A Simple Proof of an Inequality for the Complete Elliptic Integral of the First Kind
Harun Abdul Rohman
SMPN 2 Cilengkrang, Bandung Regency, West Java, Indonesia [email protected]
April 24, 2025
Abstract
This article presents a simple proof of inequalities for the complete elliptic integral of the first kind through symmetry arguments and interval reduction techniques. Fur- thermore, it provides a geometric interpretation of the bounds for this integral, offering intuitive insights into its analytic behavior.
Keywords: Complete elliptic integral, First kind elliptic integral, Symmetry argu- ments, Inequalities.
1 Introduction
In his article, Jameson [1] introduces the integral I(a, b) = 4
Z π2
0
1
pa2cos2(t) +b2sin2(t)dt (1) where a ≥ b > 0, which frequently appears alongside elliptic integrals in the context of ellipses. He establishes the inequality
π
a+b ≤I(a, b)≤ π 2√
ab (2)
using Gauss’s identityI(a, b) = AGM(a,b)2π , whereAGM(a, b) is the arithmetic-geometric mean of a and b, and satisfies√
ab≤AGM(a, b)≤ a+b2 [2–4].
The integral I(a, b) can also be written in terms of the complete elliptic integral of the first kind K(k), as
I(a, b) = 4 a
Z π2
0
1
p1−k2sin2(t)dt = 1 aK(k), where k2 = 1− ba22, so that
aI(a, b) =K(k)⇒ πa
a+b ≤K(k)≤ πa 2√
ab. (3)
This inequality can also be derived using the Cauchy-Schwarz inequality, though the proof is generally more complex than with Gauss’s identity [1].
An interesting observation is that the integrand in (1) is symmetric with respect to π2 −t, leading to the identity I(a, b) =I(b, a). Thus,
I(a, b) = 1
2[I(a, b) +I(b, a)]
= 4 Z π2
0
1 2
"
1
pa2cos2(t) +b2sin2(t) + 1
pb2cos2(t) +a2sin2(t)
# dt.
(4)
Substituting into the elliptic form, we obtain K(k) = 4
Z π2
0
1 2
"
1
p1 +k2sin2(t) + 1
p1−k2cos2(t)
#
dt. (5)
In this article, I present a simpler proof of the inequality for K(k) by modifying the limits of integration in (4) or (5) to [0,π4]. The key idea is to analyze the extremal behavior of the integrand at the endpoints of this symmetric interval. I also provide a geometric interpretation that supports the use of the symmetric limit [0,π4] in (4) or (5).
2 Results
Lemma 2.1. For a≥b >0, the integral I(a, b) = 4
Z π2
0
1
pa2cos2(t) +b2sin2(t)dt can be expressed in a symmetric form as
I(a, b) = 8 Z π4
0
1 2
"
1
pa2cos2(t) +b2sin2(t) + 1
pb2cos2(t) +a2sin2(t)
#
dt (6)
Proof. Observe that equation (6) can be simplified into a sum of two integrals:
I(a, b) = 4 Z π4
0
1
pa2cos2(t) +b2sin2(t)dt+ 4 Z π4
0
1
pb2cos2(t) +a2sin2(t)dt (7) The second term in the integral corresponds to
I(b, a) = 4 Z π4
0
1
pb2cos2(t) +a2sin2(t)dt
Since I(b, a) is symmetric with respect to the substitution u= π2 −t, we can write:
I(b, a) = 4 Z π4
π 2
1
pa2cos2(u) +b2sin2(u)(−du) (8)
Because u is a dummy variable, we can replace it back witht, yielding:
I(b, a) = 4 Z π2
π 4
1
pa2cos2(t) +b2sin2(t)dt (9) Substituting equation (9) into equation (7), we obtain:
I(a, b) = 4 Z π4
0
1
pa2cos2(t) +b2sin2(t)dt+ 4 Z π2
π 4
1
pa2cos2(t) +b2sin2(t)dt
= 4 Z π2
0
1
pa2cos2(t) +b2sin2(t)dt
This completes the proof. □
Figure 1 shows the function f(t) with
f1(t) = 1
pa2cos2(t) +b2sin2(t), f2(t) = 1
pb2cos2(t) +a2sin2(t) for a= 4 andb = 0.7 on the interval t∈[0,2π].
Figure 1: The functionf(t) is shown in red,f1(t) in black, and f2(t) in blue. The functionsf1(t) and f2(t) intersect at the minimum point off(t). If we observeRπ/2
π/4
√ 1
a2cos2(t)+b2sin2(t)dtis the reflection result of 4Rπ/4
0
√ 1
b2cos2(t)+a2sin2(t)dtaboutx=π4. This provides insight into why Lemma 2.1 holds.
The function f(t) = 12[f1(t) +f2(t)] reaches its maximum values at t= 0,π
2, π,3π 2 ,2π, and its minimum values at
t = π 4,3π
4 ,5π 4 ,7π
4 .
Meanwhile, f1(t) reaches its maximum values at t = 0, π,2π and its minimum values at t = π2,3π2 , while f2(t) reaches its maximum values at t = π2,3π2 and its minimum values at t= 0, π,2π.
Theorem 2.2. Let a≥b >0. Then the integralI(a, b) satisfies the inequality 2π
qa2+b2 2
≤I(a, b)≤ π(a+b)
ab . (10)
Proof. We know that I(a, b) = 8
Z π4
0
1 2
1
√
a2cos2t+b2sin2t + 1
√
b2cos2t+a2sin2t
dt.
Let
f(t) = 1 2
1
√
a2cos2t+b2sin2t + 1
√
b2cos2t+a2sin2t
. This function reaches its maximum and minimum values at:
f(0) = a+b 2ab , fπ
4
= 1
qa2+b2 2
.
Since f π4
≤f(0), by the Squeeze Theorem we obtain:
8 Z π4
0
1 qa2+b2
2
dt ≤I(a, b)≤8 Z π4
0
a+b 2ab dt.
Simplifying both sides, we get:
2π qa2+b2
2
≤I(a, b)≤ π(a+b) ab .
This completes the proof. □
When we observe the bounds of I(a, b), they are not tight bounds, because if I(a, b) =
2π
AGM(a,b), then:
2ab
a+b ≤AGM(a, b)≤
ra2+b2 2 As we know, the bounds for AGM(a, b) are given by:
2ab a+b ≤√
ab≤AGM(a, b)≤ a+b
2 ≤
ra2+b2 2 Corollary 2.3. Let k2 = 1− ab22 ≥0. Then, the inequality for K(k) is
2π q2−k2
2
≤K(k)≤π· 1 +√ 1−k2
√1−k2 .
Proof. Since a2−k2a2 =b2, inequality (10) becomes 2π
qa2+a2−k2a2 2
≤I(a, b)≤π· a+√
a2−k2a2 a√
a2−k2a2 .
This simplifies to
2π a
q2−k2 2
≤I(a, b)≤π·1 +√ 1−k2 a√
1−k2 . Since aI(a, b) =K(k), we obtain
2π q2−k2
2
≤K(k)≤π· 1 +√ 1−k2
√1−k2 .
This completes the proof. □
3 Discussion
The approach introduced in this article provides a novel perspective on classical inequalities involving the complete elliptic integral of the first kind, K(k), by emphasizing the role of symmetry and the efficacy of interval reduction techniques. In contrast to algebraically in- tensive methods, such as the arithmetic–geometric mean (AGM) approach, the symmetric formulation over the interval [0,π4] offers a more intuitive and geometrically motivated path- way. This methodology has also been applied in the context of the perimeter of ellipses and the complete elliptic integral of the second kind [5].
A central idea in this framework is the averaging of the reciprocal square roots of conju- gate quadratic forms, leading to the symmetric integrand
f(t) = 1 2
1
√
a2cos2t+b2sin2t + 1
√
b2cos2t+a2sin2t
,
which exploits the symmetry property I(a, b) = I(b, a). This symmetry not only facilitates the analytical process but also enables the derivation of new bounds through the investigation of extremal values of f(t) over the symmetric interval.
Moreover, the application of geometric reasoning to justify the choice of integration limits opens avenues for extending similar symmetry-based techniques to other integral inequalities.
Although the resulting bounds may not match the optimality of those obtained via the AGM method, they are derived using elementary mathematical tools and possess significant pedagogical value due to their conceptual clarity.
From a numerical standpoint, this approach contributes to more efficient and accurate computation of the complete elliptic integral of the first kind. The reduced length of the integration interval results in smaller approximation errors. Specifically, the error estimate for the trapezoidal ruleET over the interval [0,π4] is given by
ET =−
π 4
3
12n2f′′(ξ) =− π3
43·12n2f′′(ξ),
which is notably smaller than the corresponding error over [0,π2], expressed as ET =− π3
23·12n2f′′(ξ),
wherendenotes the number of subintervals andf′′(ξ) is the second derivative off evaluated at some point ξ ∈(a, b).
4 Conclusion
In this article, we have presented a simple and insightful proof of the inequality involving the complete elliptic integral of the first kind, K(k), by utilizing the symmetry of the integrand and reducing the interval of integration to
0,π4
. This approach simplifies the analysis and highlights the inherent balance in the function’s structure. The derived bounds for I(a, b) and K(k) offer a geometrically motivated perspective and extend classical results by Gauss and others. Although the resulting bounds are not as tight as those obtained from the arithmetic-geometric mean (AGM), they remain elementary, elegant, and accessible, making them a valuable alternative in studying elliptic integrals and their applications.
References
[1] G. J. O. Jameson,Inequalities for Means,The American Mathematical Monthly, vol. 107, no. 6, pp. 508–512, 2000.
[2] G. J. O. Jameson, “An Approximation to the Arithmetic-Geometric Mean,” 2014. [On- line]. Available: https://www.jstor.org/stable/24496592
[3] P. G. Almkvist and B. Berndt, “Gauss, Landen, Ramanujan, the Arithmetic-Geometric Mean, Ellipses, π, and the Ladies Diary,” The American Mathematical Monthly, vol. 95, no. 7, pp. 585–608, 1988. doi: 10.1007/978-3-319-32377-0
[4] G. Liang, “Arithmetic-Geometric Mean, π, Perimeter of Ellipse, and Beyond,” 2019.
[Online]. Available: https:http://www.math.emory.edu/~gliang7/AGM.pdf
[5] H.A. Rohman, “A Refined Symmetric Mean Integral Approach to Bounding the Perime- ter of an Ellipse,” Cambridge Open Engage, Apr. 2025. doi: 10.33774/coe-2025-5bb4j.