Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 3, Issue 4, Article 51, 2002
APPROXIMATING THE FINITE HILBERT TRANSFORM VIA AN OSTROWSKI
TYPE INEQUALITY FOR FUNCTIONS OF BOUNDED VARIATION
S.S. DRAGOMIR
S CHOOL OF C OMMUNICATIONS AND I NFORMATICS
V ICTORIA U NIVERSITY OF T ECHNOLOGY
PO B OX 14428
M ELBOURNE C ITY MC
V ICTORIA 8001, AUSTRALIA .
sever@matilda.vu.edu.au
URL: http://rgmia.vu.edu.au/SSDragomirWeb.html
Received 5 April, 2002; accepted 20 May, 2002
Communicated by B. Mond
A BSTRACT. Using the Ostrowski type inequality for functions of bounded variation, an approximation of the finite Hilbert Transform is given. Some numerical experiments are also provided.
Key words and phrases: Finite Hilbert Transform, Ostrowski’s Inequality.
2000 Mathematics Subject Classification. Primary 26D10, 26D15; Secondary 41A55, 47A99.
1. I NTRODUCTION
Cauchy principal value integrals of the form
Z t−ε
Z b
Z b
f (τ )
f (τ )
f (τ )
(1.1)
(T f ) (a, b; t) = P V
dτ := lim
dτ +
dτ
ε→0+
τ −t
a τ −t
a
t+ε τ − t
play an important role in fields like aerodynamics, the theory of elasticity and other areas of the
engineering sciences. They are also helpful tools in some methods 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 spline methods, see for example the
papers [1] – [12], [14] – [22], [24] – [33] and the references therein.
In contrast with all these methods, we point out here a new method in approximating the finite
Hilbert transform by the use of the Ostrowski inequality for functions of bounded variation
established in [13].
For a comprehensive list of papers on Ostrowski’s inequality, visit the site
http://rgmia.vu.edu.au.
ISSN (electronic): 1443-5756
c 2002 Victoria University. All rights reserved.
032-02
2
S.S. D RAGOMIR
Estimates for the error bounds and some numerical examples for the obtained approximation
are also presented.
2. S OME I NEQUALITIES ON THE I NTERVAL [a, b]
We start with the following lemma proved in [13] dealing with an Ostrowski type 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 all
x ∈ [a, b], we have the inequality:
b
Z b
1
a + b _
(2.1)
u (t) dt ≤
(b − a) + x −
(u) ,
u (x) (b − a) −
2
2 a
a
W
where ba (u) denotes the total variation of u on [a, b].
The constant 12 is the best possible one.
Proof. For the sake of completeness and since this result will be essentially used in what follows, we give here a short proof.
Using the integration by parts formula for the Riemann-Stieltjes integral we have
Z x
Z x
(t − a) du (t) = u (x) (x − a) −
u (t) dt
a
a
and
Z
b
b
Z
(t − b) du (t) = u (x) (b − x) −
x
u (t) dt.
x
If we add the above two equalities, we get
Z b
Z
(2.2)
u (x) (b − a) −
u (t) dt =
a
x
Z
b
(t − a) du (t) +
a
(t − b) du (t)
x
for any x ∈ [a, b].
If p : [c, d] → R is continuous on [c, d] and v : [c, d] → R is of bounded variation on [c, d],
then:
Z d
d
_
p (x) dv (x) ≤ sup |p (x)| (u) .
(2.3)
x∈[c,d]
c
Using (2.2) and (2.3), we deduce
Z
Z b
≤
u (x) (b − a) −
u
(t)
dt
a
a
x
c
Z b
(t − b) du (t)
(t − a) du (t) + ≤ (x − a)
x
x
_
(u) + (b − x)
a
(u)
x
"
≤ max {x − a, b − x}
b
_
x
_
a
(u) +
b
_
#
(u)
x
b
1
a + b _
=
(b − a) + x −
(u)
2
2 a
and the inequality (2.1) is proved.
J. Inequal. Pure and Appl. Math., 3(4) Art. 51, 2002
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F INITE H ILBERT T RANSFORM
3
Now, assume that the inequality (2.2) holds with a constant c > 0, i.e.,
b
Z b
a + b _
u (t) dt ≤ c (b − a) + x −
(2.4)
(u)
u (x) (b − a) −
2 a
a
for all x ∈ [a, b].
Consider the function u0 : [a, b] → R given by

 0 if x ∈ [a, b] a+b
2
u0 (x) =

1 if x = a+b
.
2
Then u0 is of bounded variation on [a, b] and
b
_
Z
b
(u0 ) = 2,
u0 (t) dt = 0.
a
a
If we apply (2.4) for u0 and choose x = a+b
, then we get 2c ≥ 1 which implies that c ≥ 12
2
showing that 12 is the best possible constant in (2.1).
The best inequality we can get from (2.1) is the following midpoint inequality.
Corollary 2.2. With the assumptions in Lemma 2.1, we have
Z b
b
_
1
a+b
(2.5)
(b − a) −
u (t) dt ≤ (b − a) (u) .
u
2
2
a
a
The constant 12 is best possible.
Using the above Ostrowski type inequality we may point out the following result in estimating
the finite Hilbert transform.
Theorem 2.3. Let f : [a, b] → R be a function such that its derivative f 0 : [a, b] → R is of
bounded variation on [a, b]. Then we have the inequality:
f
(t)
b
−
t
b
−
a
(2.6) (T f ) (a, b; t) −
ln
−
[f ; λt + (1 − λ) b, λt + (1 − λ) a]
π
t−a
π
b
1 1 1 1
a + b _ 0
≤
+ λ− (b − a) + t −
(f ) ,
π 2 2
2
2 a
for any t ∈ (a, b) and λ ∈ [0, 1), where [f ; α, β] is the divided difference, i.e.,
[f ; α, β] :=
f (α) − f (β)
.
α−β
Proof. Since f 0 is bounded on [a, b], it follows that f is Lipschitzian on [a, b] and thus the finite
Hilbert transform exists everywhere in (a, b).
As for the function f0 : (a, b) → R, f0 (t) = 1, t ∈ (a, b), we have
1
b−t
(T f0 ) (a, b; t) = ln
, t ∈ (a, b) ,
π
t−a
then obviously
(2.7)
f (t)
(T f ) (a, b; t) −
ln
π
J. Inequal. Pure and Appl. Math., 3(4) Art. 51, 2002
b−t
t−a
1
= PV
π
Z
a
b
f (τ ) − f (t)
dτ.
τ −t
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4
S.S. D RAGOMIR
Now, if we choose in (2.1), u = f 0 , x = λc + (1 − λ) d, λ ∈ [0, 1], then we get
|f (d) − f (c) − (d − c) f 0 (λc + (1 − λ) d)|
d
_ 0 1
c
+
d
(f )
≤
|d − c| + λc + (1 − λ) d −
2
2 c
where c, d ∈ (a, b), which is equivalent to
f (d) − f (c)
1 0
− f (λc + (1 − λ) d) ≤
+ λ−
d−c
2 (2.8)
d
1 _ 0 (f )
2 c
for any c, d ∈ (a, b), c 6= d.
Using (2.8), we may write
Z b
Z b
1
f
(τ
)
−
f
(t)
1
0
PV
(2.9)
dτ − P V
f (λt + (1 − λ) τ ) dτ π
τ −t
π
a
a
Z b _
t
1 1 1
0 ≤
+ λ − PV
(f ) dt
π 2 2
a τ
"
!
! #
Z t _
Z b _
t
τ
1 1 1 =
+ λ− (f 0 ) dt +
(f 0 ) dt
π 2 2
a
t
τ
t
#
"
t
b
_
_
1 1 1 ≤
+ λ − (t − a) (f 0 ) + (b − t) (f 0 )
π 2 2
a
t
b
1 1 1 1
a + b _ 0
+ λ − (b − a) + t −
(f ) .
≤
π 2
2
2
2 a
Since (for λ 6= 1)
1
PV
π
Z
a
b
f 0 (λt + (1 − λ) τ ) dτ
Z t−ε Z b 1
=
lim
+
(f 0 (λt + (1 − λ) τ ) dτ )
ε→0+
π
t+ε
" a
t−ε
b #
1
1
1
=
lim
f (λt + (1 − λ) τ ) +
f (λt + (1 − λ) τ )
π ε→0+ 1 − λ
1−λ
a
t+ε
1 f (t) − f (λt + (1 − λ) a) + f (λt + (1 − λ) b) − f (t)
·
π
1−λ
b−a
=
[f ; λt + (1 − λ) b, λt + (1 − λ) a] .
π
=
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 for λ = 12 . Thus, we
may state the following corollary.
J. Inequal. Pure and Appl. Math., 3(4) Art. 51, 2002
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F INITE H ILBERT T RANSFORM
5
Corollary 2.4. With the assumptions of Theorem 2.3, we have
b−t
f (t)
b−a
t + b a + t (2.10) (T f ) (a, b; t) −
ln
−
f;
,
π
t−a
π
2
2 b
_ 0
1 1
a
+
b
≤
(b − a) + t −
(f ) .
2π 2
2 a
The above Theorem 2.3 may be used to point out some interesting inequalities for the functions for which the finite Hilbert transforms (T f ) (a, b; t) can be expressed in terms of special
functions.
For instance, we have:
1) Assume that f : [a, b] ⊂ (0, ∞) → R, f (x) = x1 . Then
1
(b − t) a
(T f ) (a, b; t) =
ln
, t ∈ (a, b) ,
πt
(t − a) b
1
b−a
b−a
· [f ; λt + (1 − λ) b, λt + (1 − λ) a] = − ·
,
π
π [λt + (1 − λ) b] [λt + (1 − λ) a]
Z b
b
_
b 2 − a2
0
00
(f ) =
|f (t)| dt =
.
a2 b2
a
a
Using the inequality (2.6) we may write that
1
(b
−
t)
a
1
b
−
t
b
−
a
ln
−
ln
+
πt
(t − a) b
πt
t−a
π [λt + (1 − λ) b] [λt + (1 − λ) a] 2
b − a2
1 1
a
+
b
1 1 ·
+ λ − (b − a) + t −
≤
π 2
2
2
2 a2 b 2
which is equivalent to:
1
b b
−
a
− ln
(2.11) [λt + (1 − λ) b] [λt + (1 − λ) a] t
a 2
b − a2
1 1 1
a
+
b
·
≤
+ λ − (b − a) + t −
.
2
2
2
2 a2 b 2
If we use the notations
b−a
ln b − ln a
Aλ (x, y) := λx + (1 − λ) y
√
G (a, b) := ab
L (a, b)
:=
(the logarithmic mean)
(the weighted arithmetic mean)
(the geometric mean)
a+b
(the arithmetic mean)
2
then by (2.11) we deduce
1
1
1 1 1
2A (a, b)
Aλ (t, b) Aλ (t, a) − tL (a, b) ≤ 2 + λ − 2 2 (b − a) + |t − A (a, b)| G4 (a, b) ,
A (a, b)
:=
giving the following proposition:
J. Inequal. Pure and Appl. Math., 3(4) Art. 51, 2002
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6
S.S. D RAGOMIR
Proposition 2.5. With the above assumption, we have
(2.12)
|tL (a, b) − Aλ (t, b) Aλ (t, a)|
1 1
a
+
b
2A (a, b) 1 tAλ (t, b) Aλ (t, a) L (a, b)
≤ 4
+ λ − (b − a) + t −
G (a, b) 2
2
2
2 for any t ∈ (a, b), λ ∈ [0, 1).
In particular, for t = A (a, b) and λ = 12 , we get
(2.13)
(A
(a,
b)
+
a)
(A
(a,
b)
+
b)
A (a, b) L (a, b) −
4
≤
1 A2 (a, b) (A (a, b) + a) (A (a, b) + b)
·
·
L (a, b) .
2 G4 (a, b)
4
2) Assume that f : [a, b] ⊂ R → R, f (x) = exp (x). Then
exp (t)
[Ei (b − t) − Ei (a − t)] ,
π
(T f ) (a, b; t) =
where
Z
z
Ei (z) := P V
−∞
exp (t)
dt,
t
z ∈ R.
Also, we have:
b−a
[exp; λt + (1 − λ) b, λt + (1 − λ) a]
π
1 exp (λt + (1 − λ) b) − exp (λt + (1 − λ) a)
= ·
,
π
1−λ
b
_
0
Z
(f ) =
a
b
|f 00 (t)| dt = exp (b) − exp (a) .
a
Using the inequality (2.6) we may write:
b−t
(2.14) exp (t) Ei (b − t) − Ei (a − t) − ln
t−a
exp (λt + (1 − λ) b) − exp (λt + (1 − λ) a) −
1−λ
1 1 1
a
+
b
[exp (b) − exp (a)]
≤
+ λ − (b − a) + t −
2
2
2
2 for any t ∈ (a, b).
If in (2.14) we make λ =
1
2
and t =
a+b
,
2
we get
exp a + b Ei b − a − 2 exp a + 3b − exp 3a + b 2
2
4
4
≤
J. Inequal. Pure and Appl. Math., 3(4) Art. 51, 2002
1
(b − a) [exp (b) − exp (a)] ,
4
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F INITE H ILBERT T RANSFORM
7
which is equivalent to:
Ei b − a − 2 exp b − a − exp − b − a 2
4
4
b−a
b−a
1
≤ (b − a) exp
− exp −
.
4
2
2
If in this inequality we make b−a
= z > 0, then we get
2
z i 1
h
z (2.15)
− exp −
≤ z [exp (z) − exp (−z)]
Ei (z) − 2 exp
2
2
2
for any z > 0.
Consequently, we may state the following proposition.
Proposition 2.6. With the above assumptions, we have
1 (2.16)
Ei (z) − 4 sinh 2 z ≤ z sinh (z)
for any z > 0.
The reader may get other similar inequalities for special functions if appropriate examples of
functions f are chosen.
3. A Q UADRATURE F ORMULA
FOR
E QUIDISTANT D IVISIONS
The 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 all n ≥ 1,
λi ∈ [0, 1) (i = 0, . . . , n − 1) and t, τ ∈ [a, b] with t 6= τ , we have the inequality:
n−1 1 Z τ
1X
τ − t u (s) ds −
u t + (i + 1 − λi )
(3.1) τ − t t
n i=0
n _
τ
1 1
1
≤
+ max λi − (u) .
n 2 i=0,n−1
2 t
Proof. Consider the equidistant division of [t, τ ] (if t < τ ) or [τ, t] (if τ < t) given by
τ −t
En : xi = t + i ·
, i = 0, n.
n
Then the points ξi = λi t + i · τ n−t + (1 − λi ) t + (i + 1) · τ n−t
λi ∈ [0, 1] , i = 0, n − 1
are between xi and xi+1 . We observe that we may write for simplicity ξi = t + (i + 1 − λi ) τ n−t
i = 0, n − 1 . We also have
(3.2)
ξi −
xi + xi+1
τ −t
=
(1 − 2λi ) ,
2
2n
τ −t
ξi − xi = (1 − λi )
n
and
xi+1 − ξi = λi ·
τ −t
n
for any i = 0, n − 1.
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8
S.S. D RAGOMIR
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
Z xi+1
τ − t
τ −t
(3.3) u t + (i + 1 − λi )
−
u (s) ds
n
n
xi
xi+1
_
1 |τ − t| τ − t
·
+
(1 − 2λi ) (u) .
≤
2
n
2n
x
i
Summing, we get
Z
n−1 τ
τ − t τ −tX
u t + (i + 1 − λi )
u (s) ds −
t
n i=0
n x
n−1
i+1
_
|τ − t| X
≤
[1 + |1 − 2λi |] (u)
x
2n i=0
i
_
τ
1
|τ − t| 1
=
+ max λi − (u) ,
2 t
n
2 i=0,n−1
which is equivalent to (3.1).
We may now state the following theorem in approximating the finite Hilbert transform of a
differentiable function with the derivative of bounded variation on [a, b].
Theorem 3.2. Let f : [a, b] → R be a differentiable function such that its derivative f 0 is of
bounded variation on [a, b]. If λ = (λi )i=0,n−1 , λi ∈ [0, 1) i = 0, n − 1 and
(3.4)
n−1 b−t
a−t
b−aX
Sn (f ; λ, t) :=
f ; (i + 1 − λi )
+ t, (i + 1 − λi )
+t ,
πn i=0
n
n
then we have the estimate:
f
(t)
b
−
t
(T f ) (a, b; t) −
(3.5)
ln
−
S
(f
;
λ,
t)
n
π
t−a
b
b−a 1
1 1
a + b _ 0
≤
+ max λi − (b − a) + t −
(f )
nπ 2 i=0,n−1 2
2
2 a
b
≤
b−a_ 0
(f ) .
nπ a
Proof. Applying Lemma 3.1 for the function f 0 , we may write that
n−1
f (τ ) − f (t) 1 X
τ
−
t
(3.6) −
f 0 t + (i + 1 − λi )
τ −t
n i=0
n 1 1
≤
+ max λi −
n 2 i=0,n−1
τ
1 _ 0 (f )
2 t
for any t, τ ∈ [a, b], t 6= τ .
J. Inequal. Pure and Appl. Math., 3(4) Art. 51, 2002
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F INITE H ILBERT T RANSFORM
9
Consequently, we have
Z b
Z b n−1
1
f (τ ) − f (t)
1 X
τ
−
t
(3.7)
dτ −
PV
f 0 t + (i + 1 − λi )
dτ PV
π
τ −t
πn i=0
n
a
a
Z b _
τ
1 1 1
≤
+ max λi − P V
(f 0 ) dτ
2
nπ 2 i=0,n−1
a
t
_
1
b
a
+
b
1
1 1
(b − a) + t −
(f 0 ) .
≤
+ max λi − 2
2
2 a
nπ 2 i=0,n−1
On the other hand
Z b τ −t
0
PV
f t + (i + 1 − λi )
(3.8)
dτ
n
a
Z t−ε Z b τ −t
0
dτ
= lim
+
f t + (i + 1 − λi )
ε→0+
n
a
t+ε
"
t−ε
τ − t n
f t + (i + 1 − λi )
= lim
ε→0+ i + 1 − λi
n
a
b #
n
τ − t f t + (i + 1 − λi )
+
i + 1 − λi
n
t+ε
n
b−t
a−t
=
f t + (i + 1 − λi )
− f t + (i + 1 − λi )
i + 1 − λi
n
n
b−t
a−t
= (b − a) f ; t + (i + 1 − λi )
, (i + 1 − λi )
+t .
n
n
Since (see for example (2.7)),
1
(T f ) (a, b; t) = P V
π
Z
a
b
f (τ ) − f (t)
f (t)
dτ +
ln
τ −t
π
b−t
t−a
for t ∈ (a, b) , then by (3.7) and (3.8) we deduce the desired estimate (3.5).
Remark 3.3. For n = 1, we recapture the inequality (2.6).
Corollary 3.4. With the assumptions of Theorem 3.2, we have
f (t)
b−t
(3.9)
(T f ) (a, b; t) =
ln
+ lim Sn (f ; λ, t)
n→∞
π
t−a
uniformly by rapport of t ∈ (a, b) and λ with λi ∈ [0, 1) (i ∈ N).
Remark 3.5. If one needs to approximate the finite Hilbert Transform (T f ) (a, b; t) in terms of
f (t)
b−t
ln
+ Sn (f ; λ, t)
π
t−a
with the accuracy ε > 0 (ε small), then the theoretical minimal number nε to be chosen is:
"
#
b
b−a_ 0
(f ) + 1
(3.10)
nε :=
επ a
where [α] is the integer part of α.
J. Inequal. Pure and Appl. Math., 3(4) Art. 51, 2002
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10
S.S. D RAGOMIR
It is obvious that the best inequality we can get in (3.5) is for λi = 21 i = 0, n − 1 obtaining
the following corollary.
Corollary 3.6. Let f be as in Theorem 3.2. Define
n−1 b−aX
1 b−t
1 a−t
(3.11)
Mn (f ; t) :=
f; i +
+ t, i +
+t .
πn i=0
2
n
2
n
Then we have the estimate
b−t
f (t)
(3.12) (T f ) (a, b; t) −
ln
− Mn (f ; t)
π
t−a
b
_ 0
b−a 1
a
+
b
≤
(b − a) + t −
(f )
2nπ 2
2 a
for any t ∈ (a, b).
This rule will be numerically implemented in Section 5 for different choices of f and n.
4. A M ORE G ENERAL Q UADRATURE F ORMULA
We 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 and νi ∈ [µi , µi+1 ], i = 0, n − 1. Then for any t, τ ∈ [a, b] with t 6= τ , we
have the inequality:
n−1
1 Z τ
X
(4.1) u (s) ds −
(µi+1 − µi ) u [(1 − νi ) t + νi τ ]
τ − t t
i=0
τ
_
1
µ
+
µ
i
i+1 ≤ ∆n (µ) + max νi −
(u)
,
2
2
i=0,n−1
t
where ∆n (µ) := max (µi+1 − µi ) .
i=0,n−1
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 + µi τ i = 0, n .
Then the points ξi := (1 − νi ) t + νi τ i = 0, n − 1 are between xi and xi+1 . We have
xi+1 − xi = (µi+1 − µi ) (τ − t) i = 0, n − 1
and
xi + xi+1
ξi −
=
2
µi + µi+1
νi −
2
(τ − t)
i = 0, n − 1 .
Applying the inequality (2.1) on [xi , xi+1 ] with the intermediate points ξi i = 0, n − 1 , we get
Z xi+1
u (s) ds − (µi+1 − µi ) (τ − t) u [(1 − νi ) t + νi τ ]
xi
x_
i+1
1
µi + µi+1 ≤
(µi+1 − µi ) |τ − t| + |τ − t| νi −
(u)
x
2
2
i
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F INITE H ILBERT T RANSFORM
11
for any i = 0, n − 1. Summing over i, using the generalised triangle inequality and dividing by
|t − τ | > 0, we obtain
n−1
1 Z b
X
u (s) ds −
(µi+1 − µi ) u [(1 − νi ) t + νi τ ]
τ − t a
i=0
x_
n−1 i+1
X
µ
+
µ
1
i
i+1 ≤
(µi+1 − µi ) + νi −
(u)
2
2
xi
i=0
_
τ
µ
+
µ
1
i
i+1
≤ ∆n (µ) + max νi −
(u)
t
2
2
i=0,n−1
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 its derivative f 0 is of
bounded variation
on [a, b]. If 0 = µ0 < µ1 < · · · < µn−1 < µn = 1 and νi ∈ [µi , µi+1 ],
i = 0, n − 1 , then
(4.3)
f (t)
(T f ) (a, b; t) =
ln
π
b−t
t−a
+
1
Qn (µ, ν, t) + Wn (µ, ν, t)
π
for any t ∈ (a, b), where
0
(4.4) Qn (µ, ν, t) := µ1 f (t) (b − a) + (b − a)
n−2 X
(µi+1 − µi )
i=1
× [f ; (1 − νi ) t + νi b, (1 − νi ) t + νi a]
+ (1 − µn−1 ) [f (b) − f (a)]
if ν0 = 0, νn−1 = 1,
0
(4.5) Qn (µ, ν, t) := µ1 f (t) (b − a) + (b − a)
n−1
X
(µi+1 − µi )
i=1
× [f ; (1 − νi ) t + νi b, (1 − νi ) t + νi a]
if ν0 = 0, νn−1 < 1,
(4.6) Qn (µ, ν, t) := (b − a)
n−2
X
(µi+1 − µi )
i=1
× [f ; (1 − νi ) t + νi b, (1 − νi ) t + νi a] + (1 − µn−1 ) [f (b) − f (a)]
if ν0 > 0, νn−1 = 1 and
(4.7)
Qn (µ, ν, t) := (b − a)
n−1
X
(µi+1 − µi ) [f ; (1 − νi ) t + νi b, (1 − νi ) t + νi a]
i=1
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12
S.S. D RAGOMIR
if ν0 > 0, νn−1 < 1.
In all cases, the remainder satisfies the estimate:
1 1
µ
+
µ
i
i+1
(4.8)
|Wn (µ, ν, t)| ≤
∆n (µ) + max νi −
2
π 2
i=0,n−1
b
a + b _ 0
1
×
(b − a) + t −
(f )
2
2 a
b
_ 0 1
1
a
+
b
(f )
≤ ∆n (µ)
(b − a) + t −
π
2
2 a
≤
b
_
1
∆n (µ) (b − a)
π
(f 0 ) .
a
Proof. If we apply Lemma 4.1 for the function f 0 , we may write that
n−1
f (τ ) − f (t) X
−
(µi+1 − µi ) f 0 [(1 − νi ) t + νi τ ]
τ −t
i=0
τ
_
µ
+
µ
1
i
i+1 0 ≤ ∆n (µ) + max νi −
(f
)
2
2
i=0,n−1
t
for any t, τ ∈ [a, b] , t 6= τ .
Taking the P V in both sides, we may write that
(4.9)
Z b
1
f (τ ) − f (t)
PV
dτ
π
τ −t
a
! Z b X
n−1
1
− PV
(µi+1 − µi ) f 0 [(1 − νi ) t + νi τ ] dτ π
a
i=0
Z b _
τ
1 1
µi + µi+1 0 ≤
∆n (µ) + max νi −
(f ) dτ.
PV
π 2
2
i=0,n−1
a t
If ν0 = 0, νn−1 = 1, then
Z
PV
a
b
!
n−1
X
(µi+1 − µi ) f 0 [(1 − νi ) t + νi τ ] dτ
i=0
Z
= PV
b
0
µ1 f (t) dτ +
n−2
X
a
Z
Z
(µi+1 − µi ) P V
f 0 [(1 − νi ) t + νi τ ] dτ
a
i=1
b
+ (1 − µn−1 ) P V
b
f 0 (τ ) dτ
a
= µ1 f 0 (t) (b − a) + (b − a)
n−2
X
(µi+1 − µi ) [f ; (1 − νi ) t + νi b, (1 − νi ) t + νi a]
i=1
+ (1 − µn−1 ) [f (b) − f (a)] .
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F INITE H ILBERT T RANSFORM
13
If ν0 = 0, νn−1 < 1, then
Z
b
PV
a
n−1
X
!
(µi+1 − µi ) f 0 [(1 − νi ) t + νi τ ] dτ
i=0
n−1
X
0
= µ1 f (t) (b − a) + (b − a)
(µi+1 − µi ) [f ; (1 − νi ) t + νi b, (1 − νi ) t + νi a] .
i=1
If ν0 > 0, νn−1 = 1, then
Z
b
PV
a
n−1
X
!
(µi+1 − µi ) f 0 [(1 − νi ) t + νi τ ] dτ
i=0
= (b − a)
n−2
X
(µi+1 − µi ) [f ; (1 − νi ) t + νi b, (1 − νi ) t + νi a] + (1 − µn−1 ) [f (b) − f (a)] .
i=1
and, finally, if ν0 > 0, νn−1 < 1, then
Z
PV
a
b
n−1
X
!
(µi+1 − µi ) f 0 [(1 − νi ) t + νi τ ] dτ
i=0
n−1
X
= (b − a)
(µi+1 − µi ) [f ; (1 − νi ) t + νi b, (1 − νi ) t + νi a] .
i=1
Since
b
Z b _
τ
1
a + b _ 0
0 PV
(b − a) + t −
(f )
(f ) dτ ≤
2
2 a
a t
and
1
(T f ) (a, b; t) = P V
π
then by (4.9) we deduce (4.3).
Z
a
b
f (τ ) − f (t)
f (t)
dτ +
ln
τ −t
π
b−t
t−a
,
5. N UMERICAL E XPERIMENTS
For a function f : [a, b] → R, we may consider the quadrature formula
f (t)
b−t
En (f ; a, b, t) :=
ln
+ Mn (f ; t) , t ∈ [a, b] .
π
t−a
As shown above in Section 4, En (f ; a, b, t) provides an approximation for the Finite Hilbert
Transform (T f ) (a, b; t) and the error estimate fulfils the bound described in (2.3).
If we consider the function f : [−1, 1] → R, f (x) = exp(x), then the exact value of the
Hilbert transform is
exp(t)Ei(1 − t) − exp(t)Ei(−1 − t)
(T f ) (a, b; t) =
, t ∈ [−1, 1]
π
and the plot of this function is embodied in Figure 5.1.
If we implement the quadrature formula provided by En (f ; a, b, t) using Maple 6 and chose
the value of n = 100, then the error Er (f ; a, b, t) := (T f ) (a, b; t) − En (f ; a, b, t) has the
variation described in the Figure 5.2.
For n =1,000, the plot of Er (f ; a, b, t) is embodied in the following Figure 5.3.
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14
S.S. D RAGOMIR
Figure 5.1:
Figure 5.2:
Now, if we consider another function, f : [−1, 1] → R, f (x) = sin x, then the exact value
of the Hilbert transform is
(T f ) (a, b; t) =
−Si(−1 + t) cos(t) + Ci(1 − t) sin(t)
π
Si(t + 1) cos(t) − sin(t)Ci(t + 1))
+
, t ∈ [−1, 1] ;
π
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F INITE H ILBERT T RANSFORM
15
Figure 5.3:
where
Z
x
sin(t)
Si(x) =
dt,
Ci(x) = γ + ln x +
t
0
having the plot embodied in the following Figure 5.4.
Z
0
x
cos(t) − 1
dt;
t
Figure 5.4:
If we choose the value of n = 100, then the error Er (f ; a, b, t) for the function f (x) =
sin x, x ∈ [−1, 1] has the variation described in the Figure 5.5 below.
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16
S.S. D RAGOMIR
Figure 5.5:
Figure 5.6:
Moreover, for n =100,000, the behaviour of Er (f ; a, b, t) is plotted in Figure 5.6.
Finally, if we choose the function f : [−1, 1] → R, f (x) = sin (x2 ) , the Maple 6 is unable
to produce an exact value of the finite Hilbert transform. If we use our formula
f (t)
b−t
En (f ; a, b, t) :=
ln
+ Mn (f ; t) , t ∈ [a, b]
π
t−a
for n =1,000, then we can produce the plot in Figure 5.7.
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F INITE H ILBERT T RANSFORM
17
Figure 5.7:
Taking into account the bound (3.12) we know that the accuracy of the plot in Figure 5.7 is
at least of order 10−5 .
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