Applied Mathematics E-Notes, 14(2014), 193-199 c
Available free at mirror sites of http://www.math.nthu.edu.tw/ amen/
ISSN 1607-2510
Some Inequalities For Complete Elliptic Integrals
Li Yiny, Feng Qiz
Received 1 September 2014
Abstract
In this paper, by using the Lupaş integral inequality, the authors …nd some
new inequalities for the complete elliptic integrals of the …rst and second kinds.
These results improve some known inequalities.
1
Introduction
Legendre’s complete elliptic integrals of the …rst and second kind are de…ned for real
numbers 0 < r < 1 by
(r) =
Z
=2
1
p
1
0
and
"(r) =
Z
=2
0
2
r2 sin t
p
1
r2
dt =
Z
1
p
0
2
sin t d t =
Z
1
(1
0
1
r
t2 )(1
r 2 t2 )
1 r2 t2
dt
1 t2
dt
(1)
(2)
respectively. They can also de…ned by
(r; s) =
Z
=2
p
0
and
"(r; s) =
Z
0
Let r0 =
p
1
r2 . We often denote
0
=2
p
1
dt
(3)
r2 cos2 t + s2 sin2 t d t:
(4)
r2 cos2 t + s2 sin2 t
(r) = (r0 ) and "0 (r) = "(r0 ):
Mathematics Subject Classi…cations: 26D15, 33C75, 33E05.
of Mathematics, Binzhou University, Binzhou City, Shandong Province, 256603,
China
z College of Mathematics, Inner Mongolia University for Nationalities, Tongliao City, Inner Mongolia Autonomous Region, 028043, China; Department of Mathematics, College of Science, Tianjin
Polytechnic University, Tianjin City, 300387, China; Institute of Mathematics, Henan Polytechnic
University, Jiaozuo City, 454010, Henan Province, 454010, China
y Department
193
194
Some Inequalities for Complete Elliptic Integrals
These integrals are special cases of the Gauss hypergeometric function
F (a; b; c; x) =
where (a; n) =
Qn
1
k=0
(r) =
1
X
(a; n)(b; n) xn
;
(c; n)
n!
n=0
(a + k). Indeed, we have
2
1 1
; ; 1; r2
2 2
F
and "(r) =
2
1 1
; ; 1; r2 :
2 2
F
Recently, some bounds for "(r) and (r) were discovered in the paper [6]. For
example, Theorem 1 in [6] states that, for 0 < r < 1,
2
1 (1 + r)1 r
< "(r) <
ln
2 (1 r)1+r
1
2
+
1
r2 1 + r
ln
:
4r
1 r
(5)
For more information on inequalities of complete elliptic integrals, please refer to [1, 2,
3, 7, 8] and a short survey in [9, pp. 40–46].
Motivated by the double inequality (5), some estimates for "(r) in terms of rational
functions of the arithmetic, geometric, and roots square means were obtained in [5, 10,
11].
The aim of this paper is to establish some new inequalities for the complete elliptic
integrals.
2
A Lemma
In order to prove our main results, the following lemma is necessary.
LEMMA 2.1. If f 0 ; g 0 2 L2 [a; b], then
1
b
a
Z
b
1
f (t)g(t) d t
b
a
where
0
kf k2 =
Z
a
b
f
a
02
Z
b
f (t) d t
a
!1=2
dt
1
b
a
Z
b
b
g(t) d t
a
2
a
0
and kg k2 =
Z
b
g
02
dt
a
kf 0 k2 kg 0 k2 ;
!1=2
(6)
:
The inequality (6) is called the Lupaş integral inequality, see [4, p. 57].
3
Main Results
Now we are in a position to …nd some inequalities for complete elliptic integrals.
THEOREM 3.1. For r 2 (0; 1), we have
p
p
6 + 2 1 r2 3r2
p
"(r)
4 2
p
10
p
2 1 r2
p
4 2
5r2
:
(7)
L. Yin and F. Qi
195
PROOF. Taking
f (t) = g(t) =
and letting a = 0 and b =
2
Z
p
r2 sin2 t
1
in the inequality (6) yield
2
=2
4
r2 sin2 t d t
1
2
0
=
=
=
Z =2 4 2
1
r sin t cos2 t
"
(r)
dt
2
2
2 0
1 r2 sin2 t
Z =2
1
X
1
r4 sin2 t cos2 t
r2n sin2n t d t
2 0
n=0
1 Z =2
X
1
1
r2n+4 sin2n+2 t sin2n+4 t d t = h(r);
2 n=0 0
4
4
r2
2
2
where we use
Z
=2
sin2i t d t =
0
for i 2 N, and
2
A direct calculation yields
(2i 1)!!
(2i)!!
(9)
1
1
X
X
(2n + 1)!! 2n+2
(2n + 1)!! 2n+3
r
=r
r
=r
(2n + 2)!!
(2n + 2)!!
n=0
n=0
where we use
p
1
1
t2
1
X
(2i 1)!! 2n
t ;
(2i)!!
n=0
=
Hence, we have
h(r) = h(0) +
(8)
1
X
(2n + 1)!! r2n+4
:
(2n + 2)!! 2n + 4
n=0
h(r) =
h0 (r) =
"2 (r)
Z
h0 (r) d r = 1
0
1
1
r2
1 ;
jtj < 1:
p
1
r
p
(10)
r2
:
2
r2
Substituting this equality into (8) gives
4
2
" (r)
2
r2
2
2
1
4
p
r2
1
4
r2
:
8
This means the double inequality (7).
REMARK 3.1. By the well-known software Mathematica, we can show that
(i) the left-hand side inequality in (7) re…nes the corresponding one in (5);
(11)
196
Some Inequalities for Complete Elliptic Integrals
(ii) the right-hand side inequalities in (7) and (5) are not contained in each other;
(iii) when r 2 [ 41 ; 34 ], the right-hand side inequality in (7) is better than the corresponding one in (5).
THEOREM 3.2. We have that
p
32(1 r2 ) + r4
p p
for r 2 (0; 1)
8 2 4 (1 r2 )3
(r)
and
p
q
p
32(1 r2 ) r4
p p
for
x
2
0;
2
3 2
4
2
3
8 2 (1 r )
(r)
PROOF. Taking
and letting a = 0 and b =
Z
2
2
4
2
2
1
(r)
1
4 2
dt
(r)
2
r2 sin2 t
Z =2 X
1
2
r2n sin2n t d t
Z
(r)
=2
n=0
1
X
(2n 1)!! 2n
r
=
(2n)!!
n=0
4
2
2
(r)
p
1
1
r2
r4 sin2 t cos2 t
dt
1 r2 sin2 t 3
0
Z =2
1
X
1
r4 sin2 t cos2 t
(n + 2)(n + 1)r2n sin2n t d t
4 0
n=0
1
2
=
1 r2 sin2 t
in the inequality (6) lead to
0
4
=
=
2
2
4 :
1
f (t) = g(t) = p
=2
0
=
(12)
1
1X
(2n + 1)!! 1
r3
(n + 2)(n + 1)r2n+4
=
p(r);
8 n=0
(2n + 2)!! 2n + 4
32
where
p(r) =
satis…es p(0) = 0,
Z
0
r
p(r) d r =
1
X
(2n + 2)(2n + 1)!! 2n+1
r
(2n + 2)!!
n=0
1
X
(2n + 1)!! 2n+2
1
r
=p
(2n
+
2)!!
1
r2
n=0
and so
p(r) = p
(1
r
r 2 )3
:
1;
(13)
L. Yin and F. Qi
197
Consequently, we …nd
4
2
2
(r)
p
1
1
r2
The double inequality (12) follows.
r4
p
:
32 (1 r2 )3
(14)
THEOREM 3.3. For r > 0 and s > 0, we have
8
r
8rs(r2 + s2 ) (s2
rs
r2 )2
"(r; s)
8
r
8rs(r2 + s2 ) + (s2
rs
r 2 )2
:
(15)
PROOF. Taking
f (t) = g(t) =
and letting a = 0 and b =
Z
2
p
r2 cos2 t + s2 sin2 t
in the inequality (6) reveal
2
=2
4
r2 cos2 t + s2 sin2 t d t
2
0
=
4
"2 (r; s)
2
(s2
r2 )2
2
2
=
r2 )2
(s
8
"2 (r; s)
Z =2 2
r2 + s2
1
(s
r2 )2 sin2 t cos2 t
dt
2
2 0
r2 cos2 t + s2 sin2 t
Z =2
1
1
dt
4 0
r2 cos2 t + s2 sin2 t
1
s
(s2 r2 )2
=2
arctan
tan tj0
=
:
rs
r
16rs
This means the double inequality (15).
THEOREM 3.4. For s > r > 0, we have
4
2
2
(r; s)
s2 r2
32rs
1
rs
1
1
+ 2
2
s
r
PROOF. Taking
f (t) = g(t) = p
1
r2
cos2
t + s2 sin2 t
:
(16)
198
Some Inequalities for Complete Elliptic Integrals
and letting a = 0 and b =
Z
2
4
=
=
=
=
=
2
r2
8
r2
8
r2
8
r2
8
s
in the inequality (6), we acquire
=2
4 2
1
dt
(r; s)
2
2 cos2 t + s2 sin2 t
r
0
Z =2 2
s
r2 2 sin2 t cos2 t
1
1
2
(r; s)
dt
rs
2 0
r2 cos2 t + s2 sin2 t 3
Z =2
s2
1
sin t cos t d
2
2
2
0
r cos t + s2 sin2 t
"Z
Z =2
=2
s2
sin2 t
cos2 t
d
t
r2 cos2 t + s2 sin2 t 2
0
0
r2 cos2 t + s2 sin2 t
"Z
#
Z =2
=2
s2
csc2 t
sec2 t
dt
dt
r2 cot2 t + s2 2
r2 + s2 tan2 t 2
0
0
"Z
#
Z =2
1
s2
1
1
du
d
(r2 u2 + s2 )2
(r2 + s2 2 )2
0
0
2
=
2
r2
32rs
1
1
+ 2
2
s
r
2
dt
#
:
The proof is complete.
REMARK 3.2. From (16), we easily obtain
r
32r2 s2
(s4 r4 )
(r; s)
2
32 r3 s3
2
r
32r2 s2 + (s4
32 r3 s3
r4 )
:
(17)
By the software Mathematica, we can show that the double inequality in (17) and
the inequality in [6, Theorem 2] are not contained in each other.
Acknowledgements. The …rst author was supported partially by the Natural
Science Foundation of Shandong Province under grant number ZR2012AQ028, China.
The second author was partially supported by the National Natural Science Foundation
under Grant No. 11361038 of China and by the Foundation of the Research Program of
Science and Technology at Universities of Inner Mongolia Autonomous Region under
Grant No. NJZY14192, China.
The authors appreciate anonymous referees for their careful corrections to the original version of this paper.
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