On the Existence of Symmetric Bicircular Central Configurations of the 3n-Body Problem

Abstract

In this paper, we consider central configurations of the planar 3n-body problem consisting of n masses at the vertices of a regular n-gon inscribed in a circle of radius r and 2n masses at the vertices of a second (not necessarily regular) concentric 2n-gon inscribed in a circle of radius ar which are symmetric in the sense that the set of positions of the 3n masses and the set of the corresponding masses are invariant under the action of a finite subgroup of O(2). There are two different types of such configurations. In the first type, called regular bicircular central configurations of the 3n-body problem, the second 2n-gon is regular, n of the vertices of the second n-gon are aligned with the vertices of the first regular n-gon and the masses at the vertices of this 2n-gon alternate values. In the second type, called semiregular bicircular central configurations of the 3n-body problem, the second 2n-gon is semiregular and the masses at its vertices are all of them equal. A semiregular 2n-gon has n pair of vertices symmetric by a reflection of an angle $\beta$ with respect to the axis of symmetry of the first regular n-gon. Our aim is to analyze the set of values of the parameter a for the regular 2n-gon and of the parameters $(a,\beta )$ for the semiregular 2n-gon providing symmetric bicircular central configurations. In particular, for all $n\ge 2$ we prove analytically the existence of symmetric bicircular central configurations with a (respectively $(a,\beta )$) satisfying some particular conditions. Using either computer-assisted results or numerical results, we also describe the complete set of values of a (respectively $(a,\beta )$) providing symmetric bicircular central configurations for $n=2,3,4,5$ and we give numerical evidences that the pattern for $n>5$ is the same as the one for $n=5$.

Appendices

Appendix 1: Proof of Proposition 1

We state and prove some auxiliary results that will be used in the proof of Proposition 1. We need the following two propositions taken from Bang and Elmabsout (2003).

Proposition 6

(Bang and Elmabsout 2003, Proposition 7) For $0\leqslant a<1$, $\alpha \in (0,1)$ and $u\in [0,2\pi )$, we have

$$\begin{aligned} \begin{aligned}&\sum _{j=1}^n \frac{1}{\big (1+a^2-2a\cos \big (\frac{2\pi \,j}{n}+u\big )\big )^\alpha }\\&\quad = \frac{n \,\sin (\pi \alpha )}{\pi }\int _0^1 \frac{t^{\alpha -1}}{(1-t)^\alpha }\,\frac{1}{(1-a^2t)^\alpha }\,\frac{1-(at)^{2n}}{1+(at)^{2n}-2(at)^n\cos (nu)}dt. \end{aligned} \end{aligned}$$

Proposition 7

(Bang and Elmabsout 2003, Proposition 8) For $a>1$, $\alpha \in (0,1)$ and $u\in [0,2\pi )$, we have

$$\begin{aligned} \begin{aligned}&\sum _{j=1}^n \frac{1}{\big (1+a^2-2a\cos \big (\frac{2\pi \,j}{n}+u\big )\big )^\alpha }\\&\quad = \frac{n\,\sin (\pi \alpha )}{\pi }\int _0^1 \frac{t^{\alpha -1}}{(1-t)^\alpha }\,\frac{1}{(a^2-t)^\alpha }\,\frac{a^{2n}-t^{2n}}{a^{2n}+t^{2n}-2(at)^n\cos (nu)}dt. \end{aligned} \end{aligned}$$

We need the following auxiliary result.

Lemma 1

Let $u\in {\mathbb {R}}$ and let $\gamma =2\pi \,j/n+u$.

• (a) The following identities hold for all $\ell \in {\mathbb {N}}$

  $$\begin{aligned} \sum _{j=1}^n \cos \left( \frac{2\ell \pi \,j}{n}+u\right) =0, \qquad \sum _{j=1}^n \sin \left( \frac{2\ell \pi \,j}{n}+u\right) =0, \end{aligned}$$

  (18)

  when $n\ge 2$ and $n\ne \ell$.

• (b) For $n\ge 3$, we have

  $$\begin{aligned} \sum _{j=1}^n \cos ^2 \gamma =\frac{n}{2}, \end{aligned}$$

  (19)

  and for $n=2$ we have

  $$\begin{aligned} \sum _{j=1}^2 \cos ^2\big (\pi \,j+u\big ) =2\cos ^2 u. \end{aligned}$$

  (20)

• (c) For all $n\ge 1$, we get

  $$\begin{aligned}\sum _{j=1}^n\frac{a-\cos \gamma }{\Big (1+a^2-2a\cos \gamma \Big )^{3/2}}=-\frac{d}{da}\sum _{j=1}^n\frac{1}{\Big (1+a^2-2a\cos \gamma \Big )^{1/2}}. \end{aligned}$$

• (d) Let

  $$\begin{aligned} L(a,u)=\sum _{j=1}^n\frac{1-1/a\cos \gamma }{\Big (1+a^2-2a\cos \gamma \Big )^{3/2}}. \end{aligned}$$

  We have $L(a,u)=-n/2+O(a)$ when $n\ge 3$ and $L(a,u)=2-6\cos ^2u+O(a)$ when $n=2$.

Proof

Using the sum of the first n terms of a geometric series, we get

$$\begin{aligned} \sum _{j=1}^n e^{i(2\ell \pi j/n+u)}=0, \end{aligned}$$

for all $\ell \in {\mathbb {Z}}$ with $\ell \ne n$ (here $i=\sqrt{-1}$). This proves statement (a). From the formula of the cosinus of twice an angle an applying statement (a) with $\ell =2$ and 2u instead of u, we get

$$\begin{aligned} \sum _{j=1}^n \cos ^2\left( \frac{2\pi \,j}{n}+u\right) =\frac{1}{2}\sum _{j=1}^n \left( \cos \left( 2\Big (\frac{2\pi \,j}{n}+u\Big )\right) +1 \right) =\frac{n}{2}. \end{aligned}$$

This proves statement (b) for $n>2$. Statement (b) for $n=2$ and statement (c) follows from direct computations.

Expanding the function L in Laurent series around $a=0$, we have

$$\begin{aligned} L(a,u)=-\frac{1}{a} \sum _{j=1}^n \cos \big (\frac{2\pi \,j}{n}+u \big ) + \sum _{j=1}^n 1 - 3 \sum _{j=1}^n \cos ^2\big (\frac{2\pi \,j}{n}+u \big ) + O(a). \end{aligned}$$

Then, when $n\ge 3$ in view of (18) with $\ell =1$ together with (19) we obtain

$$\begin{aligned} L(a,u)=n-\frac{3n}{2} + O(a)= -\frac{n}{2} + O(a) \end{aligned}$$

and when $n=2$ in view of (18) with $\ell =1$ and (20) we obtain

$$\begin{aligned} L(a,u)=2-6\cos ^2 u + O(a). \end{aligned}$$

This completes the proof of statement (d).

$\square$

We need the following technical lemma.

Lemma 2

We have $f(v)=(1+v)(1-v^{2n})-4 n v^n (1-v) > 0$ for $n \ge 2$ and $v \in (0,1)$.

Proof

Note that

$$\begin{aligned} \begin{aligned} f(v) =&(1+v)(1-v^n)(1+v^n)-4 n v^n (1-v) \\ =&(1-v)\big ((1+v)(1+v+v^2 +\ldots + v^{n-1})(1+v^n)-4 n v^n\big ) \\ :=&(1-v) g(v) \end{aligned} \end{aligned}$$

where

$$\begin{aligned} \begin{aligned} g(v)&= (1+v)(1+v+v^2 +\ldots + v^{n-1})(1+v^n) -4 n v^n \\&= 1 +2v+2 v^2 + 2 v^3 + \ldots + 2v^{n-1}+ 2 v^n +2 v^n + 2 v^{n+1} +\ldots \\&\quad + 2 v^{2n-1} +v^{2n}-4 n v^n \\&= 1 +2 v +\ldots + 2 v^{n-1} + (2-4n) v^n + 2 v^{n+1} + \ldots + 2 v^{2n-1} + v^{2n}. \end{aligned} \end{aligned}$$

(21)

We see that

$$\begin{aligned} g(1)=0 \quad \text {and} \quad g'(1)=0. \end{aligned}$$

Indeed,

$$\begin{aligned} g(1)=1+ 2 (n-1) + 2-4n + 2 (n-1) +1 = 0, \end{aligned}$$

and since

$$\begin{aligned} \begin{aligned} g'(v)&=2 +4 v + 6v^2 + \ldots + 2(n-1) v^{n-2} + (2-4n)n v^{n-1} \\&\quad + 2 (n+1) v^n +\ldots + 2(2n-1) v^{2n-2} +2 n v^{2n-1}, \end{aligned} \end{aligned}$$

then

$$\begin{aligned} \begin{aligned} g'(1)&=2 (1+2+3 +\ldots + (n-1))+ n (2-4n) \\&\quad + 2 (n+1+n+2 + \ldots + 2n-1) +2 n\\&= 2 \Big (\sum _{j=1}^{2n-1} j - n + (1-2n) n +n \Big ) =2 \Big ( \sum _{j=1}^{2n-1} j + n(1-2n)\Big )=0. \end{aligned} \end{aligned}$$

Therefore,

$$\begin{aligned} g(v)= (1-v)^2 h(v), \quad h(v)=\sum _{j=0}^{2n-2} c_j v^j \end{aligned}$$

for some coefficients $c_j$.

We will show by induction that

$$\begin{aligned} c_j ={\left\{ \begin{array}{ll} (j+1)^2 &{} \text {for }j=0,\ldots , n-1, \\ (2n-j-1)^2 &{} \text {for }j=n,\ldots , 2n-2. \end{array}\right. } \end{aligned}$$

In view of (21), we have that the coefficients $c_j$ satisfy $c_0=1$, $c_1=2 + 2 c_0=4$,

$c_{2n-3} -2 c_{2n-2}=2$ and $c_{2n-2}=1$ (and so $c_{2n-3}= 4).$

We prove by induction the cases for $j=0,\ldots ,n-1$. It is clear for $j=0,1$ and we will prove it for some $2 \le j < n-1$. Note that by the induction hypotheses

$$\begin{aligned} c_{j+1}= -c_{j-1} + 2 c_j +2 = -j^2 +2 j^2 +4 j +2+ 2 = j^2 +4j + 4 = (j+2)^2 \end{aligned}$$

and so the induction is satisfied for $j=0,\ldots ,n-1$. For $j=n$, we have

$$\begin{aligned} c_n = 2 -4 n -(n-1)^2 + 2 n^2 =2 -4 n -n^2 +2n -1 +2 n^2 =n^2-2n +1 =(n-1)^2, \end{aligned}$$

and for $j=n+1$ we have

$$\begin{aligned} c_{n+1} = -c_{n-1}+2c_n+2=-(n-1+1)^2+2(n-1)^2+2=(2n-(n+1)-1)^2. \end{aligned}$$

For $j=n, \ldots , 2n-2$, the induction hypotheses yield $c_j=(2n-j-1)^2$. Note that it is clear for $j=n$ and $j=n+1$ and we will show it for $n+1<j \le 2n-2$. In particular, the cases $j=2n-3$ and $j=2n-2$ are also trivially satisfied. So, we only need to show it for $n+1< j < 2n-3$. By the induction hypotheses for any $n+1< j < 2n-3$, we have

$$\begin{aligned} \begin{aligned} c_j&= 2 -c_{j-2} +2 c_{j-1}= 2 - (2n-(j-2)-1)^2 + 2 (2n-(j-1)-1)^2 \\&= 2-(2n-j+1)^2 +2 (2n-j)^2 =(2n-j-1)^2 \end{aligned} \end{aligned}$$

and the induction hypotheses holds. In short, the lemma is proved. $\square$

The next result concerns properties of the functions $K_i(a)$ introduced in Sect. 3.

Lemma 3

The following statements hold for all $n\ge 2$:

• (a) $K_6(a)>K_4(a)$ for $a\in (0,1)$;

• (b) $K_6(a)<K_4(a)$ for $a>1$;

• (c) $K_5>K_1>0$.

• (d) $a^3K_4(a)=K_2(1/a)$ and $a^3K_6(a)=K_3(1/a)$.

Proof

Note that

$$\begin{aligned} \begin{aligned} aK_6(a)&=\sum _{j=1}^n \frac{a-\cos \big (\frac{2 \pi j}{n} +\frac{\pi }{n} \big )}{\big (1+a^2 -2 a \cos \big (\frac{2 \pi j}{n}+ \frac{\pi }{n} \big )\big )^{3/2}}, \\ aK_4(a)&= \sum _{j=1}^n \frac{a-\cos \big (\frac{2 \pi j}{n} \big )}{\big (1+a^2 -2 a \cos \big (\frac{2 \pi j}{n} \big )\big )^{3/2}}. \end{aligned} \end{aligned}$$

When $a \in (0,1)$, using Lemma 1(c) and Proposition 6 with $\alpha =1/2$ and $u=0$, we get

$$\begin{aligned} \begin{aligned}&a K_4(a)= -\frac{n}{\pi }\int _0^1 \frac{d}{d a}\left( \frac{t^{-1/2}}{(1-t)^{1/2}}\,\frac{1}{(1-a^2t)^{1/2}}\,\frac{1-(at)^{2n}}{1+(at)^{2n}-2(at)^n}\right) dt. \end{aligned} \end{aligned}$$

and using Lemma 1(c) and Proposition 6 with $\alpha =1/2$ and $u=\pi /n$, we get

$$\begin{aligned} \begin{aligned}&a K_6(a)= -\frac{n}{\pi }\int _0^1 \frac{d}{d a}\left( \frac{t^{-1/2}}{(1-t)^{1/2}}\,\frac{1}{(1-a^2t)^{1/2}}\,\frac{1-(at)^{2n}}{1+(at)^{2n}+2(at)^n}\right) dt. \end{aligned} \end{aligned}$$

Hence, for $a \in (0,1)$,

$$\begin{aligned} \begin{aligned} K_6(a) - K_4(a)&= -\frac{n}{\pi a} \int _0^1 \frac{d}{d a} \left( \frac{t^{-1/2}}{(1-t)^{1/2}} \, \frac{1}{(1-a^2 t)^{1/2}} \, \frac{4 (a t)^n}{((at)^{2n}-1)} \right) \, d t \\&= -\frac{n}{\pi a} \int _0^1 \frac{4 (ta)^n (t a^2 (-1+(ta)^{2n}) + n (t a^2 -1)(1+(t a)^{2n}))}{a (1-t)^{1/2} t^{1/2} (1-a^2 t)^{3/2} ((ta)^{2n}-1)^2 }\, d t >0, \end{aligned} \end{aligned}$$

because the integrand is negative for $a \in (0,1)$. Therefore, $K_6(a) > K_4(a)$ for $0< a < 1$ and so statement (a) is proved.

For $a > 1$ using Lemma 1(c) and Proposition 7 with $\alpha =1/2$ and $u=0$, we get

$$\begin{aligned}\begin{aligned}&aK_4(a)=- \frac{n}{\pi }\int _0^1 \frac{d}{d a}\left( \frac{t^{-1/2}}{(1-t)^{1/2}}\,\frac{1}{(a^2-t)^{1/2}}\,\frac{a^{2n}-t^{2n}}{a^{2n}+t^{2n}-2 a^n t^n}\right) dt. \end{aligned} \end{aligned}$$

and using Lemma 1(c) and Proposition 7 with $\alpha =1/2$ and $u=\pi /n$ and taking derivatives with respect to a, we get

$$\begin{aligned}\begin{aligned}&aK_6(a)=- \frac{n}{\pi }\int _0^1 \frac{d}{d a}\left( \frac{t^{-1/2}}{(1-t)^{1/2}}\,\frac{1}{(a^2-t)^{1/2}}\,\frac{a^{2n}-t^{2n}}{a^{2n}+t^{2n}+2 a^n t^n}\right) dt. \end{aligned} \end{aligned}$$

Hence, for $a > 1$,

$$\begin{aligned} \begin{aligned} K_6(a)- K_4(a)&= - \frac{n}{\pi a} \int _0^1 \frac{d}{d a} \left( \frac{t^{-1/2}}{(1-t)^{1/2}} \, \frac{1}{(a^2-t)^{1/2}} \, \frac{4 (a t)^n}{t^{2n}-a^{2n}} \right) \, d t \\&= -\frac{n}{\pi a} \int _0^1 \frac{4 (ta)^n (a^2(a^{2n}-t^{2n})+n(a^2-t)(a^{2n}+t^{2n}))}{a (1-t)^{1/2} t^{1/2} (a^2-t)^{3/2} (t^{2n}-a^{2n})^2 }\, d t <0, \end{aligned} \end{aligned}$$

because the integrand is positive for $a >1$. Therefore, $K_6(a) < K_4(a)$ for $a > 1$ and statement (b) is proved.

To prove statement (c), we proceed as follows. Note that

$$\begin{aligned} \begin{aligned} K_5&= \lim _{a\rightarrow 1} \frac{1}{2}\sum _{j=1}^n \frac{1}{\big (1+a^2 -2 a \cos \big (\frac{2 \pi j}{n} + \frac{\pi }{n}\big )\big )^{1/2}}:=\lim _{a\rightarrow 1} \frac{1}{2}A_0,\\ K_1&= \lim _{a\rightarrow 1}\frac{1}{2} \sum _{j=1}^{n-1} \frac{1}{\big (1+a^2 -2 a \cos \big (\frac{2 \pi j}{n} \big )\big )^{1/2}} \\&= \lim _{a\rightarrow 1} \frac{1}{2} \! \left( \sum _{j=1}^{n} \frac{1}{\big (1+a^2 -2 a \cos \big (\frac{2 \pi j}{n} \big )\big )^{1/2}}\!-\!\sum _{j=1}^1\frac{1}{(1+a^2-2a\cos (2\pi j))^{1/2}}\!\right) \\&:= \lim _{a\rightarrow 1} \frac{1}{2}(A_1-A_2), \end{aligned} \end{aligned}$$

Thus,

$$\begin{aligned} K_5-K_1=\lim _{a\rightarrow 1} \dfrac{1}{2} (A_0-A_1+A_2) \end{aligned}$$

(22)

where $A_0$, $A_1$, $A_2$ are the summations defined above. Applying Proposition 6, we have that $A_0-A_1+A_2$ when $a\in (0,1)$ is given by

$$\begin{aligned} \begin{aligned} A&= \frac{n}{\pi }\int _0^1 \frac{1-(ta)^n}{(1-t)^{1/2} t^{1/2} (1-a^2t)^{1/2}(1+(ta)^n)} \, dt\\&\quad -\frac{n}{\pi }\int _0^1 \frac{1+(ta)^n}{(1-t)^{1/2} t^{1/2} (1-a^2t)^{1/2}(1-(ta)^n)} \, dt \\&\quad + \frac{1}{\pi } \int _0^1 \frac{1+ta}{(1-t)^{1/2} t^{1/2} (1-a^2t)^{1/2}(1-ta)} \, dt\\&= \frac{n}{\pi }\int _0^1 \left( \frac{4 (ta)^n}{-1+(ta)^{2n}} + \frac{1+ta}{n(1-t a)}\right) \frac{1}{(1-t)^{1/2} t^{1/2} (1-a^2 t)^{1/2}} \, d t . \end{aligned} \end{aligned}$$

On the other hand, applying Proposition 7 we have that $A_0-A_1+A_2$ when $a>1$ is given by

$$\begin{aligned} \begin{aligned} {\overline{A}}&= \frac{n}{\pi }\int _0^1 \left( -\frac{4 (ta)^n}{a^{2n}-t^{2n}} + \frac{a+t}{n(a-t)}\right) \frac{1}{(1-t)^{1/2} t^{1/2} (a^2-t)^{1/2}} \, d t . \end{aligned} \end{aligned}$$

After doing the substitution $a\rightarrow 1/a$, the expression $\overline{A}$ can be written as $a\, A$. Thus,

$$\begin{aligned} \lim _{a\rightarrow 1^-} A=\lim _{a\rightarrow 1^+} \overline{A}=\lim _{a\rightarrow 1} A_0-A_1+A_2. \end{aligned}$$

(23)

Now we show that $A>0$. Note that taking $v=ta$, we get

$$\begin{aligned} \frac{4 (ta)^n}{-1+(ta)^{2n}} + \frac{1+ta}{n(1-t a)}=\frac{4 v^n}{-1+v^{2n}} + \frac{1}{n} \, \frac{1+v}{1-v}. \end{aligned}$$

Using Lemma 2, we get that

$$\begin{aligned} \frac{4 v^n}{-1+v^{2n}} + \frac{1}{n} \, \frac{1+v}{1-v} = \frac{(1+v)(1-v^{2n})-4 n v^n(1-v) }{n (1-v)(1-v^{2n})}> 0, \end{aligned}$$

for $v < 1$. So $A>0$ and taking the limit when $a\rightarrow 1^-$ together with (23) and (22) we get $\lim _{a\rightarrow 1^-}A=\lim _{a\rightarrow 1^+}\overline{A}=2(K_5-K_1)>0$ for all $n\ge 2$. Moreover, $K_1$ is positive by definition. So, we have proved statement (c).

Statement (d) follows from direct computations. $\square$

In the following lemma, we provide properties of the function $\varDelta (a)=-K_1-K_5+a^3(K_2(a)+K_3(a))$ that appears in the denominator $m_D$ in (4).

Lemma 4

For all $n\ge 2$, the following statements hold.

• (a) $\varDelta (a)$ is increasing for all $a \in (0,1)$;

• (b) $\varDelta (0)<0$, $\varDelta (a)\rightarrow \infty$ when $a\rightarrow 1^-$ and $\varDelta (a)\rightarrow -\infty$ when $a\rightarrow 1^+$;

• (c) $\varDelta (a)<0$ for all $a>1$;

• (d) $\varDelta (a)$ has a unique zero and it belongs to the interval (0, 1).

Proof

We first note that setting $b=1/a$ we have

$$\begin{aligned} \begin{aligned} a^3 K_2(a)&= \frac{1}{b} \sum _{j=1}^n \frac{b-\cos \big (\frac{2 \pi j}{n}\big )}{\big (1+b^2 -2 b \cos \big (\frac{2 \pi j}{n}\big )\big )^{3/2}} :=\frac{1}{b} {{\bar{K}}}_2(b) ,\\ a^3 K_3(a)&= \frac{1}{b} \sum _{j=1}^n \frac{b-\cos \big (\frac{2 \pi j}{n}+\frac{\pi }{n}\big )}{\big (1+b^2 -2 b \cos \big (\frac{2 \pi j}{n}+\frac{\pi }{n}\big )\big )^{3/2}} :=\frac{1}{b} {{\bar{K}}}_3(b). \end{aligned} \end{aligned}$$

(24)

Let ${\overline{\varDelta }}(b) =(1/b {{\bar{K}}}_2(b) +1/b {{\bar{K}}}_3(b))$. We want to show that for $a\in (0,1)$

$$\begin{aligned} \varDelta '(a)=\left. \frac{d {\overline{\varDelta }}(b)}{db}\right| _{b=1/a}\cdot \frac{db}{da}=\left. ({\overline{\varDelta }}(b))'\right| _{b=1/a}\cdot \left( -\frac{1}{a^2}\right) >0. \end{aligned}$$

So, it is sufficient to show that $\varDelta (b)'<0$ for $b>1$.

Let

$$\begin{aligned} T=\sum _{j=1}^n\frac{1}{\big (1+b^2 -2 b \cos \big (\frac{2 \pi j}{n}\big )\big )^{1/2}}+\sum _{j=1}^n \frac{1}{\big (1+b^2 -2 b \cos \big (\frac{2 \pi j}{n}+\frac{\pi }{n}\big )\big )^{1/2}}. \end{aligned}$$

Using Lemma 1(c), we get that

$$\begin{aligned} {{\bar{K}}}_2(b)+{{\bar{K}}}_3(b)=-dT/db \end{aligned}$$

(25)

and so $({\bar{\varDelta }}(b))' =\frac{1}{b^2} T_1 -\frac{1}{b} T_2$ where

$$\begin{aligned} T_1= \frac{d T}{db}, \quad T_2=\frac{d^2 T}{db^2}. \end{aligned}$$

Since from Proposition 7, we have

$$\begin{aligned} \begin{aligned} T_1&=- \frac{2n}{\pi }\int _0^1 \frac{d}{db} \left( \frac{1}{t^{1/2} (1-t)^{1/2} (b^2 -t)^{1/2}} \, \frac{t^{2n}+b^{2n}}{t^{2n}-b^{2n}}\right) \, d t \ \\&= \frac{n}{\pi }\int _0^1 \frac{1}{t^{1/2} (1-t)^{1/2}} \left( \!\frac{2 b (t^{2n}+b^{2n})}{(b^2-t)^{3/2} (t^{2n}-b^{2n})}\! - \!\frac{8 n t^{2n} b^{2n-1}}{(b^2-t)^{1/2} (t^{2n}-b^{2n})^2}\! \right) d t , \end{aligned} \end{aligned}$$

taking another derivative with respect to b we get

$$\begin{aligned} \begin{aligned} T_2&= \frac{n}{\pi }\int _0^1 \frac{1}{t^{1/2} (1-t)^{1/2}} \left( \frac{-8 n t^{2n} b^{2n-2}((2n-1)t^{2n} + (2n+1)b^{2n})}{(b^2-t)^{1/2} (t^{2n}-b^{2n})^3} \right. \\&\quad \left. +\frac{2 \left( 8 n b^{2 n} \left( b^2-t\right) t^{2 n}+\left( 2 b^2+t\right) \left( b^{4 n}-t^{4 n}\right) \right) }{\left( b^2-t\right) ^{5/2} \left( t^{2 n}-b^{2 n}\right) ^2}\right) \, d t. \end{aligned} \end{aligned}$$

Since $T_1 < 0$ (the integrand is negative) and $T_2 >0$ (the integrand is positive), we readily obtain that $({{\bar{\varDelta }}}(b))' <0$ and so $\varDelta '(a) >0$ for $a \in (0,1)$. In short, statement (a) is proved.

It is clear that $\varDelta (0)=-K_1-K_5<0$, see Lemma 3(c). Moreover,

$$\begin{aligned} \lim _{a \rightarrow 1} K_2(a)= \sum _{j=1}^{n-1} \frac{1-\cos \big (\frac{2 \pi j}{n}\big )}{\big (2 -2 \cos \big (\frac{2 \pi j}{n}\big )\big )^{3/2}} + \lim _{a \rightarrow 1} \frac{1-a}{\big (1+a^2 -2 a \big )^{3/2}}, \end{aligned}$$

so

$$\begin{aligned} \lim _{a \rightarrow 1^-} K_2(a)=\infty \quad \text {and} \quad \lim _{a \rightarrow 1^+} K_2(a)=-\infty . \end{aligned}$$

(26)

Furthermore, $\lim _{a\rightarrow 1} K_3(a)=K_5$ which is different from zero and from infinity. Hence, using (26) we get

$$\begin{aligned} \lim _{a \rightarrow 1^-} \varDelta (a)=\infty \quad \text {and} \quad \lim _{a \rightarrow 1^+} \varDelta (a)=-\infty . \end{aligned}$$

This completes the proof of statement (b).

Now we show that $\varDelta (a) < 0$ for $a > 1$. To do so, we will show that $a^3 K_2(a) +a^3 K_3(a)< 0$. Note that this is sufficient because $-K_1 -K_5 < 0$. Clearly in view of (24) and (25), we have $a^3 K_2(a) + a^3 K_3(a) = \frac{1}{b} ({{\bar{K}}}_2(b) + {{\bar{K}}}_3(b))=-\frac{1}{b}\frac{dT}{db}$ with $b < 1$. So, applying Proposition 6 we get

$$\begin{aligned} \begin{aligned} -\frac{1}{b}\dfrac{dT}{db}&=-\frac{2 n }{\pi b} \int _0^1 \frac{d}{d b} \left( \frac{1}{t^{1/2} (1-t)^{1/2} (1-b^2 t )^{1/2}} \, \frac{1+(bt)^{2n}}{1-(bt)^{2n}}\right) \, d t \\&=-\frac{2 n }{\pi b} \int _0^1 \left( \frac{4 n (t b)^{2n}}{b\, t^{1/2} (1-t)^{1/2} (1-b^2 t )^{1/2}(1-(bt)^{2n})^2}\right. \\&\quad \left. +\frac{ bt(1+(bt)^{2n})}{t^{1/2} (1-t)^{1/2} (1-b^2 t )^{3/2}(1-(bt)^{2n})} \right) \, d t \end{aligned} \end{aligned}$$

and since the integrand is positive we readily have that $a^3 K_2(a) + a^3 K_3(a)< 0$ for $a > 1$ and so $\varDelta (a) < 0$ for $a > 1$. In short, statement (c) is proved.

The proof of statement (d) is a direct consequence of the statements (a)–(c) together with the Bolzano–Cauchy theorem. $\square$

Proof of Proposition 1

We expand $m_2=a^3 m_{N,2}/{m_{D}}$ in Laurent series around $a=0$, see (4). The expansion of $m_D$ around $a=0$ is given by

$$\begin{aligned} m_{D} = (K_5-K_1)(-K_1-K_5 + O(a^3)). \end{aligned}$$

Using Lemma 1(d) with $u=0$ (respectively $u=\pi /n$) to expand $K_4$ (respectively $K_6$) in Laurent series around $a=0$, we get

$$\begin{aligned} K_4(a)=-\frac{n}{2}+O(a) \qquad \text{ and } \qquad K_6(a)=-\frac{n}{2}+O(a) \end{aligned}$$

(27)

when $n\ge 3$ and

$$\begin{aligned} K_4(a)=-4+O(a) \qquad \text{ and } \qquad K_6(a)=2+O(a) \end{aligned}$$

(28)

when $n=2$. Therefore, when $n\ge 3$ we have

$$\begin{aligned} m_2= a^3 \frac{K_1^2 + \frac{n}{2} K_1 -K_ 1 K_5 -\frac{n}{2} K_5}{(K_5-K_1)(-K_1-K_5)} + O(a^4) = a^3 \frac{K_1+ \frac{n}{2} }{K_5+K_1} + O(a^4) \end{aligned}$$

and since $K_1,K_5 > 0$, we obtain $\lim _{a \rightarrow 0^+} m_2 = 0^+$.

When $n=2$, we compute directly the quantities $K_1$ and $K_5$ and we have $K_1=1/4$, $K_5=1/\sqrt{2}$. So expanding $m_3$ in Laurent series around $a=0$, we get

$$\begin{aligned} m_2=- a^3 \Big (\frac{17}{7} + 2\sqrt{2}\Big ) + O(a^4). \end{aligned}$$

Hence, $\lim _{a \rightarrow 0^+} m_2 = 0^-$. This completes the proof of statement (a).

For statement (b), we note that by (5)

$$\begin{aligned} m_3=m_2 + \frac{a^3(K_6(a)-K_4(a))}{K_5-K_1}. \end{aligned}$$

Using the Laurent series of $K_4$ and $K_6$ around $a=0$ given in (27), we get

$$\begin{aligned} \frac{a^3(K_6-K_4(a))}{K_5-K_1} =O(a^4) \end{aligned}$$

when $n\ge 3$ and so

$$\begin{aligned} m_3= a^3 \frac{K_1+ \frac{n}{2} }{K_5+K_1} + O(a^4), \end{aligned}$$

which yields $\lim _{a \rightarrow 0^+} m_3 = 0^+$ when $n\ge 3$.

On the other hand, for $n=2$, using the Laurent series of $K_4$ and $K_6$ around $a=0$ given in (28) and the values of $K_1$ and $K_5$ for $n=2$ computed above, we get

$$\begin{aligned} \frac{a^3(K_6-K_4(a))}{K_5-K_1} =\frac{a^3}{K_5-K_1} (6 + O(a)) \end{aligned}$$

which yields

$$\begin{aligned} m_3= \frac{a^3}{7} (7+34 \sqrt{2}) + O(a^4), \end{aligned}$$

and so $\lim _{a \rightarrow 0^+} m_3 = 0^+.$ This completes the proof of statement (b).

We expand $m_3=a^3 m_{N,3}/m_{D}$ in Laurent series around $a=1$ (with $a > 1$), see (4). Note that

$$\begin{aligned} \begin{aligned} K_2(a)&= K_1 - \frac{1}{(a-1)^{2}}+O(a-1), \quad K_3(a)=K_5 + O(a-1), \\ K_4(a)&=K_1 + \frac{1}{(a-1)^{2}}-\frac{1}{a-1}+1+O(a-1), \quad K_6 =K_5 + O(a-1), \end{aligned} \end{aligned}$$

and $a^3=1+O(a-1)$. Hence,

$$\begin{aligned} m_{N,3} =\frac{1}{(a-1)^4} + O((a-1)^{-3}) \quad \text {and} \quad m_{D} =- \frac{ (K_5-K_1)}{(a-1)^{2}} + O((a-1)^{-1}). \end{aligned}$$

Therefore,

$$\begin{aligned} m_3= -\frac{1}{(a-1)^{2}(K_5-K_1)} + O((a-1)^{-1}). \end{aligned}$$

Since in view of Lemma 3(c) we have $K_5>K_1>0$, then $\lim _{a \rightarrow 1^+} m_3 = -\infty$, which completes the proof of statement (c).

For statement (d), note that taking $b=1/a$ and using Lemma 3(d) we have $K_2(a)=K_2(1/b)=b^3K_4(b)$, $K_3(a)=K_3(1/b)=b^3K_6(b)$, $K_4(a)=K_2(1/a)/a^3=b^3K_2(b)$ and $K_6(a)=K_3(1/a)/a^3=b^3K_3(b)$. Thus, using Lemma  1(d) for $n\ge 3$ we have

$$\begin{aligned} K_4(b)=-\frac{n}{2}+O(b), \quad K_6(b)=-\frac{n}{2}+O(b). \end{aligned}$$

Moreover, expanding in power series around $b=0$, we get

$$\begin{aligned} K_2(b)=n+O(b), \quad K_3(b)=n+O(b). \end{aligned}$$

Hence for $n\ge 3$

$$\begin{aligned} \begin{aligned} K_2(b)&=-\frac{n}{2} b^3+ O(b^4), \qquad K_3(b)=-\frac{n}{2} b^3+ O(b^4),\\ K_4(b)&=nb^3 + O(b^4), \qquad K_6(b)=nb^3+O(b^4), \end{aligned} \end{aligned}$$

(29)

Proceeding in the same way for $n=2$, we get

$$\begin{aligned} \begin{aligned} K_2(b)&=-4b^3+ O(b^4), \qquad K_3(b)=2b^3+ O(b^4),\\ K_4(b)&=2b^3 + O(b^4), \qquad K_6(b)=2b^3+O(b^4). \end{aligned} \end{aligned}$$

(30)

Using (29) and (30), the numerators of $m_2$ and $m_3$ (see (4)) can be written as

$$\begin{aligned} \frac{1}{b^3}(K_1^2-K_1K_5)+O(1), \end{aligned}$$

and the denominators of $m_2$ and $m_3$ (see again (4)) can be written as

$$\begin{aligned} (K_5-K_1)(-K_1-K_5-n)+O(b) \end{aligned}$$

for all $n\ge 2$. Thus, the Laurent expansion of $m_2$ and $m_3$ around $b=0$ becomes (after simplifying $K_5-K_1$)

$$\begin{aligned} \frac{1}{b^3} \frac{K_1}{K_1+K_5+n}+O(b^{-2}). \end{aligned}$$

Then taking into account that in view of Lemma 3(c) we have $K_5>K_1>0$, we conclude that

$$\begin{aligned} \lim _{a\rightarrow \infty } m_2=\lim _{b\rightarrow 0^+} m_2=\infty , \qquad \lim _{a\rightarrow \infty } m_3=\lim _{b\rightarrow 0^+} m_3=\infty . \end{aligned}$$

This proves statement (d).

From Lemma 3(a), we get $K_6-K_4(a)>0$ for $a\in (0,1)$. From Lemma 3(b), we get $K_6-K_4(a)<0$ for $a>1$, and from Lemma 3(c), we get $K_5-K_1>0$. Thus from (5) we get $m_2<m_3$ when $a\in (0,1)$ and $m_2>m_3$ when $a>1$, which proves statement (e). $\square$

Appendix 2: Proof of Proposition 3

We need the following auxiliary lemma.

Lemma 5

Let

$$\begin{aligned} E(a):=\sum _{j=0}^{[n/2]-1} \frac{-2 (1+a^2)\cos \big (\frac{2\pi \,j}{n}+ \frac{\pi }{n} \big )+5 a-a \cos \big (\frac{4\pi \,j}{n}+ \frac{2\pi }{n} \big )}{a\Big (1+a^2 -2 a \cos \big (\frac{2\pi \,j}{n}+ \frac{\pi }{n} \big )\Big )^{5/2}}. \end{aligned}$$

Then

$$\begin{aligned} \begin{aligned}&\lim _{\beta \rightarrow \pi /n} N_4(a,\beta )(K_1 -2 a^3 L_2(a,\beta ) + L_5(\beta )) =\frac{1}{4} E(a) \end{aligned} \end{aligned}$$

if n is even and

$$\begin{aligned} \begin{aligned}&\lim _{\beta \rightarrow \pi /n} N_4(a,\beta )(K_1 -2 a^3 L_2(a,\beta ) + L_5(\beta )) =\frac{1}{4} \left( E(a)+\frac{1}{a(1+a)^3}\right) \end{aligned} \end{aligned}$$

if n is odd.

Proof

First note that

$$\begin{aligned} \lim _{\beta \rightarrow \pi /n} N_4(a,\beta )(K_1 -2 a^3 L_2(a,\beta )) =0 \end{aligned}$$

because

$$\begin{aligned} \lim _{\beta \rightarrow \pi /n} N_4(a,\beta ) = \sum _{j=1}^n \frac{-1/a \sin \big (\frac{2\pi \,j}{n}+\frac{\pi }{n}\big ) }{\Big (1+a^2 -2 a\cos \big (\frac{2\pi \,j}{n}+\frac{\pi }{n}\big )\Big )^{3/2}}=0 \end{aligned}$$

and

$$\begin{aligned} \lim _{\beta \rightarrow \pi /n} L_2(a,\beta ) = \sum _{j=1}^n \frac{1-a \cos \big (\frac{2\pi \,j}{n}-\frac{\pi }{n}\big ) }{\Big (1+a^2 -2 a\cos \big (\frac{2\pi \,j}{n}-\frac{\pi }{n}\big )\Big )^{3/2}} \end{aligned}$$

is finite for all $a>0$. On the other hand,

$$\begin{aligned} \begin{aligned} L_5(\beta )&=\left( \sum _{j=0}^{n-2} \frac{1- \cos \big (\frac{2\pi \,j}{n}+2 \beta \big ) }{\Big (2 -2 \cos \big (\frac{2\pi \,j}{n}+2 \beta \big )\Big )^{3/2}}\right) + \frac{1-\cos \big (\frac{2\pi \,(n-1)}{n}+2 \beta \big ) }{\Big (2 -2 \cos \big (\frac{2\pi \,(n-1)}{n}+2 \beta \big )\Big )^{3/2}} \\&:= L_{5,1}(\beta ) + L_{5,2}(\beta ). \end{aligned} \end{aligned}$$

Clearly $\lim _{\beta \rightarrow \pi /n} N_4(a,\beta ) L_{5,1}(\beta )=0$. Therefore, we need to study $\lim _{\beta \rightarrow \pi /n} N_4(a,\beta ) L_{5,2}(\beta )$. Note that expanding $L_{5,2}$ around $\beta =\pi /n$ we have

$$\begin{aligned} L_{5,2}(\beta ) = \frac{1-\cos \big (\frac{2\pi (n-1)}{n}+2 \beta \big ) }{\Big (2 -2 \cos \big (\frac{2\pi (n-1)}{n}+2 \beta \big )\Big )^{3/2}} = \frac{1}{4 \big (\beta -\frac{\pi }{n}\big )} + \frac{1}{24} \big (\beta -\frac{\pi }{n}\big ) + O\Big (\big (\beta -\frac{\pi }{n}\big )^2\Big ). \end{aligned}$$

On the other hand, we can rewrite $N_4(a,\beta )$ as

$$\begin{aligned} \begin{aligned} {\bar{N}}_4(a,\beta )&=\sum _{j=0}^{[n/2]-1} \left( \frac{-1/a \sin \big (\frac{2\pi \,j}{n}+ \beta \big ) }{\Big (1+a^2 -2 a \cos \big (\frac{2\pi \,j}{n}+ \beta \big )\Big )^{3/2}}\right. \\&\left. \quad -\frac{1/a \sin \big (\frac{2\pi \,(-j-1)}{n}+ \beta \big ) }{\Big (1+a^2 -2 a \cos \big (\frac{2\pi \,(-j-1)}{n}+ \beta \big )\Big )^{3/2}}\right) \end{aligned} \end{aligned}$$

if n is even and as ${\bar{N}}_4(a,\beta )+{\bar{N}}_4^*(a,\beta )$ with

$$\begin{aligned} {\bar{N}}_4^*(a,\beta )= \frac{-1/a \sin \big (\frac{2\pi \,[n/2]}{n}+ \beta \big ) }{\Big (1+a^2 -2 a \cos \big (\frac{2\pi \,[n/2]}{n}+ \beta \big )\Big )^{3/2}} \end{aligned}$$

if n is odd. Expanding ${\bar{N}}_4$ around $\beta =\pi /n$, we get

$$\begin{aligned} {\bar{N}}_4(a,\beta ) =E(a) (\beta -\pi /n) + O((\beta -\pi /n)^2), \end{aligned}$$

and expanding ${\bar{N}}_4^*$ also around $\beta =\pi /n$ we get

$$\begin{aligned} {\bar{N}}_4^*(a,\beta )=\frac{1}{a(1+a)^3}(\beta -\pi /n)+ O((\beta -\pi /n)^3). \end{aligned}$$

Thus, expanding $N_4(a,\beta ) L_{5,2}(\beta )$ around $\beta =\pi /n$ we obtain

$$\begin{aligned} N_4(a,\beta ) L_{5,2}(\beta ) = \frac{1}{4} E(a) + O((\beta -\pi /n)) \end{aligned}$$

if n is even and

$$\begin{aligned} N_4(a,\beta ) L_{5,2}(\beta ) = \frac{1}{4} \left( E(a)+\frac{1}{a(1+a)^3}\right) + O((\beta -\pi /n)) \end{aligned}$$

if n is odd. Taking the limit as $\beta \rightarrow \pi /n$ we obtain the result that we wanted to prove. $\square$

Proof of Proposition 3

Note that

$$\begin{aligned} \begin{aligned} \lim _{\beta \rightarrow \pi /n}N_5(\beta )=&\sum _{j=1,j\ne n-1}^n\frac{-\sin \big (\frac{2\pi \,(j+1)}{n}\big )}{\Big (2-2\cos \big (\frac{2\pi \,(j+1)}{n}\big )\Big )^{3/2}}\\&+\lim _{\beta \rightarrow \pi /n} \frac{-\sin \big (\frac{2\pi \,(n-1)}{n}+2\beta \big )}{\Big (2-2\cos \big (\frac{2\pi \,(n-1)}{n}+2\beta \big )\Big )^{3/2}}=\infty . \end{aligned} \end{aligned}$$

In view of Lemma 5, it is clear that the sign of $F(a,\beta )$ as $\beta \rightarrow \pi /n$ is determined by the sign of the difference G given by

$$\begin{aligned} G=\lim _{\beta \rightarrow \pi /n} (K_1-L_4(a,\beta ))= K_1 - {\bar{L}}_4(a), \end{aligned}$$

where

$$\begin{aligned} {\bar{L}}_4(a)=\lim _{\beta \rightarrow \pi /n} L_4(a,\beta )= \sum _{j=1}^{n} \frac{1-1/a \cos \big (\frac{2\pi \,j}{n}+ \frac{\pi }{n} \big )}{\Big (1+a^2 -2 a \cos \big (\frac{2\pi \,j}{n}+ \frac{\pi }{n} \big )\Big )^{3/2}}. \end{aligned}$$

When $a \rightarrow \infty$, we have

$$\begin{aligned} \lim _{a \rightarrow \infty } G= K_1 -\lim _{a \rightarrow \infty }{\bar{L}}_4(a) =K_1 > 0. \end{aligned}$$

This proves statement (a) of the proposition.

Furthermore, when $a \rightarrow 1$, using Lemma 3(c), we obtain

$$\begin{aligned} \lim _{a \rightarrow 1} G = K_1 - \sum _{j=1}^{n} \frac{1- \cos \big (\frac{2\pi \,j}{n}+ \frac{\pi }{n} \big )}{\Big (2-2 \cos \big (\frac{2\pi \,j}{n}+ \frac{\pi }{n} \big )\Big )^{3/2}} =K_1 -K_5 < 0. \end{aligned}$$

So, statement (b) is proved.

Using Lemma 1(d) with $u =\pi /n$, the expansion of ${\bar{L}}_4$ around $a=0$ is given by ${\bar{L}}_4(a)= -n/2+ O(a),$ when $n\ge 3$ and ${\bar{L}}_4(a)= 2 + O(a),$ when $n=2$. Hence, when $n\ge 3$ the expansion of G around $a=0$ is $G=K_1 +n/2 + O(a)$ and so $\lim _{a \rightarrow 0} G >0.$ For $n=2$, computing the value of $K_1$, the expansion of G around $a=0$ is $G=1/4 -2 + O(a)$ and so $\lim _{a \rightarrow 0} G =-7/4<0.$ This yields statement (c) of the proposition. $\square$

Appendix 3: Proof of Propositions 4 and 5

We need the following auxiliary lemma.

Lemma 6

We have

$$\begin{aligned} \lim _{\beta \rightarrow \pi /2n} N_5(\beta ) (K_1-L_4(a,\beta ))=0. \end{aligned}$$

Proof

First note that

$$\begin{aligned} \lim _{\beta \rightarrow \pi /2n} L_4(a,\beta )= \sum _{j=1}^{n} \frac{1-1/a \cos \big (\frac{2\pi \,j}{n}+ \frac{\pi }{2n} \big )}{\Big (1+a^2 -2 a \cos \big (\frac{2\pi \,j}{n}+ \frac{\pi }{2n} \big )\Big )^{3/2}}, \end{aligned}$$

which is finite for all $a>0$. Moreover, it is easy to see that

$$\begin{aligned} \lim _{\beta \rightarrow \pi /2n}N_5(\beta )= \sum _{j=1}^{n} \frac{- \sin \big (\frac{2\pi \,j}{n}+ \frac{\pi }{n} \big )}{\Big (2 -2 \cos \big (\frac{2\pi \,j}{n}+ \frac{\pi }{n} \big )\Big )^{3/2}} =0 \end{aligned}$$

and so $\lim _{\beta \rightarrow \pi /2n} N_5(\beta )(K_1-L_4(a,\beta ))=0,$ as we wanted to prove. $\square$

Proof of Proposition 4

In view of Lemma 6 the sign of $F(a,\beta )$ around $\beta =\pi /2n$ is determined by the sign of $N_4(a,\beta )(K_1 -2 a^3 L_2(a,\beta ) + L_5(\beta ))$ unless $\lim _{\beta \rightarrow \pi /2n} N_4(a,\beta )(K_1-2a^3L_2(a,\beta )+L_5(\beta ))= 0$. From the analysis of the sign of $N_4(a,\beta )(K_1 -2 a^3 L_2(a,\beta ) + L_5(\beta ))$ around $\beta \rightarrow \pi /2n$ we will see that $\lim _{\beta \rightarrow \pi /2n} N_4(a,\beta )(K_1-2a^3L_2(a,\beta )+L_5(\beta ))\ne 0$.

In view of Proposition 2, we have that $N_4(a,\beta ) < 0$ for $\alpha \in (0,\pi /n)$ and so $N_4(a,\beta ) < 0$. So, we need to study the sign of $K_1 -2 a^3 L_2(a,\beta ) + L_5(\beta )$ as $\beta \rightarrow \pi /2n$.

Hence,

$$\begin{aligned} \lim _{\beta \rightarrow \pi /2n} K_1 -2 a^3 L_2(a,\beta ) + L_5(\beta ) =K_1-2a^3{\bar{L}}_2(a)+ K_5:=H, \end{aligned}$$

where

$$\begin{aligned} {\bar{L}}_2(a)=\lim _{\beta \rightarrow \pi /2n} L_2(a,\beta ) =\sum _{j=1}^n\frac{1-a\cos \big (\frac{2\pi \,j}{n}-\frac{\pi }{2n}\big )}{\Big (1+a^2-2a\cos \big (\frac{2\pi \,j}{n}-\frac{\pi }{2n}\big )\Big )^{3/2}}. \end{aligned}$$

When $a\rightarrow 0$, we have that ${\bar{L}}_2(a) \rightarrow n$. So $\lim _{a \rightarrow 0}H = K_1 +K_5 > 0$ in view of Lemma 3(c). This proves statement (a) of the proposition.

To study the behavior when $a \rightarrow \infty$, we first observe that making the change $b=1/a$ we get $a^3{\bar{L}}_2(a)=L(b,\pi /2n)$ where L is the function defined in Lemma 1(d). Thus, applying Lemma 1(d) with $u=\pi /2n$ and using Lemma 3(c) we have

$$\begin{aligned} \lim _{a \rightarrow \infty } H = K_1 +n + K_5 >0, \end{aligned}$$

when $n\ge 3$ and

$$\begin{aligned} \lim _{a \rightarrow \infty } H = K_1 -2\big (2-6\cos ^2\big (\tfrac{\pi }{4}\big )\big ) + K_5= K_1 +2 + K_5 >0, \end{aligned}$$

when $n=2$. This proves statement (b).

Finally, to prove statement (c) we need to study the sing of H when $a \rightarrow 1$. Note that

$$\begin{aligned} \begin{aligned} \lim _{a \rightarrow 1} H&= \frac{1}{2} \sum _{j=1}^{n-1} \frac{1}{\Big (2 -2 \cos \big (\frac{2\pi \,j}{n} \big )\Big )^{1/2}} -\sum _{j=1}^n \frac{1}{\Big (2 -2 \cos \big (\frac{2\pi \,j}{n}- \frac{\pi }{2n} \big )\Big )^{1/2}} \\&\quad + \frac{1}{2} \sum _{j=1}^n \frac{1}{\Big (2 -2 \cos \big (\frac{2\pi \,j}{n}+ \frac{\pi }{n} \big )\Big )^{1/2}}. \end{aligned} \end{aligned}$$

In order to study the sign of $\lim _{a \rightarrow 1} H$, we rewrite $\lim _{a \rightarrow 1} H$ as $\lim _{a\rightarrow 1} H_1$ where

$$\begin{aligned} \begin{aligned} H_1&= \frac{1}{2} \sum _{j=1}^{n} \frac{1}{\Big (1+a^2 -2 a \cos \big (\frac{2\pi \,j}{n} \big )\Big )^{1/2}} - \frac{1}{2} \sum _{j=1}^1\frac{1}{(1+a^2 -2 a\cos (2\pi j ) )^{1/2} } \\&\quad -\sum _{j=1}^n \frac{1}{\Big (1+a^2 -2a \cos \big (\frac{2\pi \,j}{n}- \frac{\pi }{2n} \big )\Big )^{1/2}} + \frac{1}{2} \sum _{j=1}^n \frac{1}{\Big (1+a^2 -2 a \cos \big (\frac{2\pi \,j}{n}+ \frac{\pi }{n} \big )\Big )^{1/2}}. \end{aligned} \end{aligned}$$

Now applying Proposition 6 with $\alpha =1/2$ and $u=0$; $n=1$, $\alpha =1/2$ and $u=0$; $\alpha =1/2$ and $u=-\pi /2n$; and $\alpha =1/2$ and $u=\pi /n$, respectively, we get

$$\begin{aligned} \begin{aligned} H_1&= \frac{1}{\pi } \int _0^1 \frac{1}{t^{1/2} (1-t)^{1/2} (1-a^2 t)^{1/2}}\left( \frac{4(at)^{2n}n}{1-(at)^{4n}}-\frac{1+at}{2(1-at)}\right) \, d t \end{aligned} \end{aligned}$$

Setting $v=at$ and $N=2n$ and using Lemma 2, we get

$$\begin{aligned} \begin{aligned} \frac{4(at)^{2n}n}{1-(at)^{4n}}-\frac{1+at}{2(1-at)}&=\frac{4v^{N}\frac{N}{2}}{1-v^{2N}}-\frac{1+v}{2(1-v)}\\&= \frac{4v^{N}N (1-v)-(1+v)( 1-v^{2N}) }{(1-v^{2N})2(1-v)}< 0. \end{aligned} \end{aligned}$$

Therefore $\lim _{a\rightarrow 1^-} H_1<0$.

Applying Proposition 7 to $H_1$ with $\alpha =1/2$ and $u=0$; $n=1$, $\alpha =1/2$ and $u=0$; $\alpha =1/2$ and $u=-\pi /2n$; and $\alpha =1/2$ and $u=\pi /n$, respectively, we get

$$\begin{aligned} \begin{aligned} {\bar{H}}_1&= \frac{1}{\pi } \int _0^1 \frac{1}{t^{1/2} (1-t)^{1/2} (a^2-t)^{1/2}}\left( \frac{4(at)^{2n}n}{a^{4n}-t^{4n}}-\frac{a+t}{2(a-t)}\right) \, d t. \end{aligned} \end{aligned}$$

After doing the substitution $a\rightarrow 1/a$, we get that ${\bar{H}}_1$ can be written as $a\, H_1$. Thus, $\lim _{a\rightarrow 1^+} H_1<0$. In short, we have that $H < 0$ when $a\rightarrow 1$ and statement (c) is proved. $\square$

Proof of Proposition 5

We start proving statement (a) for $n=2$. Computing directly the quantities $K_1$, $L_2(a,\beta )$, $L_4(a,\beta )$, $N_4(a,\beta )$, $L_5(\beta )$, and $N_5(\beta )$ for a fixed $\beta$ and expanding F around $a=0$ we get

$$\begin{aligned} F(a,\beta )=-\frac{1}{16} \tan \beta \sec \beta \left( 18 \cos \beta +6 \cos (3 \beta )+17 \cot ^3\beta +7\right) +O(a^2). \end{aligned}$$

So ${\bar{F}}(\beta )<0$ for all $\beta \in (\pi /2n,\pi /n)$ and $n=2$.

Now we will show that for any $\beta \in (\pi /2n,\pi /n)$ and $n\ge 3$ we have ${\bar{F}}(\beta ) >0$. We fix $\beta$ and we expand F around $a=0$. First, expanding $N_4$ around $a=0$, we get

$$\begin{aligned} N_4(a,\beta )= - \frac{1}{a} \sum _{j=1}^n \sin \big (\frac{2\pi \,j}{n}+\beta \big ) - \frac{3}{2} \sum _{j=1}^n \sin \big (\frac{4\pi \,j}{n}+2\beta \big ) + O(a) \end{aligned}$$

and in view of (18) we have $N_4(a,\beta )=O(a)$. So expanding $N_4(a,\beta )(K_1 -2 a^3 L_2(a,\beta ) + L_5(\beta ))$ around $a=0$ we obtain

$$\begin{aligned} N_4(a,\beta )(K_1 -2 a^3 L_2(a,\beta ) + L_5(\beta )) =O(a). \end{aligned}$$

On the other hand, using Lemma 1(d) with $u=\beta$ and $n\ge 3$, the expansion of $L_4$ around $a=0$ is given by $L_4(a,\beta )=-n/2+O(a)$. Therefore, around $a=0$ we get

$$\begin{aligned} F(a,\beta )= (K_1 + \frac{n}{2}) N_5(\beta ) + O(a). \end{aligned}$$

(31)

The sign of $N_5(\beta )$ for any $\beta \in (0,\pi /n)$ was studied in Proposition 2, and we obtained that $N_5(\beta )$ is negative if $\beta \in (0,\pi /2n)$ and positive if $\beta \in (\pi /2n,\pi /n)$. Therefore, from (31) we have that for any $\beta \in (\pi /2n,\pi /n)$ and $n\ge 3$, ${\bar{F}}(\beta )>0$ which completes the proof of statement (a).

Now we consider the case in which $a \rightarrow \infty$. Fixed $\beta \in (\pi /2n,\pi /n)$ we have that $N_5(\beta )$ is positive. Moreover,

$$\begin{aligned} \lim _{a \rightarrow \infty } L_4(a,\beta )= \lim _{a \rightarrow \infty } N_4(a,\beta )=0. \end{aligned}$$

Making the change $b=1/a$ we get $a^3L_2(a,\beta )=L(b,\beta )$, where L is the function defined in Lemma 1(d). Thus, applying Lemma 1(d) with $u=\beta$, we get that $\lim _{a\rightarrow \infty } a^3 L_2(a,\beta )=\lim _{b\rightarrow 0}-n/2+O(b)$ when $n\ge 3$ and $\lim _{a\rightarrow \infty } a^3 L_2(a,\beta )=\lim _{b\rightarrow 0}2-6\cos ^2(\beta )+O(b)$ when $n=2$. So,

$$\begin{aligned} \begin{aligned}&\lim _{a \rightarrow \infty } (K_1 -2 a^3 L_2(a,\beta ) + L_5(\beta ))N_4(a,\beta )=0 , \\&\quad \lim _{a \rightarrow \infty } N_5(\beta )(K_1-L_4(a,\beta )) = N_5(\beta ) K_1 >0, \end{aligned} \end{aligned}$$

for all $n\ge 2$. In short, for any $\beta \in (\pi /2n,\pi )$ and $n\ge 2$, ${\tilde{F}}(\beta )>0$. This concludes the proof of the proposition. $\square$