Sharp Approximations for the Generalized Elliptic Integral of the First Kind

Theorem 1.1

Let a ∈ (0, 1/2], c ∈ (−∞, +∞),

and

Then we have the following conclusions:

1. The function F_c(r) is strictly decreasing if and only if c ≤ λ_a, in which case the range of F_c(r) on (0, 1) is (0,eπ/[2sin⁡(πa)]−eR(a)/2−c). Consequently, the inequality

   1.12 sin⁡(πa)log⁡(cr′+eR(a)/2r′)<Ka(r)<sin⁡(πa)log⁡(cr′+eπ/[2sin⁡(πa)]−cr′)

   holds for all r ∈ (0, 1);

2. If c > λ_a, then there exists an r₀ ∈ (0, 1), such that F_c(r) is strictly increasing on (0, r₀) and strictly decreasing on (r₀, 1). Moreover, the inequality

   1.13 Ka(r)>sin⁡(πa)log⁡(cr′+eR(a)/2r′)

   holds for all r ∈ (0, 1) if and only if c≤c0∗=eπ/[2sin⁡(πa)]−eR(a)/2.

Theorem 1.2

Let a ∈ (0, 1/2] and

Then G(r) is strictly increasing from (0, 1) onto ((45a⁴ − 90a³ + 96a² − 51a + 11)/11, 2 sin(πa)/π + 6a² − 6a + 1). Consequently, the inequality

holds for all r ∈ (0, 1).

Remark 1

If we take a = 1/2, then the inequalities (1.13) and (1.14) reduce to (1.7) and (1.10), respectively.

Lemma 2.1. (cf. [7: Theorem 1.25])

Let −∞ < a < b < ∞, f, g: [a, b] → ℝ be continuous on [a, b] and differentiable on (a, b), and g′(x) ≠ 0 on (a, b). If f′(x)/g′(x) is increasing (decreasing) on (a, b), then so are the functions

If f′(x)/g′(x) is strictly monotone, then the monotonicity in the conclusion is also strict.

Lemma 2.2. (cf. [16] or [23: Lemma 2.1])

Suppose that the power series f(x)=∑n=0∞anxn and g(x)=∑n=0∞bnxn both converge for |x| < R, and b_n > 0 for all n ∈ {0, 1, 2,…}. If the non-constant sequence {an/bn}n=0∞ is increasing (decreasing), then h(x) = f(x)/g(x) is strictly increasing (decreasing) on (0, R).

Lemma 2.3

Let a ∈ (0, 1/2]. Then the following statements are true:

1. The function ϕ₁(r) = (ℰ_a − r′²𝒦_a/r² is strictly increasing from (0, 1) onto (πa/2, sin(πa)/[2(1 − a)]);

2. The function ϕ₂(r) = r′^c𝒦_a is strictly decreasing from (0, 1) onto (0, π/2) if and only if c ≥ 2a(1 − a). Moreover, r↦r′Ka is strictly decreasing for each a ∈ (0, 1/2];

3. The function ϕ₃(r) = 𝒦_a/sin(πa) + log r′ is strictly decreasing from (0, 1) onto (R(a)/2, π/(2 sin(πa)));

4. The function ϕ4(r)=[(1+r′2)sin⁡(πa)−2(1−a)(Ea−r′2Ka)]/r′2(1+a)(2−a)/3 is strictly decreasing from (0, 1) onto (0, 2 sin(πa)).

Proof

Parts (1), (2) and (3) can be found in [4: Lemmas 5.2(1), 5.4(1) and 5.5(1)].

For part (4), let ϕ41(r)=(1+r′2)sin⁡(πa)−2(1−a)(Ea−r′2Ka) and ϕ₄₂(r) = r′^{2(1 + a)(2 − a)/3}. Simple computations lead to that

and

It is not difficult to verify that r′^{2(a2 − a + 1)/3}𝒦_a is strictly decreasing on (0, 1) from part (2), so that ϕ41′(r)/ϕ42′(r) is strictly decreasing on (0, 1) by (2.3). Combining with (2.2) and applying Lemma 2.1, we derive that ϕ₄(r) is strictly decreasing on (0, 1).

Part (1) implies that ϕ₄(0⁺) = 2 sin(πa), and by (2.2), (2.3) and L’Hospital’s rule we get ϕ₄(1⁻) = 0. Therefore, part (4) is valid. □

Lemma 2.4

For a ∈ (0, 1/2], define ψ on (0, 1) by

Then φ(r) is strictly increasing from (0, 1) onto (πa2(1−a)[π+sin⁡(πa)]/4,[sin2⁡(πa)]/[4(1−a)]).

Proof

Let ψ1(r)=(1−a)(Ea−r′2Ka)2+sin⁡(πa)r′2[ar2Ka−(Ea−r′2Ka)]. Then differentiating ψ₁ gives

so that

Let ψ2(r)=r′23(1+a)(2−a)[ar2Ka−(Ea−r′2Ka)]/r4. Then from (1.1) and (2.1) one has

Since the function x ↦ (1 − x)^d F(a, b; c; x) with a, b, c > 0 and d ∈ ℝ is decreasing on (0, 1) if and only if d ≥ max{a + b − c, ab/c}(cf. [27: Lemma 2.15] and also [23]), then ψ₂(r) is strictly decreasing on (0, 1).

Obviously,

Furthermore, it follows from (2.5), (2.6) together with Lemma 2.3 (1) and (4) that ψ′(r) > 0 for all r ∈ (0, 1), so that ψ(r) is strictly increasing on (0, 1).

Using (2.6) and Lemma 2.3(1), we get

Therefore, Lemma 2.4 follows from (2.7) and the monotonicity of ψ(r). □

Lemma 2.5

For a ∈ (0, 1/2], define φ on (0, 1) by

Then φ(r) is strictly increasing from (0, 1) onto (e^{π/[2 sin(πa)]}[sin(πa) − a(1 − a)π], +∞).

Proof

Differentiation gives

where ψ(r), defined in (2.4), is strictly increasing on (0, 1).

By (1.1), (2.1) and Lemma 2.3 (1), (3) together with L’Hospital’s rule, we obtain that

and

Therefore, Lemma 2.5 follows from (2.10), (2.11) and the monotonicity of φ(r). □

Lemma 2.6

For a ∈ (0, 1/2], define the function h on (0, 1) by

Then h(r) is strictly decreasing from (0, 1) onto (0, a(1 − a)(7a² − 7a + 4)/[3(5a² − 5a + 2)]).

Proof

Let

and

Then h(r) = h₁(r)/h₂(r) and tedious computations lead to that

and

Then

where

and

for all n ∈ ℕ. Let C_n = A_n/B_n, one has

and

It is not difficult to verify that 61a⁴ − 122a³ + 113a² − 52a + 12 > 0 for all a ∈ (0, 1/2]. Thus, C_{n + 1} − C_n > 0, i.e. C_n is strictly increasing for all n ∈ ℕ, and h(r) is strictly decreasing on (0, 1) by Lemma 2.2 and (2.12).

Note that

Therefore, Lemma 2.6 follows from (2.13) and the monotonicity of h(r). □

Lemma 2.7

For a ∈ (0, 1/2] and c ∈ (−∞, +∞), define g_c on (0, 1) by

Then the following statements are true:

1. g_c(r) is strictly decreasing on (0, 1) if and only if c ≥ 2a(1 − a)(7a² − 7a + 4)/[3(5a² − 5a + 2)];

2. g_c(r) is strictly increasing on (0, 1) if and only if c ≤ 0.

Proof

Differentiating g_c gives

where h(r) is defined in Lemma 2.6. The proof of Lemma 2.6 shows that the function r↦(1−a)r2Ea−a(1−2a)r2(Ea−r′2Ka)−r′2(Ka−Ea) is positive on (0, 1). Combining with (2.15), we obtain that gc′(r)<0 for r ∈ (0, 1) if and only if c ≥ 2a(1 − a)(7a² − 7a + 4)/[3(5a² − 5a + 2)], and gc′(r)>0 for r ∈ (0, 1) if and only if c ≤ 0. Therefore, Lemma 2.7 follows. □