Journal of Computational Mathematics
Vol.xx, No.x, 200x, 1–17.
http://www.global-sci.org/jcm
doi:10.4208/jcm.1703-m2016-0544
APPROXIMATIONS OF HYPERSINGULAR INTEGRALS FOR
NEGATIVE FRACTIONAL EXPONENT*
Chaolang Hu
College of Mathematics, Sichuan University, Chengdu, Sichuan 610064, China
Email: [email protected]
Tao Lü
College of Mathematics, Sichuan University, Chengdu, Sichuan 610064, China
Email: [email protected]
Abstract
Rb
This article presents approximations of the hypersingular integrals a g(x)(x − t)α dx
Rb
and a g(x)|x − t|αdx with arbitrary singular point t ∈ (a, b) and negative fraction number
α < −1. These general expansions are applicable to a large range of hypersingular integrals, including both popular hypersingular integrals discussed in the literature and other
important ones which have not been addressed yet. The corresponding mid-rectangular
formulas and extrapolations, which can be calculated in fairly straightforward ways, are
investigated. Numerical examples are provided to illustrate the features of the numerical
methods and verify the theoretical conclusions.
Mathematics subject classification: 65N06, 65B99.
Key words: Hypersingular integral, Negative fractional exponent, Mid-rectangular quadrature formula, Extrapolation.
1. Introduction
In recent years, it has attracted attention to approximate hypersingular integrals [34, 41]
arising from different research areas, such as oil engineering,acoustics, electromagnetics, fracture
mechanics, fluid dynamics, heat conduction and elasticity [8, 11, 15, 17, 18, 20–22, 24, 26, 31, 43,
45, 46, 49, 50, 53, 54]. Furthermore, approximations for hypersingular integrals are often needed
in order to construct the numerical algorithms for solving hypersingular integral equations [30],
which have been utilized to study many real world problems, see [3–7, 28, 29, 32, 33, 37–40, 52]
and the references cited therein.
In order to evaluate the hypersingular integrals, different numeric techniques have been
developed, such as interpolation polynomials, splines, segments of orthogonal series, etc. [9, 23,
25, 27, 44, 47, 48]. Furthermore, A. Sidi discussed the Euler-Maclaurin expansions for integrals
with arbitrary algebraic endpoint singularities in [42]. I. V. Boykov etc. presented optimal,
asymtotically optimal and optimal in order algorithms for numerical evaluation of hypersingular
integrals with fixed and varying singularities [1, 2]. The direct method [11, 14], the method of
refinable operators [12] and other methods [46, 51] for the evaluation of singular integrals have
also been developed.
*
Received February 18, 2016 / Revised version received January 27, 2017 / Accepted March 9, 2017 /
Published online xxxxxx xx, 20xx /
2
C.L. HU AND T. LÜ
The Euler-Maclaurin expansions,mid-rectangular quadrature formulas and extrapolations
of the following hypersingular integrals are discussed deliberately in [16].
Z b
I(f ) =
g(x)(x − t)m dx
(1.1)
a
I(f ) =
Z
b
g(x)|x − t|m dx
(1.2)
a
where m is a negative integer and t ∈ (a, b).
Nevertheless, we have seldom found literatures which discuss the integrals with fractional
singularity as follows.
Z b
I(f ) =
g(x)(x − t)α dx
(1.3)
a
I(f ) =
Z
b
g(x)|x − t|α dx
(1.4)
a
where α is a negative fraction and t ∈ (a, b).
In most cases, Euler-Maclaurin expansions of Hypersingular integrals with negative fractional exponent concludes negative exponent of h, so that many quadrature formulas including
the mid-rectangular formula are divergent. We have to use Richardson extrapolations to get
convergent results.
In this article, we will derive Euler-Maclaurin expansions of the general hypersingular inteRb
Rb
grals a g(x)|x − t|α dx and a g(x)(x − t)α dx for negative fraction α and arbitrary singularity
point t ∈ (a, b). Then the corresponding mid-rectangular formulas and extrapolations will be
constructed. The rest of this paper is organized as follows: in section 2, a general definition of
Hadamard finite-part integrals is recalled; in section 3, we present the asymptotic expansions,
quadrature formulas and their extrapolations for (1.3) and (1.4); in section 4, numerical examples are provided to illustrate the features of the numerical methods and verify the theoretical
conclusions.
2. Definition of Hadamard Finite-Part Integrals
For a function f (x), which could be hypersingular nearby the origin of coordinates, the
Hadamard finite-part integral is defined as follows.
Definition 1 [36]. Let f (x) be integrable over (ǫ, b) for any ǫ satisfying 0 < ǫ < b < ∞.
Suppose that there exists a strictly monotonic increasing sequence α0 < α1 < α2 < · · · , and a
non-negative integer J such that the following expansion
Z
ε
b
f (x)dx =
∞ X
J
X
Ii,j εαi lnj (ε)
(2.1)
i=0 j=0
converges for any ε ∈ (0, h) with some h > 0. Then the Hadamard finite-part integral of (2.1)
is defined as follows:
Z b
0
if αi =
6 0 f or all i,
f.p.
f (x)dx =
(2.2)
I
if
α
= 0 f or some i.
i,0
i
0
Approximations of Hypersingular Integrals for Negative Fractional Exponent
3
Rb
Remark. The right side of the equation (2.1) is an expansion of the integral ε f (x)dx in
term of ε where Ii,j are the coefficients of this expansion. In the following, a simple example is
provided to illustrate the definition of the Hadamard finite-part integral:
Z b t
Z b P∞ 1 i
e
i=0 i! t
dt =
dt
1.5
t1.5
ε t
ε
Z b
Z b
∞ Z b
X
1 i−1.5
=
t−1.5 dt +
t−0.5 dt +
t
dt
i!
ε
ε
i=2 ε
= 2ε
−0.5
− 2b
−0.5
+ 2b
= (−2b−0.5 + 2b0.5 +
0.5
− 2ε
0.5
∞
X
1
1
+
·
· (bi−0.5 − εi−0.5 )
i!
i
−
0.5
i=2
∞
∞
X
X
1
1
1
1
·
· bi−0.5 )ε0 + 2ε−0.5 −
·
· εi−0.5 .
i!
i
−
0.5
i!
i
−
0.5
i=2
i=2
Then Definition 1 leads to
Z b t
∞
X
e
1
1
−0.5
0.5
f.p.
dt
=
−2b
+
2b
+
·
· bi−0.5 .
1.5
i!
i
−
0.5
0 t
i=2
Based on Definition 1, we recall the following lemma for the fundamental properties of
the Hadamard finite-part integral, which will be needed to calculate the exact value of the
hypersingular integral in Section 4.
Lemma 1 [36]. For any b > 0, We have
f.p.
Z
0
b
xα dx =
ln(b) (α = −1),
bα+1 /(α + 1) (α 6= −1).
Assume α < −1 and m > −α − 2 and g(x) ∈ C m+1 [0, b). Then for any b > 0, we have
"
#
Z b
Z b
Z b
m
m (k)
X
X
g (0)
(k)
k
α
α
f.p.
xα+k dx.
f.p.
g(x)x dx =
x g(x) −
g (0)x /k! dx +
k!
0
0
0
k=0
k=0
3. Euler-Maclaurin Expansions and Approximation Formulas
In this section, we will prove the asymptotic expansions for the hypersingular integrals
(1.3) and (1.4). These two types of integrals cover not only the popular hypersingular integral
discussed in the literature but also other important ones which have not been addressed yet.
Based on the general asymptotic expansions obtained in the theorems, mid-rectangular formulas
and their high accuracy extrapolations will be presented in the corollaries for different important
hypersingular integrals with different specific values of α.
Let ζ(p, β) denote the Riemann-zeta function [10], Γ(z) denote the Gamma function [13] ,
and [x] denote the largest integer ≤ x. Define ψ(z) = ΓΓ(z)
′ (z) .
First, we recall the following conclusion from the theorem (4.1) and the formula (5.2) in [36]
as the start of the analysis in this section.
Theorem 1 [36]. Assume α is a negative fraction, N is a positive integer, g(x) ∈ C n+1 [0, 1],
f (x) = xα g(x), and h = N1 . Then for any β ∈ (0, 1), we have the following Euler-Maclaurin
4
C.L. HU AND T. LÜ
expansion.
N −1
1 X β+k
f(
)
N
N
Z
1
1
2πi
Z
= f.p.
f (x)dx +
0
k=0
+
n
X
ζ(−k − α, β) g (k) (0)
N k+1+α
k=0
c′ +i∞
k!
−
n
X
ζ(−k, β) f (k) (1)
k=0
N k+1
N p−1 (F0 (p) − F2 (p))ζ(p, β)dp,
k!
(3.1)
c′ −i∞
where c′ ∈ (−n − α − 2, −n − α − 1), F0 (p) is Mellin transform [13] of f (x) , and F2 (p) is
Mellin transform of f (x + 1).
Since h =
1
N
and
1
2πi
′
Z
c′ +i∞
N p−1 (F0 (p) − F2 (p))ζ(p, β)dp
c′ −i∞
∞
=
N c −1
2π
=
O(hα+n+2 ),
Z
′
′
′
N is (F0 (c + is) − F2 (c + is))ζ(c + is, β)ds
−∞
then
f.p.
Z
1
f (x)dx
=
h
0
N
−1
X
+
k=0
n
X
f ((β + k)h) −
n
X
ζ(−k − α, β)hk+1+α
k=0
ζ(−k, β)hk+1
k=0
f (k) (1)
− O(hα+n+2 ).
k!
g (k) (0)
k!
(3.2)
By using (3.2), we will prove the following two theorems, which are critical to prove the
asymptotic expansions for the hypersingular integrals (1.3) and (1.4).
Theorem 2. Assume α is a negative fraction, N is a positive integer G(x) = g(x)(x − a)α , h =
b−a
n+1
[a, b]. Then for any β ∈ (0, 1), we have the following Euler-Maclaurin
N , and g(x) ∈ C
expansion:
Z b
N
−1
n
X
X
g (k) (a) k+1+α
f.p.
G(x)dx = h
G(a + (β + k)h) −
ζ(−k − α, β)
h
k!
a
+
k=0
n
X
k=0
(k)
ζ(−k, β)
k=0
G
(b) k+1
h
− O(hα+n+2 ).
k!
Proof. Without loss of generality, let l = b − a > 1, M =
α
x g(x + a) = G(x + a). There holds:
1
h,
(3.3)
y = x − a, and f (x) =
f ((β + k)h) = G(a + (β + k)h)(k = 0, 1, ..., N )
f (k) (y)|y=l = G(k) (y)|y=b (k ≤ n).
(3.4)
(3.5)
Then
f.p.
Z
b
G(x)dx
=
f.p.
a
=
=
f.p.
f.p.
Z
Z
Z
b
(x − a)α g(x)dx
a
l
y α g(y + a)dy
0
1
f (y)dy +
0
Z
1
l
f (y)dy,
(3.6)
5
Approximations of Hypersingular Integrals for Negative Fractional Exponent
f.p.
M−1
X
1
Z
y α g(y + a)dy
=
h
0
+
k=0
n
X
f ((β + k)h) −
n
X
k=0
ζ(−k, β)hk+1
f
(k)
k=0
l
Z
y α g(y + a)dy
=
h
1
N −M−1
X
+
k=0
g (k) (y + a)|y=0
k!
(y)|y=1
− O(hα+n+2 ),
k!
f (1 + (β + k)h) −
k=0
n
X
ζ(−k − α, β)hk+1+α
n
X
ζ(−k, β)hk+1
k=0
(3.7)
f (k) (y)|y=1
k!
f (k) (y)|y=l
ζ(−k, β)hk+1
+ O(hn+1 ),
k!
(3.8)
Substitute (3.7), (3.8), (3.4), and (3.5) into (3.6) to obtain
f.p.
= h
Z
b
G(x)dx
a
M−1
X
−
k=0
n
X
f ((β + k)h) + h
N −M−1
X
ζ(−k − α, β)hk+1+α
= h
−
k=0
n
X
n
f (k) (y)|y=l
g (k) (y + a)|y=0 X
+
ζ(−k, β)hk+1
− O(hα+n+2 )
k!
k!
k=0
k=0
N
−1
X
f (1 + (β + k)h)
k=0
G(a + (β + k)h)
n
ζ(−k − α, β)hk+1+α
g (k) (a) X
G(k) (b)
+
ζ(−k, β)hk+1
− O(hα+n+2 ).
k!
k!
k=0
k=0
Theorem 3. Assume α is a negative fraction, N is a positive integer G(x) = g(x)(b − x)α , h =
b−a
n+1
[a, b]. Then for any β ∈ (0, 1), we have the following Euler-Maclaurin
N , and g(x) ∈ C
expansion:
f.p.
Z
b
G(x)dx
a
=
h
N
−1
X
+
k=0
n
X
k=0
G(a + (β + k)h) −
n
X
ζ(−k − α, 1 − β)
k=0
(−1) G(k) (a) k+1
h
− O(hα+n+2 ).
k!
k
ζ(−k, 1 − β)
(−1)k g (k) (b) k+1+α
h
k!
(3.9)
′
Proof. Let y = b − x, F (y) = y α g(b − y), and β = 1 − β ∈ (0, 1). Then
G(u) = G(b − y) = F (y),
(3.10)
which implies
F (k) (y) = (−1)k G(k) (u),
F (k) (b − a) = (−1)k G(k) (a).
(3.11)
6
C.L. HU AND T. LÜ
Hence
N
−1
X
′
G(b − (1 − β + k)h)
=
k=0
N
−1
X
′
G(a + (N − 1 − k + β )h)
k=0
=
N
−1
X
′
′
G(a + (k + β )h)
k′ =0
=
N
−1
X
′
G(a + (k + β )h).
(3.12)
k=0
Using (3.3), (3.10), (3.11) and (3.12), we obtain
Z b
f.p.
G(x)dx
a
= f.p.
= h
Z
y α g(b − y)dy
0
N
−1
X
F (0 + (β + k)h) −
+
k=0
n
X
k=0
= h
b−a
N
−1
X
+
k=0
n
X
= h
+
k=0
n
X
ζ(−k − α, β)hk+1+α
k=0
g (k) (b − y)|y=0
k!
F (k) (y)|y=b−a
ζ(−k, β)hk+1
− O(hα+n+2 )
k!
G(b − (β + k)h) −
n
X
ζ(−k − α, β)hk+1+α
k=0
ζ(−k, β)hk+1
k=0
N
−1
X
n
X
g (k) (b) · (−1)k
k!
(−1) G(k) (a)
− O(hα+n+2 )
k!
′
k
G(a + (β + k)h) −
n
X
′
ζ(−k − α, 1 − β )hk+1+α
k=0
′
ζ(−k, 1 − β )hk+1
k=0
g (k) (b) · (−1)k
k!
(−1)k G(k) (a)
− O(hα+n+2 ).
k!
Based on the above two theorems, we will prove Theorem 4, Theorem 5 and Theorem 6 for
the Euler-Maclaurin expansions of the hypersingular integrals (1.3) and (1.4). Three cases will
be discussed in these three theorems respectively. In the following, we first prove the expansion
for the hypersingular integral (1.3).
Theorem 4. Assume α = − pq < 0, p 6= 1, p & q are odd numbers and coprime, N is a positive
n+1
integer, h = b−a
[a, b], xi = a + ih (i = 0, 1, · · · , N ), t ∈ {xi |i = 1, 2, · · · , N − 1},
N , g(x) ∈ C
q
α
and G(x) = g(x)(x − t) = g(x)(x − t)− p . Then we have the following Euler-Maclaurin
expansion:
f.p.
Z
a
b
G(x)dx
=
h
N
−1
X
k=0
[ n+1
2 ]
+
X
l=1
n+1
[ 2 ]
X
1
1 2g (2l−1) (t) 2l+α
G(a + ( + k)h) −
ζ(1 − 2l − α, )
h
2
2 (2l − 1)!
l=1
1 G(2l−1) (b) − G(2l−1) (a) 2l
ζ(1 − 2l, )
h + O(hn+2+α ).
2
(2l − 1)!
(3.13)
7
Approximations of Hypersingular Integrals for Negative Fractional Exponent
Proof. First we have
Z
Z b
Z t
f.p.
g(x)(x − t)α dx = −f.p.
g(x)(t − x)α dx + f.p.
a
a
b
g(x)(x − t)α dx.
(3.14)
t
Let
F (x) = g(x)(t − x)α = −G(x).
(3.15)
F k (a) = −Gk (a).
(3.16)
tnen
Since ζ(−2k, 1/2) = 0, k = 0, 1, · · · , then by substituting (3.3) and (3.9) into (3.14) and
letting β = 12 , t = xi (i = 1, 2, · · · , N − 1), we obtain
f.p.
= −h
b
Z
g(x)(x − t)α dx = −f.p.
a
i−1
X
k=0
−
n
X
k=0
+h
n
k=0
= h
i−1
X
+
k=0
n
X
k=0
= h
i−1
X
+
G(x)dx
t
k=0
1 (−1) F (k) (a) k+1
ζ(−k, 1 − )
h
+ O(hn+2+α )
2
k!
n
X
1 g (k) (t) k+1+α
1
G(t + ( + k)h) −
ζ(−k − α, )
h
2
2
k!
k=0
(k)
1 G (b) k+1
ζ(−k, )
h
− O(hn+2+α )
2
k!
k=0
(k)
1 g (t)
ζ(−k − α, )
((−1)k − 1)hk+1+α
2
k!
1 G(k) (b) + (−1)k G(k) (a) k+1
ζ(−k, )
h
− O(hn+2+α )
2
k!
n+1
k=1
k=i
l=1
1 G(k) (b) + (−1)k G(k) (a) k+1
ζ(−k, )
h
− O(hn+2+α )
2
k!
N
−1
X
k=0
X
l=1
n+1
[ 2 ]
X
1
1 2g (2l−1) (t) 2l+α
G(a + ( + k)h) −
ζ(1 − 2l − α, )
h
2
2 (2l − 1)!
[ n+1
2 ]
+
b
[ 2 ]
N
−1
X
X
1
1
1 2g (2l−1) (t) 2l+α
G(a + ( +k)h) + h
G(a + ( + k)h)−
ζ(1−2l − α, )
h
2
2
2 (2l − 1)!
k=0
n
X
= h
a
Z
NX
−i−1
1
1
G(a + ( + k)h) + h
G(t + ( + k)h)
2
2
k=0
n
X
+
F (x)dx + f.p.
k
k=0
+
t
X
1
1 g (k) (t)
F (a + ( + k)h) +
ζ(−k − α, 1 − )
(−1)k hk+1+α
2
2
k!
NX
−i−1
n
X
Z
l=1
1 G(2l−1) (b) − G(2l−1) (a) 2l
ζ(1 − 2l, )
h − O(hn+2+α ).
2
(2l − 1)!
Specifically, choosing −2 < α < −1 in Theorem 4, we can obtain a mid-rectangular formula
in Corollary 4.1 and choosing −4 < α < −1 we can obtain a mid-rectangular extrapolation
8
C.L. HU AND T. LÜ
formula in Corollary 4.2.
Corollary 4.1. Assume −2 < α = − pq < −1, p 6= 1, p & q are odd numbers and coprime,
n+1
[a, b], n ≥ 1, xi = a + ih (i = 0, 1, · · · , N ),
N is a positive integer, h = b−a
N , g(x) ∈ C
q
t ∈ {xi |i = 1, 2, · · · , N − 1}, and G(x) = g(x)(x − t)α = g(x)(x − t)− p . Let the mid-rectangular
formula
N
−1
X
1
(3.17)
Ih = h
G(a + ( + k)h).
2
k=0
Then
f.p.
b
Z
g(x)(x − t)α dx = Ih + O(h2+α ).
(3.18)
a
The conclusion is obvious.
Corollary 4.2. Assume −2 < α = − pq < −1, p 6= 1, p & q are odd numbers and coprime,
n+1
[a, b], n ≥ 1, xi = a + ih (i = 0, 1, · · · , N ),
N is a positive integer, h = b−a
N , g(x) ∈ C
q
t ∈ {xi |i = 1, 2, · · · , N − 1}, and G(x) = g(x)(x − t)α = g(x)(x − t)− p . Let the mid-rectangular
formula
N
−1
X
1
Ih = h
G(a + ( + k)h).
(3.19)
2
k=0
Qh =
22+α I h − Ih
2
Then
f.p.
2
22+α − 1
.
(3.20)
b
Z
g(x)(x − t)α dx = Q h + O(h2 ).
2
a
Proof. Using (3.13), we have the following formula.
Z b
f.p.
g(x)(x − t)α dx = Ih + ch2+α + O(h4+α ) + O(h2 ).
(3.21)
(3.22)
a
and
f.p.
Z
b
g(x)(x − t)α dx = I h + c
2
a
h2+α
+ O(h4+α ) + O(h2 ).
22+α
(3.23)
′
where c = −2ζ(−1 − α)g (t).
Using (3.22) and (3.23), we get (3.21) easily.
− pq
Theorem 5. Assume α =
< 0, p 6= 1, p is an odd number, q is an even number,p and q
n+1
are coprime, N is a positive integer, h = b−a
[a, b], xi = a + ih (i = 0, 1, · · · , N ),
N , g(x) ∈ C
q
t ∈ {xi |i = 1, 2, · · · , N − 1}, and G(x) = g(x)(x − t)α = g(x)(x − t)− p . Then we have the
following Euler-Maclaurin expansion:
f.p.
Z
a
b
G(x)dx
=
h
N
−1
X
k=0
[ n+1
2 ]
+
X
l=1
[n]
2
X
1
1 2g (2l) (t) 2l+1+α
G(a + ( + k)h) −
ζ(−2l − α, )
h
2
2 (2l)!
l=0
1 G(2l−1) (b) − G(2l−1) (a) 2l
ζ(1 − 2l, )
h − O(hn+2+α ). (3.24)
2
(2l − 1)!
9
Approximations of Hypersingular Integrals for Negative Fractional Exponent
Proof. First we have
Z b
Z t
Z
f.p.
g(x)(x − t)α dx = f.p.
g(x)(t − x)α dx + f.p.
a
a
b
g(x)(x − t)α dx.
(3.25)
t
Let
F (x) = g(x)(t − x)α = G(x).
(3.26)
F k (a) = Gk (a).
(3.27)
then
Since ζ(−2k, 1/2) = 0, k = 0, 1, · · · , then by substituting (3.3) and (3.9) into (3.25) and
letting β = 12 , t = xi (i = 1, 2, · · · , N − 1), we obtain
f.p.
Z
b
α
g(x)(x − t) dx = f.p.
a
= h
i−1
X
n
+
k=0
n
X
k=0
= h
−
k=0
n
X
k=0
= h
i−1
X
1 (−1) F (k) (a) k+1
ζ(−k, 1 − )
h
− O(hn+2+α )
2
k!
+
k=0
1 G(k) (b) k+1
ζ(−k, )
h
− O(hn+2+α )
2
k!
k=0
1 g (k) (t)
ζ(−k − α, )
((−1)k + 1)hk+1+α
2
k!
1 G(k) (b) + (−1)k G(k) (a) k+1
ζ(−k, )
h
− O(hn+2+α )
2
k!
n
k=1
k=i
(k)
1 G
ζ(−k, )
2
N
−1
X
k=0
X
l=1
k
l=0
(k)
(b) + (−1) G
k!
(a)
hk+1 − O(hn+2+α )
[n]
2
X
1
1 2g (2l) (t) 2l+1+α
G(a + ( + k)h) −
ζ(−2l − α, )
h
2
2 (2l)!
[ n+1
2 ]
+
n
X
1 g (k) (t) k+1+α
1
G(t + ( + k)h) −
ζ(−k − α, )
h
2
2
k!
[2]
N
−1
X
X
1
1
1 2g (2l) (t) 2l+1+α
G(a + ( + k)h) −
h
G(a + ( + k)h) + h
ζ(−2l − α, )
2
2
2 (2l)!
k=0
n
X
= h
G(x)dx
t
NX
−i−1
1
1
G(a + ( + k)h) + h
G(t + ( + k)h)
2
2
k=0
n
X
+
a
b
k=0
NX
−i−1
i−1
X
F (x)dx + f.p.
Z
k
k=0
+
t
X
1
1 g (k) (t)
ζ(−k − α, 1 − )
(−1)k hk+1+α
F (a + ( + k)h) −
2
2
k!
k=0
n
X
+h
Z
l=0
1 G(2l−1) (b) − G(2l−1) (a) 2l
ζ(1 − 2l, )
h − O(hn+2+α ).
2
(2l − 1)!
Specifically, choosing −3 < α < −1 in Theorem 5, we can obtain a mid-rectangular formula
in Corollary 5.1 .
10
C.L. HU AND T. LÜ
Corollary 5.1. Assume −3 < α = − pq < −1, p 6= 1, p is an odd number, q is an even
n+1
number,p and q are coprime, N is a positive integer, h = b−a
[a, b], xi = a+ih (i =
N , g(x) ∈ C
q
α
0, 1, · · · , N ), t ∈ {xi |i = 1, 2, · · · , N − 1}, and G(x) = g(x)(x − t) = g(x)(x − t)− p .
Let
N
−1
X
1
Ih = h
G(a + ( + k)h).
(3.28)
2
k=0
Qh =
Rh =
Then
Z
b
Z
b
f.p.
21+α I h − Ih
2
21+α − 1
.
23+α Q h − Qh
2
23+α − 1
(3.29)
.
(3.30)
g(x)(x − t)α dx = Ih + O(h1+α ).
(3.31)
g(x)(x − t)α dx = Qh + O(h3+α ).
(3.32)
a
f.p.
a
f.p.
b
Z
g(x)(x − t)α dx = Rh + O(h2 ).
(3.33)
a
Proof. Using (3.24),we have the following formula.
f.p.
Z
b
g(x)(x − t)α dx = Ih + c1 h1+α + c2 h3+α + O(h2 ).
(3.34)
a
Taking similar proving strategy to corollary 4.2, we get the above conclusions easily.
Theorem 6. Assume α = − pq < 0, p 6= 1, p and q are coprime, N is a positive integer,
n+1
[a, b], xi = a + ih (i = 0, 1, · · · , N ), t ∈ {xi |i = 1, 2, · · · , N − 1}, and
h = b−a
N , g(x) ∈ C
q
α
G(x) = g(x)|x − t| = g(x)|x − t|− p . Then we have the following Euler-Maclaurin expansion:
f.p.
Z
b
G(x)dx
=
a
h
N
−1
X
k=0
[ n+1
2 ]
+
X
l=1
[n]
2
X
1
1 2g (2l) (t) 2l+1+α
ζ(−2l − α, )
G(a + ( + k)h) −
h
2
2 (2l)!
l=0
1 G(2l−1) (b) − G(2l−1) (a) 2l
ζ(1 − 2l, )
h − O(hn+2+α ). (3.35)
2
(2l − 1)!
Proof. First we have
f.p.
Z
b
α
g(x)|x − t| dx = f.p.
a
Z
t
α
g(x)(t − x) dx + f.p.
a
Z
b
g(x)(x − t)α dx.
(3.36)
t
Let
F (x) = g(x)(t − x)α = G(x).
(3.37)
F k (a) = Gk (a).
(3.38)
then
11
Approximations of Hypersingular Integrals for Negative Fractional Exponent
Since ζ(−2k, 1/2) = 0, k = 0, 1, · · · , then by substituting (3.3) and (3.9) into (3.36) and
letting β = 12 , t = xi (i = 1, 2, · · · , N − 1), we obtain
f.p.
Z
b
g(x)(x − t)α dx = f.p.
a
= h
i−1
X
k=0
n
X
k=0
−
k=0
n
X
k=0
= h
i−1
X
1 (−1) F (k) (a) k+1
ζ(−k, 1 − )
h
− O(hn+2+α )
2
k!
+
k=0
(k)
1 G (b) k+1
ζ(−k, )
h
− O(hn+2+α )
2
k!
k=0
(k)
1 g (t)
ζ(−k − α, )
((−1)k + 1)hk+1+α
2
k!
1 G(k) (b) + (−1)k G(k) (a) k+1
ζ(−k, )
h
− O(hn+2+α )
2
k!
n
k=1
k=i
(k)
1 G
ζ(−k, )
2
N
−1
X
k=0
X
l=1
l=0
(b) + (−1)k G(k) (a) k+1
h
− O(hn+2+α )
k!
[n]
2
X
1
1 2g (2l) (t) 2l+1+α
h
G(a + ( + k)h) −
ζ(−2l − α, )
2
2 (2l)!
[ n+1
2 ]
+
n
X
1 g (k) (t) k+1+α
1
ζ(−k − α, )
h
G(t + ( + k)h) −
2
2
k!
[2]
N
−1
X
X
1 2g (2l) (t) 2l+1+α
1
1
ζ(−2l − α, )
G(a + ( + k)h) −
G(a + ( + k)h) + h
h
2
2
2 (2l)!
k=0
n
X
= h
G(x)dx
t
NX
−i−1
1
1
G(a + ( + k)h) + h
G(t + ( + k)h)
2
2
k=0
n
X
+
b
k=0
k=0
= h
Z
k
NX
−i−1
i−1
X
F (x)dx + f.p.
a
n
+
+
t
X
1
1 g (k) (t)
F (a + ( + k)h) −
ζ(−k − α, 1 − )
(−1)k hk+1+α
2
2
k!
k=0
n
X
+h
Z
l=0
1 G(2l−1) (b) − G(2l−1) (a) 2l
ζ(1 − 2l, )
h − O(hn+2+α ).
2
(2l − 1)!
Specifically, choosing −3 < α < −1 in Theorem 6, we can obtain a mid-rectangular formula
in Corollary 6.1.
Corollary 6.1. Assume −3 < α = − pq < −1, p 6= 1, p and q are coprime, N is a positive
n+1
integer, h = b−a
[a, b], xi = a + ih (i = 0, 1, · · · , N ), t ∈ {xi |i = 1, 2, · · · , N − 1},
N , g(x) ∈ C
q
α
and G(x) = g(x)|x − t| = g(x)|x − t|− p .
Let
N
−1
X
1
Ih = h
G(a + ( + k)h).
(3.39)
2
k=0
Qh =
21+α I h − Ih
2
21+α − 1
.
(3.40)
12
C.L. HU AND T. LÜ
Rh =
Then
Z
b
Z
b
f.p.
23+α Q h − Qh
2
23+α − 1
.
(3.41)
g(x)|x − t|α dx = Ih + O(h1+α ).
(3.42)
g(x)|x − t|α dx = Qh + O(h3+α ).
(3.43)
a
f.p.
f.p.
a
Z
b
g(x)|x − t|α dx = Rh + O(h2 ).
(3.44)
a
Proof. Using (3.35),we have the following formula.
f.p.
Z
b
g(x)|x − t|α dx = Ih + c1 h1+α + c2 h3+α + O(h2 ).
(3.45)
a
Taking similar proving strategy to corollary 4.2, we get the above conclusions easily.
4. Numerical Examples
In this section, we will provide several examples to illustrate the features of the numerical
methods and verify the theoretical conclusions.
Example 1. Consider the hypersingular integral
I(y) =
Z
1
g(x)
5
0
(x − y) 3
dx, y ∈ (0, 1).
When g(x) = x2 , the exact solution is
I(y) =
Z
0
1
x2
(x − y)
5
3
dx =
−2
−2
1
1
4
4
3 2
3
y [(1+y) 3 −(1−y) 3 ]+6y[(1−y) 3 +(1+y) 3 ]+ [(1−y) 3 −(1+y) 3 ].
2
4
Using the mid-rectangular formula (3.17) and extrapolation formula (3.20), we obtain the
(1)
2h |
numerical results in Table 1 and Table 2. The rate rh = log2 |I−I
|I−Ih | ≈0.3331 in Table 1 shows
(2)
2h |
that (3.17) has the order O(h2− 3 ) and the rate rh = log2 |I−Q
|I−Qh | ≈ 1.9937 in Table 1 shows
that (3.20) has the second order accuracy. This numerical performance is consistent with the
theoretical conclusions in (3.18) and (3.21).
5
Table 1. Numerical results for I =
h
Ih
|I − Ih |
(1)
rh
Qh
|I − Qh |
(2)
rh
1/23
1.5434565948
0.9036050369
R1
0
x2
5
dx ≈ 2.4470616318 at y = 0.25.
(x−y) 3
1/24
1.7287364976
0.7183251341
0.3310553165
2.4415679627
0.0054936690
1/25
1.8766385277
0.5704231041
0.3326045790
2.4456652766
0.0013963551
1.9761040960
1/26
1.9942441842
0.4528174476
0.3331028691
2.4467110257
0.0003506060
1.9937431911
13
Approximations of Hypersingular Integrals for Negative Fractional Exponent
Table 2. Numerical results for I =
h
Ih
|I − Ih |
(1)
rh
Qh
|I − Qh |
(2)
rh
1/24
0.8843452915
0.3592284430
R1
0
x2
5
dx ≈ 1.2435737344 at y = 0.125.
(x−y) 3
1/25
0.9583455958
0.2852281386
0.3327852690
1.2430486022
0.0005251323
1/26
1.0171608565
0.2264128779
0.3331602969
1.2434421090
0.0001316255
1.9962419135
1/27
1.0638629211
0.1797108133
0.3332787988
1.2435408065
0.0000329280
1.9990529100
Example 2. Consider the hypersingular integral
Z 1
g(x)
I(y) =
8 dx, y ∈ (0, 1).
0 (x − y) 3
When g(x) = x2 , the exact solution is
Z 1
−5
−5
−2
−2
1
1
x2
−3 2
I(y) =
y [(1 − y) 3 + y 3 ] − 3y[(1 − y) 3 − y 3 ] + 3[(1 − y) 3 + y 3 ].
8 dx =
5
0 (x − y) 3
Based on formulas (3.28), (3.29) and (3.30), we obtained the results in Table 3 and Table
(1)
2h |
4. Obviously, Ih in (3.28) is divergent. Furthermore, the rate rh = log2 |I−Q
|I−Qh | ≈ 0.3333
(2)
2h |
in Table 3 shows that Qh has the order O(h3− 3 ) and the rate rh = log2 |I−R
|I−Rh | ≈ 1.9914 in
Table 3 shows that Rh has the second order accuracy, which numerically verify the conclusions
of Corollary 5.1.
8
Table 3: Numerical results for I =
R1
0
x2
8
(x−y) 3
dx ≈ 5.1583367411 at y = 0.25.
h
1/25
1/26
1/27
1/28
1/29
Ih
281.2302286825 883.7881379560 2796.3456476958 8867.9842113956 28143.9548478132
Qh
4.1669003995
4.3713366396
4.5336696266
4.6625319066
|I − Qh |
0.9914363416
0.7870001015
0.6246671145
0.4958048345
(1)
rh
0.3331563207
0.3332762404
0.3333152425
Rh
5.1578687046
5.1582169154
5.1583066046
|I − Rh |
0.0004680365
0001198257
0.0000301365
(2)
rh
1.9656836145
1.9913518799
Table 4. Numerical results for I =
R1
0
x2
8
(x−y) 3
dx ≈ 5.1477711257 at y = 0.125.
h
1/26
1/27
1/28
1/29
1/210
Ih
224.2660840396 702.5166832538 2220.5146021397 7039.5773280917 22338.9253621600
Qh
4.3607426072
4.5230974749
4.6519646935
4.7542486140
|I − Qh |
0.7870285186
0.6246736508
0.4958064322
0.3935225117
(1)
rh
0.3333132366
0.3333256892
0.3333308197
Rh
5.1477289460
5.1477583917
5.1477678021
|I − Rh |
0.0000421797
0.0000127340
0.0000033236
(2)
rh
1.7278608079
1.9378741864
Example 3. Consider the hypersingular integral
Z 1
g(x)
I(y) =
5 dx, y ∈ (0, 1).
0 |x − y| 2
14
C.L. HU AND T. LÜ
When g(x) = x2 , the exact solution is
I(y) =
Z
0
1
x2
|x − y|
5
2
dx =
−3
−1
16 1
2
1
y 2 − y 2 (1 − y) 2 − 4y(1 − y) 2 + 2(1 − y) 2 .
3
3
Using formulas (3.39), (3.40) and (3.41), we obtain the numerical results in Table 5 and
Table 6. As predicted in theory, Ih in (3.39) is divergent. However, the straightforward ex(1)
trapolations Qh and Rh from (3.40) and (3.41) are convergent. Furthermore, the rate rh =
(2)
|I−Q2h |
|I−R2h |
log2 |I−Qh | ≈ 0.4999 and the rate rh = log2 |I−Rh | ≈ 1.9917 in Table 5 , which shows Qh
5
has order O(h3− 2 ) and Rh has order O(h2 ) , are consistent with Corollary 6.1.
Table 5. Numerical results for
R1
0
x2
5
|x−y| 2
dx ≈ 3.1798669059 at y=0.25.
h
1/25
1/26
1/27
1/28
1/29
Ih
144.3220894460 402.8437505591 1133.9207607613 3201.6249012144 9049.9089524430
Qh
2.9318980546
3.0044568253
3.0558154384
3.0921447938
|I − Qh |
0.2479688514
0.1754100807
0.1240571259
0.0877235016
(1)
rh
0.4994272471
0.4997928582
0.4999263475
Rh
3.1796291936
3.1798060988
3.1798516162
|I − Rh |
0.0002377124
0.0000608071
0.0000152898
(2)
rh
1.9669044639
1.9916726441
Table 6. Numerical results for
R1
0
x2
5
|x−y| 2
dx ≈ 3.2091975669 at y=0.125.
h
1/26
1/27
1/28
1/29
1/210
Ih
103.0117536928 285.8142232513 802.7637472161 2264.8513695386 6400.2126812477
Qh
3.0337632745
3.0851404410
3.1214740653
3.1471671996
|I − Qh |
0.1754100807
0.1240514675
0.0877221122
0.0620303673
(1)
rh
0.4999261751
0.4999693015
0.4999887314
Rh
3.2091758934
3.2091911939
3.2091959127
|I − Rh |
0.0000216735
0.0000063730
0.0000016542
(2)
rh
1.7658892315
1.9458312418
5. Conclusions
Rb
This paper presents Euler-Maclaurin expansions for the hypersingular integrals a g(x)|x −
Rb
t|α dx and a g(x)(x−t)α dx with arbitrary singular point t ∈ (a, b) and negative fraction number
α < −1, which are critical for many kinds of real world problems. Mid-rectangular formulas
and their high accuracy extrapolation formulas are also constructed. These formulas can be
calculated in a fairly straightforward way. Hence the implementation of these methods for real
world problems has a low cost.
Acknowledgments. This work was supported by Major Research Plan of National Natural
Science Foundation of China (91430105)
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