+2)+12(n+3))=2(34+nk=112k+12(n+1)+12(n+2)+12(n+3))=2(34+nk=112k+Cn), (3.14)

    where

    Cn=12(n+1)+12(n+2)+12(n+3).

    Consequently,

    An=2105Tn2105(34+nk=112k+Cn)=2105(nk=112k1nk=112k+34Cn)=2105(nk=1(1)k+1k+34Cn)=2105(Dn+34Cn), (3.15)

    where

    Dn=nk=1(1)k+1k.

    Based on relations (3.12) and (3.15), we obtain

    Sn=542+2105(Dn+34Cn)+Bn=11105+2105(DnCn)+Bn. (3.16)

    Furthermore, we know that for all real numbers x1 such that |x|1, we have

    limn+nk=0(1)k+1kxk=ln(1+x). (3.17)

    Consequently,

    limn+Dn=ln(2). (3.18)

    Moreover, it is easy to verify that

    limn+Bn=0,limn+Cn=0. (3.19)

    Finally, we have

    limn+Sn=11105+2105ln(2). (3.20)

    By grouping relations (3.7) and (3.20), we deduce that

    3F2(12,1,3;52,4;1)=9(11105+2105ln(2))=3335635ln(2). (3.21)

    The proof is complete.

    This section explores specific subfamilies of 3F2(1) and is structured into two subsections. The first subsection presents the explicit forms of the subfamilies F(1) and F(2), while the second subsection provides the explicit form of the subfamily F(3). This part aims to find the explicit form of the function H(p), pN. To do this, we will follow the same approach as that used in Section 3.3.1 to calculate H(1).

    In Lemma 3.5, we give the explicit expression of (1α)k1+α when α=2p+12.

    Lemma 3.4. For all pN,kN, we have

    (12p+12)k(1+2p+12)k=(12p)(12p+1))(1+2p)(1+2p)(2k+12p)(2k+32p)(2k1+2p)(2k+1+2p). (3.22)

    Proof of Lemma 3.5. If α=2p+12, then we have

    (1α)k=(12p)(32p)(52p)(k32p)(k12p),(α+1)k=(32+p)(52+p)(k12+p)(k+12+p). (3.23)

    Note that when k2p+2, there are j=k2p1 terms in common between (1α)k and (α+1)k. Indeed, we observe the first number in common if k12p=32+p, that is k=2p+2. Therefore, if k=2p+3, then there are two terms in common, and so on. Thus, if k2p+2, then (1α)k and (1+α)k are written as follows:

    (1α)k=(12p)(12p+2p)(32+p)(k12p),(α+1)k=(32+p)(52+p)(kp12)(kp+12)(k+12+p). (3.24)

    Therefore, their quotient can be simplified as follows:

    (1α)k(α+1)k=(12p)(12+1p)(12+2pp)(kp+12)(kp+1+12)(kp+2p+12)=(12(p0))(12(p1))(12(p2p))(k+12(p0))(k+12(p1))(k+12(p2p))=22p+1(12(p0))(12(p1))(12(p2p))22p+1(2k+12(p0))(2k+12(p1))(2k+12(p2p))=(12p)(12p+2))(1+2p)(1+2p)(2k+12p)(2k+32p)(2k1+2p)(2k+1+2p)=(12p)(32p))(1+2p)(1+2p)(2k+12p)(2k+32p)(2k1+2p)(2k+1+2p). (3.25)

    When k2p+1, we will show that the quotient (1α)k(α+1)k can be reduced to the form given by the last line of relation (3.25).

    When k=0, we have

    (1α)0(α+1)0=11=(12p)(32p)(1+2p)(1+2p)(12p)(32p))(1+2p)(1+2p)=(12p)(32p)(1+2p)(1+2p)(2×0+12p)(2×0+32p))(2×01+2p)(2×0+1+2p). (3.26)

    When k=1, we have

    (1α)1(α+1)1=12p(3+2p)=(12p)[(32p)(1+2p)(1+2p)][(32p)(1+2p)(1+2p)](3+2p)=(12p)(32p)(1+2p)(1+2p)(2×1+12p)(2×1+32p)(2×11+2p)(2×1+1+2p). (3.27)

    More generally, when k is an integer such that 0k2p+1, we have

    (1α)k(α+1)k=(12p)(32p)(2k32p)(2k12p)(3+2p)(5+2p)(2k1+2p)(2k+1+2p)=(12p)(32p)(2k12p)[(2k+12p)(1+2p)][(2k+12p)(1+2p)](3+2p)(5+2p)(2k+1+2p)=(12p)(32p)(1+2p)(1+2p)(2k+12p)(1+2p)(3+2p)(5+2p)(2k+1+2p). (3.28)

    Thus, we have shown that (1α)k(α+1)k is written in the form (3.22) for all kN. The proof is complete.

    Remark 3.1. For α=2p+12, we deduce from Lemma 3.5 that

    (1α)k(α+1)k(k+2p+1)=Kpuk, (3.29)

    where

    Kp=(12p)(32p)(1+2p)(1+2p)uk=1(2k+12p)(2k+32p)(2k+1+2p)(k+2p+1). (3.30)

    From the above remark, the decomposition into partial fractions of uk gives

    uk=a02k+12p+a12k+32p++a2p2k+1+2p+a2p+1k+2p+1=a02k+12(p0)+a12k+12(p1)++a2p2k+12(p2p)+a2p+1k+2p+1=2pi=0ai2k+12(pi)+a2p+1k+2p+1. (3.31)

    For all i=0,,2p, to find ai, simply multiply uk by (2k+12(pi)), Then, evaluate the resulting expression at k=2(pi)12. For now, let us calculate only the first three coefficients a0,a1, and a2.

    For a0, we evaluate the resulting expression at k=2p12 (i.e., 2k=2p1), so we obtain

    1a0=(2p1+12(p1))(2p1+12(p2p)12(2p1+4p+2)=(2×1)(2×2)(2×2p)12(6p+1),=22p(2p)!12(6p+1)=12(1)0(0!)22p(2p0)!(6p2×0+1). (3.32)

    Consequently,

    a0=2(1)0(0!)22p(2p0)!(6p2×0+1). (3.33)

    For a1, we evaluate the resulting expression at k=2p32 (i.e., 2k=2p3), so we obtain

    1a1=(2p3+12(p0))(2p3+12(p2p))12(2p3+4p+2)=(2)(2×1)(2×3)(2×(2p1))12(6p1)=(1)12122p1(2p1)!12(6p1)=12(1)1(1!)(2)2p(2p1)!(6p2×1+1). (3.34)

    Consequently,

    a1=2(1)1(1!)(2)2p(2p1)!(6p2×1+1). (3.35)

    For a2, we evaluate the resulting expression at k=2p52 (i.e., 2k=2p5), so we obtain

    1a2=(2p5+12(p0))(2p5+12(p2p))12(2p5+4p+2)=(22)(21)(2×1)(2×3)(2×(2p2))12(6p3)=(1)222(2!)22p2(2p2)!12(6p3)=12(1)2(2!)(2)2p(2p2)!(6p2×2+1). (3.36)

    Consequently,

    a2=2(1)2(2)2p(2p2)!(6p2×2+1). (3.37)

    The calculation of the first three coefficients a0,a1,a2 allowed us to guess the general expression of ai for all i=0,,2p, which is stated in the following Lemma 3.5.

    Lemma 3.5. Let pN. Then, for all i=0,,2p, we have

    ai=2(1)i(2)2p(i)!(2pi)!(6p2i+1). (3.38)

    Proof of Lemma 3.5. We will confirm this expression by a direct calculation of ai. To do this, we evaluate the resulting expression at k=2p(2i+1)2 (i.e., 2k=2p(2i+1)), so we obtain

    1ai=(2p(2i+1)+12(p0))(2p(2i+1)+12(p1))(2p(2i+1)+12(p(i1)))(2p(2i+1)+12(p(i+1)))(2p(2i+1)+12(p2p))12(2p(2i+1)+4p+2)=(2(i0))(2(i1))(2(i(i1)))(2×1)(2×2)(2(2pi))12(6p2i+1)=12(2)i(i)!22pi(2pi)!(6p2i+1)=12(1)i(2)i(i)!22pi(2pi)!(6p2i+1)=12(1)i(2)2p(i)!(2pi)!(6p2i+1). (3.39)

    Thus, we find the expression of ai stated in relation (3.38). The proof is then complete.

    All that remains is to determine the expression of a2p+1. This will be the subject of Lemma 3.6.

    Lemma 3.6. The last coefficient of the decomposition into partial fractions (3.31) is given by

    a2p+1=22p(2p)!(3p)!(6p+1)!(p)!. (3.40)

    Proof of Lemma 3.6. To find a2p+1, multiply uk by (k+2p+1). Then, evaluate the resulting expression at k=2p1 (i.e., 2k=4p2). We then obtain

    1a2p+1=(4p22p+1)(4p22p+3)(4p2+2p1)(4p2+2p+1)=(6p1)(6p+1)(6p+3)(2p3)(2p1)=(6p1)(6p)(6p+1)(6p+2)(6p+3)(2p3)(2p2)(2p1)(6p)(6p+2)(2p4)(2p2)=(1)2p+1(6p+1)(6p)(6p1)(6p2)(6p3)(2p+3)(2p+2)(2p+1)(1)2p(6p)(6p2)(2p+4)(2p+2)=(2p+1)(2p+2)(6p+1)2(p+1)2(p+2)2(3p)=[1×2××2p](2p+1)(2p+2)(6p+1)[1×2××p][1×2××2p]23p(p+1)+1[1×2××p](p+1)(p+2)(3p)=(6p+1)!(p)!22p(2p)!(3p)!. (3.41)

    The proof is complete.

    The following remark shows the validation of formulas (3.38) and (3.40) when p=1.

    Remark 3.2. For p=1, we will check the concordance of the coefficients a0,,a3 given by relations (3.40) and (3.38) with those presented in relation (3.9). When p=1, relations (3.38) and (3.40) give

    a0=2(1)0(2)2(0)!(2)!(7)=128a1=2(1)1(2)2(1)!(1)!(5)=110a2=2(1)2(2)2(2)!(0)!(3)=112a3=22(2)!(3)!(7)!(1)!=1105. (3.42)

    Thus, we find the same coefficients of the decomposition into partial fractions of uk given by (3.9).

    Lemma 3.7 presents a relation between the coefficient a2p+1 and the coefficients (ai)0i2p.

    Lemma 3.7. For all pN, we have

    2pi=0ai=2a2p+1. (3.43)

    Proof of Lemma 3.7. We have shown in relation (3.31) that, for all k>0, the term uk is written as a partial fraction (3.31), where (ai)0i2p and a2p+1 are given by (3.38) and (3.40).

    By reducing all the partial fractions to the same denominator and identifying the numerators, we then obtain 2p+2 equations with unknowns a0, , a2p, a2p+1. It is easy to see that the equation that relates the coefficients of the monomial k2p+1 is written in the following form 2pi=0k(2k)2pai+(2k)2p+1a2p+1=0, or in the equivalent form

    2pi=0ai=2a2p+1. (3.44)

    The proof is complete.

    In the following Lemma 3.8, we establish a supporting result that arises from the calculations performed in the previous results of this section. This result provides an explicit form of a finite sum.

    Lemma 3.8. For all pN, we have

    2pi=0(1)i(2)2p(i)!(2pi)!(6p2i+1)=22p(2p)!(3p)!(6p+1)!(p)!. (3.45)

    Proof of Lemma 3.8. Based on formula (3.43), we have

    2pi=0ai=2a2p+1. (3.46)

    By replacing in the previous equality the coefficients (ai)0i2p+1 by their expressions presented in relations (3.38) and (3.40), we directly obtain relation (3.45). The proof is complete.

    In the following section, we will prove that the series uk converges and we calculate its sum, where uk is defined by (3.30).

    Convergence of the series uk

    Let n be a fixed positive integer, and (Sn)n0 and (Tn)n1 be the sequences defined by

    Sn=nk=0uk,Tn=nk=012k2p+1. (3.47)

    We will write Sn as a function of Tn, prove that the sequence Sn converges, and determine its limit.

    For all j=1,,2p, we have

    nk=012k2(pj)+1=nk=012(k+j)2p+1=n+jk=j12k2p+1=nk=012k2p+1+n+jk=n+112k2p+1j1k=012k2p+1=Tn+Aj,nBj, (3.48)

    where

    Aj,n=n+jk=n+112k2p+1,Bj=j1k=012k2p+1. (3.49)

    Note that for j=0, we also have

    nk=012k2(p0)+1=nk=012k2p+1=Tn+A0,nB0, (3.50)

    where

    A0,n=B0=0. (3.51)

    Furthermore, we also have

    nk=01k+2p+1=2nk=012(k+2p+1)=2n+2p+1k=2p+112k=2nk=112k+2n+2p+1k=n+112k22pk=112k=2nk=112k+2Dn2Ep, (3.52)

    where

    Dn=n+2p+1k=n+112k,Ep=2pk=112k. (3.53)

    Moreover, we have

    Tn=nk=012k2p+1=p1k=012k2p+1+n+pk=p12k2p+1=p1k=012k2p+1+nk=012k+1=Bp+nk=012k+1, (3.54)

    where

    Bp=p1k=012k2p+1. (3.55)

    After writing Sn as a function of Tn, we will inject the expression of Tn given by (3.54) into Sn to prove that Sn converges and deduce its limit.

    Using relations (3.48), (3.54) (3.52), and (3.43), the partial sum Sn equals

    Sn=2pj=0ajnk=012k2(pj)+1+a2p+1nk=01k+2p+1=2pj=0aj(Tn+Aj,nBj)+a2p+1(2nk=112k+2Dn2Ep)=Tn2pj=0aj+2a2p+1nk=112k+2pj=1aj(Aj,nBj)+2a2p+1(DnEp)=2a2p+1Tn+2a2p+1nk=112k+2pj=1ajAj,n2pj=1ajBj+2a2p+1Dn2a2p+1Ep=2a2p+1nk=012k+12a2p+1Bp+2a2p+1nk=112k+2pj=1ajAj,n2pj=1ajBj+2a2p+1Dn2a2p+1Ep=2a2p+1(nk=012k+1nk=112k)2a2p+1Ep2pj=1ajBj2a2p+1Bp+2pj=1ajAj,n+2a2p+1Dn=2a2p+1nk=1(1)k+1k2a2p+1Ep2pj=1ajBj2a2p+1Bp+2pj=1ajAj,n+2a2p+1Dn. (3.56)

    \vartriangle Since

    \begin{equation} \begin{array}{lll} \lim\limits_{n\rightarrow +\infty}\sum\limits_{k = 1}^{n} \dfrac{(-1)^{k+1}}{k}& = &\ln(2)\\ \lim\limits_{n\rightarrow +\infty}A_{j, n}& = &0, \; \forall\, j = 1, \ldots, 2p\\ \lim\limits_{n\rightarrow +\infty}D_n& = &0, \end{array} \end{equation} (3.57)

    we deduce that S_n converges, and that its limit verifies

    \begin{equation} \lim\limits_{n\rightarrow +\infty}S_n = -2a_{2p+1}\ln(2)- 2a_{2p+1}(E_p+B_p)-\sum\limits_{j = 1}^{2p} a_jB_j. \end{equation} (3.58)

    \blacksquare The following Lemma 3.9 gives the sum of the series \sum u_k .

    Lemma 3.9. For all p\in {{ \mathbb{N}}}^* , we have

    \begin{equation} \sum\limits_{k = 0}^{+\infty}\dfrac{(\tfrac{1}{2}-p)_k}{(\tfrac{1}{2}+p)_k (k+2p+1)} = -K_p\left(2a_{2p+1}\ln(2)+ 2a_{2p+1}(E_p+B_p)+\sum\limits_{j = 1}^{2p} a_jB_j\right), \end{equation} (3.59)

    where K_p , a_{2p+1} , E_p , B_j , and a_j are defined in (3.5).

    Proof of Lemma 3.9. By grouping relations (3.29), (3.47), and (3.58), we obtain

    \begin{equation} \begin{array}{lll} \sum\limits_{k = 0}^{+\infty}\dfrac{(\tfrac{1}{2}-p)_k}{(\tfrac{1}{2}+p)_k (k+2p+1)}& = & K_p\lim\limits_{n\rightarrow +\infty}S_n\\ & = & -K_p\left(2a_{2p+1}\ln(2)+ 2a_{2p+1}(E_p+B_p)+\sum\limits_{j = 1}^{2p} a_jB_j\right). \end{array} \end{equation} (3.60)

    The proof is complete.

    \blacksquare The following remark gives the validation of formula (3.59) when p = 1 .

    Remark 3.3. For p = 1 , we will check the concordance of the limit given by relation (3.20) with that presented in relation (3.58). When p = 1 , we easily obtain E_1 = \tfrac{3}{4} , B_1 = -1 , and B_2 = 0 . Moreover, the values of (a_{i})_{0\leq i\leq 3} are given in (3.42). Thus, relation (3.58) gives

    \begin{equation} \begin{array}{lll} \lim\limits_{n\rightarrow +\infty}S_n& = &-2(-\dfrac{1}{105})\ln(2)- 2(-\dfrac{1}{105})(\dfrac{3}{4}-1)-[-\dfrac{1}{10}(-1)+\dfrac{1}{12}( 0)]\\ & = & \dfrac{2}{105}\ln(2)+\dfrac{1}{105}(\dfrac{3}{2}-2)-\dfrac{1}{10}\\ & = & \dfrac{2}{105}\ln(2)-\dfrac{1}{210}-\dfrac{21}{210}\\ & = & \dfrac{2}{105}\ln(2)-\dfrac{22}{210}\\ & = & \dfrac{2}{105}\ln(2)-\dfrac{11}{105}. \end{array} \end{equation} (3.61)

    Thus, we find the same limit as that found in relation (3.20).

    \blacksquare With the intermediate results from Section 3.3 now found, we can prove Theorem 3.3 stated at the beginning of this section.

    Proof of Theorem 3.3. Based on relation (2.26) and taking \alpha = \tfrac{2p+1}{2} , we obtain

    \begin{equation} \begin{array}{lll} \sum\limits_{k = 0}^{+\infty}\dfrac{(\tfrac{1}{2}-p)_k}{(\tfrac{1}{2}+p)_k (k+2p+1)}& = &\dfrac{{ _3F_2(1-\alpha, 1, 2\alpha;\alpha+1, 2\alpha+1;1) }}{2\alpha} \\ & = & \dfrac{{ _3F_2(\tfrac{1}{2}-p, 1, 1+2p;\tfrac{3}{2}+p, 2p+2;1) }}{2p+1}. \end{array} \end{equation} (3.62)

    By identifying the right-hand sides of relations (3.59) and (3.62), we obtain

    \begin{equation*} \label{Lem7Eq3bis} -K_p\left(2a_{2p+1}\ln(2)+ 2a_{2p+1}(E_p+B_p)+\sum\limits_{j = 1}^{2p} a_jB_j\right) = \dfrac{{ _3F_2(\tfrac{1}{2}-p, 1, 1+2p;\tfrac{3}{2}+p, 2p+2;1) }}{2p+1}, \end{equation*}

    that is,

    \begin{equation*} -(2p+1)K_p\left(2a_{2p+1}\ln(2)+ 2a_{2p+1}(E_p+B_p)+\sum\limits_{j = 1}^{2p} a_jB_j\right) = { _3F_2(\tfrac{1}{2}-p, 1, 1+2p;\tfrac{3}{2}+p, 2p+2;1) }. \end{equation*}

    The proof is complete.

    \blacksquare We now state and prove the following two results of Theorem 3.3 corresponding to the special cases p = 2 and p = 3 .

    \vartriangle For p = 2 , Theorem 3.3 yields the first new special result.

    Corollary 3.1.

    \begin{equation} {_3F_2(-\frac{3}{2}, 1, 5;\frac{7}{2}, 6;1)} = \frac{4045}{6006}+\frac{10}{1001}\ln(2). \end{equation} (3.63)

    Proof of corollary 3.1. Let p = 2 . Then, from Theorem 3.3, we easily compute the following quantities:

    \begin{eqnarray*} && K_2 = 45, \ E_2 = \frac{25}{24}, \ B_1 = -\frac{1}{3}, \ B_2 = -\frac{4}{3}, \ B_3 = -\frac{1}{3}, \ B_4 = 0, \ a_1 = -\frac{1}{528}, \\ &&a_2 = \frac{1}{288}, \ a_3 = -\frac{1}{336}, \ a_4 = \frac{1}{960}, \ a_5 = -\frac{1}{45045}, \ \sum\limits_{j = 1}^{4} a_jB_j = -\frac{25}{8316}. \end{eqnarray*}

    Thus, to derive relation (3.63), we simply substitute the values above into Theorem 3.3. The proof is complete.

    \vartriangle Also, for p = 3 , Theorem 3.3 yields the second new special result.

    Corollary 3.2.

    \begin{equation} {_3F_2(-\frac{5}{2}, 1, 7;\frac{9}{2}, 8;1)} = \frac{221158}{415701}-\frac{70}{138567}\ln(2). \end{equation} (3.64)

    Proof of Corollary 3.2. The proof follows similar lines of argument to that of Corollary 3.1. Let p = 3 . Then, from Theorem 3.3, we easily compute the following quantities:

    \begin{equation} \begin{array}{cccccc} B_1 = -\dfrac{1}{5}, \ B_2 = -\dfrac{8}{15}, \ B_3 = -\dfrac{23}{15}, \ B_4 = -\dfrac{8}{15}, \ B_5 = -\dfrac{3}{15}, \ B_6 = 0, \\ a_1 = -\dfrac{1}{65280}, \ a_2 = \dfrac{1}{23040}, \ a_3 = -\dfrac{1}{14976}, \ a_4 = \dfrac{1}{16896}, \ a_5 = -\dfrac{1}{34560}, \\a_6 = \dfrac{1}{161280}, \ a_7 = -\dfrac{1}{43648605}, \ K_3 = -1575, \ E_3 = \dfrac{49}{40}, \ \sum\limits_{j = 1}^{6} a_jB_j = \dfrac{371}{6563700}. \end{array} \end{equation} (3.65)

    Thus, to derive relation (3.64), we simply substitute the values above into Theorem 3.3. The proof is complete.

    This section is devoted to giving our new evaluation of a certain class of fractional integrals whose values are written in terms of hypergeometric functions _3F_2 which we obtained in the previous sections.

    Theorem 4.1. For all \tau\le s\le t\le \eta , and \alpha > 1/2, the following integral representations for the Gauss hypergeometric function hold true.

    (1)

    \begin{eqnarray} I^\alpha_1(t)&: = &\int_{\tau}^{t} (t-s)^{\alpha-1} (\eta-s)^{\alpha-1} \ ds\\ & = &\left[\dfrac{(\eta-\tau)^{\alpha-1}(t-\tau)^{\alpha}}{\alpha}\right] {_2F_1(1-\alpha, 1;\alpha+1;g(t))}, \end{eqnarray} (4.1)

    where g(t): = \dfrac{t-\tau}{\eta-\tau}.

    (2)

    \begin{eqnarray} \int_{\tau}^{\eta} I^\alpha_1(t) dt& = &\left[\dfrac{(\eta-\tau)^{2\alpha}}{\alpha(\alpha+1)} \right] {_3F_2(1-\alpha, 1, \alpha+1;\alpha+1, \alpha+2;1) } \end{eqnarray} (4.2)
    \begin{eqnarray} & = &\dfrac{(\eta-\tau)^{2\alpha}}{2\alpha^2}. \end{eqnarray} (4.3)

    Proof of Theorem 4.1.

    (1) Let s = \tau+x(t-\tau) . By changing the integration variable from s to x , the integral I^\alpha_1(t) becomes

    \begin{eqnarray} I^\alpha_1(t)& = & (t-\tau)\int_{0}^{1} ((t-\tau)-x(t-\tau))^{\alpha-1} \left((\eta-\tau)-x(t-\tau)\right)^{\alpha-1} dx \\ & = & (t-\tau)(t-\tau)^{\alpha-1} \int_{0}^{1} (1-x)^{\alpha-1} \left((\eta-\tau)-\frac{(\eta-\tau)(t-\tau)}{(\eta-\tau)} x\right)^{\alpha-1} dx\\ & = &(\eta-\tau)^{\alpha-1}(t-\tau)^{\alpha}\int_{0}^{1} (1-x)^{\alpha-1} \left(1-\frac{(t-\tau)}{(\eta-\tau)} x\right)^{\alpha-1} dx. \end{eqnarray}

    Above, we have the Euler integral representation of _2F_1(a, b; c;z) with a = 1-\alpha , b = 1 , c = \alpha+1 , and z = g(t) = \frac{t-\tau}{\eta-\tau} . Thus,

    \begin{eqnarray} I^\alpha_1(t)& = &\frac{(\eta-\tau)^{\alpha-1}(t-\tau)^{\alpha}}{\alpha} {_2F_1(1-\alpha, 1;\alpha+1;g(t)) }. \end{eqnarray}

    This completes the proof of (4.1).

    (2) Now we are in a position to evaluate the integral (4.2), so denoting the left-hand side of (4.2) by \lambda_1 , we have

    \begin{equation} \lambda_1 : = \frac{(\eta-\tau)^{\alpha-1}}{\alpha} \int_{\tau}^{\eta} (t-\tau)^{\alpha} {_2F_1(1-\alpha, 1;\alpha+1;g(t))} dt. \end{equation} (4.4)

    Now, expressing _2F_1 as a series and changing the order of integration and summation, which is justified by Lemma 2.2, we have

    \begin{eqnarray} \lambda_1 &: = &\left[\frac{(\eta-\tau)^{\alpha-1}}{\alpha} \right] \sum\limits_{k = 0}^{\infty} \dfrac{(1-\alpha)_k}{(\alpha+1)_k} \int_\tau^\eta (t-\tau)^{\alpha}\left(\dfrac{t-\tau}{\eta-\tau}\right)^k dt\\ & = &\left[\frac{(\eta-\tau)^{\alpha-1}}{\alpha} \right] \sum\limits_{k = 0}^{\infty} \dfrac{(1-\alpha)_k}{(\alpha+1)_k} \left(\dfrac{1}{\eta-\tau}\right)^k \int_\tau^\eta \left(t-\tau\right)^{k+\alpha} dt\\ & = &\left[\frac{(\eta-\tau)^{2\alpha}}{\alpha} \right] \sum\limits_{k = 0}^{\infty} \dfrac{(1-\alpha)_k}{(\alpha+1)_k} \dfrac{1}{(k+\alpha+1)}. \end{eqnarray}

    Now, by Lemma 3.1, we obtain (4.2), that is

    \begin{equation*} \lambda_1 : = \left[\dfrac{(\eta-\tau)^{2\alpha}}{\alpha(\alpha+1)} \right] {_3F_2(1-\alpha, 1, \alpha+1;\alpha+1, \alpha+2;1) }, \end{equation*}

    and Theorem 3.1 gives (4.3). This completes the proof.

    Theorem 4.2. For all \tau\le s\le t\le \eta and \alpha > 1/2, the following integral representations for the Gauss hypergeometric function hold true.

    (1)

    \begin{eqnarray} I^\alpha_2(t)&: = &\int_{\tau}^{t} (t-\tau) (t-s)^{\alpha-1} (\eta-s)^{\alpha-1} \ ds\\ & = &\left[\frac{(\eta-\tau)^{\alpha-1}(t-\tau)^{\alpha+1}}{\alpha}\right] {_2F_1(1-\alpha, 1;\alpha+1;g(t))}, \end{eqnarray} (4.5)

    where g(t): = \frac{t-\tau}{\eta-\tau}.

    (2)

    \begin{eqnarray} \int_{\tau}^{\eta} I^\alpha_2(t) dt& = &\left[\frac{(\eta-\tau)^{2\alpha+1}}{\alpha(\alpha+2)} \right] {_3F_2(1-\alpha, 1, \alpha+2;\alpha+1, \alpha+3;1) } \end{eqnarray} (4.6)
    \begin{eqnarray} & = &\frac{(\eta-\tau)^{2\alpha+1}}{\alpha} \left[ \dfrac{1}{\alpha+1}+\dfrac{1}{2\alpha}-\dfrac{2}{2\alpha+1}\right]. \end{eqnarray} (4.7)

    Proof of Theorem 4.2.

    (1) The proof is similar to the proof of Theorem 4.1, and so we just sketch the basic idea. In exactly the same manner, the integral I^\alpha_2(t) can be obtained.

    (2) Now denoting the left-hand side of (4.6) by \lambda_2 , we have

    \begin{equation} \lambda_2 : = \frac{(\eta-\tau)^{\alpha-1}}{\alpha} \int_{\tau}^{\eta} (t-\tau)^{\alpha+1} {_2F_1(1-\alpha, 1;\alpha+1;g(t))} dt. \end{equation} (4.8)

    Now, expressing _2F_1 as a series and changing the order of integration and summation, which is justified by Lemma 2.2, we have

    \begin{eqnarray} \lambda_2 &: = &\left[\frac{(\eta-\tau)^{\alpha-1}}{\alpha} \right] \sum\limits_{k = 0}^{\infty} \dfrac{(1-\alpha)_k}{(\alpha+1)_k} \int_\tau^\eta (t-\tau)^{\alpha+1}\left(\dfrac{t-\tau}{\eta-\tau}\right)^k dt\\ & = &\left[\frac{(\eta-\tau)^{\alpha-1}}{\alpha} \right] \sum\limits_{k = 0}^{\infty} \dfrac{(1-\alpha)_k}{(\alpha+1)_k} \left(\dfrac{1}{\eta-\tau}\right)^k \int_\tau^\eta \left(t-\tau\right)^{k+\alpha+1} dt\\ & = &\left[\frac{(\eta-\tau)^{2\alpha+1}}{\alpha} \right] \sum\limits_{k = 0}^{\infty} \dfrac{(1-\alpha)_k}{(\alpha+1)_k} \dfrac{1}{(k+\alpha+2)}. \end{eqnarray}

    Now, by Lemma 3.2, we obtain (4.6), that is

    \begin{eqnarray} \lambda_2 &: = &\left[\frac{(\eta-\tau)^{2\alpha+1}}{\alpha(\alpha+2)} \right] {_3F_2(1-\alpha, 1, \alpha+2;\alpha+1, \alpha+3;1) }, \end{eqnarray}

    and Theorem 3.2 gives (4.7). This completes the proof.

    Theorem 4.3. For all \tau\le s\le t\le \eta , and \alpha > 1/2, the following integral representations for the Gauss hypergeometric function hold true.

    (1)

    \begin{eqnarray} I^\alpha_3(t)&: = &\int_{\tau}^{t} (t-\tau)^{\alpha-1} (t-s)^{\alpha-1} (\eta-s)^{\alpha-1} \, ds\\ & = &\left[\frac{(\eta-\tau)^{\alpha-1}(t-\tau)^{2\alpha-1}}{\alpha}\right] {_2F_1(1-\alpha, 1, \alpha+1;g(t)) }, \end{eqnarray} (4.9)

    where g(t): = \frac{t-\tau}{\eta-\tau} .

    (2)

    \begin{equation} \int_{\tau}^{\eta} I^\alpha_3(t) dt = \left[\frac{(\eta-\tau)^{3\alpha-1}}{2\alpha^2}\right] { _3F_2(1-\alpha, 1, 2\alpha;\alpha+1, 2\alpha+1;1) }. \end{equation} (4.10)

    Proof of Theorem 4.3.

    (1) The proof is similar to the proof of Theorem 4.1, and so we just sketch the basic idea. In exactly the same manner, the integral I^\alpha_3(t) can be obtained.

    (2) Now, denoting the left-hand side of (4.10) by \lambda_3 , we have

    \begin{equation} \lambda_3 : = \frac{(\eta-\tau)^{\alpha-1}}{\alpha} \int_{\tau}^{\eta} (t-\tau)^{2\alpha-1} {_2F_1(1-\alpha, 1;\alpha+1;g(t))} dt. \end{equation} (4.11)

    By expressing _2F_1 as a series and change the order of integration and summation, which is justified by justified by Lemma 2.2, we have

    \begin{eqnarray} \lambda_3 & = &\left[\frac{(\eta-\tau)^{\alpha-1}}{\alpha} \right] \sum\limits_{k = 0}^{\infty} \dfrac{(1-\alpha)_k}{(\alpha+1)_k} \left(\dfrac{1}{\eta-\tau}\right)^k \int_\tau^\eta \left(t-\tau\right)^{k+2\alpha-1} dt\\ & = &\left[\frac{(\eta-\tau)^{3\alpha-1}}{\alpha} \right] \sum\limits_{k = 0}^{\infty} \dfrac{(1-\alpha)_k}{(\alpha+1)_k} \dfrac{1}{(k+2\alpha)}. \end{eqnarray} (4.12)

    Thus, the result of Lemma 2.5 gives (4.10). This completes the proof.

    Remark 4.1. Note that when \alpha = 2 , the results of Theorems 4.2 and 4.3 coincide. Therefore, the right-hand terms of relations (4.6) and (4.10) are equal when \alpha = 2 , which can be justified by Lemma 2.1. Thus,

    \begin{eqnarray} \int_{\tau}^{\eta} I^2_2(t) dt = \int_{\tau}^{\eta} I^2_3(t) dt & = & \frac{11(\eta-\tau)^{5}}{120}. \end{eqnarray} (4.13)

    \blacksquare For the choices \alpha = \tfrac{2p+1}{2} and p\in {{ \mathbb{N}}}^*, we have the following new explicit evaluation of a certain class of integrals as a special case from our Theorem 4.3.

    Theorem 4.4. For all \tau\le s\le t\le \eta and p\in {{ \mathbb{N}}}^*, the following integral holds true.

    (1)

    \begin{eqnarray} \int_{\tau}^{\eta} I^{\tfrac{2p+1}{2}}_3(t) dt & = & \left[\frac{2(\eta-\tau)^{\frac{6p+1}{2}}}{(2p+1)^2}\right] {_3F_2(\tfrac{1}{2}-p, 1, 1+2p;\tfrac{3}{2}+p, 2p+2;1)} \end{eqnarray} (4.14)
    \begin{eqnarray} & = &-K_p\left[\frac{2(\eta-\tau)^{\frac{6p+1}{2}}}{(2p+1)}\right] \left( 2a_{2p+1}\ln(2)+ 2a_{2p+1}(E_p+B_p)+\sum\limits_{j = 1}^{2p} a_jB_j\right ), \end{eqnarray} (4.15)

    where I^{\tfrac{2p+1}{2}}_3 is defined in (4.9) with \alpha = \tfrac{2p+1}{2} , and K_p, a_{2p+1}, E_p, B_j , and a_j are defined in (3.5).

    Proof of Theorem 4.4. By letting \alpha = \tfrac{2p+1}{2} and p\in {{ \mathbb{N}}}^* in the integral (4.10), we obtain (4.14) and (4.15) is obtained by replacing the explicit evaluation of the fucntion {_3F_2(\tfrac{1}{2}-p, 1, 1+2p;\tfrac{3}{2}+p, 2p+2;1)} given by Theorem 3.3. The proof is complete.

    \blacksquare The following result presents the integrals of I^{\alpha}_3(t) for some \alpha when \alpha = 3/2 , \alpha = 5/2 , and \alpha = 7/2 .

    Corollary 4.1. For all \tau\le s\le t\le \eta, the following integrals hold true.

    (1)

    \begin{eqnarray} \int_{\tau}^{\eta} I^{3/2}_3(t) dt & = & \left[\frac{2(\eta-\tau)^{\frac{7}{2}}}{3^2}\right] \left( \dfrac{33}{35}-\dfrac{6}{35}\ln(2)\right), \end{eqnarray} (4.16)

    where I^{3/2}_3 is defined in (4.9) with \alpha = 3/2.

    (2)

    \begin{eqnarray} \int_{\tau}^{\eta} I^{5/2}_3(t) dt & = & \left[\frac{2(\eta-\tau)^{\frac{13}{2}}}{5^2}\right] \left( \frac{4045}{6006}+\frac{10}{1001}\ln(2)\right), \end{eqnarray} (4.17)

    where I^{5/2}_3 is defined in (4.9) with \alpha = 5/2.

    (3)

    \begin{eqnarray} \int_{\tau}^{\eta} I^{7/2}_3(t) dt & = & \left[\frac{2(\eta-\tau)^{\frac{19}{2}}}{7^2}\right] \left( \frac{221158}{415701}-\frac{70}{138567}\ln(2)\right), \end{eqnarray} (4.18)

    where I^{7/2}_3 is defined in (4.9) with \alpha = 7/2.

    Proof of corollary 4.1.

    (1) Let p = 1 . Then, from Theorem 4.4 (4.14), we obtain

    \begin{equation} \int_{\tau}^{\eta} I^{3/2}_3(t) dt = \left[\frac{2(\eta-\tau)^{\frac{7}{2}}}{3^2}\right] {_3F_2(-\dfrac{1}{2}, 1, 3;\dfrac{5}{2}, 4;1)}. \end{equation} (4.19)

    From Lemma 3.3, we have

    \begin{equation} _3F_2(-\dfrac{1}{2}, 1, 3;\dfrac{5}{2}, 4;1) = \dfrac{33}{35}-\dfrac{6}{35}\ln(2), \end{equation} (4.20)

    and so substituting (4.20) into (4.19), we obtain (4.31). This completes the proof of (4.31).

    (2) Similarly, let p = 2 . Then, from Theorem 4.4 (4.14), we obtain

    \begin{equation} \int_{\tau}^{\eta} I^{5/2}_3(t) dt = \left[\frac{2(\eta-\tau)^{\frac{13}{2}}}{5^2}\right] {_3F_2(-\frac{3}{2}, 1, 5;\frac{7}{2}, 6;1)}. \end{equation} (4.21)

    From Corollary 3.1, we have

    \begin{equation} {_3F_2(-\frac{3}{2}, 1, 5;\frac{7}{2}, 6;1)} = \frac{4045}{6006}+\frac{10}{1001}\ln(2), \end{equation} (4.22)

    and so substituting (4.22) into (4.21), we obtain (4.32). This completes the proof of (4.32).

    (3) Also, if we let p = 2 , then from Theorem 4.4 (4.14), we obtain

    \begin{equation} \int_{\tau}^{\eta} I^{7/2}_3(t) dt = \left[\frac{2(\eta-\tau)^{\frac{19}{2}}}{7^2}\right]{_3F_2(-\frac{5}{2}, 1, 7;\frac{9}{2}, 8;1)}. \end{equation} (4.23)

    From Corollary 3.2, we have

    \begin{equation} {_3F_2(-\frac{5}{2}, 1, 7;\frac{9}{2}, 8;1)} = \frac{221158}{415701}-\frac{70}{138567}\ln(2), \end{equation} (4.24)

    and so substituting (4.24) into (4.23), we obtain (4.33). This completes the proof.

    Theorem 4.5. For all \tau\le s\le t\le \eta and \alpha > 1, the following integral representations for the Gauss hypergeometric functions hold true.

    (1)

    \begin{eqnarray} I^\alpha_4(t) &: = &\int_{\tau}^{t} (t-\tau)^{\alpha-1} (t-s)^{\alpha-1} (\eta-s)^{\alpha-2} \, ds\\ & = &\left[\frac{(\eta-\tau)^{\alpha-2}(t-\tau)^{2\alpha-1}}{\alpha}\right] {_2F_1(2-\alpha, 1;\alpha+1;g(t)) }, \end{eqnarray} (4.25)

    where g(t): = \frac{t-\tau}{\eta-\tau} .

    (2)

    \begin{equation} \int_{\tau}^{\eta} I^\alpha_4(t) dt = \dfrac{(\eta-\tau)^{3\alpha-2}}{(1-\alpha)(2\alpha-1)}+\left[ \dfrac{(1-3\alpha)(\eta-\tau)^{3\alpha-2}}{2\alpha^2(1-\alpha)} \right] {_3F_2(1-\alpha, 1, 2\alpha;\alpha+1, 2\alpha+1;1)}. \end{equation} (4.26)

    Proof of Theorem 4.5.

    (1) The proof follows similar lines of argument to that of the above theorems and so we just sketch the basic idea. In exactly the same manner, we have

    \begin{eqnarray} I^\alpha_4(t)& = & (\eta-\tau)^{\alpha-2}(t-\tau)^{2\alpha-1}\int_{0}^{1} (1-x)^{\alpha-1} \left(1-\frac{(t-\tau)}{(\eta-\tau)} x\right)^{\alpha-2} dx. \end{eqnarray}

    Above, we have the Euler integral representation of _2F_1(a, b; c;z) with a = 2-\alpha , b = 1 , c = \alpha+1 , and z = g(t) = \frac{t-\tau}{\eta-\tau} , thus

    \begin{equation} I^\alpha_4(t) = \frac{(\eta-\tau)^{\alpha-2}(t-\tau)^{2\alpha-1}}{\alpha} {_2F_1(2-\alpha, 1;\alpha+1;g(t)) }. \end{equation} (4.27)

    This completes the proof of (4.25).

    (2) Now denoting the left-hand side of (4.26) by \lambda_4 , we have

    \begin{equation} \lambda_4 : = \frac{(\eta-\tau)^{\alpha-2}}{\alpha} \int_{\tau}^{\eta} (t-\tau)^{2\alpha-1} {_2F_1(2-\alpha, 1;\alpha+1;g(t))} dt, \end{equation} (4.28)

    and by expressing _2F_1 as a series and change the order of integration and summation, which is justified by justified by Lemma 2.2, we have

    \begin{eqnarray} \lambda_4 & = &\left[\frac{(\eta-\tau)^{\alpha-2}}{\alpha} \right] \sum\limits_{k = 0}^{\infty} \dfrac{(2-\alpha)_k}{(\alpha+1)_k} \left(\dfrac{1}{\eta-\tau}\right)^k \int_\tau^\eta \left(t-\tau\right)^{k+2\alpha-1} dt\\ & = &\left[\frac{(\eta-\tau)^{3\alpha-2}}{\alpha} \right] \sum\limits_{k = 0}^{\infty} \dfrac{(2-\alpha)_k}{(\alpha+1)_k} \dfrac{1}{(k+2\alpha)}. \end{eqnarray} (4.29)

    Thus, the result of Lemma 2.1 gives (4.26). This completes the proof.

    \blacksquare For the choices \alpha = \tfrac{2p+1}{2} and p\in {{ \mathbb{N}}}^*, we have also the following new explicit evaluation of a certain class of Integrals as special case from our Theorem 4.5.

    Theorem 4.6. For all \tau\le s\le t\le \eta and p\in {{ \mathbb{N}}}^*, the following integral holds true.

    \begin{equation} \begin{array}{lll} \int_{\tau}^{\eta} I^{\tfrac{2p+1}{2}}_4(t) dt & = & \dfrac{(\eta-\tau)^{\frac{6p-1}{2}}}{p(1-2p)}+\left[ \dfrac{(6p+1)(\eta-\tau)^{\frac{6p-1}{2}}}{(2p-1)(2p+1)^2} \right] {_3F_2(\tfrac{1}{2}-p, 1, 1+2p;\tfrac{3}{2}+p, 2p+2;1)}\\ & = &\dfrac{(\eta-\tau)^{\frac{6p-1}{2}}}{p(1-2p)}-\left[ \dfrac{K_p(6p+1)(\eta-\tau)^{\frac{6p-1}{2}}}{(2p-1)(2p+1)} \right]\left( 2a_{2p+1}\ln(2)+ 2a_{2p+1}(E_p+B_p)+\sum\limits_{j = 1}^{2p} a_jB_j\right ), \end{array} \end{equation} (4.30)

    where I^{\tfrac{2p+1}{2}}_4 is defined in (4.25) with \alpha = \tfrac{2p+1}{2} , and K_p, a_{2p+1}, E_p, B_j, and a_j are defined in (3.5)

    Proof of Theorem 4.6. The proof is similar to the proof of Theorem 4.4, so we omit the proof for brevity.

    \blacksquare The following result presents the values of I^{\alpha}_4 for some \alpha that are when \alpha = 3/2, \alpha = 5/2 , and \alpha = 7/2.

    Corollary 4.2. For all \tau\le s\le t\le \eta, the following integrals hold true.

    (1)

    \begin{eqnarray} \int_{\tau}^{\eta} I^{3/2}_4(t) dt & = & -(\eta-\tau)^{\frac{5}{2}}+\frac{7(\eta-\tau)^{\frac{5}{2}}}{9} \left( \dfrac{33}{35}-\dfrac{6}{35}\ln(2)\right), \end{eqnarray} (4.31)

    where I^{3/2}_3 is defined in (4.25) with \alpha = 3/2.

    (2)

    \begin{eqnarray} \int_{\tau}^{\eta} I^{5/2}_4(t) dt & = & -\dfrac{(\eta-\tau)^{\frac{11}{2}}}{6}+\frac{12(\eta-\tau)^{\frac{11}{2}}}{75} \left( \frac{4045}{6006}+\frac{10}{1001}\ln(2)\right), \end{eqnarray} (4.32)

    where I^{5/2}_3 is defined in (4.25) with \alpha = 5/2.

    (3)

    \begin{eqnarray} \int_{\tau}^{\eta} I^{7/2}_4(t) dt & = & -\dfrac{(\eta-\tau)^{\frac{17}{2}}}{15}+\frac{19(\eta-\tau)^{\frac{17}{2}}}{245} \left( \frac{221158}{415701}-\frac{70}{138567}\ln(2)\right), \end{eqnarray} (4.33)

    where I^{7/2}_3 is defined in (4.25) with \alpha = 7/2.

    Proof of Corollary 4.2. The proof is similar to the proof of Theorem 4.1, so we omit the proof for brevity.

    In this study, we expressed four families of fractional integrals, denoted as {{ \mathcal{F} }}_1^{\alpha} = \{I^\alpha_1(\alpha), I^\alpha_2(\alpha)\; ;\; \alpha > \tfrac{1}{2}\} , {{ \mathcal{F} }}_2^{\alpha} = \{I^\alpha_3(\alpha), \; ;\; \alpha > \tfrac{1}{2}\} and {{ \mathcal{F} }}_3^{\alpha} = \{ I^\alpha_4(\alpha)\; ;\; \alpha > 1\} , using the class of hypergeometric functions _3F_2(1) . The series representation of the hypergeometric functions allowed us to derive explicit forms for the integrals of the family {{ \mathcal{F} }}_1^{\alpha} for all \alpha > \tfrac{1}{2} , as well as for the integrals of the subfamily {{ \mathcal{F} }}_2^{\tfrac{2p+1}{2}} and {{ \mathcal{F} }}_3^{\tfrac{2p+1}{2}} for all p\in {{ \mathbb{N}}}^* .

    For the integrals of the family {{ \mathcal{F} }}_2^{\alpha} and {{ \mathcal{F} }}_3^{\alpha} , we have only calculated the explicit forms of those of the subfamily {{ \mathcal{F} }}_2^{\tfrac{2p+1}{2}} and {{ \mathcal{F} }}_3^{\tfrac{2p+1}{2}} . However, by examining other subfamilies of {{ \mathcal{F} }}_2^{\alpha} and {{ \mathcal{F} }}_3^{\alpha} , we could derive more interesting formulas relating fractional integrals to the family of functions _3F_2(1) .

    Saleh S. Almuthaybiri: Conceptualization, supervision, validation, investigation, writing original draft preparation, formal analysis, writing review and editing. Abdelhamid Zaidi: Investigation, validation, writing original draft preparation, methodology, formal analysis, writing review and editing, doing the revision, funding acquisition. All authors have read and agreed to the published version of the manuscript.

    The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.

    The Researchers would like to thank the Deanship of Graduate Studies and Scientific Research at Qassim University for financial support (QU-APC-2025).

    The authors declare that they have no conflicts of interest.



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