Numerical Integration for Multivariable Functions with Point Singularities Yaun Yang and Kendall E. Atkinson y October 199 Abstract We consider the numerical integration of functions with point singularities over a planar wedge S using isoparametric piecewise polynomial interpolation of the function and the wedge. Such integrals often occur in solving boundary integral equations using the collocation method. In order to obtain the same order of convergence as is true with uniform meshes for smooth functions, we introduce an adaptive renement of the triangulation of S. Error analyses and several examples are given for a certain type of adaptive renement. Key Words: numerical integration, quadratic interpolation, adaptive renement. AMS(MOS) Subect Classication: 65D30, 65D3 Department of Mathematics, The University of Iowa, Iowa City, IA 54 y Department of Mathematics, The University of Iowa, Iowa City, IA 54 1
1 Introduction We consider the problem of approximating an integral of the form S f(x; y)ds where S is a closed bounded connected set in R. When the integrand has singularities within the integration region, the use of a standard quadrature method may be very inef- cient. There are several ways to deal with this problem. One approach is to use a change of variables [10] to transform a singular integrand into a well behaved function over a new region. A second approach is to use adaptive numerical integration, to place more node points near where the integrand is badly behaved in order to improve the performance of a standard quadrature method. A third approach is to use extrapolation methods to construct new and more accurate integration formulas based on the asymptotic expansion for the quadrature error in the original quadrature rule. The generalization of the classical Euler-Maclaurin expansion to functions having a particular type of singularity, as obtained by Lyness [8], [9], provides a basis for extrapolation methods in the same way as the Euler-Maclaurin expansion is used as a basis for Romberg integration. For the one dimensional case, a standard discussion of these methods and others can be found in Atkinson [3]. In this paper we discuss adaptive numerical integration for the two dimensional case. The process of placing node points with variable spacing so as to better reect the integrand is called adaptive renement or grading the mesh. We propose a type of adaptive renement for which the order of convergence is the same as for smooth functions, but with the integrand function having a particular type of singularity, specied below. In the case of smooth integrand functions, Chien [6],[7] obtained that for integrals over a piecewise smooth surface in R 3, the numerical integration using isoparametric piecewise quadratic interpolation for both the surface and the integrand leads to the order of convergence O( 4 ). In this, = max 1KN ( K ); K = max p;q4k p? q
and f4 K K = 1; ; Ng is a quasi-uniform triangulation of the surface. In terms of the number N of triangles, the order of convergence is O(1=N ). We extend these results to singular integrands. To simplify our discussion, we study only the following class of problems. The integration region S is a wedge, i.e. S = f(x; y) R 0 r 1; 0 g ; 0 < < : The integration function is singular at only the origin, and it is of the form f(x; y) = r '()h(r)g(x; y); >? (1:1) or the form f(x; y) = r ln r '()h(r)g(x; y); >? (1:) Here (x; y) are the Cartesian coordinates with the corresponding polar coordinates (r; ). The functions '(); h(r); g(x; y) are assumed to be suciently smooth. This is also the class of functions considered in Lyness [8]. In section, we describe the triangulation of the wedge S and the adaptive renement scheme we use. The interpolation-based quadrature formulas are given in section 3. Section 4 contains the error analyses and a discussion of the results. Numerical examples are given in section 5. This paper presents detailed results for only the use of quadratic interpolation. The method being used generalizes to other degrees of piecewise polynomial interpolation, and the results are consistent with the kind of results we have obtained for the quadratic case. This generalization for other degrees of interpolation is given in section 6. Because integrals of non-smooth functions often occur in solving boundary integral equations using the numerical method, we expect that the error analysis of the present numerical integration methods eventually will lead to better numerical methods for the solution of boundary integral equations. 3
The triangulation and adaptive renement In this section, we describe the triangulation of the wedge S to a ner mesh. Let and discuss its renement n = f4 K;n 1 K N n g (:1) be the triangulation mesh for a sequence n = 1; ; : : :. In referring to the element 4 K;n, the reference to n will be omitted, but understood implicitly. The initial triangulation 1 of S is obtained by connecting the midpoints of the sides of S using straight lines. The sequence of triangulations n of (.1) will be obtained by successive adaptive and uniform renements based on the initial triangulation. We construct this sequence as follows, and we call it an `La + u' renement with L a positive integer. Given f4 K;n 1 K N n g, we divide the triangle containing the origin into four new triangular elements. For the resulting triangulation, repeat the preceding subdivision. After doing this L times, we divide simultaneously every triangle into four new triangles. The nal triangulation is denoted by f4 K;n+1 1 K N n+1 g. In other words, at level n, we perform L times an adaptive subdivision on the triangle containing the origin, and then we do one simultaneous subdivision of all triangles. An advantage of this form of renement is that each set of mesh points contains those at the preceding level. As an example, we illustrate the `a + u' renement for n = 1, in Figures 1,. When n =, there are three dierent triangular elements in size. Let = dene sin, and 1?cos B 0 B 1 B = f(x; y) S x + y =g; = f(x; y) S =4 x + y =g; = f(x; y) S 0 x + y =4g: The set B l is the union of the triangles of the same size in S, and therefore it is uniformly divided by the triangulation. The diameter of triangles in B l is O(?(+l) ). 4
Figure 1: n = 1 Figure : n = 5
Moreover, functions in (1.1) and (1.) are smooth on B 0 [ B 1. More generally, by examining the structure of the `La + u' renement, we can calculate the total number N n of triangles at level n: this is (L + 1) n? 4L = O( n ). There are Ln? (L? 1) dierent triangular elements in size. The closer the triangle is to the origin, the smaller it is. As the triangles vary in size from large to small, we name the region containing triangles of the same size to be The diameter of triangles in B l, denoted by l, is O(?(n+l) ). Let N l B 0 ; B 1 ; : : : ; B Ln?L, respectively. be the number of triangles in B l. Then N l is proportional to 4 n?i where l = il+i 1 for 0 i 1 L?1. The distance from the origin to B l, denoted by r l, is O( 1 (L+1)i ). In each B l (l = 1; : : : ; Ln?L), the triangular elements 4 K are true triangles and all are congruent. The triangular elements in B 0 are nearly congruent. More importantly, any symmetric pair of triangles, as shown in Figure 3, have the following property: v 1? v =?(v 1? v 4 ); v 1? v 3 =?(v 1? v 5 ): The total number of symmetric pairs of triangles in B l triangles in B l is O( p N l ). is O(N l ) and the remaining If let L = 0, then the renement is quasi-uniform. The analysis given in [6] indicates that a quasi-uniform renement is a better scheme to use with smooth integrands. 3 Interpolation Let denote the unit simplex in the st? plane = f(s; t) 0 s; t; s + t 1g ; Let 1 ; : : : ; 6 according to Figure 4. denote the three vertices and three midpoints of the sides of, numbered 6
v3 v q q q v1 q q v4 v5 Figure 3: A symmetric pair of triangles t 6 r 4 r r 5 r r r 1 6 3 - s Figure 4: The unit simplex 7
To dene interpolation, introduce the basis functions for quadratic interpolation on. Letting u = 1? (s + t), dene l 1 (s; t) = u(u? 1); l (s; t) = t(t? 1); l 3 (s; t) = s(s? 1); l 4 (s; t) = 4tu; l 5 (s; t) = 4st; l 6 (s; t) = 4su: We give the corresponding set of basis functions fl ;K (q)g on 4 K by using its parametrization over. As a special case of piecewise smooth surfaces in R 3, discussed in [5]-[7], there is a mapping 1{1 m K : - onto 4 K (3:1) with m K C 6 (). Introduce the node points for 4 K by v ;K = m K ( ); = 1; ; 6. The rst three are the vertices and the last three are the midpoints of the sides of 4 K. Dene l ;K (m K (s; t)) = l (s; t) ; 1 6; 1 K N: Given a function f, dene P N f(q) = =1 f(v ;K )l ;K (q) ; q 4 K ; for K = 1; : : : ; N. This is called the piecewise quadratic isoparametric function interpolating f on the nodes of the mesh f4 K g for S. 4 Numerical integration and error analyses 1{1 With the triangulation f4 K g and the mapping m K : - onto 4 K ; we have 4 K f(q)ds = f(m K (s; t)) D s m K (s; t) D t m K (s; t) dsdt (4:1) D s and D t denote dierentiation with respect to s; t, respectively. The quantity D s m K (s; t) D t m K (s; t) is the Jacobian determinant of the mapping m K (s; t) used in transforming surface integrals over 4 K into integrals over. When 4 K is a triangle, the Jacobian is twice the area of 4 K. 8
The numerical integration formula used here is g(s; t)dsdt 1 6 [ g( 4) + g( 5 ) + g( 6 ) ] (4:) which is based on integrating the quadratic polynomial interpolating g on at 1 ; : : : ; 6. This integration has degree of precision. Applying (4.) to the right side of (4.1), we have 4 K f(q)ds 1 6 f(m K ( )) D s m K (s; t) D t m K (s; t) (4:3) A maor problem with (4.3) is that D s m K and D t m K are inconvenient to compute for some elements 4 K terms of only its values at 1 ; : : : ; 6. Dene on many surfaces S. Therefore, we approximate m K (s; t) in fm K (s; t) = =1 m K ( )l (s; t) = =1 v ;K l (s; t): For the case of the wedge of a circle, in this paper, only the outer triangles will be aected by this approximation. Then where 4 K f(q)ds 1 6 = f(m K ( )) D s fm K (s; t) D t fm K (s; t)! ;K f(v ;K )! ;K = 1 6 D sfm K (s; t) D t fm K (s; t) : LEMMA 1. Let f4 K g be the irregular triangulation dened by the `La+u' renement, and let N be its total number of triangles. Assume that f is of form (1.1). Then B Ln?L X 4 K B Ln?L! ;K f(v ;K ) O( 1 ) (4:4) N p 1 where p 1 = (L+1)(+) : Proof. Let Ln?L = 1 n+(ln?l) : 9
The partition of B Ln?L is shown in Figure 4. Let f4 K K = 1; : : : ; 16g denote the sixteen triangles in B Ln?L, and let 4 1 contain the origin. Then the area of 4 K is Ln?L sin(=). y 6 LL?? L L L L??? L?? L L L L L??? L??? L? L L L L L L L??? L? L??? L??? L L L L L L L L L??? L??? L??? L??? L - x 0 Ln?L Ln?L 3Ln?L 4Ln?L Figure 5: The partition of B Ln?L Dene (a) We rst estimate the error of 4 1! ;1 f(v ;1 ) : g 1 (x; y) = '()h(r)g(x; y); and note that g 1 (x; y) is integrable over 4 1. Since r 0, by the integral mean value theorem we have where 4 1 f(q)ds = 4 1 r ds inf g 1 (x; y) sup g 1 (x; y): 4 1 4 1 10
Also, is bounded: sup S sup S = max 0 M < 1 g 1 (x; y) '() sup S '() max 0r1 h(r) sup S h(r) max S g(x; y) g(x; y) Therefore, 4 1 f(q)ds M = M Ln?L 4 1 r ds 0 = M + + Ln?L: 0 r +1 ddr On the other hand,! ;1 f(v ;1 ) 1 6 + Ln?L sin (1? + cos(=) ) M This inequality comes from the following: and! ;1 = 1 6 D sfm 1 (s; t) D t fm 1 (s; t) : = 1 6 (Area of 4 1) = 1 6 + Ln?L sin ; for = 4; 5; 6 f(v ;1 ) v ;1 M = 8 >< >: ( Ln?L ) M if = 4; 6 Ln?L cos(=) M if = 5 Hence, 4 1! ;1 f(v ;1 ) O( + Ln?L): 11
1 Notice that Ln?L = O( ). It follows that the error over 4 N L+1 1 is O( l ). N p 1 (b) The error over 4 K (K = ; : : : ; 16) can be obtained by using Taylor's error 1{1 formula. Since f is smooth on 4 K and m K : - onto 4 K is also smooth, we have where H f;k (s; t; ; ) = 1 3! f(m K (s; t))? =1 4 (s s + t t )3 f(m K (; ))? f(m K ( ))l (s; t) = H f;k (s; t; ; ) (4:5) =1 (s s + t t )3 f(m K ( ; ))l (s; t) 5 : In this, = (s ; t ), (; ) is on the line segment from (0,0) to (s; t), and ( ; ) is on the line segment from (0,0) to (s ; t ). Notice that (; ) and ( ; ) belong to. For (s; t) ; m K (s; t) 4 K, and 3 0 < cos( ) Ln?L r(m K (s; t)) 4 Ln?L where r(m K (s; t)) = m K (s; t) ; the distance from the point m K (s; t) to (0,0). We would like to examine one term in H f;k, that associated with s 3, and it will show the general behavior of H f;k. 3 s f(m K(; )) = 3 3 s 3 (r (m K (s; t)) g 1 (m K (s; t))) = g 1 (m K (; )) 3 s 3 (r (m K (s; t))) s = t = s = t = (4.6) + 3 ( s g 1(m K (s; t)) s (r (m K (s; t)))) s = t = ( + 3 s g 1(m K (s; t)) s s (r (m K (s; t)))) = + r (m K (; )) 3 s 3g 1(m K (s; t)) s = t = t = (4.7) (4.8) (4.9) 1
The magnitudes of (4.6), : : :, (4.9) vary from O( Ln?L ) to O(+3 ), and Ln?L O( ) Ln?L is the dominant term in (4.6), : : :, (4.9). It follows that the coecient of s 3 is of O( Ln?L). Consequently, in H f;k for any (s; t). Therefore, H f;k (s; t; ; ) O( Ln?L) 4 K O( Ln?L ) O( Ln?L) O( + Ln?L):! ;K f(v ;K ) f(m K (s; t))? =1 H f;k (s; t; ; ) ds f(m K ( ))l (s; t) ds Hence, (4.4) holds. LEMMA. Under the assumptions in Lemma 1, let = + +?. Then L Bl X 4 K B l! ;K f(v ;K ) 8 >< >: O(?4n?l ) if < 3 O(?4n? l L?5l ) if 3 (4:10) for l = 1; : : : ; Ln? L? 1. Proof. There are two types of triangles in B l. Those triangles that are part of symmetric pairs of triangles (cf. Figure 3) are of the rst type and remaining triangles are of the second type. By analysis of the `La + u' renement in section, the number of triangles of the rst type is O(N l ), where N l is the number of triangles in B l. Then N l is proportional to 4 n?i, where l is decomposed as l = il + i 1 for 0 i 1 L? 1. Since f is smooth on B l, then by Taylor's error formula we have f(m K (s; t))? =1 f(m K ( ))l (s; t) = H f;k (s; t) + G f;k (s; t; ; ) (4:11) 13
where H f;k (s; t) = 1 3! G f;k (s; t; ; ) = 1 4! 4 (s s + t t )3 f(m K (0; 0))? =1 4 (s s + t t )4 f(m K (; ))? We examine (4.11) and we can nd the following. (s s + t t )3 f(m K (0; 0))l (s; t) 5 ; =1 (s s + t t )4 f(m K ( ; ))l (s; t) 5 : First, H f;k (s; t) is a polynomial of degree three. Second, the coecients of H f;k (s; t) are combinations of (v ;K? v 1;K ) and (v 3;K? v 1;K ). For instance, the coecient of s 3 is 1 3 3! s 3f(m K(0; 0)) 3 3 = 1 3! + 3 " 3 x 3f(m K( 1 ))(v 3;x? v 1;x ) 3 + 3 3 x y f(m K( 1 ))(v 3;x? v 1;x ) (v 3;y? v 1;y ) # 3 x y f(m K( 1 ))(v 3;x? v 1;x )(v 3;y? v 1;y ) + 3 y 3f(m K( 1 ))(v 3;y? v 1;y ) 3 ) ; (4.1) where v ;K = (v ;x ; v ;y ) for = 1; ; 3: For every symmetric pair of triangles, say 4 1 and 4, in B l (see gure 3), let m 1 (s; t) = (v 3? v 1 )s + (v? v 1 )t + v 1 m (s; t) = (v 5? v 1 )s + (v 4? v 1 )t + v 1 v = (v ;x ; v ;y ): Then v 1? v =?(v 1? v 4 ) v 1? v 3 =?(v 1? v 5 ) : We now have for H f;1 and H f;, that the coecient of s 3 in H f;1 is " 1 3 3 3! x 3f(m 1( 1 ))(v 3;x? v 1;x ) 3 + 3 x y f(m 1( 1 ))(v 3;x? v 1;x ) (v 3;y? v 1;y ) 14
+ 3 3 x y f(m 1( 1 ))(v 3;x? v 1;x )(v 3;y? v 1;y ) + 3 y 3f(m 1( 1 ))(v 3;y? v 1;y ) 3 # ; (4.13) and the coecient of s 3 in H f; is 1 3! + 3 " 3 x 3f(m ( 1 ))(v 5;x? v 1;x ) 3 + 3 3 x y f(m ( 1 ))(v 5;x? v 1;x ) (v 5;y? v 1;y ) 3 x y f(m ( 1 ))(v 5;x? v 1;x )(v 5;y? v 1;y ) + 3 y 3f(m ( 1 ))(v 5;y? v 1;y ) 3 # : (4.14) Adding (4.13) and (4.14) gives us zero, and an analogous argument holds for the remaining coecients. This means that cancellation happens on any symmetric pair of triangles. It follows that 4 1 [4 X K=1! ;K f(v ;K ) O( l ) X K=1 G f;k (s; t; ; ) dsdt where l is the diameter of 4 1 and 4. This argument is based on Chien [6]. We now bound G f;k (s; t; ; ) on 4 K B l, for K = 1;. Since 1 r(m K (s; t)) r l > 0 for (s; t), and r l = O(?(L+1)i ), we have for < 4, G f;k (s; t; ; ) O( 4 l )fo(r(m K (; ))?4 ) + O( 4 l )O(r?4 l ) O( 4 l )O((4?)(L+1)i ): =1 O(r(m K ( ; ))?4 )g The rst inequality arises from bounding the individual terms of on the kind of expansion done in (4.1). Otherwise, for 4, 4 f ; a + b = 4, based a s b t G f;k (s; t; ; ) O( 4 l ): So, 4 1 [4! ;K f(v ;K ) 8 >< >: O( 6 l )O( (4?)(L+1)i ) if < 4 O( 6 l ) otherwise 15
The error contributed by triangles of the rst type is as follows. If < 4, X 4 K of f irst type For 4, ( X 4 K of f irst type 4 K ( 4 K! ;K f(v ;K ) ) O(N l )O( 6 l )O( (4?)(L+1)i ) O( n?i 1 6n+6l )O((4?)(L+1)i ) O(?4n?l ): (4.15)! ;K f(v ;K ) ) O(N l )O( 6 l )?4n? l O( L?6l ): (4.16) For the remaining triangles in B l, those of the second type, use (4.5), the fact that r l r(m K (s; t)) 1 for 4 K B l, and that their number is O( p N l ). Then H f;k (s; t; ; ) 8 >< >: O( 3 l )O(r?3 l ) if < 3 O( 3 l ) otherwise and X 4 K of second type ( 4 K! ;K f(v ;K ) ) 8 >< >: O(?4n?l ) if < 3 O(?4n? l L?5l ) otherwise (4:17) The total error over B l is given by (4.10). LEMMA3. Under the assumptions in Lemma 1, B0 X! ;K f(v ;K ) O( 1 ): (4:18) N 4 K B 0 Proof. The function f is smooth on B 0, and B 0 is uniformly divided by triangular elements. By the results in [6], (4.18) follows. Combining the above lemmas, we get the following result, which gives the total error of integrating over S. THEOREM 1. Let f S be of the form (1.1). Then NX K=1! ;K f(v ;K ) 16 8 >< >: O( lnn N ) if p 1 = O( 1 N p ) otherwise
where N is the total number of triangles in the triangulation, p 1 = (+)(L+1), and p = minfp 1 ; g. Proof. We rst add all errors contributed by each 4 K B 1 [ [ B Ln?L?1. For < 3, if p 1 6= (i.e. 6= 0), while Ln?L?1 X l=1 Ln?L?1 X l=1 X Ln?L?1 Bl X 4 K B l O(?4n?l ) l=1! = O?4n???(Ln?L) 1?? O(?pn ) = O( 1 N p ) Bl X 4 K B l! ;K f(v ;K )! ;K f(v ;K ) O( n ) O(lnN ) for = 0. n N When 3, is nonzero. By a similar argument, Ln?L?1 X Bl X! ;K f(v ;K ) O( 1 ) for 3. p N l=1 4 K B l Combining with Lemma 1 and Lemma 3, we have S NX K=1 B Ln?L Ln?L?1 X + l=1! ;K f(v ;K ) X 4 K B Ln?L Bl X + B0 X 4 K B 0 4 K B l O( 1 ) + O( 1 N p 1 N ) + O( 1 p N ) O( 1 ) for 6= 0: N p 17! ;K f(v ;K )! ;K f(v ;K )! ;K f(v ;K )
And it is minfp 1 ; g. O( lnn N ) for = 0. The quantity p 1 is dened in Lemma 1, and p = 4 COROLLARY Let L be any positive integer greater than? 1. Then the error for + evaluating the integral over S is O(1=N ). In particular, a `1a + u' renement gives an error of O(1=N ), for any > 0. THEOREM. Let f where N S be of the form (1.). Then NX K=1! ;K f(v ;K ) O( lnn ) + O( 1 N p 1 N ) is the total number of triangles in the triangulation, p 1 = (+)(L+1). Proof. The proof is analogous to that of Theorem 1. COROLLARY Let L be any positive integer greater than 4? 1. Then the error for + evaluating the integral over S is O( 1 N ). For > 0, a `1a + u' renement still gives an error of O(1=N ). 5 Numerical examples We give numerical examples using the method analyzed in section 4. The method was implemented with a package of programs written by K. Atkinson, which is described in [1], [4]. All examples were computed on a Hewlett-Packard workstation in double precision arithmetic. EXAMPLE 1. Let S = f(x; y) R 0 r 1; 0 =g, f(x; y) = r ; >?: (5:1) The results are given in Table 1 for = 0:1; 0:5 with L = 1. The column labeled Order gives the value p n = ln E n=e n+1 ln(n n+1 =N n ) 18
where E n in the error at level n. Since the theoretical result shows that the error is O(1=N ), we expect that p n will converge to. Table gives the results for =?1 with L = 1 and L = 3. The empirical orders of convergence p n approach 1 and, respectively, as expected. Table 1: Evaluation with L = 1 n N = :1 = :5 Error Order Error Order 1 4 -.36E-4-9.47E-4 8-9.00E-6 1.68.90E-6 3.10 3 14 3.97E-7.10 3.E-6 -.3 4 508 7.13E-8 1..91E-7 1.71 5 044 6.95E-9 1.67.10E-8 1.89 6 8188 5.70E-10 1.81 1.40E-9 1.95 Table : Evaluation =?1 n N L = 1 N L = 3 Error Order Error Order 1 4 1.49E-1 4 1.49E-1 8 4.01E- 0.68 5 9.51E-3 1.08 3 14 1.07E- 0.93 44 5.61E-4 1.83 4 508.53E-3 0.98 101 3.35E-5 1.98 5 044 6.3E-4 1.00 4084.00E-6.0 6 8188 1.58E-4 1.00 19
6 Generalization We have presented results for only the planar wedge, while using polynomial interpolation of degree two to approximate the integrand and the integration region. Any other degree of interpolation could also have been used. In such cases, the denition of the nodes will change appropriately. But the denition of the triangulation will remain the same, and we will use the `La + u' renement. In addition to using quadratic interpolation, we have also examined the use of linear, cubic and quartic interpolation. For linear and cubic interpolation, the function value at the origin is needed in the numerical integration. We simply let f(0; 0) = 0. The results are consistent with the kind of results we have obtained for the quadratic case, and they are as follows. Suppose that we use interpolation of degree d to approximate both the integrand and the wedge S. Let fq 1 ; : : : ; q v g be the node points in the unit simplex and let fl 1 ; : : : ; l v g be the basis functions in the Lagrange form, where v = (d+1)(d+). The points fq 1 ; : : : ; q v g will be equally spaced over and of the form ( i ; ); 0 i + d: d d Dene the interpolating operation P N h(s; t) = vx =1 h(q )l (s; t) The numerical integration we use is based on integrating this interpolation polynomial, i.e. h(s; t)dsdt = vx =1 P N h(s; t)dsdt! (d) h(q ) Assume that the integrand is of the form (1.1). Then with the `La + u' renement, we have S XN n vx K=1 =1! (d) ;K f(v ;K) 0 8 >< >: O( lnnn N d n ) if p 1 = d O( 1 N p n ) otherwise
where N n the number of triangular elements in the triangulation. The order p = minfp 1 ; d g, p 1 = (+)(L+1), d = d+ when d is an even number, and d = d+1 when d is an odd number. The proof is completely analogous to that given earlier for the quadratic case. In addition, we need the results for smooth integrands as stated in [6]. The results can be generalized to other integration regions S, for example, a wedge with the central angle larger than, triangles, squares, regions containing the origin as an interior point, etc. We also can generalize this to curved surfaces in R 3, with the integrand having a point singularity of the type given in (1.1). We omit statements of these results, as they are straightforward. 1
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