For \(r\in(0,1)\), Legendre’s complete elliptic integral [27] of the second kind is given by
$$ \mathcal{E}(r)= \int_{0}^{\pi/2}\sqrt{1-r^{2}\sin ^{2}(t)}\,dt. $$
It is well known that \(\mathcal{E}(0^{+})=\pi/2\), \(\mathcal {E}(1^{-})=1\), and \(\mathcal{E}(r)\) is the particular case of the Gaussian hypergeometric function
$$ F(a,b;c;x)=\sum_{n=0}^{\infty}\frac{(a)_{n}(b)_{n}}{(c)_{n}} \frac{x^{n}}{n!}\quad(-1< x< 1), $$
where \((a)_{n}=\Gamma(a+n)/\Gamma(a)\) and \(\Gamma(x)=\int_{0}^{\infty}t^{x-1}e^{-t}\,dt\) (\(x>0\)) is the gamma function. Indeed, we have
$$ \mathcal{E}(r)=\frac{\pi}{2}F \biggl( -\frac{1}{2}, \frac{1}{2};1;r^{2} \biggr) =\frac{\pi}{2}\sum _{n=0}^{\infty}\frac{ ( -\frac{1}{2} ) _{n} ( \frac{1}{2} ) _{n}}{(n!)^{2}}r^{2n}. $$
(3.1)
Recently, the bounds for the complete elliptic integral \(\mathcal {E}(r)\) of the second kind have been the subject of intensive research. In particular, many remarkable inequalities for \(\mathcal{E}(r)\) can be found in the literature [28–41]. Vuorinen [42] conjectured that the inequality
$$ \mathcal{E}(r)\geq\frac{\pi}{2} \biggl( \frac {1+{r^{\prime}}^{3/2}}{2} \biggr) ^{2/3} $$
(3.2)
holds for all \(r\in(0, 1)\), where, and in what follows, \(r^{\prime }=(1-r^{2})^{1/2}\). Inequality (3.2) was proved by Barnard et al. in [43].
Very recently, the accurate bounds for \(\mathcal{E}(r)\) in terms of the Stolarsky mean \(S_{p, q}(1, r^{\prime})\) were given in [44, 45]:
$$\begin{aligned}& \frac{\pi}{2}S_{11/4, 7/4}\bigl(1, r^{\prime}\bigr)< \mathcal{E}(r)< \frac{11}{7}S_{11/4, 7/4}\bigl(1, r^{\prime}\bigr), \end{aligned}$$
(3.3)
$$\begin{aligned}& \frac{25}{16}S_{5/2, 2}\bigl(1, r^{\prime}\bigr)< \mathcal{E}(r)< \frac{\pi}{2}S_{5/2, 2}\bigl(1, r^{\prime}\bigr), \end{aligned}$$
(3.4)
where \(S_{p, q}(a,b)=[q(a^{p}-b^{p})/(p(a^{q}-b^{q}))]^{1/(p-q)}\).
In this section, we shall use Theorem 2.1 to present new bounds for the complete elliptic integral \(\mathcal{E}(r)\) of the second kind. In order to prove our main result, we need three lemmas, which we present in this section.
Lemma 3.1
see [46], Lemma 7
Let
\(n\in\mathbb{N}\)
and
\(m\in\mathbb{N}\cup\{0\}\)
with
\(n>m\), \(a_{i}\geq0\)
for all
\(0\leq i\leq n\), \(a_{n}a_{m}>0\)
and
$$ P_{n}(t)=-\sum_{i=0}^{m}a_{i}t^{i}+ \sum_{i=m+1}^{n}a_{i}t^{i}. $$
Then there exists
\(t_{0}\in(0, \infty)\)
such that
\(P_{n}(t_{0})=0\), \(P_{n}(t)<0\)
for
\(t\in(0, t_{0})\)
and
\(P_{n}(t)>0\)
for
\(t\in(t_{0}, \infty)\).
Lemma 3.2
see [47, 48]
The double inequality
$$ \frac{1}{(x+a)^{1-a}}< \frac{\Gamma(x+a)}{\Gamma(x+1)}< \frac{1}{x^{1-a}} $$
holds for all
\(x>0\)
and
\(a\in(0, 1)\).
Lemma 3.3
Let
\(n\in\mathbb{N}\), \(p_{1}(n)\), \(p_{2}(n)\), \(p_{3}(n)\)
and
\(w_{n}\)
be defined by
$$\begin{aligned}& p_{1}(n)=125n^{4}+1\text{,}482n^{3}+1\text{,}601n^{2}-7\text{,}532n+1\text{,}944, \end{aligned}$$
(3.5)
$$\begin{aligned}& p_{2}(n)=4\text{,}375n^{6}+213\text{,}247n^{5}+1\text{,}696\text{,}697n^{4}-4\text{,}433\text{,}167n^{3} \\& \qquad \quad {}-1\text{,}571\text{,}832n^{2}+4\text{,}662\text{,}360n-1\text{,}555\text{,}200, \end{aligned}$$
(3.6)
$$\begin{aligned}& p_{3}(n)=625n^{6}+22\text{,}583n^{5}-304\text{,}413n^{4}+234\text{,}181n^{3} \\& \qquad \quad {}+1\text{,}906\text{,}328n^{2}-2\text{,}918\text{,}064n+675\text{,}360, \end{aligned}$$
(3.7)
$$\begin{aligned}& w_{n}=-648np_{1}(n) \biggl( \frac{1}{4} \biggr) _{n-2}+p_{2}(n) \biggl( \frac{1}{2} \biggr) _{n-3}+\frac{5np_{3}(n)}{2n-5} \biggl( \frac{3}{4} \biggr) _{n-3}, \end{aligned}$$
(3.8)
respectively. Then
\(w_{n}\geq0\)
for all
\(n\geq3\).
Proof
Let \(p_{0}(n)\), \(p_{4}(n)\), \(p_{5}(n)\), \(\alpha_{n}\) and \(\beta_{n}\) be defined by
$$\begin{aligned}& p_{0}(n)=625n^{4}-2\text{,}018n^{3}+3\text{,}985n^{2}-4\text{,}032n+1\text{,}080, \end{aligned}$$
(3.9)
$$\begin{aligned}& p_{4}(n)=875n^{7}-81\text{,}128n^{6}-2\text{,}341\text{,}894n^{5}-10\text{,}120\text{,}928n^{4}+19\text{,}839\text{,}719n^{3} \\& \qquad \quad {}-22\text{,}615\text{,}904n^{2}+73\text{,}667\text{,}340n-71\text{,}971\text{,}200, \end{aligned}$$
(3.10)
$$\begin{aligned}& p_{5}(n)=250n^{6}-1\text{,}834n^{5}+391\text{,}275n^{4}+975\text{,}000n^{3}-7\text{,}770\text{,}415n^{2} \\& \quad \qquad {}+7\text{,}980\text{,}844n+298\text{,}140, \end{aligned}$$
(3.11)
$$\begin{aligned}& \alpha_{n}=\frac{p_{4}(n)}{n(n+1)} \biggl( \frac{1}{2} \biggr) _{n-3}, \end{aligned}$$
(3.12)
$$\begin{aligned}& \beta_{n}=\frac{10(n-1)p_{5}(n)}{(2n-3)(2n-5)} \biggl( \frac{3}{4} \biggr)_{n-3}, \end{aligned}$$
(3.13)
respectively.
Then from (3.5)-(3.13) and elaborated computations we get
$$\begin{aligned}& w_{3}=w_{4}=w_{5}=w_{6}=0, \end{aligned}$$
(3.14)
$$\begin{aligned}& w_{7}=\frac{2\text{,}848\text{,}355}{16}>0, \end{aligned}$$
(3.15)
$$\begin{aligned}& \frac{w_{n+1}}{(n+1)p_{1}(n+1)}-\frac{(n-7/4)w_{n}}{np_{1}(n)} \\& \quad = \frac{p_{2}(n+1)}{(n+1)p_{1}(n+1)} \biggl( \frac{1}{2} \biggr) _{n-2}- \frac{(n-7/4)p_{2}(n)}{np_{1}(n)} \biggl( \frac{1}{2} \biggr) _{n-3} \\& \quad \quad {}+\frac{5p_{3}(n+1)}{(2n-3)p_{1}(n+1)} \biggl( \frac{3}{4} \biggr) _{n-2}- \frac{5(n-7/4)p_{3}(n)}{(2n-5)p_{1}(n)} \biggl( \frac{3}{4} \biggr) _{n-3} \\& \quad = \biggl[ \frac{(n-5/2)p_{2}(n+1)}{(n+1)p_{1}(n+1)}-\frac {(n-7/4)p_{2}(n)}{np_{1}(n)} \biggr] \biggl( \frac{1}{2} \biggr) _{n-3} \\& \quad \quad {}+ \biggl[ \frac{5(n-9/4)p_{3}(n+1)}{(2n-3)p_{1}(n+1)}-\frac {5(n-7/4)p_{3}(n)}{(2n-5)p_{1}(n)} \biggr] \biggl( \frac{3}{4} \biggr) _{n-3} \\& \quad = \frac{p_{0}(n+1)p_{4}(n)}{4n(n+1)p_{1}(n)p_{1}(n+1)} \biggl( \frac {1}{2} \biggr) _{n-3} \\& \quad \quad {}+\frac{5(n-1)p_{0}(n+1)p_{5}(n)}{2(2n-3)(2n-5)p_{1}(n)p_{1}(n+1)} \biggl( \frac{3}{4} \biggr) _{n-3} \\& \quad = \frac{p_{0}(n+1)}{4p_{1}(n)p_{1}(n+1)} ( \alpha_{n}+\beta_{n} ) , \end{aligned}$$
(3.16)
$$\begin{aligned}& \frac{\alpha_{n}}{\beta_{n}}=\frac {(2n-3)(2n-5)p_{4}(n)}{10n(n-1)(n+1)p_{5}(n)} \frac{ ( \frac{1}{2} ) _{n-3}}{ ( \frac{3}{4} ) _{n-3}}, \end{aligned}$$
(3.17)
$$\begin{aligned}& \frac{\alpha_{7}}{\beta_{7}}=-\frac{7\text{,}313\text{,}056}{7\text{,}313\text{,}875}, \end{aligned}$$
(3.18)
$$\begin{aligned}& \frac{10 ( \frac{3}{4} ) _{n-3}}{ ( \frac{1}{2} ) _{n-1}} \biggl( \frac{\alpha_{n+1}}{\beta_{n+1}}-\frac{\alpha_{n}}{\beta_{n}} \biggr) \\& \quad = \frac{(2n-1)(2n-3)(n-5/2)p_{4}(n+1)}{n(n+1)(n+2)(n-9/4)p_{5}(n+1)}-\frac{(2n-3)(2n-5)p_{4}(n)}{n(n-1)(n+1)p_{5}(n)} \\& \quad = -\frac{(2n-3)(2n-5)(n-3)(n-4) (125n^{4}+1\text{,}982n^{3}+6\text{,}797n^{2}+616n-2\text{,}380 )}{n(4n-9)(n-1)(n+1)(n+2)p_{5}(n)p_{5}(n+1)} \\& \quad \quad {} \times\bigl(1\text{,}750n^{8}-802\text{,}264n^{7}-54\text{,}353\text{,}513n^{6}+811\text{,}227\text{,}431n^{5}-4\text{,}748\text{,}597\text{,}075n^{4} \\& \quad \quad {}+21\text{,}568\text{,}863\text{,}989n^{3}-80\text{,}185\text{,}046\text{,}202n^{2} \\& \quad \quad {}+144\text{,}028\text{,}552\text{,}644n-85\text{,}225\text{,}499\text{,}160 \bigr) \\& \quad = -\frac{(2n-3)(2n-5)(n-3)(n-4) ( 125n^{4}+1\text{,}982n^{3}+6\text{,}797n^{2}+616n-2\text{,}380 ) }{n(4n-9)(n-1)(n+1)(n+2)p_{5}(n)p_{5}(n+1)} \\& \quad \quad {} \times\bigl[-420\text{,}331\text{,}641\text{,}120-714\text{,}515\text{,}222\text{,}844(n-7)-475\text{,}749\text{,}995\text{,}856(n-7)^{2} \\& \quad \quad {}-152\text{,}526\text{,}681\text{,}341(n-7)^{3}-25\text{,}642\text{,}525\text{,}865(n-7)^{4} \\& \quad \quad {}-2\text{,}263\text{,}535\text{,}771(n-7)^{5} \\& \quad \quad {}-91\text{,}263\text{,}449(n-7)^{6}-704\text{,}264(n-7)^{7}+1\text{,}750(n-7)^{8} \bigr]. \end{aligned}$$
(3.19)
From Lemma 3.1 and (3.19) together with the facts that \(p_{5}(n)>0\) and \(p_{5}(n+1)>0\) for \(n\geq7\), we clearly see that there exists \(n_{0}>7\) such that the sequence \(\{\alpha_{n}/\beta _{n}\}_{n=7}^{\infty}\) is increasing for \(7\leq n\leq n_{0}\) and decreasing for \(n\geq n_{0}\), which implies that
$$ \frac{\alpha_{n}}{\beta_{n}}\geq \min \biggl\{ \frac{\alpha _{7}}{\beta_{7}}, \lim _{n\rightarrow\infty}\frac{\alpha_{n}}{\beta_{n}} \biggr\} . $$
(3.20)
It follows from Lemma 3.2 that
$$ \frac{\Gamma ( \frac{3}{4} ) ( n-\frac{5}{2} ) ^{-1/4}}{\Gamma ( \frac {1}{2} ) } < \frac{ ( \frac{1}{2} ) _{n-3}}{ ( \frac{3}{4} ) _{n-3}}=\frac{\Gamma ( \frac{3}{4} ) }{\Gamma ( \frac{1}{2} ) }\frac {\Gamma ( n-\frac{13}{4}+\frac{3}{4} ) }{\Gamma ( n-\frac{13}{4}+1 ) } < \frac{\Gamma ( \frac{3}{4} ) ( n-\frac{13}{4} ) ^{-1/4}}{\Gamma ( \frac{1}{2} ) } $$
(3.21)
for all \(n\geq7\).
From (3.10), (3.11), (3.17), (3.18), (3.20) and (3.21) we get
$$ \frac{\alpha_{n}}{\beta_{n}}\geq\frac{\alpha_{7}}{\beta_{7}}>-1 $$
(3.22)
for all \(n\geq7\).
Therefore, Lemma 3.3 follows easily from (3.14)-(3.16) and (3.22) together with the facts that \(p_{1}(n)>0\), \(p_{0}(n+1)>0\) and \(p_{1}(n+1)>0\) for \(n\geq7\). □
Theorem 3.4
The double inequality
$$ \frac{40(\pi-2)}{29}J\bigl(r^{\prime}\bigr)-\frac{51\pi-160}{58}< \mathcal{E}(r)< \frac{\pi}{2}J\bigl(r^{\prime}\bigr) $$
(3.23)
holds for all
\(r\in(0, 1)\), where
$$ J\bigl(r^{\prime}\bigr)=\frac{51{r^{\prime}}^{2}+20r^{\prime}\sqrt {r^{\prime}}+50r^{\prime}+20\sqrt{r^{\prime}}+51}{16(5r^{\prime }+2\sqrt{r^{\prime}}+5)}. $$
(3.24)
Proof
Let \(r\in(0, 1)\), \(x=r^{2}\), \(P(x)\), \(f_{1}(r)\), \(f_{2}(r)\) and \(F(r)\) be defined by
$$\begin{aligned}& P(x)=1\text{,}536-1\text{,}248x, \end{aligned}$$
(3.25)
$$\begin{aligned}& f_{1}(r)= 2\text{,}592\bigl(2-r^{2}\bigr) \bigl(1-r^{2} \bigr)^{3/4}+15\bigl(425r^{4}-624r^{2}+192\bigr) \bigl(1-r^{2}\bigr)^{1/2} \\& \quad \quad {}-50\bigl(r^{4}-96r^{2}+96\bigr) \bigl(1-r^{2} \bigr)^{1/4}-105r^{4}+3\text{,}600r^{2}+2\text{,}880, \end{aligned}$$
(3.26)
$$\begin{aligned}& f_{2}(r)=625r^{4}-384r^{2}+384, \end{aligned}$$
(3.27)
$$\begin{aligned}& F(r)=\frac{1-J(r^{\prime})}{1-2\mathcal{E}(r)/\pi}, \end{aligned}$$
(3.28)
respectively.
Then from (3.1) and (3.24)-(3.28) we have
$$\begin{aligned}& f_{1}(r)= 2\text{,}880+3\text{,}600r^{2}-105r^{4}+2\text{,}592 \Biggl( 2 \sum_{n=0}^{\infty}\frac{ ( -\frac{3}{4} ) _{n}}{n!}r^{2n}- \sum_{n=1}^{\infty}\frac{ ( -\frac{3}{4} ) _{n-1}}{(n-1)!}r^{2n} \Biggr) \\& \hphantom{f_{1}(r)=}+15 \Biggl( 425\sum_{n=2}^{\infty} \frac{ ( -\frac{1}{2} ) _{n-2}}{(n-2)!}r^{2n}- 624\sum_{n=1}^{\infty} \frac{ ( -\frac{1}{2} ) _{n-1}}{(n-1)!}r^{2n} +192\sum_{n=0}^{\infty} \frac{ ( -\frac{1}{2} ) _{n}}{(n!}r^{2n} \Biggr) \\& \hphantom{f_{1}(r)=}-50 \Biggl( \sum_{n=2}^{\infty} \frac{ ( -\frac{1}{4} ) _{n-2}}{(n-2)!}r^{2n}- 96\sum_{n=1}^{\infty} \frac{ ( -\frac{1}{4} ) _{n-1}}{(n-1)!}r^{2n} +96\sum_{n=0}^{\infty} \frac{ ( -\frac{1}{4} ) _{n}}{n!}r^{2n} \Biggr) \\& \hphantom{f_{1}(r)} = 6\text{,}144-7\text{,}680r^{2}-105r^{4}+2\text{,}592\sum _{n=2}^{\infty}\frac{ ( n-\frac{7}{2} ) ( -\frac{3}{4} ) _{n-1}}{n!}r^{2n} \\& \hphantom{f_{1}(r)=}-15\sum_{n=2}^{\infty}\frac{ ( 7n^{2}-367n-720 ) ( -\frac{1}{2} ) _{n-2}}{n!}r^{2n} \\& \hphantom{f_{1}(r)=}-50\sum_{n=2}^{\infty}\frac{ ( n^{2}-121n+270 ) ( -\frac{1}{4} ) _{n-2}}{n!}r^{2n} \\& \hphantom{f_{1}(r)} = 6\text{,}144-7\text{,}680r^{2}+11\text{,}248r^{4}-1\text{,}944\sum _{n=3}^{\infty}\frac{ ( n-\frac{7}{2} ) ( \frac{1}{4} ) _{n-2}}{n!}r^{2n} \\& \hphantom{f_{1}(r)=}+\frac{15}{2}\sum_{n=3}^{\infty} \frac{ ( 7n^{2}-367n-720 ) ( \frac{1}{2} ) _{n-3}}{n!}r^{2n} \\& \hphantom{f_{1}(r)=}+\frac{25}{2}\sum_{n=3}^{\infty} \frac{ ( n^{2}-121n+270 ) ( \frac{3}{4} ) _{n-3}}{n!}r^{2n}, \\& 16f_{2}(r)-f_{1}(r) \\& \quad =1\text{,}536r^{2}-1\text{,}248r^{4}- \frac{25}{2}\sum_{n=3}^{\infty} \frac{ ( n^{2}-121n+270 ) ( \frac{3}{4} ) _{n-3}}{n!}r^{2n} \\& \quad \quad {}-\frac{15}{2}\sum_{n=3}^{\infty} \frac{ ( 7n^{2}-367n-720 ) ( \frac{1}{2} ) _{n-3}}{n!}r^{2n} +1\text{,}944\sum _{n=3}^{\infty} \frac{ ( n-\frac{7}{2} ) ( \frac{1}{4} ) _{n-2}}{n!}r^{2n} \\& \quad = 1\text{,}536r^{2}-1\text{,}248r^{4}+\sum_{n=3}^{\infty}u_{n}r^{2n}, \end{aligned}$$
where
$$\begin{aligned}& u_{n}=\frac{972(2n-7)}{n!} \biggl( \frac{1}{4} \biggr) _{n-2} \\& \quad \quad {}-\frac{15(7n^{2}-367n-720)}{2n!} \biggl( \frac{1}{2} \biggr) _{n-3} -\frac{25(n^{2}-121n+270)}{2n!} \biggl( \frac{3}{4} \biggr) _{n-3}, \\& 1-\frac{2}{\pi}\mathcal{E}(r) \\& \quad =-\sum_{n=1}^{\infty} \frac{ ( -\frac{1}{2} ) _{n} ( \frac{1}{2} ) _{n}}{(n!)^{2}}r^{2n} =\sum _{n=1}^{\infty} \frac{ ( \frac{1}{2} ) _{n-1} ( \frac{1}{2} ) _{n}}{2(n!)^{2}}r^{2n}, \\& 16f_{2}(r) \biggl( 1-\frac{2}{\pi}\mathcal{E}(r) \biggr) \\& \quad = 8 \bigl( 625r^{4}-384r^{2}+384 \bigr) \sum _{n=1}^{\infty}\frac{ ( \frac{1}{2} ) _{n-1} ( \frac{1}{2} ) _{n}}{(n!)^{2}}r^{2n} \\& \quad = 1\text{,}536r^{2}-1\text{,}248r^{4}+\sum_{n=3}^{\infty}v_{n}r^{2n}, \end{aligned}$$
(3.29)
where
$$ v_{n}=\frac{8p_{0}(n)}{(n!)^{2}} \biggl( \frac{1}{2} \biggr) _{n-3} \biggl( \frac{1}{2} \biggr) _{n-2} $$
(3.30)
and \(p_{0}(n)\) is defined by (3.9).
$$\begin{aligned}& J\bigl(r^{\prime}\bigr)={} \frac{(51{r^{\prime}}^{2}+20r^{\prime}\sqrt {r^{\prime}}+50r^{\prime}+20\sqrt{r^{\prime}}+51) (5r^{\prime}-2\sqrt{r^{\prime}}+5)(25{r^{\prime}}^{2}-46r^{\prime }+25)}{16(5r^{\prime}+2\sqrt{r^{\prime}}+5)(5r^{\prime}-2\sqrt {r^{\prime}}+5)(25{r^{\prime}}^{2}-46r^{\prime}+25)} \\& \hphantom{J\bigl(r^{\prime}\bigr) }= \frac{f_{1}(r)}{16f_{2}(r)}, \\& F(r)=\frac{16f_{2}(r)-f_{1}(r)}{16f_{2}(r) ( 1-\frac{2}{\pi}\mathcal {E}(r) ) } =\frac{P(x)+\sum_{n=2}^{\infty}u_{n+1}x^{n}}{P(x)+\sum _{n=2}^{\infty}v_{n+1}x^{n}}. \end{aligned}$$
(3.31)
It follows from (3.9), (3.29), (3.30) and elaborated computations that
$$\begin{aligned}& \frac{u_{3}}{v_{3}}=1, \end{aligned}$$
(3.32)
$$\begin{aligned}& u_{n+1}-\frac{v_{n+1}}{v_{n}}u_{n}=-\frac{15(2n-5)}{8(n+1)p_{0}(n)(n+1)!}w_{n}, \end{aligned}$$
(3.33)
where \(w_{n}\) is defined by (3.8).
It is not difficult to verify that
$$ p_{0}(n)=625n^{4}-2\text{,}018n^{3}+3\text{,}985n^{2}-4\text{,}032n+1\text{,}080>0 $$
(3.34)
for all \(n\geq3\).
From Lemma 3.3, (3.30), (3.33) and (3.34) we know that
for all \(n\geq3\), and the sequence \(\{u_{n}/v_{n}\}_{n=3}^{\infty}\) is decreasing.
Equation (3.25) implies that
for \(x\in(0, 1)\), and \(P(x)\) is decreasing on \((0, 1)\).
It follows from Theorem 2.1, (3.31) and (3.32) together with the monotonicity of the sequence \(\{u_{n}/v_{n}\} _{n=3}^{\infty}\) and the function \(P(x)\) on \((0, 1)\) that the function \(F(r)\) is strictly decreasing on \((0, 1)\) and
$$ \lim_{r\rightarrow1^{-}}F(r)< F(r)< \lim_{r\rightarrow0^{+}}F(r) $$
(3.37)
for all \(r\in(0, 1)\).
Note that (3.24), (3.28) and (3.31) lead to the conclusion that
$$ \lim_{r\rightarrow1^{-}}F(r)=\frac{1-J(0^{+})}{1-2\mathcal {E}(1^{-})/\pi}=\frac{1-51/80}{1-2/\pi}= \frac{29\pi}{80(\pi-2)}, \qquad\lim_{r\rightarrow0^{+}}F(r)=1. $$
(3.38)
Therefore, Theorem 3.4 follows from (3.28), (3.37) and (3.38). □
Remark 3.5
Let
$$\begin{aligned}& \lambda_{1}(r)=\frac{\pi}{2}S_{11/4,7/4}\bigl(1, r^{\prime}\bigr), \qquad\mu_{1}(r)=\frac{11}{7}S_{11/4,7/4} \bigl(1, r^{\prime}\bigr), \end{aligned}$$
(3.39)
$$\begin{aligned}& \lambda_{2}(r)=\frac{25}{16}S_{5/2,2}\bigl(1, r^{\prime}\bigr), \qquad\mu_{2}(r)=\frac{\pi}{2}S_{5/2,2} \bigl(1, r^{\prime}\bigr), \end{aligned}$$
(3.40)
$$\begin{aligned}& \lambda(r)=\frac{40(\pi-2)}{29}J\bigl(r^{\prime}\bigr)-\frac{51\pi-160}{58}, \qquad\mu(r)=\frac{\pi}{2}J\bigl(r^{\prime}\bigr), \end{aligned}$$
(3.41)
where \(J(r^{\prime})\) is defined by (3.24). Then simple computations lead to
$$\begin{aligned}& \lambda_{1}\bigl(1^{-}\bigr)=\frac{7\pi}{22}=0.999597 \ldots, \qquad\mu_{1}\bigl(0^{+}\bigr)= \frac{11}{7}=1.571428\ldots, \end{aligned}$$
(3.42)
$$\begin{aligned}& \lambda_{2}\bigl(0^{+}\bigr)=\frac{25}{16}=1.5625, \qquad\mu_{2}\bigl(1^{-}\bigr)=\frac{2\pi}{5}=1.256637 \ldots, \end{aligned}$$
(3.43)
$$\begin{aligned}& \lambda\bigl(0^{+}\bigr)=\frac{\pi}{2}=\mu\bigl(0^{+} \bigr)=1.5707963\ldots, \end{aligned}$$
(3.44)
$$\begin{aligned}& \lambda\bigl(1^{-}\bigr)=1, \qquad\mu\bigl(1^{-}\bigr)= \frac{51\pi}{160}=1.00138\ldots. \end{aligned}$$
(3.45)
From (3.3), (3.4), (3.23) and (3.39)-(3.45) we clearly see that there exists small enough \(\delta\in(0, 1)\) such that the lower bound given in (3.23) for \(\mathcal{E}(r)\) is better than the lower bound given in (3.3) for \(r\in(\delta, 1-\delta)\), the lower bound given in (3.23) for \(\mathcal{E}(r)\) is better than the lower bound given in (3.4) for \(r\in(0, \delta)\), the upper bound given in (3.23) for \(\mathcal{E}(r)\) is better than the upper bound given in (3.3) for \(r\in(0, \delta)\), and the upper bound given in (3.23) for \(\mathcal{E}(r)\) is better than the upper bound given in (3.4) for \(r\in(\delta, 1-\delta)\).
Corollary 3.6
Let
\(J(r^{\prime})\)
be defined by (3.24). Then the double inequality
$$ \frac{\pi}{2}J\bigl(r^{\prime}\bigr)- \biggl( \frac{51\pi}{160}-1 \biggr) < \mathcal{E}(r)< \frac{\pi}{2}J\bigl(r^{\prime}\bigr) $$
(3.46)
holds for all
\(r\in(0, 1)\).
Proof
Let \(F(r)\) be defined by (3.28) and
$$ A(r)=J\bigl(r^{\prime}\bigr)-\frac{2}{\pi}\mathcal{E}(r). $$
(3.47)
Then we clearly see that
$$\begin{aligned}& A\bigl(0^{+}\bigr)=J\bigl(1^{-}\bigr)-\frac{2}{\pi} \mathcal{E}\bigl(0^{+}\bigr)=0, \qquad A\bigl(1^{-}\bigr)=J \bigl(0^{+}\bigr)-\frac{2}{\pi}\mathcal{E}\bigl(1^{-} \bigr)=\frac{51\pi-160}{80\pi}, \end{aligned}$$
(3.48)
$$\begin{aligned}& A(r)= \biggl[ 1-\frac{2}{\pi}\mathcal{E}(r) \biggr] \biggl[ 1- \frac{1-J(r^{\prime})}{1-\frac{2}{\pi}\mathcal{E}(r)} \biggr] = \biggl[ 1-\frac{2}{\pi}\mathcal{E}(r) \biggr] \bigl[1-F(r)\bigr]. \end{aligned}$$
(3.49)
From (3.49) and the proof of Theorem 3.4 we know that \(F(r)\) is strictly decreasing on \((0, 1)\) and \(A(r)\) is strictly increasing on \((0, 1)\). Therefore, inequality (3.46) follows from (3.47) and (3.48) together with the monotonicity of \(A(r)\) on the interval \((0, 1)\). □
Corollary 3.7
Let
\(J(r^{\prime})\)
be defined by (3.24). Then the double inequality
$$ \frac{80}{51}J\bigl(r^{\prime}\bigr)< \mathcal{E}(r)< \frac{\pi}{2}J \bigl(r^{\prime}\bigr) $$
(3.50)
holds for all
\(r\in(0, 1)\).
Proof
Let \(A(r)\) be defined by (3.47) and
$$ B(r)=\frac{J(r^{\prime})}{\frac{2}{\pi}E(r)}. $$
(3.51)
Then we clearly see that
$$\begin{aligned}& B\bigl(0^{+}\bigr)=\frac{J(1^{-})}{\frac{2}{\pi}E(0^{+})}=1, \qquad B\bigl(1^{-} \bigr)=\frac{J(0^{+})}{\frac{2}{\pi}E(1^{-})}=\frac{51\pi}{160}, \end{aligned}$$
(3.52)
$$\begin{aligned}& B(r)=\frac{1}{\frac{2}{\pi}\mathcal{E}(r)} \biggl[ J\bigl(r^{\prime }\bigr)-\frac{2}{\pi} \mathcal{E}(r) \biggr] +1=\frac{A(r)}{\frac{2}{\pi}\mathcal{E}(r)}+1. \end{aligned}$$
(3.53)
From (3.53) and the proof of Corollary 3.6 we know that both \(A(r)\) and \(B(r)\) are strictly increasing on \((0, 1)\). Therefore, inequality (3.50) follows from (3.51) and (3.52) together with the monotonicity of \(B(r)\) on the interval \((0, 1)\). □
Remark 3.8
From Corollaries 3.6 and 3.7 we have
$$ \biggl\vert \mathcal{E}(r)-\frac{\pi}{2}J\bigl(r^{\prime}\bigr) \biggr\vert < \frac{51\pi}{160}-1=0.001382\ldots,\qquad\biggl\vert \frac{\mathcal{E}(r)-\frac{\pi}{2}J(r^{\prime})}{\mathcal{E}(r)} \biggr\vert < \frac{51\pi}{160}-1=0.001382\ldots\,. $$
for all \(r\in(0, 1)\), which implies that both the absolute and relative errors using \(\pi J(r^{\prime})/2\) to approximate \(\mathcal{E}(r)\) are less than 0.14%.