1. Introduction
Fractional Fredholm integrodifferential equations have several applications in sciences and engineering. The closed form of the exact solution of such problems is difficult to find and in most of the cases is not available. For this reason, researchers are looking for the numerical solutions of such problems. Irandoust-Pakchin and Abdi-mazraeh [1] used the variational iteration method for solving fractional integrodifferential equations with the nonlocal boundary conditions. Adomian decomposition method is used in [2, 3] while the homotopy perturbation method is used in [4, 5]. Wazwaz [6–8] studied the Fredholm integral equations of the form(1)ux=fx+λ∫abKx,tutdt,where a and b are constants, λ is a parameter, u(x) is a smooth function as the discussion required, and K(x,t)∈C(R×[a,b]) is the kernel. In this paper, we study the generalization of the above problem. We study the following class of fractional Fredholm integrodifferential equations of the form(2)Dαux=fx+λ∫abKx,tumtdt, 0<α≤1, x∈R, a≤t≤b,subject to(3)ua=a0.The fractional derivative in (2) is in the Caputo sense. If α=0, we do not need the initial condition (3) and we return back to the problem which is discussed by Wazwaz [8]. In the following definition and theorem, we write the definition of Caputo derivative as well as the power rule which we are using in this paper. For more details on the geometric and physical interpretation for Caputo fractional derivatives, see [9].

Definition 1.
For m to be the smallest integer that exceeds α, the Caputo fractional derivatives of order α>0 are defined as(4)Dαux=1Γm-α∫0xx-τm-α-1dmuτdτmdτ,m-1<α<m,dmuxdxm,α=m∈N.

Theorem 2.
The Caputo fractional derivative of the power function satisfies(5)Dαxp=Γp+1Γp-α+1xp-α,m-1<α<m, p>m-1, p∈R,0,m-1<α<m, p≤m-1, p∈N.

Definition 3.
The Riemann-Liouville fractional integral operator of order α of u(x)∈Cγ, γ>-1, is defined as(6)Iαux=1Γα∫0xx-τα-1uτdτ,0≤m-1<α<m,ux,α=m∈N,where Cγ is the space of real functions u(x), x∈R, such that, for each u(x), there exists a real number ρ>γ such that u(x)=xρu1(x) where u1(x)∈C(R).

In addition, u(x)∈Cγm if um(x)∈Cγ where m∈N. We present the following definition and some properties of the fractional power series which are used in this paper. More details can be found in [10].

Definition 4.
A power series expansion of the form(7)∑m=0∞cmx-x0mα=c0+c1x-x0α+c2x-x02α+⋯,where 0≤m-1<α≤m, is called fractional power series FPS about. x=x0.

Theorem 5.
Suppose that f has a fractional FPS representation at x=x0 of the form(8)gx=∑m=0∞cmx-x0mα, x0≤x<x0+β.If Dmαgx, m=0,1,2,…, are continuous on R, then cm=Dmαg(x0))/Γ(1+mα).

Theorem 6.
Let u(x)∈C([x0,x0+R)) and Diαu(x)∈C((x0,x0+R)) for i=0,1,…,m+1 where 0≤m-1<α≤m. Then,(9)Im+1αDm+1αux=Dm+1αϖΓm+1α+1x-x0m+1α+1, x0≤ϖ≤x<x0+R.

Theorem 7.
Let u(x)∈C([x0,x0+R)), Diαu(x)∈C((x0,x0+R)), and Diαu(x) can differentiate (m-1) with respect to x for i=0,1,…,m+1 where 0≤m-1<α≤m. Then,(10)ux=∑k=0mDkαϖΓkα+1x-x0kα+Dm+1αϖΓm+1α+1x-x0m+1α+1, x0≤ϖ≤x<x0+R.

Theorem 8.
Let |D(m+1)αu(x)|∈A on x0≤x<s where m-1<α≤m. Then, the reminder Rm satisfies(11)Rm≤AΓm+1α+1x-x0m+1α, x0≤x<s.

This paper is organized as follows. A description of the modified fractional power series method (MFPS) for approximating the fractional Fredholm integrodifferential equations problem (2)-(3) is presented in Section 2. Several numerical examples are discussed in Section 3. Conclusions and closing remarks are given in Section 4.

2. Algorithm of the MFPS Method
Consider the following class of fractional Fredholm integrodifferential equations of the form(12)ux=fx+λ∫abKx,tutdt, 0<α≤1subject to(13)ua=a0.Using the MFPS method, the solution problem (2)-(3) can be written the fractional power series form as(14)ux=∑n=0∞fnx-anαΓ1+nα.To obtain the approximate values of the above series (14), the kth truncated series uk(x) is written in the form(15)ukx=∑n=0kfnx-anαΓ1+nα.Since u(a)=f0=a0, we rewrite (14) as(16)ukx=a0+∑n=1kfnx-anαΓ1+nα, k=1,2,…,where u0(x)=a0 is considered to be the 0th MRPS approximate solution of u(x). To find the values of the MFPS-coefficients fk, k=1,2,3,…, we solve the fractional differential equation(17)Dk-1αReskua=0, k=1,2,3,…,where Resk(u(a)) is the kth residual function and is defined by(18)Reskux=Dαukx-fx-λ∫abKx,tuktdt.To determine the coefficient f1 in the expansion (15), we substitute the 1st RPS approximate solution(19)u1x=a0+f1x-aαΓ1+αinto (18) to get(20)Res1ua=Dαu1x-fx-λ∫abKx,tu1tdtx=a=f1-fa-λ∫aba0Ka,tdt-λf1∫abt-aαΓ1+αKx,tdt=0.Then, we solve Res1(a)=0 to get(21)f1=fa+λ∫aba0Ka,tdt1-λ∫abt-aα/Γ1+αKa,tdt.To find f2, we substitute the 2nd RPS approximate solution(22)u2x=a0+f1x-aαΓ1+α+f2x-a2αΓ1+2αinto the 2nd residual function Res2(u(x)) such that(23)Res2ux=Dαu2x-fx-λ∫abKx,tu2tdt=f1+f2x-aαΓ1+α-fx-λ∫abKx,ta0+f1t-aαΓ1+α+f2t-a2αΓ1+2αdt.Then, we solve DαRes2(u(a))=0 to get(24)f2=Dαfa+λa0∫abDxαKa,tdt+λf1∫abt-aα/Γ1+αDxαKa,tdt1-λ∫abt-a2α/Γ1+2αDxαKx,tdt.To find f3, we substitute the 3rd RPS approximate solution (25)u3x=a0+f1x-aαΓ1+α+f2x-a2αΓ1+2α+f3x-a3αΓ1+3αinto the 3rd residual function Res3(u(x)) such that(26)Res3ux=Dαu3x-fx-λ∫abKx,tu3tdt=f1+f2x-aαΓ1+α+f3x-a2αΓ1+2α-fx-λ∫abKx,ta0+f1t-aαΓ1+α+f2t-a2αΓ1+2α+f3t-a3αΓ1+3αdt.Then, we solve D2αRes3(u(a))=0 to get(27)f3=f2+D2αfa+λ∑k=02fk∫abt-akα/Γ1+kαDx2αKa,tdt1-λ∫abt-a3α/Γ1+3αDx2αKa,tdt.To find f4, we substitute the 4th RPS approximate solution(28)u4x=a0+f1x-aαΓ1+α+f2x-a2αΓ1+2α+f3x-a3αΓ1+3α+f4x-a4αΓ1+4αinto the 4th residual function Res4(u(x)) such that(29)Res4ux=Dαu4x-fx-λ∫abKx,tu4tdt=f1+f2x-aαΓ1+α+f3x-a2αΓ1+2α+f4x-a4αΓ1+4α-fx-λ∫abKx,ta0+f1x-aαΓ1+α+f2x-a2αΓ1+2α+f3x-a3αΓ1+3α+f4x-a4αΓ1+4αdt.Then, we solve D3αRes4(u(a))=0 to get(30)f4=f3+D3αfa+λ∑k=03fk∫abt-akα/Γ1+kαDx2αKa,tdt1-λ∫abt-a4α/Γ1+4αDx2αKa,tdt.Using similar argument, we find that(31)fn=fn-1+Dn-1αfa+λ∑k=0n-1fk∫abt-akα/Γ1+kαDxn-1αKa,tdt1-λ∫abt-anα/Γ1+nαDxn-1αKa,tdt, n=1,2,3,….Thus,(32)ukx=a0+∑n=1kfn-1+Dn-1αfa+λ∑k=0n-1fk∫abt-akα/Γ1+kαDxn-1αKa,tdt1-λ∫abt-anα/Γ1+nαDxn-1αKa,tdtx-anαΓ1+nα.For k=1,2,….

3. Numerical Results
In this section, we present three examples to show the efficiency of the proposed method. We use Mathematica software to generate the results in this section.

Example 1.
Consider the following fractional Fredholm integrodifferential equation:(33)D1/2ux=323πx1.5+163πx2.5-2x+∫01xtut, x∈Rsubject to(34)u0=0.The exact solution is(35)ux=4x2+5x3.Using the same argument described in the previous section, we find that(36)f0=f1=f2=f3=0,f4=8,f5=0,f6=30,fn=0, n=7,8,….Thus,(37)u6x=4x2+5x3which is the exact solution.

Example 2.
Consider the following fractional Fredholm integrodifferential equation:(38)D1/4ux=∑k=0∞xk+3/42k+1Γk-1/4-4-2ex2+∫01x2tut, x∈Rsubject to(39)u0=1.The exact solution is(40)ux=e1/2x.Using the same argument described in the previous section, we find the first few terms which are(41)f0=1,f1=f2=f3=0,f4=12,f5=f6=f7=0,f8=14.Continuing in this process, we find that(42)fn=12k,n=4k, k=0,1,2,…,0,otherwise.Thus, if n=4k for some positive integer k,(43)unx=∑m=0nfmxm/4Γ1+m/4=∑m=0k12mx4m/4Γ1+4m/4=∑m=0k12mxmm!.Hence,(44)limn→∞unx=limn→∞∑m=0k12mxmm!=∑m=0∞12mxmm!=e1/2xwhich is the exact solution.

Example 3.
Consider the following fractional Fredholm integrodifferential equation:(45)D1/3ux=Γ4/3Γ11/3-5ex+4∫01extutdt, x∈Rsubject to(46)u0=1.The exact solution is(47)ux=1+x3.Using the same argument described in the previous section, we find that the first few terms are(48)f0=1,f1=f2=f3=f4=0,f5=f6=f7=f8=0,f9=6,f9=f10=0.Continuing in this process, we find that(49)fn=1,n=0,6,n=9,0,otherwise.Thus,(50)u9x=1+x3which is the exact solution.