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N.S. BARNETT AND S.S. DRAGOMIR

Abstract. Estimates for a singular integral for which the Gr¨uss, ˇCebyˇsev,

Lupa¸s and Ostrowski inequalities fail to work are given by utilising trapezoidal type rules.

1. Introduction

Define the integral

Ip(f) :=

Z b

a

f(t)

(t−a)pdt, p∈(0,1) ;

when f : [a, b]→R is Lebesgue measurable on [a, b] and the Lebesgue integral is finite for any givenp∈(0,1).

In order to obtain estimates for this integral one may use the following refinement of the Gr¨uss inequality obtained by Cheng and Sun in [1]:

1

b−a Z b

a

f(t)g(t)dt− 1 b−a

Z b

a

f(t)dt· 1 b−a

Z b

a

g(t)dt

≤1

2(M−m) 1

b−a Z b

a

g(t)− 1 b−a

Z b

a

g(s)ds

dt,

provided thatf, g: [a, b]→Rare Lebesgue measurable and

−∞< m≤f(t)≤M <∞

for a.e. t∈[a, b].

For an extension to the general case of Lebesgue integrals on measurable spaces and for the sharpness of the constant12,see [2].

For this type of integral, the classical Gr¨uss inequality,

1

b−a Z b

a

f(t)g(t)dt− 1 b−a

Z b

a

f(t)dt· 1 b−a

Z b

a

g(t)dt

≤ 1

4(M −m) (N−n) where

−∞< m≤f(t)≤M <∞,−∞< n≤g(t)≤N <∞

for a.e. t∈[a, b],cannot be used since the second function, namelyg(t) = 1/(t−a)p,

p∈(0,1) is not essentially bounded on [a, b].

Date: August 12, 2005.

1991Mathematics Subject Classification. Primary 26D15; Secondary 41A55. Key words and phrases. Singular integral, Inegral Inequalities. Trapezoidal Rule.

(2)

The same can be said for attemps to use the ˇCebyˇsev [3, p. 297], Lupa¸s [3, p. 301] or Ostrowski [3, p. 300] inequalities that provide bounds for the quantity

1

b−a Z b

a

f(t)g(t)dt− 1 b−a

Z b

a

f(t)dt· 1 b−a

Z b

a

g(t)dt

in terms of the derivatives of the functionsf, g. The details are left as an exercise to the reader.

The main aim of this paper is to point out various estimates forIp(f) using an approach which employs the removal of the singularity and the utilisation of the trapezoidal rule.

2. Some Inequalities

The following result that establishes an estimate of the singular integralIp(f) in terms of the function values at the pointsaandb and in terms of the derivative at the pointaholds true:

Theorem 1. Let f : [a, b]→R be a differentiable function so thatf0 is absolutely continuous on [a, b], then we have the inequalities:

Ip(f)−

2f(a) + (1−p)f(b) (3−p) (1−p) (b−a)

1−p

− 1

(3−p) (2−p)f

0(a) (ba)2−p

(2.1)

≤                       

1 2 (3−p)

Rb

a(t−a) 2−p

kf00k[a,t],dt if f00∈L∞[a, b] ;

1

(3−p) (β+ 1)β1

Rb

a(t−a) 1+1

β−pkf00k

[a,t],αdt if f

00L

α[a, b],

α >1, α1 +1β = 1; 1

(3−p)

Rb

a(t−a) 1−p

kf00k

[a,t],1dt;

≤                           

1

2 (3−p)2(b−a) 3−p

kf00k[a,b], if f00∈L∞[a, b] ;

(b−a)2+1β−p

(3−p) (β+ 1)β1

2 + β1−p kf00k

[a,b],α if f

00L

α[a, b],

α >1, α1 +β1 = 1; (b−a)2−p

(2−p) (3−p)kf

00k

[a,b],1,

for allp∈(0,1). Proof. We have

Ip(f) =

Z b

a

f(t) (t−a)pdt=

Z b

a

f(a) (t−a)pdt+

Z b

a

f(t)−f(a) (t−a)p dt (2.2)

=f(a)(b−a) 1−p

1−p + Z b

a

(3)

Sincef is absolutely continuous, then we have

f(t)−f(a) =

Z t

a

f0(u)du.

Using the following elementary identity which can be obtained from the integration by parts formula

(2.3)

Z β

α

g(u)du= g(α) +g(β)

2 (β−α) +

Z β

α

α+β

2 −u

g0(u)du,

provided thatg is absolutely continuous on the interval [α, β], we may state that:

(2.4)

Z t

a

f0(u)du=f

0(t) +f0(a)

2 (t−a) +

Z t

a

a+t

2 −u

f00(u)du.

Consequently,

(2.5)

Z b

a

f(t)−f(a) (t−a)p dt=

Z b

a

(t−a)1−p

f0(t) +f0(a)

2

dt

+

Z b

a

1

(t−a)p

Z t

a

a+t

2 −u

f00(u)du

dt.

However,

Z b

a

(t−a)1−p

f0(t) +f0(a)

2

dt

= 1 2

" Z b

a

f0(a) (t−a)1−pdt+

Z b

a

(t−a)1−pf0(t)dt #

= 1 2

"

f0(a) (b−a)2−p

2−p +f(t) (t−a)

1−p

b

a

−(1−p)

Z b

a

f(t) (t−a)pdt

#

= 1 2

"

f0(a) (b−a)2−p

2−p +f(b) (b−a)

1−p

−(1−p)Ip(f)

#

= 1

2f(b) (b−a) 1−p

+1 2·

f0(a) (b−a)2−p

2−p −

1−p

2 Ip(f).

Using (2.2), we may write that:

(2.6) Ip(f) =f(a)

(b−a)1−p 1−p +

1

2f(b) (b−a) 1−p

+1 2

f0(a) (b−a)2−p

2−p −

1−p

2 Ip(f) +R[f],

where

R[f] =

Z b

a

1

(t−a)p

Z t

a

a+t

2 −u

f00(u)du

(4)

From (2.6), we get:

Ip(f)

1 +1−p 2

=f(a)(b−a) 1−p

1−p +

1

2f(b) (b−a) 1−p

+1 2

f0(a) (b−a)2−p

2−p +R[f],

which implies that

Ip(f) = 1 3−p

"

2f(a)(b−a) 1−p

1−p +f(b) (b−a)

1−p +f

0(a) (ba)2−p 2−p

#

+ 2

3−pR[f],

i.e.,

(2.7) Ip(f) =

2f(a) + (1−p)f(b) (3−p) (1−p) (b−a)

1−p +f

0(a) (ba)2−p (3−p) (2−p) +

2

3−pR[f].

On the other hand, we have

|R[f]| ≤ Z b

a

1 (t−a)p

Z t

a

a+t

2 −u

f00(u)du

dt=:J.

However,

Z t

a

a+t

2 −u

f00(u)du

≤ kf00k[a,t], Z t

a

a+t

2 −u

du

= (t−a) 2

4 kf

00k

[a,t],∞,

by H¨older’s integral inequality, we have:

Z t

a

a+t

2 −u

f00(u)du

≤ kf00k[a,t] Z t

a

a+t

2 −u

β

du !β1

= (t−a) 1+1

β

2 (β+ 1)β1

kf00k[a,t], α >1, 1 α+

1

β = 1;

and finally,

Z t

a

a+t

2 −u

f00(u)du

≤ sup u∈[a,t]

a+t

2 −u

Z t

a

|f00(u)|du

=(t−a) 2 kf

00k

(5)

and then

J ≤                         

1 4

Z b

a

(t−a)2−pkf00k

[a,t],∞dt,

1

2 (β+ 1)β1

Z b

a

(t−a)1+1β−pkf00k

[a,t],αdt,

1 2

Z b

a

(t−a)1−pkf00k

[a,t],1dt,

and the first part of (2.1) is proved. We now observe that,

b

Z

a

(t−a)2−pkf00k[a,t],dt≤ kf00k[a,b],(b−a)

3−p

3−p ,

b

Z

a

(t−a)1+1β−pkf00k

[a,t],αdt≤ kf

00k

[a,b],α

(b−a)2+1β−p

2 + 1β−p

and

b

Z

a

(t−a)1−pkf00k[a,t],1dt≤ kf00k[a,b],1(b−a)

2−p

2−p ,

which proves the last part of the theorem.

Theorem 2. Let f : [a, b]→Rbe such that the second derivative f00 is absolutely continuous on [a, b], then,

(2.8)

Ip(f)−

2f(a) + (1−p)f(b) (3−p) (1−p) (b−a)

1−p

− 1

(3−p) (2−p)f

0(a) (ba)2−p

≤                         

1 12

Z b

a

(t−a)3−pkf000k

[a,t],∞dt if f000∈L∞[a, b] ;

[B(β+ 1, β+ 1)]β1

2

Z b

a

(t−a)2+1β−pkf000k

[a,t],αdt if f

000L

α[a, b],

α >1, 1 α+

1 β = 1; 1

8

Z b

a

(t−a)2−pkf000k

[a,t],1dt;

≤                         

1

12 (4−p)(b−a) 4−p

kf000k

[a,b],∞

[B(β+ 1, β+ 1)]β1

23 +β1 −p

(b−a)3+β1−pkf000k

[a,b],α, α >1, 1 α+

1 β = 1;

1 8(b−a)

3−p

kf000k

(6)

for allp∈(0,1), whereB(·,·)is Euler’s Beta function,

B(s+ 1, q+ 1) :=

Z 1

0

ts(1−t)qdt, s, q >−1.

Proof. Using the elementary identity

Z β

α

g(u)du= g(α) +g(β)

2 (β−α) + 1 2

Z β

α

(u−α) (u−β)g00(u)du,

where g : [α, β] →R is such that the first derivative is absolutely continuous, we have,

Z t

a

f0(u)du=f

0(t) +f0(a)

2 (t−a) + 1 2

Z t

a

(u−a) (u−t)f000(u)du.

Consequently,

Z b

a

f(t)−f(a) (t−a)p dt=

Z b

a

(t−a)1−p

f0(t) +f0(a)

2

dt

+1 2

Z b

a

1

(t−a)p

Z t

a

(u−a) (u−t)f000(u)du

dt.

By a similar argument to that in the proof of the above theorem, we have (see also (2.7)) that,

Ip(f) =

2f(a) + (1−p)f(b) (3−p) (1−p) (b−a)

1−p

+ 1

(3−p) (2−p)f

0(a) (ba)2−p

+ 2

3−pR1[f].

where

R1[f] = 1 2

Z b

a

1

(t−a)p

Z t

a

(u−a) (t−u)f000(u)du

dt.

Now, we have

|R1[f]| ≤ 1 2

Z b

a

1 (t−a)p

Z t

a

(u−a) (t−u)f000(u)du

dt=:K.

However,

Z t

a

(u−a) (u−t)f000(u)du

≤ kf000k[a,t], Z t

a

(u−a) (t−u)du

=kf000k

[a,t],∞

(t−a)3

6 .

Using H¨older’s integral inequality, we have

Z t

a

(u−a) (u−t)f000(u)du

≤ kf000k[a,t] Z t

a

(u−a)β(t−u)βdu 1β

=kf000k

[a,t],α(t−a) 2+1

β[B(β+ 1, β+ 1)]

(7)

Also,

Z t

a

(u−a) (u−t)f000(u)du

≤ max u∈[a,t]{

(t−u) (u−a)} Z t

a

|f000(u)|du

= (t−a) 2

4 kf

000k

[a,t],1.

We can now state that

K≤                       

1 12

Z b

a

(t−a)3−pkf000k

[a,t],∞dt

1

2[B(β+ 1, β+ 1)]

1 β

Z b

a

(t−a)2+β1−pkf000k

[a,t],αdt

1 8

Z b

a

(t−a)2−pkf000k

[a,t],1dt

and the first part of (2.8) is proved. Now, we observe that

Z b

a

(t−a)3−pkf000k

[a,t],∞dt≤ kf000k[a,b],∞·

(b−a)4−p 4−p ,

Z b

a

(t−a)2+β1−pkf000k

[a,t],αdt≤ kf

000k

[a,b],α·

(b−a)3+β1−p

3 +β1 −p

and

Z b

a

(t−a)2−pkf000k[a,t],1dt≤ kf000k[a,b],1(b−a)

3−p

3−p

which proves the last part of the inequality (2.8).

3. Applications

A natural application of the above results is to approximate the Euler Beta function:

(3.1) B(p+ 1, q+ 1) :=

Z 1

0

tp(1−t)qdt, p, q >−1

in the case thatp∈(−1,0) andq∈(0,∞).

For this purpose, we consider the function f : [0,1]→ R, f(t) = (1−t)q with

q∈(0,∞).

We observe that

f0(t) =−q(1−t)q−1, f00(t) =q(q−1) (1−t)q−2.

We also observe that

kf00k[0,1],=q(q−1) if q≥2,

kf00k[0,1] =q(q−1)

Z 1

0

(1−t)(q−2)αdt α1

= q(q−1) [(q−2)α+ 1]1α

(8)

and

kf00k[0,1],1=q(q−1)

Z 1

0

(1−t)q−2dt=q if q >1.

Applying Theorem 1 forf as above, we have the inequalities

(3.2)

B(p+ 1, q+ 1)− 2

(3−p) (1−p)+

2 (3−p) (2−p)

≤                       

q(q−1)

2 (3−p) if p∈(−1,0), q∈(2,∞) ;

1

(3−p) (β+ 1)1β

2 + 1β−p

· q(q−1)

[(q−2)α+ 1]α1

ifp∈(−1,0) andq, α >1,with (q−2)α+ 1>0 and α1 +β1 = 1;

q

(2−p) (3−p), if p∈(−1,0) andq∈(1,∞).

A similar result can be obtained by using Theorem 2. The details are omitted.

References

[1] X.-L. CHENG and J. SUN, Note on the perturbed trapezoid inequal-ity, J. Inequal. Pure & Appl. Math., 3(2002), No. 2, Article 29. [ONLINE http://jipam.vu.edu.au/article.php?sid=181].

[2] P. CERONE and S.S. DRAGOMIR, A refinement of the Gr¨uss inequality and applications, RGMIA Res. Rep. Coll., 5(2002), No. 2, Article 14. [ONLINE http://rgmia.vu.edu.au/v5n2.html ].

[3] D.S. MITRINOVI ´C, J.E. PE ˇCARI ´C and A.M. FINK,Classical and New Inequalities in Anal-ysis, Kluwer Academic Publishers, Dordrecht/Boston/London, 1993.

School of Computer Science & Mathematics, Victoria University, PO Box 14428, MCMC 8001, Victoria, Australia.

E-mail address:{neil,sever}@csm.vu.edu.au

References

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