General Mathematics Vol. 14, No. 2 (2006), 43–54
On a class of convergent sequences defined
by integrals 1
Dorin Andrica
and
Mihai Piticari
Abstract
The main result shows that if g : [0, 1] → R is a continuous funcg(x)
exists and it is finite, then for any continuous
tion such that lim
x→0 x
x>0
function f : [0, 1] → R
Z
lim n
n→∞
1
Z
f (x)g(xn )dx = f (1)
0
0
1
g(x)
dx.
x
The order of convergence in the above relation, consequences and
some applications are given.
2000 Mathematics Subject Classification: 26D15, 65D30
Key words and phrases: quadrature rule, numerical integration, error
bounds, optimal quadrature
1
Received March 6, 2006
Accepted for publication (in revised form) April 10, 2006
43
44
Dorin Andrica and Mihai Piticari
1
Introduction
There are many important classes of sequences defined by using Riemann
integrals. We mention here only two. The first one is called the RiemannLebesgue Lemma and it asserts that if g : [0, +∞) → R is a continuous
and T -periodic function, then for any continuous function f : [a, b] → R,
0 ≤ a < b, the following relation holds:
Z b
Z
Z b
1 T
(1)
lim
f (x)g(nx)dx =
g(x)dx
f (x)dx
n→∞ a
T 0
a
For the proof we refer to [3] (in special case a = 0, b = T ) and [8]. In the
paper [1] we have proved that a similar relation as (1) holds for all continuous
and bounded functions g : [0, +∞) → R having finite Cesaro mean. The
second one was given in our paper [2] and shows that if f : [1, +∞) → R is
a continuous function such that lim xf (x) exists and it is finite, then
x→∞
Z
(2)
∞
n
f (x )dx =
lim n
n→∞
Z
a
1
1
f (x)
dx,
x
for any real number a > 1.
Z
In this paper we investigate the class of sequences defined by
1
n
f (x)g(xn )dx, where f, g : [0, 1] → R are continuous functions. The
0
main results in [6] are obtained as consequences and some applications are
given.
2
The main results
We begin with two preliminary results.
On a class of convergent sequences defined by integrals
45
Lemma 1. Let g : [0, 1] → R be a continuous function such that
g(x)
lim
exists and it is finite. Then
x→0 x
x>0
Z
(3)
Z
1
lim
1
n
g(x )dx =
n→∞
0
0
g(u)
du.
u
Proof. Define the function h : [0, 1] → R,

g(x)


if x ∈ (0, 1]

x
(4)
h(x) =

lim if x = 0

 x→0
x>0
It is clear that h is continuous and denote
Z
x
H(x) =
h(t)dt.
0
We have
Z
1
n
Z
1
n
g(x )dx = n
0
¯1 Z
¯
x h(x )dx = xH(x )¯ −
n
n
0
0
Z
1
= H(1) −
Z
1
n
H(x )dx =
0
0
If 0 < a < 1, then we can write
¯Z 1
¯ Z 1
Z
¯
¯
n
n
¯
H(x )dx¯¯ ≤
|H(x )|dx =
¯
0
(5)
0
g(x)
dx −
x
a
t∈[0,1]
Z
1
H(xn )dx.
0
Z
1
|H(x )|dx +
0
H(xn )dx
0
n
≤ a|H(αnn )| + (1 − a)M,
where αn ∈ [0, a] and M max |H(t)|.
1
n
a
|H(xn )|dx
46
Dorin Andrica and Mihai Piticari
ε
Consider ε > 0 such that a > 1 −
. Because lim |H(αnn )| = 0, it
n→∞
2M
ε
n
follows that a|H(αn )| < for all positive integers n ≥ N (ε). From (5) we
2
get
¯
¯Z 1
¯ ε
¯
ε ³
ε ´
n
¯
¯
H(x )dx¯ ≤ + (1 − a)M < + 1 − 1 +
M = ε,
¯
2
2
2M
0
Z
1
i.e. lim
n→∞
H(xn )dx = 0 and the conclusion follows.
0
Lemma 2. Let g : [0, 1] → R be a continuous function such that
g(x)
lim
exists and it is finite. Then for any function f : [0, 1] → R of
x→0 x
x>0
class C 1 ,
Z
(6)
1
n
f (x)g(x )dx = f (1)
lim n
n→∞
Z
1
0
0
Z
g(x)
dx
x
x
g(t)
dt, x ∈ [0, 1], and note that
t
0
Z 1
Z 1
g(xn )
n
n
f (x)g(x )dx = n
xn f (x) n dx
x
0
0
¯1 Z 1
¯
n
= G(x )xf (x)¯ −
(xf 0 (x) + f (x))G(xn )dx
Proof. Denote G(x) =
0
0
Z
1
= G(1)f (1) −
(xf 0 (x) + f (x))G(xn )dx
0
Z
(7)
1
= f (1)
0
g(x)
dx −
x
Z
1
(xf 0 (x) + f (x))G(xn )dx.
0
We will prove that
Z
1
lim
n→∞
0
(xf 0 (x) + f (x))G(xn )dx = 0.
On a class of convergent sequences defined by integrals
47
Indeed, by considering M = max |xf 0 (x) + f (x)| we have
x∈[0,1]
¯Z 1
¯ Z
¯
¯
0
n
¯ (xf (x) + f (x))G(x )dx¯ ≤
¯
¯
0
Z
1
≤M
|xf 0 (x) + f (x)||G(xn )|dx
0
|G(xn )|dx.
0
Z
1
Using that lim
n→∞
1
|G(xn )|dx = 0 (see the proof of Lemma 1) the desired
0
relation (6) follows from (7).
Our main results are the following.
Theorem 1. Let g : [0, 1] → R be a continuous function such that
g(x)
lim
exists and it is finite. Then for any continuous function
x→0 x
x>0
f : [0, 1] → R the relation (6) holds.
Proof. According to the well-known Weierstrass approximation theorem, consider (fm )m≥1 a sequence of polynomials uniformly convergent to f
on the interval [0, 1]. Let ε > 0 be a fixed real number. We will show that
we can find a positive integer N (ε) such that for any n ≥ N (ε) and for any
x ∈ [0, 1], we have
¯ Z
¯
¯n
(8)
¯
1
Z
1
n
f (x)g(x )dx − f (1)
0
0
µ Z
From technical reasons, take ε = ε/ 2
¯
g(x) ¯¯
dx¯ < ε
x
¶
g(x)
dx + 1 and consider
x
0
the positive integer N (ε) with the property that |fm (x) − f (x)| < ε0 for any
1
0
x ∈ [0, 1]. Because f and g are bounded it follows that we can assume that
f ≥ 0 and g ≥ 0. For m ≥ N (ε) we have
fm (x)g(xn ) − ε0 g(xn ) ≤ f (x)g(xn ) ≤ fm (x)g(xn ) + ε0 g(xn ),
48
Dorin Andrica and Mihai Piticari
hence
Z
Z
1
n
n
1
0
fm (x)g(x )dx − ε n
Z
g(x )dx ≤ n
0
0
Z
(9)
f (x)g(xn )dx
0
Z
1
≤n
1
n
n
1
0
fm (x)g(x )dx + ε n
0
g(xn )dx
0
From Lemma 2 we have
Z
lim n
n→∞
Z
1
1
n
fm (x)g(x )dx = fm (1)
0
0
and
Z
Z
1
0
lim nε
n
g(x )dx = ε
n→∞
1
0
0
0
g(x)
dx
x
g(x)
dx
x
and it follows that for any positive integer n ≥ N 0 (ε)
Z
Z
1
n
n
fm (x)g(x )dx − ε n
0
Z
−ε
1
0
0
Z
n
fm (x)g(x )dx + ε n
Z
n
g(x )dx ≤ fm (1)
0
Z
1
+ε0
0
g(x)
dx
x
1
g(x)
dx
x
0
1
0
0
1
g(x)
dx − ε0
x
Z
1
n
Z
n
g(x )dx ≥ fm (1)
0
and
1
0
0
g(x)
dx + ε0
x
0
But f (1) − ε < fm (1) < f (1) + ε0 imply for all n ≥ N 0 (ε)
¶
µZ 1
Z 1
Z 1
g(x)
g(x)
0
0
(f (1) − ε )
dx − ε
dx + 1 ≤ n
f (x)g(xn )dx
x
x
0
0
0
Z
1
0
≤ (f (1) + ε )
0
g(x)
dx + ε0
x
µZ
1
0
¶
g(x)
dx + 1 .
x
On a class of convergent sequences defined by integrals
49
The last relation is equivalent to
¯ Z 1
¯
µ Z 1
¶
Z 1
¯
¯
g(x)
g(x)
n
0
¯n
f (x)g(x )dx − f (1)
dx¯¯ < ε 2
dx + 1
¯
x
x
0
0
0
= ε, for all n ≥ N 0 (ε),
and the conclusion follows.
Remarks. 1) Consider the function h : [0, 1] → R,

g(x)


if x 6= 0



 x
h(x) =


g(x)


lim
if x = 0.

 x→0
x
x>0
g(x)
exists and it is finite, it follows that function h is
x→0 x
x>0
continuous on [0, 1]. Applying the result in Theorem 1 we obtain that for
Because lim
any continuous functions f, h : [0, 1] → R the following relation holds:
Z 1
Z 1
n
n
(10)
lim n
x f (x)h(x )dx = f (1)
h(x)dx
n→∞
0
0
Relation (10) was proved in [6] in the case when f is differentiable and
f 0 is continuous on [0, 1].
2) If u : [0, 1] → R is a continuous function such that its right derivative
at 0 exists and it is finite, then the function g(x) = u(x) − u(0) satisfies the
hypotheses in Theorem 1. From (6) it follows
Z 1
f (x)g(xn )dx = 0,
lim
n→∞
0
i.e.
Z
(11)
lim
n→∞
1
Z
1
n
f (x)u(x )dx = u(0)
0
f (x)dx
0
50
Dorin Andrica and Mihai Piticari
In the paper [6] (see also [3]) is proved that the above relation holds
even f , u are only continuous on [0, 1].
3) If f = 1, the constant function on [0, 1], from (10) we get the result
in paper [7].
The order of convergence in (10) is given in the following result.
Theorem 2. Let f : [0, 1] → R be a function of class C 1 and let
h : [0, 1] → R be a continuous function. Then
·
¸
Z 1
Z 1
n
n
lim n f (1)
h(x)dx − n
x f (x)h(x )dx
n→∞
0
0
Z
1
0
(12)
= (f (1) + f (1))
0
Z
H(x)
dx,
x
x
where H(x) =
h(t)dt.
0
Proof. We can write
Z
1
n
Z
n
1
n
x f (x)h(x )dx =
0
xf (x)(H(xn ))0 dx
0
¯1 Z 1
¯
= xf (x)H(x )¯ −
(xf (x))0 H(xn )dx.
n
0
Therefore
·
Z 1
Z
n f (1)
h(x)dx − n
0
1
0
¸
n
Z
1
n
x f (x)h(x )dx = n
0
(xf (x))0 H(xn )dx.
0
Functions x 7→ xf (x), x 7→ H(x) satisfy the hypothesis in Theorem 1,
hence we have
Z
1
lim
n→∞
Z
0
n
1
0
n(xf (x)) H(x )dx = (f (1) + f (1))
0
0
H(x)
dx
x
On a class of convergent sequences defined by integrals
51
and the desired relation follows.
g(x)
g(x)
if x 6= 0 and h(0) = lim
, where
x→0 x
x
x>0
g(x)
g : [0, 1] → R is a continuous function such that lim
exists and it is
x→0 x
x>0
finite, from (11) we derive the following relation
·
¸
Z 1
Z 1
g(x)
n
lim f (1)
dx − n
f (x)g(x )dx
n→∞
x
0
0
¶
Z 1µ Z x
g(t)
1
0
dt dx.
= (f (1) + f (1))
x 0 t
0
Remarks. 1) Writing h(x) =
This is the order of convergence in (6) when f is of class C 1 .
2) If h = 1, the constant function on [0, 1], from (10) we derive Problem
2.83.b) in [3].
3
Some applications
Application 1. 1) If f : [0, 1] → R is a continuous function, then
Z 1 n
x f (x)
π
lim n
dx = f (1).
2n
n→∞
4
0 1+x
2) If f : [0, 1] → R is a function of class C 1 , then
·
¸
Z 1 n
Z 1
π
x f (x)
arctg x
0
lim n f (1) − n
dx = (f (1) + f (1))
dx.
2n
n→∞
4
x
0 1+x
0
These results follows from (6) and (13), where
g(x) =
x
,
1 + x2
x ∈ [0, 1].
If f (x) = 1 for all x ∈ [0, 1], then we get Problem 2 of the 12th Form in
final Round of Romanian National Olympiad 2006.
52
Dorin Andrica and Mihai Piticari
Application 2. 1) (Romanian National Olympiad, County round 2001,
partial statement) If a > 0, then
Z
1
lim n
n→∞
0
xn
a+1
.
dx = ln
n
a+x
a
2) The following relation holds
µ
a+1
lim n ln
−n
n→∞
a
(14)
Z
1
0
xn
dx
a + xn
¶
=
∞
X
(−1)n+1
n=1
a n n2
.
x
we easily derive the first
a+x
relation. For the second one we use (13) for the same choosing of functions.
Indeed, taking in (6) f = 1 and g(x) =
The right hand side in (13) becomes
¶
Z 1µ Z x
Z 1
1
dt
ln(x + a) − ln a
dx =
dx
x 0 a+t
x
0
0
Z
1
=
0
Z 1 X
∞
∞
X
1 ³
x´
1
(−1)n+1 ³ x ´n
(−1)n+1
ln 1 +
dx =
dx =
.
n n2
x
a
n
a
a
0 x n=1
n=1
If a = 1, from (14) we get the interesting relation
µ
¶
Z 1
xn
π2
(15)
lim n ln 2 − n
dx =
n
n→∞
12
0 1+x
Application 3. 1) If f : [0, 1] → R is a continuous function, then
Z
(16)
1
lim n
n→∞
f (x) ln(1 + xn )dx =
0
π2
f (1)
12
2) If f : [0, 1] → R is a function of class C 1 , then
¸
· 2
Z 1
3
π
n
f (1) − n
f (x) ln(1 + x )dx = (f (1) + f 0 (1))ζ(3),
(17)
lim n
n→∞
12
4
0
where ζ is the well-known Riemann, s function.
On a class of convergent sequences defined by integrals
53
To prove (16) we take in (6), g(x) = ln(1 + x). We have
Z
1
0
g(x)
dx =
x
=
Z
1
0
ln(1 + x)
dx =
x
∞
X
(−1)n+1
n=1
n2
= ζ(2) −
Z
1
0
∞
1 X (−1)n+1 xn
dx
x n=1
n
2
1
π2
ζ(2)
=
ζ(2)
.
22
2
12
In order to prove (17) we use relation (13) and observe that in the right
hand side we obtain
!
¶
Z 1µ Z x
Z 1Ã Z xX
∞
ln(1 + t)
1
(−1)n+1 tn−1
1
dt dx =
dt dx
x 0
t
x 0 n=1
n
0
0
Z
1
=
0
Ã
∞
1 X (−1)n+1 xn
x n=1
n2
!
dx =
∞
X
(−1)n+1
n=1
n3
= ζ(3) −
2
3
ζ(3) = ζ(3).
3
2
4
References
[1] Andrica, D., Piticari, M., An extension of the Riemann-Lebesgue
lemma and some applications, Proc. International Conf. on Theory and
Applications of Mathematics and Informatics (ICTAMI 2004), Thessaloniki, Acta Universitatis Apulensis, No.8(2004), 26-39.
[2] Andrica, D., Piticari, M., On a class of sequences defined by using Riemann integral, Proc. International Conf. on Theory and Applications
of Mathematics and Informatics (ICTAMI 2005), Albac, Acta Universitatis Apulensis, No. 10 (2005), 381-385.
[3] Dumitrel, F., Problems in Mathematical Analysis (Romanian), Editura
Scribul, 2002.
54
Dorin Andrica and Mihai Piticari
[4] Lang, S., Analysis I, Addison-Wesley, 1968.
[5] Mocanu, M., Sandovici, A., On the convergence of a sequence generated
by an integral, Creative Math., 14(2005), 31-42.
[6] Muşuroia, N., On the limits of some sequences of integrals, Creative
Math., 14(2005), 43-48.
[7] Păltănea, E., O reprezentare a unui şir de integrale, G.M. No.11 (2003),
419-421.
[8] Sireţchi, Gh., Mathematical Analysis II. Advanced Problems in Differential and Integral Calculus, (Romanian), University of Bucharest,
1982.
”Babeş-Bolyai” University
Faculty of Mathematics and Computer Science
Cluj-Napoca, Romania
E-mail address: dandrica@math.ubbcluj.ro
”Dragoş-Vodă” National College
Câmpulung Moldovenesc, Romania