Abstract

Many algorithms that have been proposed for the numerical evaluation of Cauchy principal value integrals are numerically unstable. In this work we present some formulae to evaluate the known Gaussian quadrature rules for finite part integrals 𝑏𝑎𝑤(𝑥)(𝑓(𝑥)/(𝑥𝑡)𝑟)𝑑𝑥, 𝑟=1,2,, and extend Clenshow's algorithm to evaluate these integrals in a stable way.

1. Introduction

In the theory of the numerical approximation of Cauchy principal value integrals, basically two kinds of Gaussian quadrature formulae have been investigated. For the modified Gaussian formula, we refer to [17] and the literature cited therein. Results on the Gaussian quadrature formula in the strict sense can be found, for example, in [1, 3, 713]. In [14] Diethelm proposed a quadrature formula which is called Gaussian quadrature formula of the third kind. Many algorithms that have been proposed for the numerical evaluation of Cauchy principal value integrals are numerically unstable. In particular, this is true for the standard algorithm proposed in [3, Section 3] for the evaluation of the modified Gaussian formula [15, Section 4]. In this paper, we present some formulae to evaluate the known Gaussian quadrature rules for finite part integrals𝑏𝑎𝑤(𝑥)𝑓(𝑥)(𝑥𝑡)𝑟𝑑𝑥,𝑟=1,2,,(1.1) and extend Clenshow's algorithm [16] to evaluate these integrals in a stable way.

This paper is organized as follows: in Section 1, we introduce some known quadrature rules for finite part integrals and we analyze the relations among these rules. In Section 2, we extend Clenshow's algorithm [16] to evaluate any sum of derivatives of functions. In Section 3, we apply the extended algorithm to compute Lagrange interpolation polynomial and its derivatives, to evaluate some known singular integrals exactly and quadrature rules for finite part integrals in a stable way. Finally, we present a numerical example.

Some hypersingular integrals can be found by successive differentiation; in some cases evaluation of these integrals becomes tedious and the formulas are lengthily. The evaluation of the such singular integrals is a necessary step for the numerical approach to the integral equations in crack problems. One of the main tasks of the proposed algorithm is evaluate these integrals. The appendix, with closed form solutions for class of these integrals, supplements the paper.

1.1. Guassian Quadrature Rule

Let 𝑤(𝑥) be an integrable weight function on the interval [𝑎,𝑏], and let {𝜋𝑛(𝑥)} be orthogonal system of polynomials associated with 𝑤(𝑥) which satisfiesa three-term recurrence relation:𝜋𝑛+1(𝑥)=𝐴𝑛𝑥+𝐵𝑛𝜋𝑛(𝑥)𝑙𝑛𝜋𝑛1(𝑥),𝜋1(𝑥)=0,𝜋1(𝑥)=0,(1.2) and hence {𝜋(𝑟)𝑛(𝑥)} satisfy a recurrence relation:𝜋(𝑟)𝑛+1(𝑥)=𝐴𝑛𝑥+𝐵𝑛𝜋(𝑟)𝑛(𝑥)𝑙𝑛𝜋(𝑟)𝑛1(𝑥)+𝑟𝐴𝑛𝜋(𝑟1)𝑛(𝑥).(1.3) The known Gaussian quadrature rule 𝑄𝑛(𝑓) for 𝐼(𝑓)=𝑏𝑎𝑤(𝑥)𝑓(𝑥)𝑑𝑥 is defined as𝑄𝑛(𝑓)=𝑛𝑖=1𝑤𝑖𝑓𝑥𝑖,𝑤𝑖=1𝜋𝑛𝑥𝑖𝑏𝑎𝑤(𝑥)𝜋𝑛(𝑥)𝑥𝑥𝑖𝑑𝑥,𝜋𝑛𝑥𝑖=0,𝑖=1,𝑛.(1.4)

1.2. Modified Gaussian Quadrature Rules for Singular Integrals

Definition 1.1. 𝑓[𝑥,𝑦]=𝑓(𝑥)𝑓(𝑦)𝑥𝑦,𝑓[𝑥,𝑥]=𝑓(𝑥),𝑓𝑥,𝑡,,𝑡𝑟times=1(𝑟1)!𝑑𝑟1𝑑𝑡𝑟1𝑓[𝑥,𝑡].(1.5)

The finite part integral is defined as follows.

Definition 1.2. 𝐼(𝑓,𝑡,𝑟)=𝑏𝑎𝑤(𝑥)𝑓(𝑥)(𝑥𝑡)𝑟𝑑𝑥=𝑏𝑎𝑤(𝑥)𝑓𝑥,𝑡,,𝑡𝑟times𝑑𝑥+𝑟1𝑘=0𝑓(𝑘)(𝑡)𝑘!𝐼(1,𝑡,𝑟𝑘).(1.6)

The modified Gaussian quadrature rule for 𝐼(𝑓,𝑡,𝑟) can be defined as follows.

Definition 1.3. 𝑄𝑀𝑛(𝑓,𝑡,𝑟)=𝑛𝑖=1𝑤𝑖𝑓𝑥𝑖,𝑡,,𝑡𝑟times+𝑟1𝑘=0𝑓(𝑘)(𝑡)𝑘!𝐼(1,𝑡,𝑟𝑘).(1.7)

Since 𝑄𝑀𝑛(𝑓,𝑡,1)=𝑛𝑖=1𝑤𝑖𝑓[𝑥𝑖,𝑡]+𝑓(𝑡)𝐼(1,𝑡,1), then by differentiating both sides (𝑟1)-times, using Leibniz formula and Definition 1.1, we have the following the lemma.

Lemma 1.4. 𝑄𝑀𝑛(𝑓,𝑡,𝑟)=1(𝑟1)!𝑑𝑟1𝑑𝑡𝑟1𝑄𝑀𝑛(𝑓,𝑡,1).(1.8)

Lemma 1.5. Let 𝑞𝑛(𝑡)=𝐼(𝜋𝑛,𝑡,1); then 𝐼(1,𝑡,𝑚)=1(𝑚1)!𝑑𝑚1𝑑𝑡𝑚1𝑞𝑛(𝑡)𝜋𝑛(𝑡)+𝑛𝑗=1𝑤𝑗𝑥𝑗𝑡𝑚,𝑚=1,2,.(1.9)

Proof. Since the quadrature rule (1.7) is Gaussian: 𝐼𝜋𝑛,𝑡,1=𝑄𝑛𝜋𝑛,𝑡,1=𝑛𝑖=1𝑤𝑖𝜋𝑛𝑥𝑖,𝑡+𝜋𝑛(𝑡)𝐼(1,𝑡,1),(1.10) then 𝑞𝑛(𝑡)=𝜋𝑛(𝑡)𝑛𝑖=1𝑤𝑖𝑥𝑖𝑡+𝜋𝑛(𝑡)𝐼(1,𝑡,1).(1.11) This implies that 𝐼(1,𝑡,1)=𝑞𝑛(𝑡)𝜋𝑛(𝑡)+𝑛𝑗=1𝑤𝑗𝑥𝑗𝑡.(1.12) Differentiate both sides (𝑚1)-times to have (1.9).

Thus the modified Gaussian quadrature rule for 𝐼(𝑓,𝑡,𝑟) can be written as 𝑄𝑀𝑛(𝑓,𝑡,𝑟)=𝑛𝑖=1𝑤𝑖𝑓𝑥𝑖,𝑡,,𝑡𝑟times+𝑟1𝑘=0𝑓(𝑘)(𝑡)𝑘!1(𝑟𝑘1)!𝑑𝑟𝑘1𝑑𝑡𝑟𝑘1𝑞𝑛(𝑡)𝜋𝑛(𝑡)+𝑛𝑗=1𝑤𝑗𝑥𝑗𝑡𝑟𝑘.(1.13)

Special Cases
The first case: 𝑄𝑀𝑛(𝑓,𝑡,1)=𝑛𝑖=1𝑤𝑖𝑓𝑥𝑖𝑥𝑖𝑡+𝑓(𝑡)𝑞𝑛(𝑡)𝜋𝑛(𝑡),(1.14) The second case: 𝑄𝑀𝑛(𝑓,𝑡,2)=𝑛𝑖=1𝑤𝑖𝑓𝑥𝑖𝑥𝑖𝑡2+𝑑𝑑𝑡𝑓(𝑡)𝑞𝑛(𝑡)𝜋𝑛(𝑡),(1.15) which are the known formulae

1.3. The Gaussian Quadrature Formula in the Strict Sense for Singular Integrals

In the integral 𝐼(𝑓,𝑡,𝑟), if we interpolate 𝑓 by a polynomial 𝐿𝑛1𝑓 of degree 𝑛1 using the zeros of 𝜋𝑛(𝑥), that is, the nodes of the classical Gauss formula, as interpolation nodes and use 𝐼(𝐿𝑛1𝑓,𝑡,𝑟) as an approximation for 𝐼(𝑓,𝑡,𝑟), we obtain the Gaussian quadrature formula in the strict sense for singular integrals:𝑄strict𝑛(𝑓,𝑡,1)=𝐼𝐿𝑛1𝑓,𝑡,1=𝑛𝑖=1𝑤𝑖𝑞𝑛𝑥𝑖𝑞𝑛𝑥𝑖,𝑡𝑓𝑥𝑖,(1.16)𝑄strict𝑛(𝑓,𝑡,𝑟)=1(𝑟1)!𝑑𝑟1𝑑𝑡𝑟1𝑄strict𝑛(𝑓,𝑡,1),(1.17) so due to (1.16), one obtains the following lemma.

Lemma 1.6. 𝑄strict𝑛(𝑓,𝑡,𝑟)=𝑛𝑖=1𝑤𝑖𝑞𝑛𝑥𝑖𝑞𝑛𝑥𝑖,𝑡,,𝑡𝑟times𝑓𝑥𝑖.(1.18)

1.4. Gaussian Quadrature Rule of the Third Kind

Let 𝑓𝐶𝑟1[𝑎,𝑏]. According to [14], the integral 𝐼(𝑓,𝑡,𝑟) can be approximated by 𝑄𝐿3𝑛+1(𝑓,𝑡,𝑟)𝑄𝐿3𝑛+1(𝑓,𝑡,𝑟)=𝑏𝑎𝑤(𝑥)𝐿𝑛1𝑓𝑥,𝑡,,𝑡𝑟times𝑑𝑥+𝑟1𝑘=0𝑓(𝑘)(𝑡)𝑘!𝐼(1,𝑡,𝑟𝑘),(1.19) which is called Gaussian quadrature rule of the third kind:

for 𝑟=1, 𝑄𝐿3𝑛+1(𝑓,𝑡,1)=𝑏𝑎𝑤(𝑥)𝐿𝑛1𝑓[𝑥,𝑡]𝑑𝑥+𝑓(𝑡)𝑞0(𝑡).(1.20)

Lemma 1.7. 𝑄𝐿3𝑛+1(𝑓,𝑡,𝑟)=1(𝑟1)!𝑑𝑟1𝑑𝑡𝑟1𝑄𝐿3𝑛+1(𝑓,𝑡,1).(1.21)

1.5. The Connection between the Gaussian Quadrature Rules

Lemma 1.8. 𝑄𝑀𝑛(𝑓,𝑡,1)=𝑄strict𝑛(𝑓,𝑡,1)+𝑞𝑛(𝑡)𝜋𝑛(𝑡)𝑓(𝑡)𝐿𝑛𝑓(𝑡).(1.22)

Proof. From (1.16), 𝑄strict𝑛(𝑓,𝑡,1)=𝑛𝑖=1𝑤𝑖𝑞𝑛𝑥𝑖𝑞𝑛𝑥𝑖,𝑡𝑓𝑥𝑖,𝑤𝑖=𝑞𝑛𝑥𝑖𝜋𝑛𝑥𝑖=𝑛𝑖=1𝑤𝑖𝑓𝑥𝑖𝑥𝑖𝑡+𝑞𝑛(𝑡)𝜋𝑛(𝑡)𝑛𝑖=1𝑓𝑥𝑖𝜋𝑛(𝑡)𝜋𝑛𝑥𝑖𝑡𝑥𝑖=𝑛𝑖=1𝑤𝑖𝑓𝑥𝑖𝑥𝑖𝑡+𝑞𝑛(𝑡)𝜋𝑛(𝑡)𝐿𝑛𝑓(𝑡).(1.23)
Since 𝑄𝑀𝑛(𝑓,𝑡,1)=𝑛𝑖=1𝑤𝑖(𝑓(𝑥𝑖)/𝑥𝑖𝑡)+𝑓(𝑡)(𝑞𝑛(𝑡)/𝜋𝑛(𝑡)), one obtains (1.22).

Corollary 1.9. If 𝜋𝑛(𝑡)=0, then 𝑄𝑀𝑛(𝑓,𝑡,1)=𝑄strict𝑛(𝑓,𝑡,1)+𝑞𝑛(𝑡)𝜋𝑛(𝑡)𝑓(𝑡)𝐿𝑛𝑓(𝑡).(1.24)

Proof. From Definition 1.3, 𝑄𝑀𝑛(𝑓,𝑡,1)=𝑛𝑖=1𝑤𝑖𝑞𝑛𝑥𝑖𝑞𝑛𝑥𝑖,𝑡𝑓𝑥𝑖+𝑞𝑛(𝑡)𝑛𝑖=1𝑤𝑖𝑞𝑛𝑥𝑖𝑓𝑥𝑖,𝑡=𝑄strict𝑛(𝑓,𝑡,1)+𝑞𝑛(𝑡)𝑛𝑖=1𝑓𝑥𝑖𝑓(𝑡)𝜋𝑛𝑥𝑖𝑥𝑖𝑡,(1.25) and since 𝑛𝑖=1𝑓𝑥𝑖𝑓(𝑡)𝜋𝑛𝑥𝑖𝑥𝑖𝑡=1𝜋𝑛(𝑡)𝑓(𝑡)𝐿𝑛1[𝑓](𝑡),(1.26) then Lemma has been proved.

Lemma 1.10. 𝑄𝐿3𝑛+1(𝑓,𝑡,1)=𝑄strict𝑛(𝑓,𝑡,1)+𝑓(𝑡)𝐿𝑛1𝑓(𝑡)𝑞0(𝑡).(1.27)

2. Sum Series Algorithm

Theorem 2.1. The sum 𝑛𝑘=𝑚𝑎𝑘Φ𝑘(𝑥)(𝑚0) of any series of functions which satisfy a linear recurrence relation Φ𝑘+1(𝑥)+𝛿𝑘(𝑥)Φ𝑘(𝑥)+𝜂𝑘(𝑥)Φ𝑘1(𝑥)=𝑠𝑘𝑓𝑘(𝑥)(2.1) is given by 𝑛𝑘=𝑚𝑎𝑘Φ𝑘(𝑥)=𝑏𝑚(𝑥)Φ𝑚(𝑥)+𝑏𝑚+1(𝑥)Φ𝑚+1(𝑥)+𝛿𝑚(𝑥)Φ𝑚(𝑥)+𝑛1𝑘=𝑚+1𝑏𝑘+1𝑠𝑘𝑓𝑘(𝑥),(2.2) where 𝑏𝑛+1=𝑏𝑛+2=0,𝑏𝑘+𝛿𝑘𝑏𝑘+1+𝜂𝑘+1𝑏𝑘+2=𝑎𝑘,𝑘=𝑛,𝑛1,,𝑚.(2.3)

Moreover, if 𝑓𝑘(𝑥) satisfies the linear recurrence relation𝑓𝑘+1(𝑥)+𝜆𝑘(𝑥)𝑓𝑘(𝑥)+𝜇𝑘(𝑥)𝑓𝑘1(𝑥)=0,(2.4) then𝑛𝑘=𝑚𝑎𝑘Φ𝑘(𝑥)=𝑏𝑚(𝑥)Φ𝑚(𝑥)+𝑏𝑚+1(𝑥)Φ𝑚+1(𝑥)+𝛿𝑚(𝑥)Φ𝑚(𝑥)+𝑐𝑚+1(𝑥)𝑓𝑚+1(𝑥)+𝑐𝑚+2(𝑥)𝑓𝑚+2(𝑥)+𝜆𝑚+1(𝑥)𝑓𝑚+1(𝑥),(2.5) where𝑐𝑛=𝑐𝑛+1=0,𝑐𝑘+𝜆𝑘𝑐𝑘+1+𝜇𝑘+1𝑐𝑘+2=𝑠𝑘𝑏𝑘+1,𝑘=𝑛1,,𝑚+1.(2.6)

Proof. 𝑛𝑘=𝑚𝑎𝑘Φ𝑘=𝑛𝑘=𝑚𝑏𝑘+𝛿𝑘𝑏𝑘+1+𝜂𝑘+1𝑏𝑘+2Φ𝑘(𝑥)=𝑏𝑚+𝛿𝑚𝑏𝑚+1Φ𝑚(𝑥)+𝑏𝑚+1Φ𝑚+1(𝑥)+𝑛𝑘=𝑚+2𝑏𝑘Φ𝑘(𝑥)+𝛿𝑘1Φ𝑘1(𝑥)+𝜂𝑘1Φ𝑘2(𝑥).(2.7) Using (2.1), we get (2.2). Moreover, if 𝑓𝑘(𝑥) satisfies (2.4), we can similarly deduce (2.5).

In a similar way, we can prove the following theorem.

Theorem 2.2. 𝑛𝑘=𝑚𝑎𝑘Φ𝑘(𝑥)=𝜂𝑛+1(𝑥)𝑏𝑛Φ𝑛(𝑥)+𝑏𝑛1(𝑥)𝜂𝑛Φ𝑛1(𝑥)+𝛿𝑛(𝑥)Φ𝑛(𝑥)+𝑛1𝑘=𝑚+1𝑏𝑘1𝑠𝑘𝑓𝑘(𝑥),(2.8) where 𝑏𝑚2=𝑏𝑚1=0,𝜂𝑘+1𝑏𝑘+𝛿𝑘𝑏𝑘1+𝑏𝑘2=𝑎𝑘,𝑘=𝑚,𝑚+1,,𝑛.(2.9) Moreover, if 𝑓𝑘(𝑥) satisfies the linear recurrence relation (2.4), then 𝑛𝑘=𝑚𝑎𝑘Φ𝑘(𝑥)=𝜂𝑛+1𝑏𝑛Φ𝑛(𝑥)+𝑏𝑛1𝜂𝑛Φ𝑛1(𝑥)+𝛿𝑛Φ𝑛(𝑥)+𝜇𝑛𝑐𝑛1𝑓𝑛1(𝑥)+𝑐𝑛2𝜇𝑛1𝑓𝑛2(𝑥)+𝜆𝑛1𝑓𝑛1(𝑥),(2.10) where 𝑐𝑚1=𝑐𝑚=0,𝜇𝑘+1𝑐𝑘+𝜆𝑘𝑐𝑘1+𝑐𝑘2=𝑠𝑘𝑏𝑘1,𝑘=𝑚+1,,𝑛1.(2.11)

Now we extend the previous theorems to evaluate 𝑛𝑘=0𝑎𝑘Φ(𝑟1)𝑘(𝑥).

Suppose that Φ𝑘(𝑥) satisfies (2.1) with 𝑓𝑘0 and that Φ(𝑖)𝑘(𝑥) satisfies the following recurrence relation:Φ(𝑖)𝑘+1(𝑥)+𝛿𝑘(𝑥)Φ(𝑖)𝑘(𝑥)+𝜂𝑘(𝑥)Φ(𝑖)𝑘1(𝑥)=𝑠𝑖𝑘(𝑥)Φ(𝑖1)𝑘(𝑥).(2.12)

Theorem 2.3. Let 𝑀=max{𝑛,2𝑟1}; then 𝑛1𝑖=0𝑎𝑖Φ(𝑟1)𝑖=𝑟𝑖=1𝑐𝑟𝑖𝑖1Φ(𝑟𝑖)𝑖1+𝑐𝑟𝑖𝑖𝛿𝑖1Φ(𝑟𝑖)𝑖1+Φ(𝑟𝑖)𝑖,(2.13) where 𝑐𝑟𝑖𝑀𝑖+1=𝑐𝑟𝑖𝑀𝑖+2=0,𝑐𝑟𝑖𝑘+𝛿𝑘𝑐𝑟𝑖𝑘+1+𝜂𝑘+1𝑐𝑟𝑖𝑘+2=𝑎𝑟𝑖𝑘,𝑘=𝑀𝑖,,𝑖1,1𝑖𝑟,𝑎𝑟1𝑘=𝑎𝑘,0𝑘𝑛1,0,𝑛𝑘𝑀,𝑎𝑟𝑖𝑘=𝑠𝑟𝑖+1𝑘𝑐𝑟𝑖+1𝑘+1,𝑖2.(2.14)

Proof. Let 𝑌𝑚=𝑀𝑚𝑖=𝑚1𝑎𝑟𝑚𝑖Φ(𝑟𝑚)𝑖(1𝑚𝑟).(2.15) Due to Theorem 2.1, we obtain 𝑌𝑚=𝑋𝑚+𝑌𝑚+1, 𝑋𝑚=𝑐𝑟𝑚𝑚1Φ(𝑟𝑚)𝑚1+𝑐𝑟𝑚𝑚𝛿𝑚1Φ(𝑟𝑚)𝑚1+Φ(𝑟𝑚)𝑚,𝑐𝑟𝑚𝑀𝑚+1=𝑐𝑟𝑚𝑀𝑚+2=0,𝑐𝑟𝑚𝑘+𝛿𝑘𝑐𝑟𝑚𝑘+1+𝜂𝑘+1𝑐𝑟𝑚𝑘+2=𝑎𝑟𝑚𝑘,𝑘=𝑀𝑚,,𝑚1.(2.16) By induction, we have 𝑌1=𝑟1𝑖=1𝑋𝑖+𝑌𝑟, and by using Theorem 2.1, we obtain 𝑌𝑟=𝑀𝑟𝑖=𝑟1𝑎0𝑖Φ𝑖=𝑐0𝑟1Φ𝑟1+𝑐0𝑟1𝛿𝑟1Φ𝑟1+Φ𝑟,(2.17) where 𝑐0𝑀𝑟+1=𝑐0𝑀𝑟+2=0,𝑐0𝑘+𝛿𝑘𝑐0𝑘+1+𝜂𝑘+1𝑐0𝑘+2=𝑎0𝑘,𝑘=𝑀𝑟,,𝑟1.(2.18) Finally 𝑌1=𝑟𝑖=1𝑐𝑟𝑖𝑖1Φ𝑟𝑖𝑖1+𝑐𝑟𝑖𝑖𝛿𝑖1Φ(𝑟𝑖)𝑖1+Φ(𝑟𝑖)𝑖,(2.19) so 𝑛1𝑖=0𝑎𝑖Φ(𝑟1)𝑖=𝑌1=𝑟𝑖=1𝑐𝑟𝑖𝑖1Φ(𝑟𝑖)𝑖1+𝑐𝑟𝑖𝑖𝛿𝑖1Φ(𝑟𝑖)𝑖1+Φ(𝑟𝑖)𝑖.(2.20)

The alternative algorithm that incorporate 𝑎𝑖's in an upward direction can be deduced as follows.

Theorem 2.4. Let 𝑀=max{𝑛,2𝑟1}; then 𝑛1𝑖=0𝑎𝑖Φ(𝑟1)𝑖=𝑟𝑗=1𝜂𝑀𝑗+1𝑐𝑗1𝑀𝑗Φ(𝑟𝑗)𝑀𝑗+𝑐𝑗1𝑀𝑗1𝜂𝑀𝑗Φ(𝑟𝑗)𝑀𝑗1+𝛿𝑀𝑗Φ(𝑟𝑗)𝑀𝑗,(2.21) where 𝑐𝑗1𝑗3=𝑐𝑗1𝑗2=0,𝜂𝑘+1𝑐𝑗1𝑘+𝛿𝑘𝑐𝑗1𝑘1+𝑐𝑗1𝑘2=𝑎𝑗1𝑘,𝑘=𝑗1,,𝑀𝑗,1𝑗𝑟,𝑎0𝑘=𝑎𝑘,0𝑘𝑛1,0,𝑛𝑘𝑀,𝑎𝑗1𝑘=𝑠𝑗2𝑘𝑐𝑗2𝑘1,𝑘=𝑗1,,𝑀𝑗,𝑗2.(2.22)

Proof. Let 𝑌𝑚=𝑀𝑚𝑖=𝑚1𝑎(𝑚1)𝑖Φ(𝑟𝑚)𝑖(1𝑚𝑟).(2.23) Due to Theorem 2.2, we obtain 𝑌𝑚=𝑋𝑚+𝑌𝑚+1, 𝑋𝑚=𝜂𝑀𝑚+1𝑐𝑚1𝑀𝑚Φ(𝑟𝑚)𝑀𝑚+𝑐𝑚1𝑀𝑚1𝜂𝑀𝑚Φ(𝑟𝑚)𝑀𝑚1+𝛿𝑀𝑚Φ(𝑟𝑚)𝑀𝑚,𝑐𝑚1𝑚3=𝑐𝑚1𝑚2=0,𝜂𝑘+1𝑐𝑚1𝑘+𝛿𝑘𝑐𝑚1𝑘1+𝑐𝑚1𝑘2=𝑎𝑚1𝑘,𝑘=𝑚1,,𝑀𝑚,1𝑚𝑟1.(2.24) By induction, we have 𝑌1=𝑟1𝑖=1𝑋𝑖+𝑌𝑟, and by using Theorem 2.2, we obtain 𝑌𝑟=𝑀𝑟𝑖=𝑟1𝑎𝑟1𝑖Φ𝑖=𝜂𝑀𝑟+1𝑐𝑟1𝑀𝑟Φ𝑀𝑟+𝑐𝑟1𝑀𝑟1𝜂𝑀𝑟Φ𝑀𝑟1+𝛿𝑀𝑟Φ𝑀𝑟,(2.25) where 𝑐𝑟1𝑟3=𝑐𝑟1𝑟2=0,𝜂𝑘+1𝑐𝑟1𝑘+𝛿𝑘𝑐𝑟1𝑘1+𝑐𝑟1𝑘2=𝑎𝑟1𝑘,𝑘=𝑟1,,𝑀𝑟.(2.26) Finally 𝑌1=𝑟𝑗=1𝜂𝑀𝑗+1𝑐𝑗1𝑀𝑗Φ(𝑟𝑗)𝑀𝑗+𝑐𝑗1𝑀𝑗1𝜂𝑀𝑗Φ(𝑟𝑗)𝑀𝑗1+𝛿𝑀𝑗Φ(𝑟𝑗)𝑀𝑗,(2.27) so 𝑛1𝑖=0𝑎𝑖Φ(𝑟1)𝑖=𝑌1=𝑟𝑗=1𝜂𝑀𝑗+1𝑐𝑗1𝑀𝑗Φ(𝑟𝑗)𝑀𝑗+𝑐𝑗1𝑀𝑗1𝜂𝑀𝑗Φ(𝑟𝑗)𝑀𝑗1+𝛿𝑀𝑗Φ(𝑟𝑗)𝑀𝑗.(2.28)

3. Computational Aspects and Numerical Examples

In this section, we use the previous algorithm to evaluate the known Gaussian quadrature rules for the integrals in (1.6).

Lemma 3.1. Let 𝐿𝑛1(𝑓)=𝑛1𝑖=0𝑎𝑖[𝑓]𝜋𝑖; then𝑎𝑖[𝑓]=1𝑖𝑛𝑗=1𝑤𝑗𝜋𝑖𝑥𝑗𝑓𝑥𝑗,(3.1) where 𝑤𝑖 is the weight of the classical Gaussian quadrature formula corresponding to the node 𝑥𝑖.

Proof. Since 𝑛1𝑖=0𝑎𝑖[𝑓]𝜋𝑖=𝑛𝑖=1𝑓𝑥𝑖𝜋𝑛(𝑥)𝜋𝑛𝑥𝑖𝑥𝑖𝑥,(3.2) then multiply both sides by 𝑤(𝑥)𝜋𝑗(𝑥)(0𝑗𝑛1), and integrate over [𝑎,𝑏], we derive 𝑎𝑗[𝑓]=1𝑗𝑛𝑖=1𝑓𝑥𝑖𝜋𝑛𝑥𝑖𝐼𝜋𝑛𝜋𝑗,𝑥𝑖,1,(3.3) by using the orthogonality of 𝜋𝑛, we get 𝐼(𝜋𝑛𝜋𝑗,𝑥𝑖,1)=𝜋𝑗(𝑥𝑖)𝑞𝑛(𝑥𝑖) and by knowing that 𝑤𝑖=𝑞𝑛(𝑥𝑖)/𝜋𝑛(𝑥𝑖), we obtain (3.1).

Now, since 𝜋𝑛(𝑥) satisfy (2.1) which 𝑓𝑟=0and 𝜂𝑛=𝑙𝑛, the following statement follows from Theorem 2.3.

Lemma 3.2. 𝐿𝑛1[𝑓]=𝑐0+𝑐1𝛼0+𝜋1,(3.4) where 𝑐𝑛+1=𝑐𝑛+2=0,𝑐𝑘+𝛼𝑘𝑐𝑘+1+𝑙𝑘+1𝑐𝑘+2=𝑎𝑘,𝑘=𝑛1,,0,(3.5)

𝛼𝑘=𝐴𝑘𝑡+𝐵𝑘.(3.6) Since 𝜋𝑛 satisfies recurrence relation (2.12), so we can apply Theorem 2.3 to have the following lemma.

Lemma 3.3. 𝐿𝑛1[𝑓]=𝑐11𝜋1+𝑐01𝜋1+𝑐02𝛼1𝜋1+𝜋2,(3.7) where 𝑐11, 𝑐01, 𝑐02 are given in Theorem 2.3.

Since 𝑞(𝑟1)𝑖(𝑡) obeys recurrence relation (2.12), so we can evaluate the pervious Gaussian quadrature rules introduced in Sections 1.21.4 by the following theorems.

Theorem 3.4. 𝑄strict𝑛(𝑓,𝑡,𝑟)=1(𝑟1)!𝑛1𝑖=0𝑎𝑖[𝑓]𝑞(𝑟1)𝑖(𝑡),(3.8) where 𝑛1𝑖=0𝑎𝑖[𝑓]𝑞(𝑟1)𝑖(𝑡)=𝑟𝑖=1𝑐𝑟𝑖𝑖1𝑞(𝑟𝑖)𝑖1(𝑡)+𝑐𝑟𝑖𝑖𝛼𝑖1𝑞(𝑟𝑖)𝑖1(𝑡)+𝑞(𝑟𝑖)𝑖,(3.9) where 𝑐𝑘𝑖’s are given by Theorem 2.3.

Theorem 3.5. If 𝜋𝑛(𝑡)0, then 𝑄𝑀𝑛(𝑓,𝑡,𝑟)=1(𝑟1)!𝑛1𝑖=0𝑎𝑖[𝑓]𝜌(𝑟1)𝑖(𝑡)+1(𝑟1)!𝑑𝑟1𝑑𝑡𝑟1𝑞𝑛(𝑡)𝜋𝑛(𝑡)𝑓(𝑡),(3.10) where (using Theorem 2.3) 𝑛1𝑖=0𝑎𝑖[𝑓]𝜌(𝑟1)𝑖(𝑡)=𝑟𝑖=1𝑐𝑟𝑖𝑖1𝜌(𝑟𝑖)𝑖(𝑡)+𝑐𝑟𝑖𝑖𝛼𝑖1𝜌(𝑟1)𝑖1(𝑡)+𝜌(𝑟1)𝑖(𝑡),(3.11) where 𝜌𝑖(𝑡)=𝑞𝑖(𝑡)𝑞𝑛(𝑡)𝜋𝑖(𝑡)/𝜋𝑛(𝑡), which fulfill a three-term recurrence (1.2), and 𝑐𝑘𝑖 are given by Theorem 2.3.

Proof. Using Lemma 1.8, we have 𝑄𝑀𝑛(𝑓,𝑡,1)=𝑛1𝑖=0𝑎𝑖[𝑓]𝐼𝜋𝑖,𝑡,1+𝑞𝑛(𝑡)𝜋𝑛(𝑡)𝑓(𝑡)𝐿𝑛1[𝑓](𝑡)=𝑛1𝑖=0𝑎𝑖[𝑓]𝑞𝑖(𝑡)+𝑞𝑛(𝑡)𝜋𝑛(𝑡)𝑓(𝑡)𝐿𝑛1[𝑓](𝑡)=𝑛1𝑖=0𝑎𝑖[𝑓]𝜌𝑖(𝑡)+𝑞𝑛(𝑡)𝜋𝑛(𝑡)𝑓(𝑡).(3.12) Then differentiating both sides (𝑟1)-times, we obtain relation (3.10).

Theorem 3.6. If 𝜋𝑛(𝑡)=0, then 𝑄𝑀𝑛(𝑓,𝑡,𝑟)=1(𝑟1)!𝑛1𝑖=0𝑎𝑖[𝑓]𝑞(𝑟1)𝑖(𝑡)+1(𝑟1)!𝑑𝑟1𝑑𝑡𝑟1𝑞𝑛(𝑡)𝜋𝑛(𝑡)𝑓(𝑡)𝐿𝑛1[𝑓](𝑡).(3.13)

Proof. Using Corollary 1.9, we have 𝑄𝑀𝑛(𝑓,𝑡,1)=𝑛1𝑖=0𝑎𝑖[𝑓]𝐼𝜋𝑖,𝑡,1+𝑞𝑛(𝑡)𝜋𝑛(𝑡)𝑓(𝑡)𝐿𝑛1[𝑓](𝑡)=𝑛1𝑖=0𝑎𝑖[𝑓]𝑞𝑖(𝑡)+𝑞𝑛(𝑡)𝜋𝑛(𝑡)𝑓(𝑡)𝐿𝑛1[𝑓](𝑡),(3.14) and by differentiating both sides (𝑟1)-times, we obtain relation (3.13).

A direct result of Lemma 1.10 and Theorem 2.3 is the following theorem.

Theorem 3.7. 𝑄𝐿3𝑛+1(𝑓,𝑡,𝑟)=1(𝑟1)!𝑛1𝑖=0𝑎𝑖[𝑓]𝜎(𝑟1)𝑖(𝑡)+1(𝑟1)!𝑑𝑟1𝑑𝑡𝑟1𝑓(𝑡)𝑞0(𝑡),𝑄𝐿3𝑛+1(𝑓,𝑡,𝑟)=1(𝑟1)!𝑟1𝑖=0𝑐𝑟𝑖1𝑖𝜎(𝑟𝑖1)𝑖+𝑐𝑟𝑖1𝑖+1𝛼𝑖𝜎(𝑟𝑖1)𝑖+𝜎(𝑟𝑖1)𝑖+1(𝑟1)!𝑑𝑟1𝑑𝑡𝑟1𝑓(𝑡)𝑞0(𝑡),(3.15) where 𝜎𝑖=𝑞𝑖(𝑡)𝑞0(𝑡)𝜋𝑖(𝑡) and 𝑐(𝑘)𝑖 are given by Theorem 2.3.

4. A Numerical Example

Hui* and Shia [17] consider the modified Gaussian quadrature111𝑥2𝑓(𝑥)(𝑥𝑡)2𝑑𝑥𝜋𝑓(𝑡)𝑈𝑛1(𝑡)𝑇𝑛(𝑡)𝜋𝑓(𝑡)𝑊𝑈𝑛1,𝑇𝑛𝑈𝑛1(𝑡)2+𝑛𝑖=1𝑤𝑖𝑓𝑥𝑖𝑥𝑖𝑡2,(4.1) where 𝑊(𝑈𝑛1,𝑇𝑛)=𝑈𝑛1𝑇𝑛𝑈𝑛1𝑇𝑛.

In Table 1, numerical results are obtained for𝐼(𝑓,0,2)=111𝑥2𝑒𝑥𝑥2𝑑𝑥2.3395562533389707,(4.2) by using the quadrature rule 𝑄𝑀𝑛(𝑓,0,2); in this case, 𝑤(𝑥)=1𝑥2; and 𝑥𝑖,𝑛 are the zeros of Chebyshev polynomial of the second kind 𝑈𝑛; then we have𝑥𝑖,𝑛=cos𝑖+1𝑛+1𝜋,𝑤𝑖,𝑛=𝜋𝑛+11𝑥2𝑖,𝑛,𝑞𝑛(𝑡)=𝜋𝑇𝑛+1(𝑥).(4.3) Then applying the algorithm in Theorem 3.6 for 𝑟=2, 𝑙𝑖=1, 𝛼𝑖=2𝑡, we obtain the following results comparing with these of Hui* and Shia [17].

The results indicate that the algorithm is indeed stable even when 𝑡 coincides with one of the nodes of the quadrature formula (in this case, this node is located at the origin). However, in deriving the modified method [17] there are no numerical values presented for 𝑛=5 and 7 that is why it is assumed that the roots of the orthogonal polynomials are distinct from 𝑡.

Appendix

A. Evaluation for Some Finite Part Integrals Using the Proposed Algorithm

The goal of this appendix is to provide the evaluations for some integrals using the proposed algorithm with suitable 𝑛 points

A.1. 𝑤(𝑥)=1𝑥2, 𝑟=1

1𝜋11𝑤(𝑥)𝑇0(𝑥)𝑥𝑡𝑑𝑥=𝑡,1𝜋11𝑤(𝑥)𝑇1(𝑥)𝑥𝑡𝑑𝑥=12𝑡2,1𝜋11𝑤(𝑥)𝑇2(𝑥)𝑥𝑡𝑑𝑥=2𝑡2𝑡3,1𝜋11𝑤(𝑥)𝑇3(𝑥)𝑥𝑡𝑑𝑥=1+5𝑡24𝑡4,1𝜋11𝑤(𝑥)𝑇4(𝑥)𝑥𝑡𝑑𝑥=4𝑡+12𝑡38𝑡5,1𝜋11𝑤(𝑥)𝑇5(𝑥)𝑥𝑡𝑑𝑥=113𝑡2+28𝑡416𝑡6,1𝜋11𝑤(𝑥)𝑇6(𝑥)𝑥𝑡𝑑𝑥=6𝑡38𝑡3+64𝑡532𝑡7,1𝜋11𝑤(𝑥)𝑇7(𝑥)𝑥𝑡𝑑𝑥=1+25𝑡2104𝑡4+144𝑡664𝑡8,1𝜋11𝑤(𝑥)𝑇8(𝑥)𝑥𝑡𝑑𝑥=8𝑡+88𝑡3272𝑡5+320𝑡7128𝑡9,(A.1)

A.2. 𝑤(𝑥)=(1𝑥2)3/2, 𝑟=1

1𝜋11𝑤(𝑥)𝑇0(𝑥)𝑥𝑡𝑑𝑥=𝑡332𝑡,1𝜋11𝑤(𝑥)𝑇1(𝑥)𝑥𝑡𝑑𝑥=3832𝑡2+𝑡4,1𝜋11𝑤(𝑥)𝑇2(𝑥)𝑥𝑡𝑑𝑥=94𝑡4𝑡3+2𝑡5,1𝜋11𝑤(𝑥)𝑇3(𝑥)𝑥𝑡𝑑𝑥=78+6𝑡29𝑡4+4𝑡6,1𝜋11𝑤(𝑥)𝑇4(𝑥)𝑥𝑡𝑑𝑥=4𝑡+16𝑡320𝑡5+8𝑡7,1𝜋11𝑤(𝑥)𝑇5(𝑥)𝑥𝑡𝑑𝑥=114𝑡2+41𝑡444𝑡6+16𝑡8,1𝜋11𝑤(𝑥)𝑇6(𝑥)𝑥𝑡𝑑𝑥=6𝑡44𝑡3+102𝑡596𝑡7+32𝑡9,1𝜋11𝑤(𝑥)𝑇7(𝑥)𝑥𝑡𝑑𝑥=1+26𝑡2129𝑡4+248𝑡6208𝑡8+64𝑡10,1𝜋11𝑤(𝑥)𝑇8(𝑥)𝑥𝑡𝑑𝑥=8𝑡+96𝑡3360𝑡5+592𝑡7448𝑡9+128𝑡11,(A.2)

A.3. 𝑤(𝑥)=(1𝑥2)5/2, 𝑟=1

(a)
1𝜋11𝑤(𝑥)𝑇0(𝑥)𝑥𝑡𝑑𝑥=158𝑡+52𝑡3𝑡5,1𝜋11𝑤(𝑥)𝑇1(𝑥)𝑥𝑡𝑑𝑥=516158𝑡2+52𝑡4𝑡6,1𝜋11𝑤(𝑥)𝑇2(𝑥)𝑥𝑡𝑑𝑥=512𝑡254𝑡3+6𝑡52𝑡7,1𝜋11𝑤(𝑥)𝑇3(𝑥)𝑥𝑡𝑑𝑥=2532+558𝑡215𝑡4+13𝑡64𝑡8,1𝜋11𝑤(𝑥)𝑇4(𝑥)𝑥𝑡𝑑𝑥=6516𝑡+20𝑡336𝑡5+28𝑡78𝑡9,1𝜋11𝑤(𝑥)𝑇5(𝑥)𝑥𝑡𝑑𝑥=313215𝑡2+55𝑡485𝑡6+60𝑡816𝑡10,1𝜋11𝑤(𝑥)𝑇6(𝑥)𝑥𝑡𝑑𝑥=6𝑡50𝑡3+146𝑡5198𝑡7+128𝑡932𝑡11,1𝜋11𝑤(𝑥)𝑇7(𝑥)𝑥𝑡𝑑𝑥=1+27𝑡2155𝑡4+377𝑡6456𝑡8+272𝑡1064𝑡12,1𝜋11𝑤(𝑥)𝑇8(𝑥)𝑥𝑡𝑑𝑥=8𝑡+104𝑡3456𝑡5+952𝑡71040𝑡9+576𝑡11128𝑡13,(A.3)

(b)
1𝜋11𝑤(𝑥)𝑈0(𝑥)𝑥𝑡𝑑𝑥=158𝑡+52𝑡3𝑡5,1𝜋11𝑤(𝑥)𝑈1(𝑥)𝑥𝑡𝑑𝑥=58154𝑡2+5𝑡42𝑡6,1𝜋11𝑤(𝑥)𝑈2(𝑥)𝑥𝑡𝑑𝑥=258𝑡10𝑡3+11𝑡54𝑡7,1𝜋11𝑤(𝑥)𝑈3(𝑥)𝑥𝑡𝑑𝑥=1516+10𝑡225𝑡4+24𝑡68𝑡8,1𝜋11𝑤(𝑥)𝑈4(𝑥)𝑥𝑡𝑑𝑥=5𝑡+30𝑡361𝑡5+52𝑡716𝑡9,1𝜋11𝑤(𝑥)𝑈5(𝑥)𝑥𝑡𝑑𝑥=120𝑡2+85𝑡4146𝑡6+112𝑡832𝑡10,1𝜋11𝑤(𝑥)𝑈6(𝑥)𝑥𝑡𝑑𝑥=7𝑡70𝑡3+231𝑡5344𝑡7+240𝑡964𝑡11,1𝜋11𝑤(𝑥)𝑈7(𝑥)𝑥𝑡𝑑𝑥=1+34𝑡2225𝑡4+608𝑡6800𝑡8+512𝑡10128𝑡12,1𝜋11𝑤(𝑥)𝑈8(𝑥)𝑥𝑡𝑑𝑥=9𝑡+138𝑡3681𝑡5+1560𝑡71840𝑡9+1088𝑡11256𝑡13,(A.4)

A.4. 𝑤(𝑥)=(1𝑥2)5/2, 𝑟=2

(a)
1𝜋11𝑤(𝑥)𝑇0(𝑥)(𝑥𝑡)2𝑑𝑥=158+152𝑡25𝑡4,1𝜋11𝑤(𝑥)𝑇1(𝑥)(𝑥𝑡)2𝑑𝑥=154𝑡+10𝑡36𝑡5,1𝜋11𝑤(𝑥)𝑇2(𝑥)(𝑥𝑡)2𝑑𝑥=512754𝑡2+30𝑡414𝑡6,1𝜋11𝑤(𝑥)𝑇3(𝑥)(𝑥𝑡)2𝑑𝑥=554𝑡60𝑡3+78𝑡532𝑡7,1𝜋11𝑤(𝑥)𝑇4(𝑥)(𝑥𝑡)2𝑑𝑥=6516+60𝑡2180𝑡4+196𝑡672𝑡8,1𝜋11𝑤(𝑥)𝑇5(𝑥)(𝑥𝑡)2𝑑𝑥=30𝑡+220𝑡3510𝑡5+480𝑡7160𝑡9,1𝜋11𝑤(𝑥)𝑇6(𝑥)(𝑥𝑡)2𝑑𝑥=6150𝑡2+730𝑡41386𝑡6+1152𝑡8352𝑡10,1𝜋11𝑤(𝑥)𝑇7(𝑥)(𝑥𝑡)2𝑑𝑥=54𝑡620𝑡3+2262𝑡53648𝑡7+2720𝑡9768𝑡11,1𝜋11𝑤(𝑥)𝑇8(𝑥)(𝑥𝑡)2𝑑𝑥=8+312𝑡22280𝑡4+6664𝑡69360𝑡8+6336𝑡101664𝑡12,(A.5)

(b)
1𝜋11𝑤(𝑥)𝑈0(𝑥)(𝑥𝑡)2𝑑𝑥=158+152𝑡25𝑡4,1𝜋11𝑤(𝑥)𝑈1(𝑥)(𝑥𝑡)2𝑑𝑥=152𝑡+20𝑡312𝑡5,1𝜋11𝑤(𝑥)𝑈2(𝑥)(𝑥𝑡)2𝑑𝑥=25830𝑡2+55𝑡428𝑡6,1𝜋11𝑤(𝑥)𝑈3(𝑥)(𝑥𝑡)2𝑑𝑥=20𝑡100𝑡3+144𝑡564𝑡7,1𝜋11𝑤(𝑥)𝑈4(𝑥)(𝑥𝑡)2𝑑𝑥=5+90𝑡2305𝑡4+364𝑡6144𝑡8,1𝜋11𝑤(𝑥)𝑈5(𝑥)(𝑥𝑡)2𝑑𝑥=40𝑡+340𝑡3876𝑡5+896𝑡7320𝑡9,1𝜋11𝑤(𝑥)𝑈6(𝑥)(𝑥𝑡)2𝑑𝑥=7210𝑡2+1155𝑡42408𝑡6+2160𝑡8704𝑡10,1𝜋11𝑤(𝑥)𝑈7(𝑥)(𝑥𝑡)2𝑑𝑥=68𝑡900𝑡3+3648𝑡56400𝑡7+5120𝑡91536𝑡11,1𝜋11𝑤(𝑥)𝑈8(𝑥)(𝑥𝑡)2𝑑𝑥=9+414𝑡23405𝑡4+10920𝑡616560𝑡8+11968𝑡103328𝑡12,(A.6)

A.5. 𝑤(𝑥)=(1𝑥2)5/2, 𝑟=3

(a)
1𝜋11𝑤(𝑥)𝑇0(𝑥)(𝑥𝑡)3𝑑𝑥=152𝑡10𝑡3,1𝜋11𝑤(𝑥)𝑇1(𝑥)(𝑥𝑡)3𝑑𝑥=158+15𝑡215𝑡4,1𝜋11𝑤(𝑥)𝑇2(𝑥)(𝑥𝑡)3𝑑𝑥=754𝑡+60𝑡342𝑡5,1𝜋11𝑤(𝑥)𝑇3(𝑥)(𝑥𝑡)3𝑑𝑥=55890𝑡2+195𝑡4112𝑡6,1𝜋11𝑤(𝑥)𝑇4(𝑥)(𝑥𝑡)3𝑑𝑥=60𝑡360𝑡3+588𝑡5288𝑡7,1𝜋11𝑤(𝑥)𝑇5(𝑥)(𝑥𝑡)3𝑑𝑥=15+330𝑡21275𝑡4+1680𝑡6720𝑡8,1𝜋11𝑤(𝑥)𝑇6(𝑥)(𝑥𝑡)3𝑑𝑥=150𝑡+1460𝑡34158𝑡5+4608𝑡71760𝑡9,1𝜋11𝑤(𝑥)𝑇7(𝑥)(𝑥𝑡)3𝑑𝑥=27930𝑡2+5655𝑡412768𝑡6+12240𝑡84224𝑡10,1𝜋11𝑤(𝑥)𝑇8(𝑥)(𝑥𝑡)3𝑑𝑥=312𝑡4560𝑡3+19992𝑡537440𝑡7+31680𝑡99984𝑡11,(A.7)

(b)
1𝜋11𝑤(𝑥)𝑈0(𝑥)(𝑥𝑡)3𝑑𝑥=152𝑡10𝑡3,1𝜋11𝑤(𝑥)𝑈1(𝑥)(𝑥𝑡)3𝑑𝑥=154+30𝑡230𝑡4,1𝜋11𝑤(𝑥)𝑈2(𝑥)(𝑥𝑡)3𝑑𝑥=30𝑡+110𝑡384𝑡5,1𝜋11𝑤(𝑥)𝑈3(𝑥)(𝑥𝑡)3𝑑𝑥=10150𝑡2+360𝑡4224𝑡6,1𝜋11𝑤(𝑥)𝑈4(𝑥)(𝑥𝑡)3𝑑𝑥=90𝑡610𝑡3+1092𝑡5576𝑡7,1𝜋11𝑤(𝑥)𝑈5(𝑥)(𝑥𝑡)3𝑑𝑥=20+510𝑡22190𝑡4+3136𝑡61440𝑡8,1𝜋11𝑤(𝑥)𝑈6(𝑥)(𝑥𝑡)3𝑑𝑥=210𝑡+2310𝑡37224𝑡5+8640𝑡73520𝑡9,1𝜋11𝑤(𝑥)𝑈7(𝑥)(𝑥𝑡)3𝑑𝑥=341350𝑡2+9120𝑡422400𝑡6+23040𝑡88448𝑡10,1𝜋11𝑤(𝑥)𝑈8(𝑥)(𝑥𝑡)3𝑑𝑥=414𝑡6810𝑡3+32760𝑡566240𝑡7+59840𝑡919968𝑡11,(A.8)

A.6. 𝑤(𝑥)=(1𝑥2)5/2, 𝑟=4

(a)
1𝜋11𝑤(𝑥)𝑇0(𝑥)(𝑥𝑡)4𝑑𝑥=5210𝑡2,1𝜋11𝑤(𝑥)𝑇1(𝑥)(𝑥𝑡)4𝑑𝑥=10𝑡20𝑡3,1𝜋11𝑤(𝑥)𝑇2(𝑥)(𝑥𝑡)4𝑑𝑥=254+60𝑡270𝑡4,1𝜋11𝑤(𝑥)𝑇3(𝑥)(𝑥𝑡)4𝑑𝑥=60𝑡+260𝑡3224𝑡5,1𝜋11𝑤(𝑥)𝑇4(𝑥)(𝑥𝑡)3𝑑𝑥=20360𝑡2+980𝑡4672𝑡6,1𝜋11𝑤(𝑥)𝑇5(𝑥)(𝑥𝑡)4𝑑𝑥=220𝑡1700𝑡3+3360𝑡51920𝑡7,1𝜋11𝑤(𝑥)𝑇6(𝑥)(𝑥𝑡)4𝑑𝑥=50+1460𝑡26930𝑡4+10752𝑡65280𝑡8,1𝜋11𝑤(𝑥)𝑇7(𝑥)(𝑥𝑡)4𝑑𝑥=620𝑡+7540𝑡325536𝑡5+32640𝑡714080𝑡9,1𝜋11𝑤(𝑥)𝑇8(𝑥)(𝑥𝑡)4𝑑𝑥=1044560𝑡2+33320𝑡487360𝑡6+95040𝑡836608𝑡10,(A.9)

(b)
1𝜋11𝑤(𝑥)𝑈0(𝑥)(𝑥𝑡)4𝑑𝑥=5210𝑡2,1𝜋11𝑤(𝑥)𝑈1(𝑥)(𝑥𝑡)4𝑑𝑥=20𝑡40𝑡3,1𝜋11𝑤(𝑥)𝑈2(𝑥)(𝑥𝑡)4𝑑𝑥=10+110𝑡2140𝑡4,1𝜋11𝑤(𝑥)𝑈3(𝑥)(𝑥𝑡)4𝑑𝑥=100𝑡+480𝑡3448𝑡5,1𝜋11𝑤(𝑥)𝑈4(𝑥)(𝑥𝑡)4𝑑𝑥=30610𝑡2+1820𝑡41344𝑡6,1𝜋11𝑤(𝑥)𝑈5(𝑥)(𝑥𝑡)4𝑑𝑥=340𝑡2920𝑡3+6272𝑡53840𝑡7,1𝜋11𝑤(𝑥)𝑈6(𝑥)(𝑥𝑡)4𝑑𝑥=70+2310𝑡212040𝑡4+20160𝑡610560𝑡8,1𝜋11𝑤(𝑥)𝑈7(𝑥)(𝑥𝑡)4𝑑𝑥=900𝑡+12160𝑡344800𝑡5+61440𝑡728160𝑡9,1𝜋11𝑤(𝑥)𝑈8(𝑥)(𝑥𝑡)4𝑑𝑥=1386810𝑡2+54600𝑡4154560𝑡6+179520𝑡873216𝑡10.(A.10)