SHARP ERROR BOUNDS OF A QUADRATURE RULE WITH ONE MULTIPLE NODE FOR THE FINITE HILBERT TRANSFORM IN SOME CLASSES OF CONTINUOUS DIFFERENTIABLE FUNCTIONS

S.S. DRAGOMIR

Abstract. Some inequalities and approximations for the finite Hilbert transform by the use of Taylor’s formula with the integral remainder are given.

1. Introduction

Cauchy principal value integrals of the form

(1.1) (T f) (a, b; t) = P V ∫_{a}^{b} \frac{f (τ)}{τ − t}dτ := lim_{ε→0+} [∫_{a}^{t−ε} \frac{f (τ)}{τ − t}dτ + ∫_{t+ε}^{b} \frac{f (τ)}{τ − t}dτ]

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. [22]).

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], [13] – [21], [23] – [32] 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 Taylor formula with integral remainder for functions whose n-th derivatives are absolutely continuous.

For a comprehensive list of papers on generalized Taylor formulae and Ostrowski’s inequality, visit the site http://rgmia.vu.edu.au.

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

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

The following result holds. Theorem 1. Let f : [a, b] → ℝ be such that f^{(n−1)} (n ≥ 1) is absolutely continuous on [a, b]. Then we have the bounds: (2.1)

|(T f) (a, b; t) − f (t) ln (\frac{b − t}{t − a}) − ∑_{k=1}^{n−1} \frac{f^{(k)} (t)}{k!} · [\frac{(b − t)^{k} + (−1)^{k+1} (t − a)^{k}}{k} ]|

Date: May 28, 2001.

2000 Mathematics Subject Classification. Primary 26D15, 26D10; Secondary 41A55.

Key words and phrases. Hilbert Transform Taylor’s Formula, Analytic Inequalities.

≤ \{ \begin{matrix}\frac{∥f^{(n)}∥_{[a,b],∞}}{n · n!} [(b − t)^{n} + (t − a)^{n}] , & if f^{(n)} ∈ L_{∞} [a, b] ; \\ \frac{q ∥f^{(n)}∥_{[a,b],p} [(b − t)^{n−1+ \frac{1}{q}} + (t − a)^{n−1+ \frac{1}{q}} ]}{(n − 1)! [(n − 1) q + 1]^{1+ \frac{1}{q}}}, & if f^{(n)} ∈ L_{p} [a, b] , \\ & p > 1, \frac{1}{p} + \frac{1}{q} = 1; \\ \frac{∥f^{(n)}∥_{[a,b],1}}{(n − 1) · (n − 1)!} [(b − t)^{n−1} + (t − a)^{n−1}] , \end{matrix}

≤ \{ \begin{matrix}\frac{(b − a)^{n}}{n · n!} ∥f^{(n)}∥_{[a,b],∞} , & if f^{(n)} ∈ L_{∞} [a, b] ; \\ \frac{q (b − a)^{n−1+ \frac{1}{q}}}{(n − 1)! [(n − 1) q + 1]^{1+ \frac{1}{q}}} ∥f^{(n)}∥_{[a,b],p} , & if f^{(n)} ∈ L_{p} [a, b] , \\ & p > 1, \frac{1}{p} + \frac{1}{q} = 1; \\ \frac{1}{(n − 1) · (n − 1)!} (b − a)^{n−1} ∥f^{(n)}∥_{[a,b],1} , \end{matrix}

for any t ∈ (a, b).

Proof. Start with Taylor’s formula for a function g : I → ℝ (I is a compact interval) with the property that g^{(n−1)} (n ≥ 1) is absolutely continuous on I, then we have

g (x) = ∑_{k=0}^{n−1} \frac{(x − a)^{k}}{k!} g^{(k)} (a) + \frac{1}{(n − 1)!} ∫_{a}^{x} (x − t)^{n−1} g^{(n)} (t) dt,

where a, x ∈ I̊(I̊is the interior of I). This implies that

|g (x) − ∑_{k=0}^{n−1} \frac{(x − a)^{k}}{k!} g^{(k)} (a)| ≤ \frac{1}{(n − 1)!} |∫_{a}^{x} |x − t|^{n−1} |g^{(n)} (t)| dt|

= : \frac{1}{(n − 1)!} · M (x)

for any a, x ∈ I̊.

Before we estimate M (x) , let us introduce the following notations

∥h∥_{[a,x],p} := |∫_{a}^{x} |h (t)|^{p} dt|^{\frac{1}{p}} if p ≥ 1

and

∥h∥_{[a,x],∞} := ess sup_{\begin{matrix}t∈[a,x] \\ t∈([x,a]) \end{matrix}} |h (t)| ,

where a, x ∈ I̊.

It is obvious now that

M (x) ≤ sup_{\begin{matrix}t∈[a,x] \\ (t∈[x,a]) \end{matrix}} |g^{(n)} (t)| |∫_{a}^{x} |x − t|^{n−1} dt| = ∥g^{(n)}∥_{[a,x],∞} \frac{|x − a|^{n}}{n!}

for any a, x ∈ I̊.

Using Hölder’s integral inequality, we may state that

M (x) ≤ |∫_{a}^{x} |g^{(n)} (t)|^{p} dt|^{\frac{1}{p}} |∫_{a}^{x} |x − t|^{(n−1)q} dt|^{\frac{1}{q}}

= ∥g^{(n)}∥_{[a,x],p} \begin{matrix}|x − a| & n−1+ \frac{1}{q} \\ [(n − 1) q + 1] & \frac{1}{q} \end{matrix}

for any a, x ∈ I̊.

Also, we observe that

M (x) ≤ |x − a|^{n−1} |∫_{a}^{x} |g^{(n)} (t)| dt| = ∥g^{(n)}∥_{[a,x],1} |x − a|^{n−1}

for all a, x ∈ I̊.

In conclusion, we may state the following inequality which will be used in the sequel

|g (x) − ∑_{k=0}^{n−1} \frac{(x − a)^{k}}{k!} g^{(k)} (a)| (2.2)

≤ \{ \begin{matrix}\frac{|x − a|^{n}}{n!} ∥g^{(n)}∥_{[a,x],∞} & if & g^{(n)} ∈ L_{∞}(I̊); \\ \frac{|x − a|^{n−1+ \frac{1}{q}}}{(n − 1)! [(n − 1) q + 1]^{\frac{1}{q}}} ∥g^{(n)}∥_{[a,x],p} & if & g^{(n)} ∈ L_{p}(I̊), \\ & & p > 1, \frac{1}{p} + \frac{1}{q} = 1; \\ \frac{|x − a|^{n−1}}{(n − 1)!} & ∥g^{(n)}∥ & [a,x],1 \end{matrix}

for any a, x ∈I̊.

Now, let us note for the function f_{0} : [a, b] → ℝ, f_{0} (t) = 1, we have that

(T f_{0}) (a, b; t) = ln (\frac{b − t}{t − a}) , t ∈ (a, b) ,

and then

(T f) (a, b; t) = P V ∫_{a}^{b} \frac{f (τ) − f (t) + f (t)}{τ − t}dτ
= P V ∫_{a}^{b} \frac{f (τ) − f (t)}{τ − t}dτ + f (t) ln (\frac{b − t}{t − a}) ,

giving the equality

(2.3)  (T f) (a, b; t) − f (t) ln (\frac{b − t}{t − a}) = P V ∫_{a}^{b} \frac{f (τ) − f (t)}{τ − t}dτ.

4 S.S. DRAGOMIR

Writing (2.2) for g = f, x = τ, a = t, we get

|\frac{f (τ) − f (t)}{τ − t} − ∑_{k=1}^{n−1} \frac{(τ − t)^{k−1}}{k!} f^{(k)} (t)| (2.4)

≤ \{ \begin{matrix}\frac{|τ − t|^{n−1}}{n!} ∥f^{(n)}∥_{[t,τ],∞} & if & f^{(n)} ∈ L_{∞} [a, b] ; \\ \frac{|τ − t|^{n−2+ \frac{1}{q}}}{(n − 1)! [(n − 1) q + 1]^{\frac{1}{q}}} ∥f^{(n)}∥_{[t,τ],p} & if & f^{(n)} ∈ L_{p} [a, b] , \\ & & p > 1, \frac{1}{p} + \frac{1}{q} = 1; \\ \frac{|τ − t|^{n−2}}{(n − 1)!} & ∥f^{(n)}∥ & [t,τ],1 \end{matrix}

for any t, τ ∈ (a, b) , t ≠ τ.

If we take the PV in (2.4), then we may write

|P V ∫_{a}^{b} \frac{f (τ) − f (t)}{τ − t}dτ − ∑_{k=1}^{n−1} \frac{f^{(k)} (t)}{k!}P V ∫_{a}^{b} (τ − t)^{k−1} dτ| (2.5)

≤ P V ∫_{a}^{b} |\frac{f (τ) − f (t)}{τ − t} − ∑_{k=1}^{n−1} \frac{(τ − t)^{k−1}}{k!} f^{(k)} (t)| dτ
≤ \{ \begin{matrix}\frac{1}{n!}P V ∫_{a}^{b} |τ − t|^{n−1} ∥f^{(n)}∥_{[t,τ],∞} dτ \\ \frac{1}{(n − 1)! [(n − 1) q + 1]^{\frac{1}{q}}} P V ∫_{a}^{b} |τ − t|^{n−2+ \frac{1}{q}} ∥f^{(n)}∥_{[t,τ],p} dτ \\ \frac{1}{(n − 1)!}P V ∫_{a}^{b} |τ − t|^{n−2} ∥f^{(n)}∥_{[t,τ],1} dτ \end{matrix}

≤ \{ \begin{matrix}\frac{1}{n!} ∥f^{(n)}∥_{[a,b],∞} P V ∫_{a}^{b} |τ − t|^{n−1} dτ \\ \frac{1}{(n − 1)! [(n − 1) q + 1]^{\frac{1}{q}}} ∥f^{(n)}∥_{[a,b],p} P V ∫_{a}^{b} |τ − t|^{n−2+ \frac{1}{q}} dτ \\ \frac{1}{(n − 1)!} ∥f^{(n)}∥_{[a,b],1} P V ∫_{a}^{b} |τ − t|^{n−2} dτ. \end{matrix}

However,

P V ∫_{a}^{b} |τ − t|^{n−1} dτ = \frac{1}{n} [(b − t)^{n} + (t − a)^{n}] ,

P V ∫_{a}^{b} |τ − t|^{n−2+ \frac{1}{q}} dτ = \frac{q}{[(n − 1) q + 1]} [(b − t)^{n−1+ \frac{1}{q}} + (t − a)^{n−1+ \frac{1}{q}} ] ,

P V ∫_{a}^{b} |τ − t|^{n−2} dτ = \frac{1}{n − 1} [(b − t)^{n−1} + (t − a)^{n−1}]

and

P V ∫_{a}^{b} (τ − t)^{k−1} dτ = \frac{1}{k} [(b − t)^{k} + (−1)^{k+1} (t − a)^{k}]

and then by (2.5) we deduce the desired result (2.1).

It is obvious that the best inequality one would deduce from (2.1) is the one for t = \frac{a+b}{2}, getting the following corollary. Corollary 1. With the assumptions of Theorem 1, we have

|(T f) (a, b; \frac{a + b}{2}) − ∑_{k=1}^{n−1} \frac{(b − a)^{k}}{2^{k} · k · k!} [1 + (−1)^{k+1}] f^{(k)} (\frac{a + b}{2})| (2.6)

≤ \{ \begin{matrix}\frac{(b − a)^{n}}{2^{n−1} · n · n!} ∥f^{(n)}∥_{[a,b],∞} , & if f^{(n)} ∈ L_{∞} [a, b] ; \\ \frac{q (b − a)^{n−1+ \frac{1}{q}}}{2^{n−2+ \frac{1}{q}} (n − 1)! [(n − 1) q + 1]^{1+ \frac{1}{q}}} ∥f^{(n)}∥_{[a,b],p} , & if f^{(n)} ∈ L_{p} [a, b] , \\ & p > 1, \frac{1}{p} + \frac{1}{q} = 1; \\ \frac{(b − a)^{n−1}}{2^{n−2} · (n − 1) · (n − 1)!} ∥f^{(n)}∥_{[a,b],1} . \end{matrix}

It is important to note that for small intervals, we basically have the following representation: Corollary 2. Assume that f ∈ C^{∞} [a, b] and 0 < b − a ≤ 1. Then

(T f) (a, b; t) = f (t) ln (\frac{b − t}{t − a}) + ∑_{k=1}^{∞} \frac{f^{(k)} (t)}{k!} · [\frac{(b − t)^{k} + (−1)^{k+1} (t − a)^{k}}{k} ]

and the convergence is uniform by rapport of t ∈ [a, b].

3. The Composite Case

The following lemma holds. Lemma 1. Let g : [a, b] → ℝ be such that g^{(n−1)} (n ≥ 1) is absolutely continuous on [a, b] . Then for any m ∈ ℕ, m ≥ 1, we have the inequality:

|\frac{1}{b − a} ∫_{a}^{b} g (u) du − ∑_{i=0}^{m−1} ∑_{k=1}^{n} \frac{(b − a)^{k−1}}{m^{k}k!} · g^{(k−1)} (a + i · \frac{b − a}{m})| (3.1)

≤ \{ \begin{matrix}\frac{(b − a)^{n}}{m^{n} (n + 1)!} ∥g^{(n)}∥_{[a,b],∞} , & if g^{(n)} ∈ L_{∞} [a, b] ; \\ \frac{(b − a)^{n−1+ \frac{1}{q}}}{m^{n}n! (nq + 1)^{\frac{1}{q}}} ∥g^{(n)}∥_{[a,b],p} , & if g^{(n)} ∈ L_{p} [a, b] , \\ & p > 1, \frac{1}{p} + \frac{1}{q} = 1; \\ \frac{(b − a)^{n−1}}{m^{n}n!} ∥g^{(n)}∥_{[a,b],1} . \end{matrix}

Proof. Write Taylor’s formula with the integral remainder for φ (x)=∫_{α}^{x} g (u) du and then choose x = β, to get:

|∫_{α}^{β} g (u) du − ∑_{k=1}^{n} \frac{(β − α)^{k}}{k!} g^{(k−1)} (α)| (3.2)

≤ \{ \begin{matrix}\frac{|β − α|^{n+1}}{(n + 1)!} ∥g^{(n)}∥_{[α,β],∞} , & if g^{(n)} ∈ L_{∞} [a, b] ; \\ \frac{|β − α|^{n+ \frac{1}{q}}}{n! (nq + 1)^{\frac{1}{q}}} ∥g^{(n)}∥_{[α,β],p} , & if g^{(n)} ∈ L_{p} [a, b] , \\ & p > 1, \frac{1}{p} + \frac{1}{q} = 1; \\ \frac{|β − α|^{n}}{n!} & ∥g^{(n)}∥ & [α,β],1 \end{matrix}

for any α, β ∈ [a, b] .

Now, if we consider the division

I_{n} : x_{i} = a + i · \frac{b − a}{m}, i = \overline{0, m},

and apply (3.2) on the intervals [x_{i,}x_{i+1}] (i = \overline{0, m − 1}) , we can write:

|∫_{x_{i,}}^{x_{i+1}} g (u) du − ∑_{k=1}^{n} \frac{(b − a)^{k}}{m^{k}k!} · g^{(k−1)} (a + i · \frac{b − a}{m})|

≤ \{ \begin{matrix}\frac{(b − a)^{n+1}}{m^{n+1} (n + 1)!} ∥g^{(n)}∥_{[x_{i,}x_{i+1}],∞} , \\ \frac{(b − a)^{n+ \frac{1}{q}}}{m^{n+ \frac{1}{q}} n! (nq + 1)^{\frac{1}{q}}} ∥g^{(n)}∥_{[x_{i,}x_{i+1}],p} , \\ \frac{(b − a)^{n}}{m^{n}n!} ∥g^{(n)}∥_{[x_{i,}x_{i+1}],1} . \end{matrix}

Summing over i from 0 to m − 1 and using the generalized triangle inequality, we deduce (3.2).

The following main result holds.

Theorem 2. Let f : [a, b] → ℝ be such that f^{(n)} (n ≥ 0) is absolutely continuous on [a, b]. Then for any m ∈ ℕ, m ≥ 1, we have:

(T f) (a, b; t) = f (t) ln (\frac{b − t}{t − a}) + A_{n,m} (f, t) + R_{n,m} (f, t) ,

FINITE HILBERT TRANSFORM 7

where A_{n,m} (f, t) (3.3)

= ∑_{k=1}^{n} \frac{f^{(k)} (t)}{m^{k}k!} · [\frac{(b − t)^{k} + (−1)^{k+1} (t − a)^{k}}{k} ] + (b − a) ∑_{i=1}^{m−1} ∑_{k=1}^{n} \frac{1}{m^{k}k!}

× \{∑_{ν=1}^{k−1} (−1)^{ν−1} (k − 1) ⋯ (k − ν) (\frac{m}{i})^{ν−1}

× [f^{(k−ν)}; t + \frac{i}{m} (b − t) , t − \frac{i}{m} (t − a)]

+ (−1)^{k−1} (\frac{m}{i})^{k−1} (k − 1)! [f; t + \frac{i}{m} (b − t) , t − \frac{i}{m} (t − a)]\}

and the remainder R_{n,m} (f, t) satisfies the estimate

(3.4)  |R_{n,m} (f, t)|

≤ \{ \begin{matrix}\frac{∥f^{(n+1)}∥_{[a,b],∞}}{m^{n} (n + 1)! · (n + 1)} [(b − t)^{n+1} + (t − a)^{n+1}] , & if f^{(n+1)} ∈ L_{∞} [a, b] ; \\ \frac{q [(b − t)^{n+ \frac{1}{q}} + (t − a)^{n+ \frac{1}{q}} ]}{m^{n}n! (nq + 1)^{1+ \frac{1}{q}}} ∥f^{(n+1)}∥_{[a,b],p} , & if f^{(n+1)} ∈ L_{p} [a, b] , \\ & p > 1, \frac{1}{p} + \frac{1}{q} = 1; \\ \frac{∥f^{(n+1)}∥_{[a,b],1}}{m^{n}n! · n} [(b − t)^{n} + (t − a)^{n}] ; \end{matrix}

≤ \{ \begin{matrix}\frac{(b − a)^{n+1}}{m^{n} (n + 1) · (n + 1)!}, ∥f^{(n+1)}∥_{[a,b],∞} & if f^{(n+1)} ∈ L_{∞} [a, b] ; \\ \frac{q (b − a)^{n+ \frac{1}{q}}}{m^{n}n! (nq + 1)^{1+ \frac{1}{q}}} ∥f^{(n+1)}∥_{[a,b],p} , & if f^{(n+1)} ∈ L_{p} [a, b] , \\ & p > 1, \frac{1}{p} + \frac{1}{q} = 1; \\ \frac{(b − a)^{n}}{m^{n}n! · n} ∥f^{(n+1)}∥_{[a,b],1} ; \end{matrix}

Proof. We have (see (2.3)) that:

(T f) (a, b; t) − f (t) ln (\frac{b − t}{t − a}) = P V ∫_{a}^{b} \frac{f (τ) − f (t)}{τ − t}dτ.

If we write the inequality (3.1) for g = f′, we get

(3.5)  |\frac{f (t) − f (τ)}{τ − t} − ∑_{i=0}^{m−1} ∑_{k=1}^{n} \frac{(τ − t)^{k−1}}{m^{k}k!} · f^{(k)} (t + i · \frac{τ − t}{m})|

(3.6)  ≤ \{ \begin{matrix}\frac{|τ − t|^{n}}{m^{n} (n + 1)!} ∥f^{(n+1)}∥_{[t,τ],∞} , & if f^{(n+1)} ∈ L_{∞} [a, b] ; \\ \frac{|τ − t|^{n−1+ \frac{1}{q}}}{m^{n}n! (nq + 1)^{\frac{1}{q}}} ∥f^{(n+1)}∥_{[t,τ],p} , & if f^{(n+1)} ∈ L_{p} [a, b] , \\ & p > 1, \frac{1}{p} + \frac{1}{q} = 1; \\ \frac{|τ − t|^{n−1}}{m^{n}n!} ∥f^{(n+1)}∥_{[t,τ],1} . \end{matrix}

If we apply P V to (3.5), we may write that:

|P V ∫_{a}^{b} \frac{f (t) − f (τ)}{τ − t}dτ (3.7)

− ∑_{i=0}^{m−1} ∑_{k=1}^{n} P V ∫_{a}^{b} \frac{(τ − t)^{k−1}}{m^{k}k!} · f^{(k)} (t + i · \frac{τ − t}{m}) dτ|

≤ \{ \begin{matrix}\frac{1}{m^{n} (n + 1)!}P V ∫_{a}^{b} |τ − t|^{n} ∥f^{(n+1)}∥_{[t,τ],∞} dτ, \\ \frac{1}{m^{n}n! (nq + 1)^{\frac{1}{q}}} P V ∫_{a}^{b} |τ − t|^{n−1+ \frac{1}{q}} ∥f^{(n+1)}∥_{[t,τ],p} dτ, \\ \frac{1}{m^{n}n!}P V ∫_{a}^{b} |τ − t|^{n−1} ∥f^{(n+1)}∥_{[t,τ],1} dτ, \end{matrix}

≤ \{ \begin{matrix}\frac{1}{m^{n} (n + 1)!} ∥f^{(n+1)}∥_{[a,b],∞} P V ∫_{a}^{b} |τ − t|^{n} dτ, \\ \frac{1}{m^{n}n! (nq + 1)^{\frac{1}{q}}} ∥f^{(n+1)}∥_{[a,b],p} P V ∫_{a}^{b} |τ − t|^{n−1+ \frac{1}{q}} dτ, \\ \frac{1}{m^{n}n!} ∥f^{(n+1)}∥_{[a,b],1} P V ∫_{a}^{b} |τ − t|^{n−1} dτ, \end{matrix}

≤ \{ \begin{matrix}\frac{1}{m^{n} (n + 1)!} ∥f^{(n+1)}∥_{[a,b],∞} [\frac{(b − t)^{n+1} + (t − a)^{n+1}}{n + 1} ] , \\ \frac{1}{m^{n}n! (nq + 1)^{\frac{1}{q}}} ∥f^{(n+1)}∥_{[a,b],p} [\frac{(b − t)^{n+ \frac{1}{q}} + (t − a)^{n+ \frac{1}{q}}}{n + \frac{1}{q}} ] , \\ \frac{1}{m^{n}n!} ∥f^{(n+1)}∥_{[a,b],1} [\frac{(b − t)^{n} + (t − a)^{n}}{n} ] . \end{matrix}

Now, let us denote

I_{i,k} := P V ∫_{a}^{b} (τ − t)^{k−1} f^{(k)} (t + \frac{i}{m} (τ − t)) dτ,

where i = 0, …, m − 1, k = 1, …, n.

For i = 0, we have

I_{0,k} := P V ∫_{a}^{b} (τ − t)^{k−1} f^{(k)} (t) dτ = f^{(k)} (t) · \frac{(b − t)^{k} + (−1)^{k+1} (t − a)^{k}}{k}

for any k = 1, …, n.

For k = 1, …, n and i = 1, …, m − 1, we have

I_{i,k} = lim_{ε→0+} [∫_{a}^{t−ε} (τ − t)^{k−1} f^{(k)} [t + \frac{i}{m} (τ − t)] dτ (3.8)

+ ∫_{t+ε}^{b} (τ − t)^{k−1} f^{(k)} [t + \frac{i}{m} (τ − t)] dτ]

= lim_{ε→0+} [\frac{m}{i}f^{(k−1)} [t + \frac{i}{m} (τ − t)] (τ − t)^{k−1}|_{a}^{t−ε}

−\frac{m}{i} ∫_{a}^{t−ε} (k − 1) (τ − t)^{k−2} f^{(k−1)} [t + \frac{i}{m} (τ − t)] dτ

+\frac{m}{i}f^{(k−1)} [t + \frac{i}{m} (τ − t)] (τ − t)^{k−1}|_{t+ε}^{b}

− \frac{m}{i} ∫_{t+ε}^{b} (k − 1) (τ − t)^{k−2} f^{(k−1)} [t + \frac{i}{m} (τ − t)] dτ]

= \frac{m}{i} [f^{(k−1)} [t + \frac{i}{m} (b − t)] − f^{(k−1)} [t − \frac{i}{m} (t − a)]]

−\frac{m}{i} (k − 1) P V ∫_{a}^{b} (τ − t)^{k−2} f^{(k−1)} (t + \frac{i}{m} (τ − t)) dτ

= (b − a) [f^{(k−1)}; t + \frac{i}{m} (b − t) , t − \frac{i}{m} (t − a)] − \frac{m}{i} (k − 1) I_{i,k−1}

for any k = 2, …, n.

For k = 1, we have

I_{i,1} = P V ∫_{a}^{b} f^{(1)} (t + \frac{i}{m} (τ − t)) dτ
= \frac{m}{i} [f [t + \frac{i}{m} (b − t)] − f [t − \frac{i}{m} (t − a)]]
= (b − a) [f; t + \frac{i}{m} (b − t) , t − \frac{i}{m} (t − a)] .

Using the recursive relation (3.8), we may write

I_{i,k} (3.9)

= (b − a) [f^{(k−1)}; t + \frac{i}{m} (b − t) , t − \frac{i}{m} (t − a)] − (\frac{m}{i}) (k − 1) I_{i,k−1}
= (b − a) [f^{(k−1)}; t + \frac{i}{m} (b − t) , t − \frac{i}{m} (t − a)] − (\frac{m}{i}) (k − 1)

× [(b − a) [f^{(k−2)}; t + \frac{i}{m} (b − t) , t − \frac{i}{m} (t − a)] − (\frac{m}{i}) (k − 2) I_{i,k−2}]

10 S.S. DRAGOMIR

= (b − a) [f^{(k−1)}; t + \frac{i}{m} (b − t) , t − \frac{i}{m} (t − a)]

− (b − a) (\frac{m}{i}) (k − 1) [f^{(k−2)}; t + \frac{i}{m} (b − t) , t − \frac{i}{m} (t − a)]

+ (\frac{m}{i})^{2} (k − 1) (k − 2) I_{i,k−2}

= · · · · · · · · · · · · =
= (b − a) ∑_{ν=1}^{k−1} (−1)^{ν−1} (k − 1) ⋯ (k − ν) (\frac{m}{i})^{ν−1}

× [f^{(k−ν)}; t + \frac{i}{m} (b − t) , t − \frac{i}{m} (t − a)] + (−1)^{k−1} (\frac{m}{i})^{k−1} (k − 1)!I_{i,1}

= (b − a) [∑_{ν=1}^{k−1} (−1)^{ν−1} (k − 1) ⋯ (k − ν) (\frac{m}{i})^{ν−1}

× [f^{(k−ν)}; t + \frac{i}{m} (b − t) , t − \frac{i}{m} (t − a)]

+ (−1)^{k−1} (\frac{m}{i})^{k−1} (k − 1)! [f; t + \frac{i}{m} (b − t) , t − \frac{i}{m} (t − a)] .

Replacing I_{i,k} in (3.7), we deduce the estimate (3.4) with A_{m,n} as defined by (3.3).

4. A Numerical Example

Consider the function f : [1, 2] → R, f (t) = sin t. The exact finite Hilbert transform of f provided by Maple is

(4.1) (T f) (1, 2; t) = − Si(t − 2) cos t + Ci(2 − t) sin t + Si(t − 1) cos t − Ci(t − 1)

where

Si(x) := ∫_{0}^{x} \frac{sin u}{u}du, Ci(x) := γ + ln x + ∫_{0}^{x} \frac{cos u − 1}{u}du.

The plot of the finite Hilbert transform is incorporated in Figure 1.

FINITE HILBERT TRANSFORM 11

Figure 1.

Define by A_{n} (1, 2; t) the approximation for the finite Hilbert transform provided by the formula (2.1), i.e.,

A_{n} (1, 2; t) = sin t · ln (\frac{2 − t}{t − 1}) + ∑_{k=1}^{n−1} \frac{[sin t]^{(k)}}{k!} · [\frac{(b − t)^{k} + (−1)^{k+1} (t − a)^{k}}{k} ] .

Figure 2 contains the plot of the error

Er_{5} (1, 2; t) := (T f) (1, 2; t) − A_{5} (1, 2; t) , t ∈ [1, 2] . Figure 2.

Figure 3 contains the plot of the error

Er_{10} (1, 2; t) := (T f) (1, 2; t) − A_{10} (1, 2; t) , t ∈ [1, 2] .

Figure 3.

Acknowledgement 1. The author would like to thank the anonymous referee for some valuable comments that have been incorporated in the present version of the paper.

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