In order to prove our main results, we need several formulas and lemmas, which we present in this section.
For \(r\in(0, 1)\) and \(r'=\sqrt{1-r^{2}}\), the well-known complete elliptic integrals of the first and second kinds are defined by
$$ \textstyle\begin{cases} \mathcal{K}=\mathcal{K}(r)=\int^{\pi/2}_{0}(1-r^{2}\sin^{2}\theta )^{-1/2}\,d\theta,\\ \mathcal{K}'=\mathcal{K}'(r)=\mathcal{K}(r'),\\ \mathcal{K}(0)=\pi/2,\qquad\mathcal{K}(1)=+\infty, \end{cases} $$
and
$$ \textstyle\begin{cases} \mathcal{E}=\mathcal{E}(r)=\int^{\pi/2}_{0}(1-r^{2}\sin^{2}\theta )^{1/2}\,d\theta,\\ \mathcal{E}'=\mathcal{E}'(r)=\mathcal{E}(r')\\ \mathcal{E}(0)=\pi/2,\qquad\mathcal{E}(1)=1, \end{cases} $$
respectively, and the following formulas were presented in [18], Appendix E, pp.474-475:
$$\begin{aligned} \begin{aligned} & \frac{d\mathcal{K}}{dr}=\frac{\mathcal{E}-r^{\prime 2}\mathcal {K}}{rr^{\prime 2} },\qquad \frac{d\mathcal{E}}{dr}=\frac{\mathcal {E}-\mathcal{K}}{r}, \\ & \frac{d(\mathcal{E}-r^{\prime 2}\mathcal{K})}{dr}=r\mathcal{K},\qquad \frac {d(\mathcal{K}-\mathcal{E})}{dr}=\frac{r\mathcal{E}}{r^{\prime 2}}, \end{aligned} \end{aligned}$$
(2.1)
$$\begin{aligned} & \mathcal{E} \biggl(\frac{2\sqrt{r}}{1+r} \biggr)=\frac{2\mathcal {E}-r^{\prime 2}\mathcal{K}}{1+r}. \end{aligned}$$
(2.2)
In what follows, four special values \(\mathcal{E}(\sqrt {2}/2),\mathcal{K}(\sqrt{2}/2)\) and \(\mathcal{E}(0.9),\mathcal {K}(0.9)\) will be used. By numerical computations, these are given by
$$\begin{aligned}& \mathcal{E}(\sqrt{2}/2) =1.35064\cdots,\qquad\mathcal{K}(\sqrt {2}/2)=1.85407 \cdots, \end{aligned}$$
(2.3)
$$\begin{aligned}& \mathcal{E}(0.9)=1.1717\cdots,\qquad\mathcal{K}(0.9)=2.28055\cdots. \end{aligned}$$
(2.4)
Lemma 2.1
See [18], Theorem 1.25
For
\(-\infty< a < b <\infty\), let
\(f,g:[a, b]\rightarrow\mathbb{R}\)
be continuous on
\([a, b]\)
and be differentiable on
\((a, b)\), let
\(g'(x)\neq0\)
on
\((a, b)\). If
\(f'(x)/g'(x)\)
is increasing (decreasing) on
\((a, b)\), then so are
$$ \frac{f(x)-f(a)}{g(x)-g(a)}\quad\textit{and}\quad\frac {f(x)-f(b)}{g(x)-g(b)}. $$
If
\(f '(x)/g'(x)\)
is strictly monotone, then the monotonicity in the conclusion is also strict.
Lemma 2.2
(1) The function
\(r\rightarrow(\mathcal{E}-r^{\prime 2}\mathcal{K})/r^{2}\)
is strictly increasing from
\((0, 1)\)
onto
\((\pi/4, 1)\);
(2) The function
\(r\rightarrow2\mathcal{E}-r^{\prime 2}\mathcal{K}\)
is increasing and log-convex from
\((0,1)\)
onto
\((\pi/2,2)\);
(3) The function
\(\mathcal{K}/\log(e^{2}/r')\)
is strictly increasing from
\((0,1)\)
onto
\((\pi/4, 1)\);
(4) The function
\((\mathcal{K}-\mathcal{E})/r^{2}\)
is strictly increasing on
\((0,1)\); in particular, \(\mathcal{K}-\mathcal{E}>(\pi /4)r^{2}\)
for all
\(r\in(0,1)\).
Proof
Parts (1) and (2) follow from [18], Theorem 3.21(1) and Exercise 3.43(13). □
Lemma 2.3
The equation
$$(1+p)^{1/p}=\frac{\pi}{2} $$
has a unique solution
\(p=p_{0}=3.15295\cdots\) .
Proof
Let
$$ \varphi(p)= \textstyle\begin{cases} (1+p)^{1/p}-\pi/2,& p\in(-1,0)\cup(0,+\infty),\\ e-\pi/2,&p=0. \end{cases} $$
It is easy to verify that the function φ is continuous and strictly decreasing from \((-1,+\infty)\) onto \((1,+\infty)\). Therefore, Lemma 2.3 easily follows from the continuity and monotonicity of φ together with the facts that \(\varphi(3.15295)=6.14999\times10^{-7}\) and \(\varphi (3.15296)=-4.35155\times10^{-7}\). □
Lemma 2.4
The function
$$f(r)=\frac{2(2\mathcal{E}-r^{\prime 2}\mathcal{K})/\pi-1-r^{2}/4}{r^{4}} $$
is strictly increasing from
\((0,1)\)
onto
\((1/64,4/\pi-5/4)\).
Proof
Let \(f_{1}(r)=2(2\mathcal{E}-r^{\prime 2}\mathcal{K})/\pi-1-r^{2}/4\) and \(\widehat{f}_{1}(r)=r^{4}\), then \(f_{1}(0)=\widehat{f}_{1}(0)=0\) and \(f(r)=f_{1}(r)/\widehat{f}_{1}(r)\).
A simple calculation yields
$$\begin{aligned}& \frac{f'_{1}(r)}{\widehat{f}'_{1}(r)}=\frac{4(\mathcal{E}-r^{\prime 2}\mathcal {K})-\pi r^{2}}{8\pi r^{4}}\triangleq\frac{f_{2}(r)}{\widehat{f}_{2}(r)}, \end{aligned}$$
(2.5)
$$\begin{aligned}& f_{2}(0)=\widehat{f}_{2}(0), \end{aligned}$$
(2.6)
$$\begin{aligned}& \frac{f'_{2}(r)}{\widehat{f}'_{2}(r)}=\frac{2\mathcal{K}-\pi}{16\pi r^{2}}\triangleq\frac{f_{3}(r)}{\widehat{f}_{3}(r)}, \end{aligned}$$
(2.7)
$$\begin{aligned}& f_{3}(0)=\widehat{f}_{3}(0), \end{aligned}$$
(2.8)
$$\begin{aligned}& \frac{f'_{3}(r)}{\widehat{f}'_{3}(r)}=\frac{1}{16\pi}\cdot\frac {\mathcal{E}-r^{\prime 2}\mathcal{K}}{r^{2}}\cdot \frac{1}{r^{\prime 2}}. \end{aligned}$$
(2.9)
Following from Lemma 2.2(1) and (2.9) together with the monotonicity of \(1/r^{\prime 2}\), we clearly see that \(f'_{3}(r)/\widehat{f}'_{3}(r)\) is strictly increasing on \((0,1)\). Equations (2.5)-(2.8) and Lemma 2.1 lead to the conclusion that \(f(r)\) is strictly increasing on \((0,1)\).
Therefore, Lemma 2.4 follows from the monotonicity of \(f(r)\) together with the facts that \(f(0^{+})=1/64\) and \(f(1^{-})=4/\pi-5/4\). □
The following double inequalities can be obtained from Lemma 2.4 immediately.
Corollary 2.5
Inequalities
$$1+\frac{r^{2}}{4}+\frac{r^{4}}{64}< \frac{2}{\pi}\bigl(2\mathcal {E}-r^{\prime 2}\mathcal{K}\bigr)< 1+\frac{r^{2}}{4}+ \biggl( \frac{4}{\pi}-\frac {5}{4} \biggr)r^{4} $$
hold for
\(0< r<1\).
Lemma 2.6
The inequality
$$ \biggl[\frac{(1+r)^{7/2}-(1-r)^{7/2}}{7r} \biggr]^{2/5}< 1+\frac{r^{2}}{4} $$
(2.10)
holds for
\(0< r<1\).
Proof
In order to prove inequality (2.10), it suffices to prove that
$$\begin{aligned} g(r) =& \bigl[(1+r)^{7/2}-(1-r)^{7/2} \bigr]^{2}-49r^{2} \biggl(1+\frac{r^{2}}{4} \biggr)^{5} \\ =& (1+r)^{7}+(1-r)^{7}-49r^{2} \biggl(1+ \frac{r^{2}}{4} \biggr)^{5}-2\bigl(1-r^{2} \bigr)^{7/2} \\ =& g_{1}(r)-g_{2}(r)< 0 \end{aligned}$$
(2.11)
for \(0< r<1\), where
$$\begin{aligned}& g_{1}(r)=(1+r)^{7}+(1-r)^{7}-49r^{2} \biggl(1+\frac{r^{2}}{4} \biggr)^{5}, \\& g_{2}(r)=2\bigl(1-r^{2}\bigr)^{7/2}. \end{aligned}$$
Observe that
$$\begin{aligned}& g'_{1}(r)=-\frac{7r}{256} \bigl[32 \bigl(4-5r^{2}\bigr)^{2}+2{,}848r^{4} + 2{,}240r^{6} + 350 r^{8} + 21 r^{10} \bigr]< 0, \end{aligned}$$
(2.12)
$$\begin{aligned}& g_{1}(0.56)=0.0755\cdots>0,\qquad g_{1}(0.57)=-0.00966 \cdots< 0, \end{aligned}$$
(2.13)
we conclude, from (2.12) and (2.13), that there exists \(r_{0}\in (0.56,0.57)\) such that \(g_{1}(r)>0\) for \(r\in(0,r_{0})\) and \(g_{1}(r)<0\) for \(r\in(r_{0},1)\).
In order to prove (2.11), we divide it into two cases.
Case A \(r\in[r_{0},1)\). In this case, we clearly see that \(g_{1}(r)\leq0\) and \(g_{2}(r)>0\). This implies that \(g(r)=g_{1}(r)-g_{2}(r)<0\).
Case B \(r\in(0,r_{0})\). In this case, \(g_{1}(r)>0\). Let \(g_{3}(r)=2-7r^{2}+\frac{35}{4}r^{4}-6r^{6}\), the difference between \(g_{1}(r)\) and \(g_{3}(r)\) yields
$$ g_{1}(r)-g_{3}(r)=-\frac{r^{6}(10{,}880 + 7{,}840 r^{2} + 980 r^{4} + 49 r^{6})}{1{,}024}< 0. $$
(2.14)
We know from (2.14) that \(g_{3}(r)>g_{1}(r)>0\). Moreover,
$$ g_{3}^{2}(r)-g_{2}^{2}(r)=- \frac{r^{6}}{16}\bigl[\bigl(2-4r^{2}\bigr) \bigl(4+r^{2} \bigr)+r^{2}\bigr] \biggl[16 \biggl(r^{2}-\frac{5}{8} \biggr)+\frac{27}{4} \biggr]< 0, $$
this in conjunction with \(g_{3}(r)>0\) implies that
$$ g_{3}(r)-g_{2}(r)< 0. $$
(2.15)
Therefore, we clearly see that \(g(r)=[g_{1}(r)-g_{3}(r)]+[g_{3}(r)-g_{2}(r)]<0\) from (2.14) and (2.15). □
Lemma 2.7
Let
\(\eta(r)=[(1+r)^{p_{0}+1}-(1-r)^{p_{0}+1}]/r\)
and
\(\omega (r)=[(1-r)^{p_{0}}(1+p_{0}r)-(1+r)^{p_{0}}(1-p_{0}r)]/r^{2}\), then the functions
\(\eta(r)\)
and
\(\omega(r)\)
both are strictly increasing on
\((0,1)\).
Proof
We assume that
$$\begin{aligned}& \eta_{1}(r)=(1+r)^{p_{0}+1}-(1-r)^{p_{0}+1},\qquad \eta_{2}(r)=r, \\& \omega_{1}(r)=(1-r)^{p_{0}}(1+p_{0}r)-(1+r)^{p_{0}}(1-p_{0}r), \qquad\omega_{2}(r)=r^{2}, \end{aligned}$$
then \(\eta(r)=\eta_{1}(r)/\eta_{2}(r)\) and \(\omega(r)=\omega _{1}(r)/\omega_{2}(r)\).
A simple calculation yields
$$\begin{aligned}& \eta_{1}(0)=\eta_{2}(0)=\omega_{1}(0)= \omega_{2}(0)=0, \end{aligned}$$
(2.16)
$$\begin{aligned}& \frac{\eta'_{1}(r)}{\eta'_{2}(r)}=(1+p_{0})\bigl[(1+r)^{p_{0}}-(1-r)^{p_{0}} \bigr], \end{aligned}$$
(2.17)
$$\begin{aligned}& \frac{\omega'_{1}(r)}{\omega'_{2}(r)}=\frac {p_{0}(p_{0}+1)[(1+r)^{p_{0}-1}-(1-r)^{p_{0}-1}]}{2}. \end{aligned}$$
(2.18)
Lemma 2.1 and (2.16)-(2.18) lead to the conclusion that \(\eta(r)\) and \(\omega(r)\) are strictly increasing on \((0,1)\). □
Lemma 2.8
Let
$$ \phi_{p}(r)=\frac{2}{\pi} \bigl(2\mathcal{E}-r^{\prime 2} \mathcal{K} \bigr)- \biggl[\frac{(1+r)^{p+1}-(1-r)^{p+1}}{2(p+1)r} \biggr]^{1/p}, $$
then
\(\phi_{p}(r)>0\)
for
\(0< r<1\)
if and only if
\(p\leq5/2\); \(\phi _{p}(r)<0\)
for
\(0< r<1\)
if and only if
\(p\geq p_{0}\).
Proof
It is well known that \(L_{p}(a,b)\) is strictly increasing with respect to \(p\in\mathbb{R}\) for fixed \(a,b>0\) with \(a\neq b\), then \(\phi_{p}(r)\) is strictly decreasing with respect to \(p\in\mathbb{R}\). In order to prove Lemma 2.8, we divide it into three cases.
Case 1 \(p=5/2\).
From Corollary 2.5 and Lemma 2.6, we clearly see that
$$\begin{aligned} \phi_{5/2}(r) =& \frac{2}{\pi} \bigl(2\mathcal{E}-r^{\prime 2} \mathcal {K} \bigr)- \biggl[\frac{(1+r)^{7/2}-(1-r)^{7/2}}{7r} \biggr]^{2/5} \\ >& 1+\frac{r^{2}}{4}+\frac{r^{4}}{64}- \biggl[\frac {(1+r)^{7/2}-(1-r)^{7/2}}{7r} \biggr]^{2/5} \\ >& \frac{r^{4}}{64}>0 \end{aligned}$$
for \(0< r<1\).
Case 2 \(p=p_{0}\).
We divide it into two subcases.
Subcase A \(\phi_{p_{0}}(r)<0\) for \(r\in(0,0.9)\).
Since \(\phi_{p}(r)\) is strictly decreasing with respect to \(p\in\mathbb {R}\), we clearly see that \(\phi_{p_{0}}(r)<\phi_{3}(r)\). It suffices to prove that \(\phi_{3}(r)<0\) for \(r\in(0,0.9)\).
For \(r\in(0,\sqrt{2}/2]\), it follows from Corollary 2.5 that
$$\begin{aligned} \phi_{3}(r) =& \frac{2}{\pi} \bigl(2\mathcal{E}-r^{\prime 2} \mathcal{K} \bigr)- \biggl[\frac{(1+r)^{4}-(1-r)^{4}}{8r} \biggr]^{1/3} \\ < & 1+\frac{r^{2}}{4}+ \biggl(\frac{4}{\pi}-\frac{5}{4} \biggr)r^{4}- \biggl(1+\frac{r^{2}}{3}-\frac{r^{4}}{9} \biggr)^{3} \\ = & -\frac{r^{2}}{12} \biggl[1- \biggl(\frac{48}{\pi}-\frac{41}{3} \biggr)r^{2} \biggr] \\ \leq & -\frac{r^{2}}{12} \biggl[1-\frac{1}{2} \biggl(\frac{48}{\pi }- \frac{41}{3} \biggr) \biggr] \\ =& -\frac{(47\pi-144)r^{2}}{72\pi}< 0, \end{aligned}$$
where the first inequality easily follows from
$$ \frac{(1+r)^{4}-(1-r)^{4}}{8r}- \biggl(1+\frac{r^{2}}{3}-\frac {r^{4}}{9} \biggr)^{3}=\frac{r^{6}}{729}\bigl[126+9\bigl(1-r^{4} \bigr)+r^{6}\bigr]>0. $$
For \(r\in(\sqrt{2}/2,0.9)\), taking the derivative of \(\phi_{3}(r)\) yields
$$ \phi'_{3}(r)=\frac{2(\mathcal{E}-r^{\prime 2}\mathcal{K})}{\pi r}-\frac {2r}{3(1+r^{2})^{2/3}}= \mu_{1}(r)+\mu_{2}(r), $$
(2.19)
where
$$ \mu_{1}(r)=\frac{2(\mathcal{E}-r^{\prime 2}\mathcal{K})}{\pi r}-\frac {r}{2},\qquad \mu_{2}(r)=\frac{r}{2}-\frac{2r}{3(1+r^{2})^{2/3}}. $$
From Lemma 2.2(4), we clearly see that
$$ \frac{d\mu_{1}(r)}{dr}=\frac{2}{\pi r^{2}} \biggl(\mathcal{K}-\mathcal {E}- \frac{\pi}{4}r^{2} \biggr)>0 $$
(2.20)
for \(r\in(0,1)\) and
$$\begin{aligned} \frac{d\mu_{2}(r)}{dr} =& \frac {9(1+r^{2})^{5/3}-12+4r^{2}}{18(1+r^{2})^{5/3}}> \frac {9[1+(5/3)r^{2}]-12+4r^{2}}{18(1+r^{2})^{5/3}} \\ =& \frac{19r^{2}-3}{18(1+r^{2})^{5/3}}>0 \end{aligned}$$
(2.21)
for \(r\in(\sqrt{2}/2,0.9)\). Equations (2.19)-(2.21) lead to the conclusion that \(\phi'_{3}(r)\) is strictly increasing on \((\sqrt {2}/2,0.9)\). This in conjunction with (2.3) implies that
$$ \phi'_{3}(r)>\phi'_{3}( \sqrt{2}/2)=0.02163\cdots>0 $$
(2.22)
for \(r\in(\sqrt{2}/2,9/10)\). Therefore, from (2.22) we clearly see that \(\phi_{3}(r)\) is strictly increasing on \((\sqrt{2}/2,0.9)\). This in conjunction with (2.4) yields \(\phi_{3}(r)<\phi _{3}(0.9)=-0.002687\cdots<0\) for \(r\in(\sqrt{2}/2,0.9)\).
Subcase B \(\phi_{p_{0}}(r)<0\) for \(r\in[0.9,1)\).
For \(0.9\leq r<1\), taking the derivation of \(\phi_{p_{0}}(r)\) yields
$$ \phi'_{p_{0}}(r)=\frac{2(\mathcal{E}-r^{\prime 2}\mathcal{K})}{\pi r}-\frac {\omega(r)}{p_{0}(p_{0}+1)2^{1/p_{0}}\eta(r)^{1-1/p_{0}}}, $$
(2.23)
where \(\omega(r)\) and \(\eta(r)\) are defined as in Lemma 2.7. From Lemma 2.2(1), we clearly see that \((\mathcal{E}-r^{\prime 2}\mathcal{K})/r\) is strictly increasing on \((0,1)\). Lemma 2.7 and (2.4), (2.20) lead to the conclusion that
$$\begin{aligned} \phi'_{p_{0}}(r) \geq & \frac{2[\mathcal{E}(0.9)-(1-0.9^{2})\mathcal {K}(0.9)]}{0.9\pi}-\frac{\omega(1)}{p_{0}(p_{0}+1)2^{1/p_{0}}\eta (0.9)^{1-1/p_{0}}} \\ =& 0.522306\cdots-0.46787\cdots=0.054436\cdots>0 \end{aligned}$$
for \(0.9\leq r<1\).
Therefore, it follows from the monotonicity of \(\phi'_{p_{0}}(r)\) on \((9/10,1)\) that \(\phi_{p_{0}}(r)<\phi_{p_{0}}(1)=2 [2/\pi -1/(1+p_{0})^{1/p_{0}} ]=0\) for \(0< r<1\).
Case 3 \(5/2< p< p_{0}\).
Taking the Taylor series of \(\phi_{p}(r)\) at \(r=0\) yields
$$ \phi_{p}(r)= \biggl(\frac{5}{12}-\frac{p}{6} \biggr)r^{2}+\frac{(149 - 144 p + 24 p^{2} + 16 p^{3}) r^{4}}{2{,}880}+o\bigl(r^{4}\bigr). $$
(2.24)
From (2.24) we clearly see that there exists a sufficiently small \(\delta_{1}>0\) such that \(\phi_{p}(r)<0\) for \(r\in(0,\delta_{1})\) if \(p>5/2\). If \(p< p_{0}\), then \(\phi_{p}(1)=2[2/\pi-1/(1+p)^{1/p}]>0\). By the continuity of \(\phi_{p}(r)\) with respect to r, there exists a sufficiently small \(\delta_{2}>0\) such that \(\phi_{p}(r)>0\) for \(r\in (\delta_{2},1)\). □