journal of inequalities in pure and applied...

volume 3, issue 4, article 51,2002.

Received 5 April, 2002;accepted 20 May, 2002.

Communicated by: B. Mond

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Journal of Inequalities in Pure andApplied Mathematics

APPROXIMATING THE FINITE HILBERT TRANSFORM VIA ANOSTROWSKI TYPE INEQUALITY FOR FUNCTIONS OFBOUNDED VARIATION

S.S. DRAGOMIRSchool of Communications and InformaticsVictoria University of TechnologyPO Box 14428Melbourne City MC8001 Victoria, AustraliaEMail : [email protected]: http://rgmia.vu.edu.au/SSDragomirWeb.html

c©2000Victoria UniversityISSN (electronic): 1443-5756032-01

Please quote this number (032-01) in correspondence regarding this paper with the Editorial Office.

mailto:[email protected]

http://jipam.vu.edu.au/


http://rgmia.vu.edu.au/SSDragomirWeb.html

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Approximating the Finite HilbertTransform via an Ostrowski

Type Inequality for Functions ofBounded Variation

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J. Ineq. Pure and Appl. Math. 3(4) Art. 51, 2002

http://jipam.vu.edu.au

Abstract

Using the Ostrowski type inequality for functions of bounded variation, an ap-proximation of the finite Hilbert Transform is given. Some numerical experi-ments are also provided.

2000 Mathematics Subject Classification: Primary 26D10, 26D15; Secondary41A55, 47A99.Key words: Finite Hilbert Transform, Ostrowski’s Inequality.

Contents1 Introduction. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32 Some Inequalities on the Interval[a, b] . . . . . . . . . . . . . . . . . . . 43 A Quadrature Formula for Equidistant Divisions. . . . . . . . . . 154 A More General Quadrature Formula. . . . . . . . . . . . . . . . . . . . 215 Numerical Experiments. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .28References




http://www.ams.org/msc/



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1. IntroductionCauchy principal value integrals of the form

(Tf) (a, b; t) = PV

∫ b

a

f (τ)

τ − tdτ(1.1)

:= limε→0+

[∫ t−ε

a

f (τ)

τ − tdτ +

∫ b

t+ε

f (τ)

τ − tdτ

]play an important role in fields like aerodynamics, the theory of elasticity andother areas of the engineering sciences. They are also helpful tools in somemethods for the solution of differential equations (cf., e.g. [23]).

For different approaches in approximating the finite Hilbert transform (1.1)including: interpolatory, noninterpolatory, Gaussian, Chebychevian and splinemethods, see for example the papers [1] – [12], [14] – [22], [24] – [33] and thereferences therein.

In contrast with all these methods, we point out here a new method in ap-proximating the finite Hilbert transform by the use of the Ostrowski inequalityfor functions of bounded variation established in [13].

For a comprehensive list of papers on Ostrowski’s inequality, visit the sitehttp://rgmia.vu.edu.au .

Estimates for the error bounds and some numerical examples for the obtainedapproximation are also presented.




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2. Some Inequalities on the Interval[a, b]

We start with the following lemma proved in [13] dealing with an Ostrowskitype inequality for functions of bounded variation.

Lemma 2.1. Let u : [a, b] → R be a function of bounded variation on[a, b].Then, for allx ∈ [a, b], we have the inequality:

(2.1)

∣∣∣∣u (x) (b− a)−∫ b

a

u (t) dt

∣∣∣∣ ≤ [1

2(b− a) +

∣∣∣∣x− a + b

2

∣∣∣∣] b∨a

(u) ,

where∨b

a (u) denotes the total variation ofu on [a, b].The constant1

2is the best possible one.

Proof. For the sake of completeness and since this result will be essentially usedin what follows, we give here a short proof.

Using the integration by parts formula for the Riemann-Stieltjes integral wehave ∫ x

a

(t− a) du (t) = u (x) (x− a)−∫ x

a

u (t) dt

and ∫ b

x

(t− b) du (t) = u (x) (b− x)−∫ b

x

u (t) dt.

If we add the above two equalities, we get

(2.2) u (x) (b− a)−∫ b

a

u (t) dt =

∫ x

a

(t− a) du (t) +

∫ b

x

(t− b) du (t)






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for anyx ∈ [a, b].If p : [c, d] → R is continuous on[c, d] andv : [c, d] → R is of bounded

variation on[c, d], then:

(2.3)

∣∣∣∣∫ d

c

p (x) dv (x)

∣∣∣∣ ≤ supx∈[c,d]

|p (x)|d∨c

(u) .

Using (2.2) and (2.3), we deduce∣∣∣∣u (x) (b− a)−∫ b

a

u (t) dt

∣∣∣∣ ≤ ∣∣∣∣∫ x

a

(t− a) du (t)

∣∣∣∣+ ∣∣∣∣∫ b

x

(t− b) du (t)

∣∣∣∣≤ (x− a)

x∨a

(u) + (b− x)b∨x

(u)

≤ max {x− a, b− x}

[x∨a

(u) +b∨x

(u)

]

=

[1

2(b− a) +

∣∣∣∣x− a + b

2

∣∣∣∣] b∨a

(u)

and the inequality (2.1) is proved.Now, assume that the inequality (2.2) holds with a constantc > 0, i.e.,

(2.4)

∣∣∣∣u (x) (b− a)−∫ b

a

u (t) dt

∣∣∣∣ ≤ [c (b− a) +

∣∣∣∣x− a + b

2

∣∣∣∣] b∨a

(u)

for all x ∈ [a, b].






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Consider the functionu0 : [a, b] → R given by

u0 (x) =

0 if x ∈ [a, b]

∖{a+b2

}1 if x = a+b

2.

Thenu0 is of bounded variation on[a, b] and

b∨a

(u0) = 2,

∫ b

a

u0 (t) dt = 0.

If we apply (2.4) for u0 and choosex = a+b2

, then we get2c ≥ 1 which impliesthatc ≥ 1

2showing that1

2is the best possible constant in (2.1).

The best inequality we can get from (2.1) is the following midpoint inequal-ity.

Corollary 2.2. With the assumptions in Lemma2.1, we have

(2.5)

∣∣∣∣u(a + b

2

)(b− a)−

∫ b

a

u (t) dt

∣∣∣∣ ≤ 1

2(b− a)

b∨a

(u) .

The constant12

is best possible.

Using the above Ostrowski type inequality we may point out the followingresult in estimating the finite Hilbert transform.






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Theorem 2.3. Let f : [a, b] → R be a function such that its derivativef ′ :[a, b] → R is of bounded variation on[a, b]. Then we have the inequality:

(2.6)

∣∣∣∣(Tf) (a, b; t)− f (t)

πln

(b− t

t− a

)−b− a

π[f ; λt + (1− λ) b, λt + (1− λ) a]

∣∣∣∣≤ 1

π

[1

2+

∣∣∣∣λ− 1

2

∣∣∣∣] [1

2(b− a) +

∣∣∣∣t− a + b

2

∣∣∣∣] b∨a

(f ′) ,

for anyt ∈ (a, b) andλ ∈ [0, 1), where[f ; α, β] is the divided difference, i.e.,

[f ; α, β] :=f (α)− f (β)

α− β.

Proof. Sincef ′ is bounded on[a, b], it follows thatf is Lipschitzian on[a, b]and thus the finite Hilbert transform exists everywhere in(a, b).

As for the functionf0 : (a, b) → R, f0 (t) = 1, t ∈ (a, b), we have

(Tf0) (a, b; t) =1

πln

(b− t

t− a

), t ∈ (a, b) ,

then obviously

(2.7) (Tf) (a, b; t)− f (t)

πln

(b− t

t− a

)=

1

πPV

∫ b

a

f (τ)− f (t)

τ − tdτ.






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Now, if we choose in (2.1), u = f ′, x = λc + (1− λ) d, λ ∈ [0, 1], then we get

|f (d)− f (c)− (d− c) f ′ (λc + (1− λ) d)|

≤[1

2|d− c|+

∣∣∣∣λc + (1− λ) d− c + d

2

∣∣∣∣]∣∣∣∣∣

d∨c

(f ′)

∣∣∣∣∣wherec, d ∈ (a, b), which is equivalent to

(2.8)

∣∣∣∣f (d)− f (c)

d− c− f ′ (λc + (1− λ) d)

∣∣∣∣ ≤ [1

2+

∣∣∣∣λ− 1

2

∣∣∣∣]∣∣∣∣∣

d∨c

(f ′)

∣∣∣∣∣for anyc, d ∈ (a, b), c 6= d.

Using (2.8), we may write∣∣∣∣ 1πPV

∫ b

a

f (τ)− f (t)

τ − tdτ − 1

πPV

∫ b

a

f ′ (λt + (1− λ) τ) dτ

∣∣∣∣(2.9)

≤ 1

π

[1

2+

∣∣∣∣λ− 1

2

∣∣∣∣]PV

∫ b

a

∣∣∣∣∣t∨τ

(f ′)

∣∣∣∣∣ dt

=1

π

[1

2+

∣∣∣∣λ− 1

2

∣∣∣∣][∫ t

a

(t∨τ

(f ′)

)dt +

∫ b

t

(τ∨t

(f ′)

)dt

]

≤ 1

π

[1

2+

∣∣∣∣λ− 1

2

∣∣∣∣][(t− a)

t∨a

(f ′) + (b− t)b∨t

(f ′)

]

≤ 1

π

[1

2+

∣∣∣∣λ− 1

2

∣∣∣∣] [1

2(b− a) +

∣∣∣∣t− a + b

2

∣∣∣∣] b∨a

(f ′) .






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Since (forλ 6= 1)

1

πPV

∫ b

a

f ′ (λt + (1− λ) τ) dτ

=1

πlim

ε→0+

[∫ t−ε

a

+

∫ b

t+ε

](f ′ (λt + (1− λ) τ) dτ)

=1

πlim

ε→0+

[1

1− λf (λt + (1− λ) τ)

∣∣∣∣t−ε

a

+1

1− λf (λt + (1− λ) τ)

∣∣∣∣bt+ε

]

=1

π· f (t)− f (λt + (1− λ) a) + f (λt + (1− λ) b)− f (t)

1− λ

=b− a

π[f ; λt + (1− λ) b, λt + (1− λ) a] .

Using (2.9) and (2.7), we deduce the desired result (2.6).

It is obvious that the best inequality we can get from (2.6) is the one forλ = 1

2. Thus, we may state the following corollary.

Corollary 2.4. With the assumptions of Theorem2.3, we have

(2.10)

∣∣∣∣(Tf) (a, b; t)− f (t)

πln

(b− t

t− a

)− b− a

π

[f ;

t + b

2,a + t

2

]∣∣∣∣≤ 1

2π

[1

2(b− a) +

∣∣∣∣t− a + b

2

∣∣∣∣] b∨a

(f ′) .






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The above Theorem2.3 may be used to point out some interesting inequal-ities for the functions for which the finite Hilbert transforms(Tf) (a, b; t) canbe expressed in terms of special functions.

For instance, we have:

1) Assume thatf : [a, b] ⊂ (0,∞) → R, f (x) = 1x. Then

(Tf) (a, b; t) =1

πtln

[(b− t) a

(t− a) b

], t ∈ (a, b) ,

b− a

π· [f ; λt + (1− λ) b, λt + (1− λ) a]

= − 1

π· b− a

[λt + (1− λ) b] [λt + (1− λ) a],

b∨a

(f ′) =

∫ b

a

|f ′′ (t)| dt =b2 − a2

a2b2.

Using the inequality (2.6) we may write that∣∣∣∣ 1

πtln

[(b− t) a

(t− a) b

]− 1

πtln

(b− t

t− a

)+

b− a

π [λt + (1− λ) b] [λt + (1− λ) a]

∣∣∣∣≤ 1

π

[1

2+

∣∣∣∣λ− 1

2

∣∣∣∣] [1

2(b− a) +

∣∣∣∣t− a + b

2

∣∣∣∣] · b2 − a2

a2b2






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which is equivalent to:

(2.11)

∣∣∣∣ b− a

[λt + (1− λ) b] [λt + (1− λ) a]− 1

tln

(b

a

)∣∣∣∣≤[1

2+

∣∣∣∣λ− 1

2

∣∣∣∣] [1

2(b− a) +

∣∣∣∣t− a + b

2

∣∣∣∣] · b2 − a2

a2b2.

If we use the notations

L (a, b) :=b− a

ln b− ln a(the logarithmic mean)

Aλ (x, y) := λx + (1− λ) y (the weighted arithmetic mean)

G (a, b) :=√

ab (the geometric mean)

A (a, b) :=a + b

2(the arithmetic mean)

then by (2.11) we deduce∣∣∣∣ 1

Aλ (t, b) Aλ (t, a)− 1

tL (a, b)

∣∣∣∣≤[1

2+

∣∣∣∣λ− 1

2

∣∣∣∣] [1

2(b− a) + |t− A (a, b)|

]2A (a, b)

G4 (a, b),

giving the following proposition:






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Proposition 2.5. With the above assumption, we have

(2.12) |tL (a, b)− Aλ (t, b) Aλ (t, a)|

≤ 2A (a, b)

G4 (a, b)

[1

2+

∣∣∣∣λ− 1

2

∣∣∣∣] [1

2(b− a) +

∣∣∣∣t− a + b

2

∣∣∣∣]× tAλ (t, b) Aλ (t, a) L (a, b)

for anyt ∈ (a, b), λ ∈ [0, 1).

In particular, fort = A (a, b) andλ = 12, we get

(2.13)

∣∣∣∣A (a, b) L (a, b)− (A (a, b) + a) (A (a, b) + b)

4

∣∣∣∣≤ 1

2· A2 (a, b)

G4 (a, b)· (A (a, b) + a) (A (a, b) + b)

4L (a, b) .

2) Assume thatf : [a, b] ⊂ R → R, f (x) = exp (x). Then

(Tf) (a, b; t) =exp (t)

π[Ei (b− t)− Ei (a− t)] ,

where

Ei (z) := PV

∫ z

−∞

exp (t)

tdt, z ∈ R.

Also, we have:

b− a

π[exp; λt + (1− λ) b, λt + (1− λ) a]

=1

π· exp (λt + (1− λ) b)− exp (λt + (1− λ) a)

1− λ,






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b∨a

(f ′) =

∫ b

a

|f ′′ (t)| dt = exp (b)− exp (a) .

Using the inequality (2.6) we may write:

(2.14)

∣∣∣∣exp (t)

[Ei (b− t)− Ei (a− t)− ln

(b− t

t− a

)]− exp (λt + (1− λ) b)− exp (λt + (1− λ) a)

1− λ

∣∣∣∣≤[1

2+

∣∣∣∣λ− 1

2

∣∣∣∣] [1

2(b− a) +

∣∣∣∣t− a + b

2

∣∣∣∣] [exp (b)− exp (a)]

for anyt ∈ (a, b).

If in (2.14) we makeλ = 12

andt = a+b2

, we get∣∣∣∣exp

(a + b

2

)Ei

(b− a

2

)− 2

[exp

(a + 3b

4

)− exp

(3a + b

4

)]∣∣∣∣≤ 1

4(b− a) [exp (b)− exp (a)] ,

which is equivalent to:∣∣∣∣Ei

(b− a

2

)− 2

[exp

(b− a

4

)− exp

(−b− a

4

)]∣∣∣∣≤ 1

4(b− a)

[exp

(b− a

2

)− exp

(−b− a

2

)].






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If in this inequality we makeb−a2

= z > 0, then we get

(2.15)∣∣∣Ei (z)− 2

[exp

(z

2

)− exp

(−z

2

)]∣∣∣≤ 1

2z [exp (z)− exp (−z)]

for anyz > 0.

Consequently, we may state the following proposition.

Proposition 2.6. With the above assumptions, we have

(2.16)

∣∣∣∣Ei (z)− 4 sinh

(1

2z

)∣∣∣∣ ≤ z sinh (z)

for anyz > 0.

The reader may get other similar inequalities for special functions if appro-priate examples of functionsf are chosen.






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3. A Quadrature Formula for Equidistant DivisionsThe following lemma is of interest in itself.

Lemma 3.1. Let u : [a, b] → R be a function of bounded variation on[a, b].Then for alln ≥ 1, λi ∈ [0, 1) (i = 0, . . . , n− 1) andt, τ ∈ [a, b] with t 6= τ ,we have the inequality:

(3.1)

∣∣∣∣∣ 1

τ − t

∫ τ

t

u (s) ds− 1

n

n−1∑i=0

u

[t + (i + 1− λi)

τ − t

n

]∣∣∣∣∣≤ 1

n

[1

2+ max

i=0,n−1

∣∣∣∣λi −1

2

∣∣∣∣]∣∣∣∣∣

τ∨t

(u)

∣∣∣∣∣ .Proof. Consider the equidistant division of[t, τ ] (if t < τ ) or [τ, t] (if τ < t)given by

(3.2) En : xi = t + i · τ − t

n, i = 0, n.

Then the pointsξi = λi

[t + i · τ−t

n

]+ (1− λi)

[t + (i + 1) · τ−t

n

](λi ∈ [0, 1] , i = 0, n− 1

)are betweenxi andxi+1. We observe that we may

write for simplicity ξi = t + (i + 1− λi)τ−tn

(i = 0, n− 1

). We also have

ξi −xi + xi+1

2=

τ − t

2n(1− 2λi) ,

ξi − xi = (1− λi)τ − t

n






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andxi+1 − ξi = λi ·

τ − t

n

for anyi = 0, n− 1.If we apply the inequality (2.1) on the interval[xi, xi+1] and the intermediate

point ξi

(i = 0, n− 1

), then we may write that

(3.3)

∣∣∣∣τ − t

nu

(t + (i + 1− λi)

τ − t

n

)−∫ xi+1

xi

u (s) ds

∣∣∣∣≤[1

2· |τ − t|

n+

∣∣∣∣τ − t

2n(1− 2λi)

∣∣∣∣]∣∣∣∣∣xi+1∨xi

(u)

∣∣∣∣∣ .Summing, we get∣∣∣∣∣

∫ τ

t

u (s) ds− τ − t

n

n−1∑i=0

u

[t + (i + 1− λi)

τ − t

n

]∣∣∣∣∣≤ |τ − t|

2n

n−1∑i=0

[1 + |1− 2λi|]

∣∣∣∣∣xi+1∨xi

(u)

∣∣∣∣∣=|τ − t|

n

[1

2+ max

i=0,n−1

∣∣∣∣λi −1

2

∣∣∣∣]∣∣∣∣∣

τ∨t

(u)

∣∣∣∣∣ ,which is equivalent to (3.1).

We may now state the following theorem in approximating the finite Hilberttransform of a differentiable function with the derivative of bounded variationon [a, b].






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Theorem 3.2. Let f : [a, b] → R be a differentiable function such that itsderivativef ′ is of bounded variation on[a, b]. If λ = (λi)i=0,n−1, λi ∈ [0, 1)(i = 0, n− 1

)and

(3.4) Sn (f ; λ, t)

:=b− a

πn

n−1∑i=0

[f ; (i + 1− λi)

b− t

n+ t, (i + 1− λi)

a− t

n+ t

],

then we have the estimate:∣∣∣∣(Tf) (a, b; t)− f (t)

πln

(b− t

t− a

)− Sn (f ; λ, t)

∣∣∣∣(3.5)

≤ b− a

nπ

[1

2+ max

i=0,n−1

∣∣∣∣λi −1

2

∣∣∣∣] [1

2(b− a) +

∣∣∣∣t− a + b

2

∣∣∣∣] b∨a

(f ′)

≤ b− a

nπ

b∨a

(f ′) .

Proof. Applying Lemma3.1for the functionf ′, we may write that

(3.6)

∣∣∣∣∣f (τ)− f (t)

τ − t− 1

n

n−1∑i=0

f ′[t + (i + 1− λi)

τ − t

n

]∣∣∣∣∣≤ 1

n

[1

2+ max

i=0,n−1

∣∣∣∣λi −1

2

∣∣∣∣]∣∣∣∣∣

τ∨t

(f ′)

∣∣∣∣∣






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for anyt, τ ∈ [a, b], t 6= τ .Consequently, we have∣∣∣∣ 1πPV

∫ b

a

f (τ)− f (t)

τ − tdτ(3.7)

− 1

πn

n−1∑i=0

PV

∫ b

a

f ′[t + (i + 1− λi)

τ − t

n

]dτ

∣∣∣∣∣≤ 1

nπ

[1

2+ max

i=0,n−1

∣∣∣∣λi −1

2

∣∣∣∣]PV

∫ b

a

∣∣∣∣∣τ∨t

(f ′)

∣∣∣∣∣ dτ

≤ 1

nπ

[1

2+ max

i=0,n−1

∣∣∣∣λi −1

2

∣∣∣∣] [1

2(b− a) +

∣∣∣∣t− a + b

2

∣∣∣∣] b∨a

(f ′) .

On the other hand

PV

∫ b

a

f ′[t + (i + 1− λi)

τ − t

n

]dτ(3.8)

= limε→0+

[∫ t−ε

a

+

∫ b

t+ε

](f ′[t + (i + 1− λi)

τ − t

n

]dτ

)= lim

ε→0+

[n

i + 1− λi

f

(t + (i + 1− λi)

τ − t

n

)∣∣∣∣t−ε

a

+n

i + 1− λi

f

(t + (i + 1− λi)

τ − t

n

)∣∣∣∣bt+ε

]






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=n

i + 1− λi

[f

(t + (i + 1− λi)

b− t

n

)−f

(t + (i + 1− λi)

a− t

n

)]= (b− a)

[f ; t + (i + 1− λi)

b− t

n, (i + 1− λi)

a− t

n+ t

].

Since (see for example (2.7)),

(Tf) (a, b; t) =1

πPV

∫ b

a

f (τ)− f (t)

τ − tdτ +

f (t)

πln

(b− t

t− a

)for t ∈ (a, b) , then by (3.7) and (3.8) we deduce the desired estimate (3.5).

Remark 3.1. For n = 1, we recapture the inequality (2.6).

Corollary 3.3. With the assumptions of Theorem3.2, we have

(3.9) (Tf) (a, b; t) =f (t)

πln

(b− t

t− a

)+ lim

n→∞Sn (f ; λ, t)

uniformly by rapport oft ∈ (a, b) andλ with λi ∈ [0, 1) (i ∈ N).

Remark 3.2. If one needs to approximate the finite Hilbert Transform(Tf) (a, b; t)in terms of

f (t)

πln

(b− t

t− a

)+ Sn (f ; λ, t)






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with the accuracyε > 0 (ε small), then the theoretical minimal numbernε to bechosen is:

(3.10) nε :=

[b− a

επ

b∨a

(f ′)

]+ 1

where[α] is the integer part ofα.

It is obvious that the best inequality we can get in (3.5) is for λi = 12(

i = 0, n− 1)

obtaining the following corollary.

Corollary 3.4. Letf be as in Theorem3.2. Define

(3.11) Mn (f ; t) :=b− a

πn

n−1∑i=0

[f ;

(i +

1

2

)b− t

n+ t,

(i +

1

2

)a− t

n+ t

].

Then we have the estimate

(3.12)

∣∣∣∣(Tf) (a, b; t)− f (t)

πln

(b− t

t− a

)−Mn (f ; t)

∣∣∣∣≤ b− a

2nπ

[1

2(b− a) +

∣∣∣∣t− a + b

2

∣∣∣∣] b∨a

(f ′)

for anyt ∈ (a, b).

This rule will be numerically implemented in Section5 for different choicesof f andn.






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4. A More General Quadrature FormulaWe may state the following lemma.

Lemma 4.1. Let u : [a, b] → R be a function of bounded variation on[a, b],0 = µ0 < µ1 < · · · < µn−1 < µn = 1 andνi ∈ [µi, µi+1], i = 0, n− 1. Thenfor anyt, τ ∈ [a, b] with t 6= τ , we have the inequality:

(4.1)

∣∣∣∣∣ 1

τ − t

∫ τ

t

u (s) ds−n−1∑i=0

(µi+1 − µi) u [(1− νi) t + νiτ ]

∣∣∣∣∣≤[1

2∆n (µ) + max

i=0,n−1

∣∣∣∣νi −µi + µi+1

2

∣∣∣∣]∣∣∣∣∣

τ∨t

(u)

∣∣∣∣∣ ,where∆n (µ) := max

i=0,n−1(µi+1 − µi) .

Proof. Consider the division of[t, τ ] (if t < τ ) or [τ, t] (if τ < t) given by

(4.2) In : xi := (1− µi) t + µiτ(i = 0, n

).

Then the pointsξi := (1− νi) t + νiτ(i = 0, n− 1

)are betweenxi andxi+1.

We havexi+1 − xi = (µi+1 − µi) (τ − t)

(i = 0, n− 1

)and

ξi −xi + xi+1

2=

(νi −

µi + µi+1

2

)(τ − t)

(i = 0, n− 1

).






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Applying the inequality (2.1) on [xi, xi+1] with the intermediate pointsξi

(i = 0, n− 1

), we get∣∣∣∣∫ xi+1

xi

u (s) ds− (µi+1 − µi) (τ − t) u [(1− νi) t + νiτ ]

∣∣∣∣≤[1

2(µi+1 − µi) |τ − t|+ |τ − t|

∣∣∣∣νi −µi + µi+1

2

∣∣∣∣]∣∣∣∣∣xi+1∨xi

(u)

∣∣∣∣∣for anyi = 0, n− 1. Summing overi, using the generalised triangle inequalityand dividing by|t− τ | > 0, we obtain∣∣∣∣∣ 1

τ − t

∫ b

a

u (s) ds−n−1∑i=0

(µi+1 − µi) u [(1− νi) t + νiτ ]

∣∣∣∣∣≤

n−1∑i=0

[1

2(µi+1 − µi) +

∣∣∣∣νi −µi + µi+1

2

∣∣∣∣]∣∣∣∣∣xi+1∨xi

(u)

∣∣∣∣∣≤[1

2∆n (µ) + max

i=0,n−1

∣∣∣∣νi −µi + µi+1

2

∣∣∣∣]∣∣∣∣∣

τ∨t

(u)

∣∣∣∣∣and the inequality (4.1) is proved.

The following theorem holds.

Theorem 4.2. Let f : [a, b] → R be a differentiable function such that itsderivativef ′ is of bounded variation on[a, b]. If 0 = µ0 < µ1 < · · · < µn−1 <






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µn = 1 andνi ∈ [µi, µi+1],(i = 0, n− 1

), then

(4.3) (Tf) (a, b; t) =f (t)

πln

(b− t

t− a

)+

1

πQn (µ, ν, t) + Wn (µ, ν, t)

for anyt ∈ (a, b), where

(4.4) Qn (µ, ν, t) := µ1f′ (t) (b− a) + (b− a)

n−2∑i=1

{(µi+1 − µi)

× [f ; (1− νi) t + νib, (1− νi) t + νia]

}+ (1− µn−1) [f (b)− f (a)]

if ν0 = 0, νn−1 = 1,

(4.5) Qn (µ, ν, t) := µ1f′ (t) (b− a) + (b− a)

n−1∑i=1

(µi+1 − µi)

× [f ; (1− νi) t + νib, (1− νi) t + νia]

if ν0 = 0, νn−1 < 1,

(4.6) Qn (µ, ν, t) := (b− a)n−2∑i=1

(µi+1 − µi)

× [f ; (1− νi) t + νib, (1− νi) t + νia]

+ (1− µn−1) [f (b)− f (a)]






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if ν0 > 0, νn−1 = 1 and

(4.7) Qn (µ, ν, t)

:= (b− a)n−1∑i=1

(µi+1 − µi) [f ; (1− νi) t + νib, (1− νi) t + νia]

if ν0 > 0, νn−1 < 1.In all cases, the remainder satisfies the estimate:

|Wn (µ, ν, t)| ≤ 1

π

[1

2∆n (µ) + max

i=0,n−1

∣∣∣∣νi −µi + µi+1

2

∣∣∣∣](4.8)

×[1

2(b− a) +

∣∣∣∣t− a + b

2

∣∣∣∣] b∨a

(f ′)

≤ 1

π∆n (µ)

[1

2(b− a) +

∣∣∣∣t− a + b

2

∣∣∣∣]∣∣∣∣∣

b∨a

(f ′)

∣∣∣∣∣≤ 1

π∆n (µ) (b− a)

b∨a

(f ′) .

Proof. If we apply Lemma4.1for the functionf ′, we may write that∣∣∣∣∣f (τ)− f (t)

τ − t−

n−1∑i=0

(µi+1 − µi) f ′ [(1− νi) t + νiτ ]

∣∣∣∣∣≤[1

2∆n (µ) + max

i=0,n−1

∣∣∣∣νi −µi + µi+1

2

∣∣∣∣]∣∣∣∣∣

τ∨t

(f ′)

∣∣∣∣∣






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for anyt, τ ∈ [a, b] , t 6= τ .Taking thePV in both sides, we may write that

(4.9)

∣∣∣∣ 1πPV

∫ b

a

f (τ)− f (t)

τ − tdτ

− 1

πPV

∫ b

a

(n−1∑i=0

(µi+1 − µi) f ′ [(1− νi) t + νiτ ]

)dτ

∣∣∣∣∣≤ 1

π

[1

2∆n (µ) + max

i=0,n−1

∣∣∣∣νi −µi + µi+1

2

∣∣∣∣]PV

∫ b

a

∣∣∣∣∣τ∨t

(f ′)

∣∣∣∣∣ dτ.

If ν0 = 0, νn−1 = 1, then

PV

∫ b

a

(n−1∑i=0

(µi+1 − µi) f ′ [(1− νi) t + νiτ ]

)dτ

= PV

∫ b

a

µ1f′ (t) dτ +

n−2∑i=1

(µi+1 − µi) PV

∫ b

a

f ′ [(1− νi) t + νiτ ] dτ

+ (1− µn−1) PV

∫ b

a

f ′ (τ) dτ

= µ1f′ (t) (b− a) + (b− a)

n−2∑i=1

(µi+1 − µi)

× [f ; (1− νi) t + νib, (1− νi) t + νia] + (1− µn−1) [f (b)− f (a)] .






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If ν0 = 0, νn−1 < 1, then

PV

∫ b

a

(n−1∑i=0

(µi+1 − µi) f ′ [(1− νi) t + νiτ ]

)dτ

= µ1f′ (t) (b− a) + (b− a)

n−1∑i=1

(µi+1 − µi)

× [f ; (1− νi) t + νib, (1− νi) t + νia] .

If ν0 > 0, νn−1 = 1, then

PV

∫ b

a

(n−1∑i=0

(µi+1 − µi) f ′ [(1− νi) t + νiτ ]

)dτ

= (b− a)n−2∑i=1

(µi+1 − µi) [f ; (1− νi) t + νib, (1− νi) t + νia]

+ (1− µn−1) [f (b)− f (a)] .

and, finally, ifν0 > 0, νn−1 < 1, then

PV

∫ b

a

(n−1∑i=0

(µi+1 − µi) f ′ [(1− νi) t + νiτ ]

)dτ

= (b− a)n−1∑i=1

(µi+1 − µi) [f ; (1− νi) t + νib, (1− νi) t + νia] .






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Since

PV

∫ b

a

∣∣∣∣∣τ∨t

(f ′)

∣∣∣∣∣ dτ ≤[1

2(b− a) +

∣∣∣∣t− a + b

2

∣∣∣∣] b∨a

(f ′)

and

(Tf) (a, b; t) =1

πPV

∫ b

a

f (τ)− f (t)

τ − tdτ +

f (t)

πln

(b− t

t− a

),

then by (4.9) we deduce (4.3).






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5. Numerical ExperimentsFor a functionf : [a, b] → R, we may consider the quadrature formula

En(f ; a, b, t) :=f (t)

πln

(b− t

t− a

)+ Mn (f ; t) , t ∈ [a, b] .

As shown above in Section 4,En(f ; a, b, t) provides an approximation for theFinite Hilbert Transform(Tf) (a, b; t) and the error estimate fulfils the bounddescribed in(2.3).

If we consider the functionf : [−1, 1] → R, f (x) = exp(x), then the exactvalue of the Hilbert transform is

(Tf) (a, b; t) =exp(t)Ei(1− t)− exp(t)Ei(−1− t)

π, t ∈ [−1, 1]

and the plot of this function is embodied in Figure1.

If we implement the quadrature formula provided byEn(f ; a, b, t) usingMaple 6 and chose the value ofn = 100, then the errorEr (f ; a, b, t) :=(Tf) (a, b; t)− En(f ; a, b, t) has the variation described in the Figure2.

Forn =1,000, the plot ofEr (f ; a, b, t) is embodied in the following Figure3.

Now, if we consider another function,f : [−1, 1] → R, f(x) = sin x, then






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Figure 1:

the exact value of the Hilbert transform is

(Tf) (a, b; t) =−Si(−1 + t) cos(t) + Ci(1− t) sin(t)

π

+Si(t + 1) cos(t)− sin(t)Ci(t + 1))

π, t ∈ [−1, 1] ;

where

Si(x) =

∫ x

0

sin(t)

tdt, Ci(x) = γ + ln x +

∫ x

0

cos(t)− 1

tdt;

having the plot embodied in the following Figure4.If we choose the value ofn = 100, then the errorEr (f ; a, b, t) for the

functionf(x) = sin x, x ∈ [−1, 1] has the variation described in the Figure5below.






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Figure 2:

Moreover, forn =100,000, the behaviour ofEr (f ; a, b, t) is plotted in Fig-ure6.

Finally, if we choose the functionf : [−1, 1] → R, f(x) = sin (x2) , theMaple 6 is unable to produce an exact value of the finite Hilbert transform. Ifwe use our formula

En(f ; a, b, t) :=f (t)

πln

(b− t

t− a

)+ Mn (f ; t) , t ∈ [a, b]

for n =1,000, then we can produce the plot in Figure7.Taking into account the bound(3.12) we know that the accuracy of the plot

in Figure7 is at least of order10−5.






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Figure 3:

Figure 4:






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Figure 5:

Figure 6:






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Figure 7:






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