SOME INEQUALITIES FOR A SINGULAR INTEGRAL

  J. Indones. Math. Soc. (MIHMI) Vol. 12, No. 2 (2006), pp. 175–183.

  

SOME INEQUALITIES FOR A SINGULAR

  

INTEGRAL

_Abstract.

N.S. Barnett and S.S. Dragomir

Estimates for a singular integral for which the Gr¨ uss, ˇ Cebyˇ sev, Lupa¸s and Ostrowski inequalities fail to work are given by utilising trapezoidal type rules.

  1. INTRODUCTION Define the integral _{Z b} f (t)

I p (f ) := p dt, p ∈ (0, 1) ; a (t − a) when f : [a, b] → R is Lebesgue measurable on [a, b] and the Lebesgue integral is

  finite for any given p ∈ (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 [2]: _{Z b Z b Z b}

  1

  1

  1 f (t) g (t) dt − f (t) dt · g (t) dt b − a b − a b − a _{a a a Z b Z b}

  1

  1

  1 ≤ (M − m) g (t) − g (s) ds dt, 2 b − a b − a _{a a} Received 13 September 2005, Accepted 16 December 2005. _{2000 Mathematics Subject Classification Key words and Phrases} : Primary 26D15; Secondary 41A55.

  

: singular integral, integral inequalities, trapezoidal rule.

  N.S. Barnett and S.S. Dragomir

  176 provided that f, g : [a, b] → R are 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

  1 spaces and for the sharpness of the constant , see [1].

  2 For this type of integral, the classical Gr¨ uss inequality, _{Z b Z b Z b}

  1

  1

  1

  1 f (t) g (t) dt − f (t) dt · g (t) dt ≤ (M − m) (N − n) b − a b − a b − a _{a a a}

  4 where −∞ < m ≤ f (t) ≤ M < ∞, −∞ < n ≤ g (t) ≤ N < ∞ for a.e. t ∈ [a, b] , cannot be used since the second function, namely g (t) = 1/(t − _p a) , p ∈ (0, 1) is not essentially bounded on [a, b] .

  The same can be said for attempts to use the ˇ Cebyˇsev [3, p. 297], Lupa¸s [3, p. 301] or Ostrowski [3, p. 300] inequalities that provide bounds for the quantity _{Z b Z b Z b}

  1

  1

  1 f (t) g (t) dt − f (t) dt · g (t) dt b − a a b − a a b − a a in terms of the derivatives of the functions f, g.

  The details are left as an exercise to the reader. The main aim of this paper is to point out various estimates for I p (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 integral I p (f ) in terms of the function values at the points a and b and in terms of the derivative at the point a holds true: _′ Theorem 1. Let is absolutely f : [a, b] → R be a differentiable function so that f continuous on

  [a, b], then we have the inequalities: 2f (a) + (1 − p) f (b)

  1

  1−p 2−p _′

  I p (f ) − (b − a) − f (a) (b − a) (1) (3 − p) (1 − p) (3 − p) (2 − p)

• 1 _β

  Proof.

  (3 − p) (β + 1) ^{1 β} 2 +

  1 _β

  − p ^{kf ′′} k

  [a,b],α

  if f ^′′ ∈ L α [a, b] , α > 1,

  1 _α

  = 1; (b − a)

  2−p

  (2 − p) (3 − p) kf ^′′ k

  [a,b],1

  , for all p ∈ (0, 1).

  We have, I p (f ) = ^{Z b} _a f (t)

  if f ^′′ ∈ L ∞ [a, b] ; (b − a)

  (t − a) ^{p dt = Z b a} f (a) (t − a) ^{p dt + Z b a} f (t) − f (a)

  (t − a) ^{p dt (2)} = f (a)

  (b − a)

  1−p

  1 − p

  (t − a) ^p dt. Since f is absolutely continuous, then, f (t) − f (a) = ^{Z t} _a f ^′ (u) du.

  Using the following elementary identity which can be obtained from the integration by parts formula, _{Z β α} g (u) du = g (α) + g (β)

  2 (β − α) + ^{Z β} _α

  α + β

  2 − u g ^′ (u) du, (3) provided that g is absolutely continuous on the interval [α, β], we may state that: _{Z t} f ^′ (u) du = f ^′ (t) + f ^′ (a)

  2 (t − a) + ^{Z t} a + t

  2+ ¹ _β ^{− p}

  [a,b],∞

  2 − u f ^′′ (u) du. (4)

  [a,t],α

  Inequalities for a singular integral 177

  ≤ ^{            } _{           }

  1 2 (3 − p) ^{Z b} _a (t − a)

  2−p

  kf ^′′ k

  [a,t],∞

  dt if f ^′′ ∈ L ∞ [a, b] ;

  1 (3 − p) (β + 1) ^{1 β Z b a}

  (t − a)

  1+ ¹ _β

^{− p}

  kf ^′′ k

  dt if f ^′′ ∈ L α [a, b] , α > 1,

  kf ^′′ k

  1 _α

  = 1;

  1 (3 − p) ^{Z b} _a

  (t − a)

  1−p

  kf ^′′ k

  [a,t],1

  dt; ≤ ^{             } _{            }

  1 2 (3 − p)

  2

  (b − a)

  3−p

• 1 _β
• ^{Z b} _a f (t) − f (a)
• ^{Z b} _a

• f (t) (t − a)
• f (b) (b − a)

  (6)

  2 − u f ^′′ (u) du dt.

  1 (t − a) ^{p Z t} _a a + t

  2 I p (f ) + R [f ] , where R [f ] = ^{Z b} _a

  2 − p − 1 − p

  2−p

  2 f ^′ (a) (b − a)

  1−p

  2 = f (a)

  2 f (b) (b − a)

  1 − p

  1−p

  (b − a)

  2 Using (2), we may write that: I p (f ) = f (a)

  2 − p 1 − p

  From (6), we get: I p (f ) 1 + 1 − p

  (b − a)

  2 f ^′ (a) (b − a)

  (b − a)

  2 − p ^#

  2−p

  (a) (b − a)

  1−p

  1 − p

  1−p

  1 3 − p ^" 2f (a)

  1−p

  I p (f ) =

  2 − p

  2−p

  2 f ^′ (a) (b − a)

  1−p

  2 f (b) (b − a)

  1 − p

  2−p

  1

  R [f ] ,

  2 − u f ^′′ (u) du dt.

  2 ^{" Z b a} f ^′ (a) (t − a)

  1

  2 dt =

  f ^′ (t) + f ^′ (a)

  1−p

  However, _{Z b a} (t − a)

  1 (t − a) ^{p Z t} _a a + t

  dt + ^{Z b} _a (t − a)

  2 dt (5)

  f ^′ (t) + f ^′ (a)

  1−p

  (t − a)

  Consequently, _{Z b a} f (t) − f (a) (t − a) ^{p dt = Z b} _a

  N.S. Barnett and S.S. Dragomir

  178

  1−p

  1−p

  1−p

  2 ^" f ^′ (a) (b − a)

  2

  1

  − (1 − p) I p (f ) ^#

  1−p

  2 − p

  2−p

  1

  f ^′ (t) dt ^# =

  =

  − (1 − p) ^{Z b} _a f (t) (t − a) ^{p dt #}

  1−p ^b _a

  2 − p

  2−p

  2 ^" f ^′ (a) (b − a)

  1

• 1
• 1
• 1
• 1
• R [f ] , which implies that,
• f (b) (b − a)
• f ^′
• 2

  Inequalities for a singular integral 179

  i.e., _{′ 2−p} 2f (a) + (1 − p) f (b) f (a) (b − a)

  2

  

1−p

  I p (f ) = (b − a) R [f ] . (7) + + (3 − p) (1 − p) (3 − p) (2 − p) 3 − p

  On the other hand, _{Z b Z t} 1 a + t _′′ |R [f ]| ≤ p − u f (u) du dt =: J. _{a (t − a) a}

  2 However, _{Z t Z t} a + t a + t _{′′ ′′} − u f (u) du ≤ kf k − u du _a [a,t],∞

  2 a

  2

  2

  (t − a) _′′ = kf k ,

  [a,t],∞

  4 by H¨older’s integral inequality, we have: _{Z t Z t β ! 1 β} a + t _{′′ ′′} a + t − u f (u) du ≤ kf k − u du

  [a,t],α _{a a}

  2

  2

  1+ _β

  (t − a) _′′

  1

  1 = kf k , α > 1, = 1; + _{1 β} [a,t],α

  α β 2 (β + 1) and finally, _{Z t Z t} a + t _{′′ ′′} a + t

  − u f (u) du ≤ sup − u |f (u)| du _{a u∈ [a,t] a}

  2

  2 (t − a) _′′

  = kf k

  [a,t],1

  2 and then, _{ Z b   2−p ′′}

  1 _ (t − a) kf k dt, _{ } [a,t],∞ _{  Z b     }

  4 _{a 1} 1 − p

  1+ ′′ _β

  (t − a) kf k dt, J ≤ _{1  β a} [a,t],α _{       } 2 (β + 1) _{Z b   1−p ′′  (t − a) kf k dt,}

  1

  [a,t],1

  2 _a and the first part of (1) is proved. We now observe that, _{Z b 3−p}

  (b − a)

  2−p ′′ ′′

  (t − a) kf k dt ≤ kf k ,

  [a,t],∞ [a,b],∞

  3 − p

  

N.S. Barnett and S.S. Dragomir

  180 _{Z b 1 − p 1 − (b − a)} 2+ _β 1+ p ′′ ′′ _β (t − a) kf k dt ≤ kf k

  [a,t],α [a,b],α

  1 _a 2 + − p _β and _{Z b 2−p}

  (b − a)

  1−p ′′ ′′

  (t − a) kf k dt ≤ kf k , _{a 2 − p} [a,t],1 [a,b],1 which proves the last part of the theorem. _′′ ✷ Theorem 2. Let is absolutely f : [a, b] → R be such that the second derivative f continuous on [a, b] , then,

  2f (a) + (1 − p) f (b)

  1

  1−p ′ 2−p

  I p (f ) − (b − a) − f (a) (b − a) (8) _{ Z b} (3 − p) (1 − p) (3 − p) (2 − p) _{ 3−p ′′′ ′′′ }

  1 _ (t − a) kf k dt if f ∈ L ∞ [a, b] ; _{ } [a,t],∞ _{  β     }

  12 _{a 1 Z b 1} [B (β + 1, β + 1)] _{β ′′′}

  ≤ (t − a) kf k dt if f ∈ L α [a, b] , _ [a,t],α _{ }

  2 _{a  }

  1

  1 _{    Z b}

  α > 1, + = 1; _{α β   2−p ′′′}

  1 _ (t − a) kf k dt;

  [a,t],1 _

  8 _a

  1 _{ } 4−p ′′′ _{  [a,b],∞} (b − a) kf k _{       } 12 (4 − p) _{1 β 1} [B (β + 1, β + 1)] − p

  1

  1

  

3+ ′′′

  (b − a) kf k , α > 1, = 1; ^β + ≤

  [a,b],α

  1 _

  α β _{        } 2 3 + − p _β

  1 _{ } 3−p ′′′ (b − a) kf k ;

  [a,b],1

  8 for all p ∈ (0, 1), where B (·, ·) is Euler’s Beta function, _Z

  1

_s

_q B (s + 1, q + 1) := t (1 − t) dt, s, q > −1.

  Proof. _{Z β Z β} Using the elementary identity g (α) + g (β) 1 _′′ g (u) du = (β − α) + (u − α) (u − β) g (u) du,

  2

  2 Inequalities for a singular integral 181

  where g : [α, β] → R is such that the first derivative is absolutely continuous, we have, _{Z t ′ ′ Z t ′} f (t) + f (a)

  1 _′′′ f (u) du = (t − a) + (u − a) (u − t) f (u) du. _{a a}

  2

  2 Consequently, _{Z b Z b ′ ′} f (t) − f (a) f (t) + f (a) _{a (t − a) a p dt = (t − a) dt}

1−p

_{Z b Z t}

  2

  1

  1 _′′′ + (u − a) (u − t) f (u) du dt. _p 2 (t − a) _{a a} By a similar argument to that in the proof of the above theorem, we have (see also (7)) that,

  2f (a) + (1 − p) f (b)

  1−p

  I (f ) = (b − a) _p (3 − p) (1 − p)

  1 _{′ 2−p}

  2

• f (a) (b − a) R [f ] . +

  1

  (3 − p) (2 − p) 3 − p where _{Z b Z t}

  1

  1 _′′′ R

1 [f ] = p (u − a) (t − u) f (u) du dt.

  2 a (t − a) a Now, _{Z b Z t}

  1

  1 _′′′ |R [f ]| ≤ (u − a) (t − u) f (u) du dt =: K.

  1 p

  2 (t − a) _{a a} However, _{Z t Z t ′′′ ′′′}

  (u − a) (u − t) f (u) du ≤ kf k (u − a) (t − u) du _{a a} [a,t],∞

  3 _′′′ (t − a) = kf k .

  [a,t],∞

  6 Using H¨older’s integral inequality, we have _{Z t ′′′ ′′′ Z t β β β 1} (u − a) (u − t) f (u) du ≤ kf k (u − a) (t − u) du _{a ′′′ 2+} [a,t],α _{a β β 1 1}

  = kf k (t − a) [B (β + 1, β + 1)]

  [a,t],α

  1

  1 _{α β} + for α > 1, = 1.

  N.S. Barnett and S.S. Dragomir

  182 _{Z t} Also, _{′′′ ′′′ Z t} (u − a) (u − t) f (u) du ≤ max {(t − u) (u − a)} |f (u)| du _{u∈ a} [a,t] _a

  2

  (t − a) _′′′ = kf k .

  [a,t],1

  4 We can now state that _{ Z b}

  1 _{ } 3−p ′′′ _ (t − a) kf k dt _ [a,t],∞ _{      a }

  12 _{1 Z b 1} 1 − p _{β β} 2+ ′′′ K ≤ [B (β + 1, β + 1)] (t − a) kf k dt _ [a,t],α _{       Z b }

  2 _a

  1 _{  (t − a) kf k dt} 2−p ′′′ [a,t],1 8 _a and the first part of (8) is proved.

  Now, we observe that _{Z b}

  4−p

  (b − a)

  3−p ′′′ ′′′

  (t − a) kf k dt ≤ kf k · , _{Z b}

_{a 4 − p}

[a,t],∞ [a,b],∞ _{1 − p 1 − p (b − a)} 3+ _β 2+ ′′′ ′′′ _β (t − a) kf k dt ≤ kf k ·

_{a 3 + − p}

[a,t],α [a,b],α _β

  1

  and _{Z b 3−p}

  2−p (b − a)

_{′′′ ′′′}

  (t − a) kf k dt ≤ kf k

  [a,t],1 [a,b],1 _a 3 − p which proves the last part of the inequality (8).

  ✷

  3. APPLICATIONS A natural application of the above results is to approximate the Euler Beta function: _Z

  1

_p

_q

  B (p + 1, q + 1) := t (1 − t) dt, p, q > −1 (9) for the case that p ∈ (−1, 0) and q ∈ (0, ∞) . _q with For this purpose, we consider the function f : [0, 1] → R, f (t) = (1 − t) q ∈ (0, ∞) . We observe that _{′ q− q−}

  1 ′′

  2 f (t) = −q (1 − t) , f (t) = q (q − 1) (1 − t) . Inequalities for a singular integral 183

  We also observe that _′′ kf k = q (q − 1) if q ≥ 2,

  [0,1],∞ _{Z α 1 ′′ (q−2)α}

  1

  kf k = q (q − 1) (1 − t) dt

  [0,1],α

  q (q − 1) = if (q − 2) α + 1 > 0, α > 1, q > 1 _{α 1}

  [(q − 2) α + 1] and _{Z ′′}

  1 _q−

  2 kf k = q (q − 1) (1 − t) dt = q if q > 1. [0,1],1

  Applying Theorem 1 for f as above, we have the inequalities

  2

  2 (10) + B (p + 1, q + 1) − _ (3 − p) (1 − p) (3 − p) (2 − p) _{ } q (q − 1) _{ } if p ∈ (−1, 0) , q ∈ (2, ∞) ; _{    } 2 (3 − p) _{   } 1 q (q − 1) _{β 1 · 1}

  1 α

  [(q − 2) α + 1] (3 − p) (β + 1) 2 + − p _β

  ≤ _

  1

  1 _{  } ^{ } _{  }

• if p ∈ (−1, 0) and q, α > 1, with (q − 2) α + 1 > 0 and = 1; _{α β  }

  q _ , if p ∈ (−1, 0) and q ∈ (1, ∞) . (2 − p) (3 − p) A similar result can be obtained by using Theorem 2. The details are omitted. _REFERENCES

  

1. P. Cerone and S.S. Dragomir, “A refinement of the Gr¨ uss inequality and applica-

tions”, RGMIA Res. Rep. Coll. 5(2002), No. 2, Article 14.

  [ONLINE: http://rgmia.vu.edu.au/v5n2.html].

  

2. X.-L. Cheng and J. Sun, “Note on the perturbed trapezoid inequality”, J. Inequal.

  Pure & Appl. Math. 3 (2002), No. 2, Article 29. [ONLINE: http://jipam.vu.edu.au/article.php?sid=181].

  3. D.S. Mitrinovi´

  c, J.E. Peˇ cari´ c and A.M. Fink , Classical and New Inequalities in Analysis

, Kluwer Academic Publishers, Dordrecht/Boston/London, 1993.

  N.S. Barnett : School of Computer Science & Mathematics, Victoria University, PO Box 14428, MCMC 8001, Victoria, Australia. E-mail: neil@csm.vu.edu.au S.S. Dragomir

  : School of Computer Science & Mathematics, Victoria University, PO Box 14428, MCMC 8001, Victoria, Australia. E-mail: sever@csm.vu.edu.au

  184

  

N.S. Barnett and S.S. Dragomir