Approximating $\ln 2$ by numbers in the field ${\mathbb{Q}}(\sqrt{2}\,)$

The object of this paper is to prove the following result.

Theorem 1.  Let $\mu{\kern1pt}{>}{\kern1pt}7.98101$, $p_1,p_2,p_3,p_4 {\kern1pt}{\in}{\kern1pt}\mathbb{Z}$, $(p_3,p_4){\kern1pt}{\ne}{\kern1pt}(0,0)$, $P{\kern1pt}{=}\max_{1 {\kern.7pt}{\leq}{\kern.7pt} i {\kern.7pt}{\leq}{\kern.7pt} 4}|p_i|$, and $P>P_0(\mu)$. Then the following inequality holds:

$$\begin{equation} \biggl|\ln 2-\frac{p_1\sqrt{2}+p_2}{p_3\sqrt{2}+p_4} \biggr|>P^{-\mu}. \end{equation} \tag{ 1 }$$

A similar bound for $\mu=12.4288$ was obtained in [1], and this result was improved in [2]; the corresponding value of $\mu$ was $9.307$.

The derivation of the new bound (1) is related to an application of the following integral construction. Let $h,j,k,l,m,q\in\mathbb{Z^+}$, $h+j+q=k+l+m$, $h+j-k\geq 0$, $k+l-j\geq0$ and $k+m-h\geq 0$. Let $x\in\mathbb{C}$, $\operatorname{Re}x> 0$ and $x\ne 1$. Consider the integral

$$\begin{equation} J=\frac{1}{2\pi i}\int_0^{-\infty}ds\int_{-i\infty}^{i\infty} \frac{s^{h}t^{j}\,dt}{\sqrt{{s}/{(s-1)}}(1-s)^{k+l-j+1} (s-t)^{h+j-k+1}(t-x)^{k+m-h+1}}. \end{equation} \tag{ 2 }$$

The result in Theorem 1 is obtained by taking

$$\begin{equation} x=\frac{3+2\sqrt{2}}{4\sqrt{2}}, \end{equation} \tag{ 3 }$$

$$\begin{equation} \begin{gathered} \, \begin{gathered} \, h=7n,\quad j=38n,\quad k=25n,\quad l=32n,\quad m=38n,\quad q=50n, \\ n\in\mathbb{N},\quad n\to\infty. \end{gathered}\end{gathered} \end{equation} \tag{ 4 }$$

The integral (2) differs from the integral introduced by Marcovecchio in [3], p. 148, formula (5), only by the factor $\sqrt{{s}/{(s-1)}}$ in the denominator of the integrand. The integral (2) was considered for the first time in [4]. To make the picture complete, we describe a brief scheme of some transformations of this integral (see [4], Russian pp. 484, 485). We denote the integrand of (2) by $G(s,t)$. Then

$$\begin{equation} J=-\int_0^{-\infty}\operatorname{res}_{t=x}G(s,t)\,ds. \end{equation} \tag{ 5 }$$

Evaluating the residue in (5) and making the change $s={z^2}/{(z^2-1)}$ in the integral with respect to the variable $s$, we obtain the relation

$$\begin{equation} \operatorname{res}_{t=x}G(s,t)\, ds\equiv R(z)\, dz,\qquad s=\frac{z^2}{z^2-1}, \end{equation} \tag{ 6 }$$

where

$$\begin{equation} \begin{gathered} \, R(z)=2(-1)^{j-k}\sum_{l_1=\max(0,q-l)}^{k+m-h}(-1)^{l_1}{\binom{j}{k+m-h-l_1}} \frac{x^{l-q+l_1}}{(x-1)^{h+j-k+l_1+1}} \\ \qquad\times{\binom{h+j-k+l_1}{l_1}}R_{l_1}(z),\end{gathered} \end{equation} \tag{ 7 }$$

$$\begin{equation} R_{l_1}(z)=\frac{z^{2h}(1-z^2)^{l+l_1}}{({x}/{(x-1)}-z^2)^{h+j-k+l_1+1}} =\frac{z^{2h}(1-z^2)^{l+l_1}}{((3+2\sqrt{2})^2-z^2)^{h+j-k+l_1+1}}. \end{equation} \tag{ 8 }$$

By choosing for $s\in[0,-\infty)$ a value $z\in[0,1)$, we see from (5) and (6) that

$$\begin{equation} J=-\int_0^1R(z)\, dz. \end{equation} \tag{ 9 }$$

We write

$$\begin{equation} \omega(l_1)={\binom{h+j-k+l_1}{l_1}}\int_0^1R_{l_1}(z)\, dz, \end{equation} \tag{ 10 }$$

let $K$ be the ring of numbers of the form $a+b\sqrt{2}$, where $a, b\in\mathbb{Z}$, and for positive integers $M\in\mathbb{N}$ write $q_M=\operatorname{\text{lcm}}(1,2,\dots,M)$ with $q_0= 1$.

Lemma 1.  Let $M_0=\max(2k+2l-2j,\, h+j-k,\, k+m-h)$, $m\geq q$. Then the following representation holds for all $l_1\leq k+m-h$:

$$\begin{equation} q_{M_0}\omega(l_1)=2^{-5(m-q)/2-2}\bigl(a(l_1)\ln 2+b(l_1)\bigr), \end{equation} \tag{ 11 }$$

where the $a(l_1), b(l_1)$ belong to $\mathbb{K}$.

Proof.  For $N\in\mathbb{Z^+}$ we write

$$\begin{equation*} D_N(f(z))=\frac{1}{N!} f^{(N)}(3+2\sqrt{2}). \end{equation*}$$

Since the integrand (8) of the integral in (10) is even, we have the following expansion into a sum of simplest fractions:

$$\begin{equation} \begin{aligned} \, R_{l_1}(z)&=\frac{(-1)^{q-m-1}z^{2h}(z^2-1)^{l+l_1}}{(z^2-(3+2\sqrt{2})^2)^{h+j-k+l_1+1}} \\ &=P(z)+\sum_{\nu=1}^{h+j-k+l_1+1}\biggl(\frac{(-1)^{\nu}k_{\nu}}{(z-3-2\sqrt{2})^{\nu}} +\frac{k_{\nu}} {(z+3+2\sqrt{2})^{\nu}}\biggr), \end{aligned} \end{equation} \tag{ 12 }$$

where $P(z)\in\mathbb{K}[z]$, $\deg{P(z)}=2k+2l-2j-2$, and, moreover,

$$\begin{equation*} (-1)^{\nu}k_{\nu}=D_{h+j-k+l_1+1-\nu}\bigl(R_{l_1}(z)(z-3-2\sqrt{2})^{h+j-k+l_1+1}\bigr). \end{equation*}$$

By Leibniz' formula, we see from (12) that

$$\begin{equation*} \begin{aligned} \, k_{\nu}&=(-1)^{q-m+\nu-1}D_{h+j-k+l_1+1 -\nu}\biggl(\frac{z^{2h}(z-1)^{l+l_1}(z+1)^{l+l_1}} {(z+3+2\sqrt{2})^{h+j-k+l_1+1}}\biggr) =(-1)^{q-m+\nu-1} \\ &\qquad\times\sum_{\overline{m}\in M_{\nu}}D_{m_1}(z^{2h}) D_{m_2}\bigl((z-1)^{l+l_1}\bigr) D_{m_3}\bigl((z+1)^{l+l_1}\bigr) \\ &\qquad\qquad\times D_{m_4}\bigl((z+3+2\sqrt{2})^{-(h+j-k+l_1+1)}\bigr), \end{aligned} \end{equation*}$$

where

$$\begin{equation*} \begin{aligned} \, M_{\nu}&\,{=}\,\bigl\{\overline{m}\,{=}\,(m_1, m_2, m_3, m_4)\,{\in}\,(\mathbb{Z^+})^4\bigm| m_1\,{+}\,m_2\,{+}\,m_3\,{+}\,m_4\, {=}\,h\,{+}\,j\,{-}\,k\,{+}\,l_1\,{+}\,1\,{-}\,\nu, \\ &\qquad m_1\leq 2h;\, m_2, m_3\leq l+l_1\bigr\}. \end{aligned} \end{equation*}$$

Therefore,

$$\begin{equation*} \begin{aligned} \, k_{\nu}&=(-1)^{q-m+\nu-1}\sum_{\overline{m}\in M_{\nu}}{\binom{2h}{m_1}} {\binom{l+l_1}{m_2}} {\binom {l+l_1}{m_3}} {\binom{h+j-k+l_1+m_4}{m_4}}(-1)^{m_4} \\ &\qquad\times(3+2\sqrt{2})^{2h-m_1}(2+2\sqrt{2})^{l+l_1-m_2}(4+2\sqrt{2})^{l+l_1-m_3} \\ &\qquad\times(2(3+2\sqrt{2}))^{-(h+j-k+l_1+1+m_4)} \\ &=\sum_{\overline{m}\in M_{\nu}}k_{\nu}(\overline{m})2^{l+l_1-m_2}(2\sqrt{2})^{l+l_1-m_3} 2^{-(h+j-k+l_1+1+m_4)}, \end{aligned} \end{equation*}$$

where the $k_{\nu}(\overline{m})$ belong to $\mathbb{K}$.

However, $m_2+1.5m_3+m_4\leq 1.5(h+j-k+l_1+1-\nu)$. Hence,

$$\begin{equation*} \begin{aligned} \, &l+l_1-m_2+1.5(l+l_1-m_3)-(h+j-k+l_1+1+m_4) \\ &\qquad\geq 2.5(l+l_1)-2.5(h+j-k+l_1)+1.5(\nu-1)-1 \\ &\qquad=-2.5(m-q)+1.5(\nu-1)-1, \end{aligned} \end{equation*}$$

that is,

$$\begin{equation} k_{\nu}=2^{-2.5(m-q)+1.5(\nu-1)-1}k_{\nu}',\qquad k_{\nu}'\in\mathbb{K},\quad \nu=1, \dots, h+j-k+l_1+1. \end{equation} \tag{ 13 }$$

By (12), we have

$$\begin{equation} \begin{aligned} \, \int_0^1R_{l_1}(z)\, dz=&\int_0^1P(z)\, dz \\ &\qquad+\sum_{\nu=2}^{h+j-k+l_1+1}\biggl(-\frac{k_{\nu}}{\nu-1}\biggl(\frac{1}{(4+2\sqrt{2})^{\nu-1}} -\frac{1}{(2+2\sqrt{2})^{\nu-1}}\biggr)\biggr) \\ &\qquad+k_1\ln{\biggl(\frac{3+2\sqrt{2}+z}{3+2\sqrt{2}-z}\biggr)}\bigg|^1_0. \end{aligned} \end{equation} \tag{ 14 }$$

Obviously,

$$\begin{equation*} \begin{gathered} \, \frac{1}{(4+2\sqrt{2})^{\nu-1}}=2^{-1.5(\nu-1)}(\sqrt{2}-1)^{\nu-1},\qquad \frac{1}{(2+2\sqrt{2})^{\nu-1}}=2^{-(\nu-1)}(\sqrt{2}-1)^{\nu-1}, \\ \ln\biggl(\frac{3+2\sqrt{2}+z}{3+2\sqrt{2}-z}\biggr)\bigg|^1_0 =\ln\biggl(\frac{4+2\sqrt{2}}{2+2\sqrt{2}}\biggr)=0.5\ln 2. \end{gathered} \end{equation*}$$

Further, it follows from the definition of $M_0$ that $q_{M_0}\int_0^1P(z)\, dz\equiv A_1\in\mathbb{K}$. It can also readily be seen that

$$\begin{equation*} q_{M_0}\binom{h+j-k+l_1}{l_1}\frac{1}{\nu-1}\equiv A_{\nu}\in\mathbb{N} \end{equation*}$$

for all $\nu=2,\dots,h+j-k+l_1+1$. Then it follows from (13) and (14) that

$$\begin{equation*} \begin{aligned} \, q_{M_0}\omega(l_1)&=q_{M_0}{\binom{h+j-k+l_1}{l_1}}\int_0^1R_{l_1}(z)\, dz \\ &= {\binom {h+j-k+l_1}{l_1}} A_1+2^{-2.5(m-q)-1} \\ &\qquad\times\sum_{\nu=2}^{h+j-k+l_1+1}\bigl(A_{\nu}k_{\nu}'((2-\sqrt{2})^{\nu-1} -(\sqrt{2}-1)^{\nu-1})\bigr) \\ &\qquad+2^{-2.5(m-q)-2}k_{1}'\ln{2}, \end{aligned} \end{equation*}$$

whence, since $m\geq q$, (11) follows. $\square$

Corollary 1.  The integral (2) admits the following representation for $m\geq q$:

$$\begin{equation} 2^{-2.5q}q_{M_0}J=a\ln 2+b,\qquad a, b\in\mathbb{K}. \end{equation} \tag{ 15 }$$

Proof.  We have

$$\begin{equation*} \begin{aligned} \, \frac{x^{l-q+l_1}}{(x-1)^{h+j-k+l_1+1}}&=\frac{(3+2\sqrt{2})^{l-q+l_1}}{(4\sqrt{2})^{l-q+l_1}} (3+2\sqrt{2})^{h+j-k+l_1+1}(4\sqrt{2})^{h+j-k+l_1+1} \\ &=2^{2.5(m+1)}C(l_1), \end{aligned} \end{equation*}$$

where $C(l_1)\in\mathbb{K}$. Applying Lemma 1, we see from (7) and (9) that

$$\begin{equation*} \begin{aligned} \, 2^{-2.5q}q_{M_0}J&\,{=}\,2(-1)^{j-k}\sum _{l_1=\max(0, l-q)}^{k+m-h}(-1)^{l_1} {\binom{j}{k\,{+}\,m\,{-}\,h\,{-}\,l_1}} 2^{2.5(m+1)} C(l_1)\,2^{-5q/2} \\ &\qquad\times2^{-2.5(m-q)-2}(a(l_1)\ln 2+b(l_1))=a\ln 2 +b, \end{aligned} \end{equation*}$$

where $a,b\in\mathbb{K}$, and this proves Corollary 1. $\square$

Along with the family of parameters (4), we shall use a more general situation in which

$$\begin{equation} (h, j, k, l, m, q)=n(h', j', k', l', m', q'), { } \end{equation} \tag{ 16 }$$

where $h', j', k', l', m', q'\in\mathbb{Z^+}$. It is convenient to denote the integral (2) for parameters of the form (16) and for $x$ of the form (3) as follows:

$$\begin{equation} J\equiv J_n\equiv J_n(h', j', k', l', m', q'). \end{equation} \tag{ 17 }$$

For the family of parameters (16) we write

$$\begin{equation} Mn=\max\{2(k+l-j),\, 2h,\, 2k,\, h+j-k,\, k+m-h,\, l,\, m,\, j,\, q\}. \end{equation} \tag{ 18 }$$

Let $p$ be a prime, $p>\sqrt{Mn}$, and $\omega=\{{n}/{p}\}$ the fractional part of the number ${n}/{p}$. Consider the inequalities

$$\begin{equation} \begin{aligned} \, &[2k'\omega]+[(l'+k'-j')\omega]+[m'\omega]+[l'\omega]-[k'\omega]-[2(l'+k'-j')\omega] \\ &\qquad-[(h'+j'-k')\omega]-[(k'+m'-h')\omega]>0, \\ &[2h'\omega]+[(l'+k'-j')\omega]+[j'\omega]+[q'\omega]-[h'\omega]-[2(l'+k'-j')\omega] \\ &\qquad-[(h'+j'-k')\omega]-[(k'+m'-h')\omega]>0, \\ &[j'\omega]+[m'\omega]-[(h'+j'-k')\omega]-[(k'+m'-h')\omega]>0, \\ &[2k'\omega]+[(l'+k'-j')\omega]+[q'\omega]-[k'\omega]-[2(l'+k'-j')\omega] \\ &\qquad-[(k'+m'-h')\omega]>0, \\ &[2h'\omega]+[(l'+k'-j')\omega]+[l'\omega]-[h'\omega]-[2(l'+k'-j')\omega] \\ &\qquad-[(h'+j'-k')\omega]>0. \end{aligned} \end{equation} \tag{ 19 }$$

We denote by $\Delta_n$ the product of all primes $\smash{p>\sqrt{Mn}}$ such that $\omega=\{{n}/{p}\}$ satisfies at least one of the inequalities (19). The following lemma sharpens the result obtained in Corollary 1.

Lemma 2.  When $m\geq q$ the integral (17) admits the representation

$$\begin{equation} \frac{q_{Mn}}{\Delta_n}\, 2^{-2.5q}J_n= A_n\ln 2+B_n, \end{equation} \tag{ 20 }$$

where $A_n, B_n\in\mathbb{K}$, $n\in\mathbb{N}$.

Proof.  The representation (20) follows from (15) by the standard procedure of refining the denominator (see, for example, Lemma 3 in [6]). The inequalities (19) were obtained for the integral (2) for the first time in [4], Russian p. 491, inequalities (11). The inequalities (19) differ somewhat from those considered, for the same purpose, by Marcovecchio in [3], inequalities (31). $\square$

The following lemma contains the final version of a linear form of the form (20) which we shall use to prove Theorem 1.

Lemma 3.  The following representation holds ( see (17)):

$$\begin{equation} \begin{aligned} \, L_n&\equiv (3-2\sqrt{2})^{140n} 2^{-125n}\frac{q_{56n}}{\Delta_n}J_n(7, 38, 25, 32, 38, 50) \\ &\equiv(\Lambda_1(n)\sqrt{2}+\Lambda_2(n))\ln 2+(\Lambda_3(n)\sqrt{2}+\Lambda_4(n)), \end{aligned} \end{equation} \tag{ 21 }$$

where the $\Lambda_i(n)$ belong to $\mathbb{Z}$ and $\Delta_n$ is defined by the inequalities (19) for the family of parameters (4).

Proof.  It was proved in [4], equation (10), that the equation $J_n(h', j', k', l', m', q')=x^{l-q}J_n(k', m', h', q', j', l')$ holds (see (17)), that is,

$$\begin{equation*} J_n(7, 38, 25, 32, 38, 50)=\frac{(3+2\sqrt{2})^{-18n}}{(4\sqrt{2})^{-18n}}J_n(25, 38, 7, 50, 38 ,32). \end{equation*}$$

For the family of parameters (4) we have from (18) that

$$\begin{equation*} Mn=n \max\{38, 14, 50, 20, 56, 32, 38, 38, 50\}=56n. \end{equation*}$$

However, then

$$\begin{equation*} L_n=(3-2\sqrt{2})^{158n} 2^{-80n} \frac{q_{56n}}{\Delta_n} J_n(25, 38, 7, 50, 38, 32). \end{equation*}$$

By Lemma 2,

$$\begin{equation*} 2^{-80n}\frac{q_{56n}}{\Delta_n}J_n(25, 38, 7, 50, 38, 32)=A_n\ln 2+B_n, \end{equation*}$$

where $A_n, B_n\in\mathbb{K}$, which implies (21). $\square$

We conclude §1 by proving the following important lemma.

Lemma 4.  Let $n,d\in\mathbb{N}$, $\theta\in\mathbb{R}$, $\sqrt{d}\notin\mathbb{N}$, and

$$\begin{equation*} L_n=(\Lambda_1(n)\sqrt{d}+\Lambda_2(n))\theta+\Lambda_3(n)\sqrt{d}+\Lambda_4(n), \end{equation*}$$

where the $\Lambda_i(n)$ belong to $\mathbb{Z}$, and let $\Lambda(n)=\max_{1 \leq i \leq 4}|\Lambda_i(n)|$. Let

$$\begin{equation*} \lim_{n\to\infty}\frac{1}{n}\ln{\bigl|\Lambda_1(n)\sqrt{d}+\Lambda_2(n)\bigr|}=\gamma_1, \qquad\lim_{n\to\infty}\sup\frac{1}{n}\ln{|\Lambda(n)|}\leq \gamma_2, \end{equation*}$$

and for some constant $\gamma_3>\gamma_2$ and every $\varepsilon_1, \varepsilon_2> 0$ let there be an $N=N(\varepsilon_1, \varepsilon_2)$ such that the following inequalities hold for every $n\geq N$ and at least one of the values $m\in\{n, n+1\}$:

$$\begin{equation} e^{-(\gamma_3+\varepsilon_1)m}\leq |L_m|\leq e^{-(\gamma_3-\varepsilon_2)m}. \end{equation} \tag{ 22 }$$

Further, let $\gamma_1+\gamma_2\,{>}\,0$, $\mu\,{>}\,2(\gamma_1+\gamma_3)/(\gamma_3-\gamma_2)$, $p_1,p_2,p_3,p_4\,{\in}\,\mathbb{Z}$, $(p_3, p_4)\,{\ne}\,(0, 0)$, $P=\max_{1 \leq i \leq4}|p_i|$ and $P>P_0(\mu)$. Then

$$\begin{equation} \biggl|\theta-\frac{p_1\sqrt{d}+p_2}{p_3\sqrt{d}+p_4}\biggr|>P^{-\mu}. \end{equation} \tag{ 23 }$$

Proof.  We fix a $\tau\,{\in}\,(0,(\gamma_3-\gamma_2)/2)$ such that $2(\gamma_1+\gamma_3)/(\gamma_3-\gamma_2-2\tau)\,{<}\,\mu$ and write

$$\begin{equation} \varepsilon=\biggl(\frac{\gamma_3-\gamma_2}{2}-\tau \biggr) \biggl(\mu-\frac{2(\gamma_1+\gamma_3)}{\gamma_3-\gamma_2-2\tau}\biggr)>0. \end{equation} \tag{ 24 }$$

Let us choose $\varepsilon_1={\varepsilon}/{3}$, $\varepsilon_2={\tau}/{2}$, $N=N({\varepsilon}/{3},{\tau}/{2})$ in the inequalities (22). We introduce an $n_0\in\mathbb{N}$, where $n_0\geq N$ and the following family of conditions holds: for all $n\in\mathbb{N}$, $n>n_0$,

$$\begin{equation} \Lambda_1(n)\sqrt{d}+\Lambda_2(n)\ne 0, \end{equation} \tag{ 25 }$$

$$\begin{equation} 2 (2d+2)(\sqrt{d}+1)^2<e^{\tau n/3}, \end{equation} \tag{ 26 }$$

$$\begin{equation} \Lambda(n)<e^{(\gamma_2+{\tau}/{3})n}, \end{equation} \tag{ 27 }$$

$$\begin{equation} |\Lambda_1(n)\sqrt{d}+\Lambda_2(n)|<e^{(\gamma_1+\min(\tau,\varepsilon)/3)n}, \end{equation} \tag{ 28 }$$

$$\begin{equation} \gamma_3-\gamma_2-2\tau<\frac{1}{3}\,\tau n, \end{equation} \tag{ 29 }$$

$$\begin{equation} \gamma_1+\gamma_3+\frac{2\varepsilon}{3}<\frac{1}{3}\,\varepsilon n, \end{equation} \tag{ 30 }$$

$$\begin{equation} \gamma_1+\gamma_2+\tau<\tau n. \end{equation} \tag{ 31 }$$

Let $P_0(\mu)=\exp\bigl(({(\gamma_3-\gamma_2)}/{2}-\tau)n_0\bigr)$ and $P>P_0(\mu)$. We define $n\in\mathbb{N}$, $n> n_0$, by the inequalities

$$\begin{equation} \exp{\biggl(\biggl(\frac{\gamma_3-\gamma_2}{2}-\tau\biggr)n\biggr)}\leq P<\exp{\biggl(\biggl(\frac{\gamma_3-\gamma_2}{2}-\tau\biggr)(n+1)\biggr)}. \end{equation} \tag{ 32 }$$

Then the inequalities (22) hold for at least one of the values $m\in\{n, n+1\}$. We write (see (25))

$$\begin{equation*} \omega_1=\frac{p_1\sqrt{d}+p_2}{p_3\sqrt{d}+p_4},\qquad\! \omega_2=-\frac{\Lambda_3(m)\sqrt{d}+\Lambda_4(m)}{\Lambda_1(m)\sqrt{d}+\Lambda_2(m)},\qquad\! \beta_1=\theta-\omega_1,\qquad\!\beta_2=\theta-\omega_2. \end{equation*}$$

Let us first consider the case in which the inequalities (22) hold for $m=n$. Two situations are possible: $\omega_1=\omega_2$, $\omega_1\ne\omega_2$.

1) Let $\omega_1=\omega_2$. We have

$$\begin{equation*} |\beta_1|=|\beta_2|=\frac{|L_n|}{|\Lambda_1(n)\sqrt{d}+\Lambda_2(n)|}. \end{equation*}$$

We have $|L_n|>e^{-(\gamma_3+{\varepsilon}/{3})n}$ by (22), and $|\Lambda_1(n)\sqrt{d}+\Lambda_2(n)|<e^{(\gamma_1+{\varepsilon}/{3})n}$ by (28). Hence, by (32), the condition $\gamma_1+\gamma_3>\gamma_1+\gamma_2> 0$ and the inequality (24) yield that

$$\begin{equation*} |\beta_1|>\frac{1}{e^{(\gamma_1+\gamma_3+\varepsilon)n}}\geq P^{-\frac{\gamma_1+\gamma_3+\varepsilon}{(\gamma_3-\gamma_2)/2-\tau}}=P^{-\mu}, \end{equation*}$$

which coincides with the inequality (23).

2) Let $\omega_1\ne \omega_2$. We consider

$$\begin{equation*} O\ne \Omega\equiv |\omega_1-\omega_2|= \frac{|C\sqrt{d}+D|}{|p_3\sqrt{d}+p_4|\, |\Lambda_1(n)\sqrt{d}+\Lambda_2(n)|}, \end{equation*}$$

where

$$\begin{equation*} \begin{alignedat}{2} C&=\Lambda_1(n)p_2+\Lambda_2(n)p_1+\Lambda_3(n)p_4+\Lambda_4(n)p_3, &\qquad |C|&\leq 4P \Lambda(n), \\ D&=\Lambda_1(n)p_1d+\Lambda_2(n)p_2+\Lambda_3(n)p_3d+\Lambda_4(n)p_4,&\qquad |D|&\leq (2d+2)P\Lambda(n). \end{alignedat} \end{equation*}$$

Obviously,

$$\begin{equation*} \begin{gathered} \, O\ne |C\sqrt{d}+D|=\frac{|C^{2}d-D^2|}{|C\sqrt{d}-D|}\geq \frac{1}{(\sqrt{d}+1)\max(|C|, |D|)}, \\ |p_3\sqrt{d}+p_4|\leq P(\sqrt{d}+1). \end{gathered} \end{equation*}$$

Therefore,

$$\begin{equation} \Omega\geq \frac{1}{(2d+2)(\sqrt{d}+1)^{2}P^{2}\Lambda(n)|\Lambda_1(n)\sqrt{d}+\Lambda_2(n)|}. \end{equation} \tag{ 33 }$$

We claim that

$$\begin{equation*} |\beta_2|=\frac{|L_n|}{|\Lambda_1(n)\sqrt{d}+\Lambda_2(n)|}<\frac{\Omega}2. { } \end{equation*}$$

By the inequality (33), it suffices to prove that

$$\begin{equation} |L_n|<\frac{1}{2(2d+2)(\sqrt{d}+1)^{2}P^{2}\Lambda(n)}. \end{equation} \tag{ 34 }$$

Applying the inequalities (22), (26), (32), (27) and (29), we obtain that

$$\begin{equation*} \begin{aligned} \, |L_n|\cdot 2(2d+2)(\sqrt{d}+1)^{2}P^{2}\Lambda(n)&<e^{(-\gamma_3+{\tau}/{2})n} e^{\tau n/3} e^{(\gamma_3-\gamma_2-2\tau)(n+1)}e^{(\gamma_2+{\tau}/{3})n} \\ &=e^{\gamma_3-\gamma_2-2\tau}e^{-5\tau n/6}<e^{-\tau n/2}<1. \end{aligned} \end{equation*}$$

This proves the inequality (34), and therefore $|\beta_2|<\Omega/2$. However, $\Omega=|\beta_1-\beta_2|\leq |\beta_1|+|\beta_2|$. Hence, $|\beta_1|\geq \Omega-|\beta_2|>\Omega/2$. Using the inequalities (33), (26), (27), (28) and (32) in succession and taking into account that $\gamma_1+\gamma_2> 0$, we have

$$\begin{equation*} \begin{aligned} \, |\beta_1|&>\frac{1}{2(2d+2)(\sqrt{d}+1)^{2}P^{2}\Lambda(n)|\Lambda_1(n)\sqrt{d}+\Lambda_2(n)|} \\ &>P^{-2}e^{-\tau n/3}e^{-(\gamma_2+{\tau}/{3})n}e^{-(\gamma_1+{\tau}/{3})n} \\ &>e^{-(\gamma_1+\gamma_2+2\tau)n}P^{-2}>P^{-{(\gamma_1+\gamma_2+2\tau)}/ {((\gamma_3-\gamma_2)/2-\tau)}-2}>P^{-\mu}, \end{aligned} \end{equation*}$$

which coincides with the inequality (23).

Let us now consider the case in which the inequalities (22) hold for $m=n+ 1$. Again, two situations are possible: $\omega_1=\omega_2$, $\omega_1\ne\omega_2$.

3) Let $\omega_1=\omega_2$. As in part 1), we have

$$\begin{equation*} |\beta_1|=\frac{|L_{n+1}|}{|\Lambda_1(n+1)\sqrt{d}+\Lambda_2(n+1)|}. \end{equation*}$$

Applying the inequalities (22), (28) and (30), we obtain that

$$\begin{equation*} |\beta_1|>e^{-(\gamma_1+\gamma_3+{2\varepsilon}/{3})(n+1)}>e^{-(\gamma_1+\gamma_3+\varepsilon)n} >P^{-\mu}. \end{equation*}$$

4) Let $\omega_1\ne \omega_2$. The arguments here are similar to that in part 2), where $n\to n+1$, in particular, in the inequalities (33) and (34). Applying the inequalities (22), (26), (32) and (27), we obtain that

$$\begin{equation*} \begin{aligned} \, &|L_{n+1}|2(2d+2)(\sqrt{d}+1)^{2}P^{2}\Lambda(n+1) \\ &\qquad<e^{(-\gamma_3+{\tau}/{2})(n+1)}e^{\tau n/3}e^{(\gamma_3-\gamma_2-2\tau)(n+1)} e^{(\gamma_2+{\tau}/{3})(n+1)} =e^{\tau n/3}e^{-7\tau(n+1)/6}<1, \end{aligned} \end{equation*}$$

which proves the inequality (34) with $n\to n+1$, and simultaneously the inequality $|\beta_1|>\Omega/2$. Then, applying the inequalities (26), (27), (28), (31) and (32) in succession, we obtain

$$\begin{equation*} \begin{aligned} \, |\beta_1|&>\frac{1}{2(2d+2)(\sqrt{d}+1)^{2}P^{2}\Lambda(n+1)|\Lambda_1(n+1)\sqrt{d} +\Lambda_2(n+1)|} \\ &>P^{-2}e^{-\tau n/3}e^{-(\gamma_2+{\tau}/{3})(n+1)} e^{-(\gamma_1+{\tau}/{3})(n+1)}>P^{-2}e^{-(\gamma_1+\gamma_2+\tau)(n+1)} \\ &>P^{-2}e^{-(\gamma_1+\gamma_2+2\tau)n}>P^{-\mu}, \end{aligned} \end{equation*}$$

which coincides with the inequality (23).

This completes the proof of the lemma. $\square$

Remark 1.  Assertions similar to Lemma 4 were used in [1], [2] and [5].

To prove Theorem 1, we shall apply Lemma 4 to the linear form (21). We need to evaluate the constants $\gamma_1$, $\gamma_2$ and $\gamma_3$. In this section, we evaluate $\gamma_1$ and $\gamma_3$ and, in the next section, the constant $\gamma_2$. To evaluate both the constants $\gamma_1$ and $\gamma_3$, we apply the saddle-point method. We have (see (17) and (21))

$$\begin{equation*} J_n\equiv J_n(7, 38, 25, 32, 38, 50)\equiv \frac{1}{2\pi i}\int_0^{-\infty} ds\int_{-i\infty}^{i\infty}G(s,t)\, dt, \end{equation*}$$

where

$$\begin{equation} \begin{aligned} \, G(s,t)&=\varphi(s, t)(f(s, t))^n, \\ f(s,t)&=\frac{s^7t^{38}}{(1-s)^{19}(s-t)^{20}(t-x)^{56}}, \\ \varphi(s, t)&=\frac{1}{\sqrt{{s}/{(s-1)}}(1-s)(s-t)(t-x)}. \end{aligned} \end{equation} \tag{ 35 }$$

The saddle points are the solutions of the system $f_s'(s, t)= 0$, $f_t'(s, t)= 0$ that differ from the zeros of the function $f(s, t)$. In [4], Russian p. 492, equations (12), this system was solved in the general case for the integral (17). For the function $f(s, t)$ written above we have three saddle points:

$$\begin{equation} (s_1, t_1)\approx(1.006013, 1.012356), \end{equation} \tag{ 36 }$$

$$\begin{equation} (s_2, t_2)\approx(-0.013064+0.334437i, -0.665181-0.286087i), \end{equation} \tag{ 37 }$$

$(s_3, t_3)=(\overline{s}_2, \overline{t}_2)$, the complex conjugate of $(s_2, t_2)$. We write $\xi=(s, t)\in\mathbb{C}^2$.

Lemma 5.  Let $\xi^0$ be a non-degenerate saddle point of the function $S(\xi)$, let $\gamma$ be a two-dimensional smooth complex manifold with boundary, let $\xi^0$ be an interior point of $\gamma$, let the functions $\varphi(\xi)$ and $S(\xi)$ be holomorphic at the point $\xi^0$, and let also $\max_{\xi\in\gamma}\operatorname{Re} S(\xi)$ be attained only at the point $\xi^0$, let

$$\begin{equation*} \begin{gathered} \, F(\lambda)=\int_{\gamma}\varphi(\xi) \exp(\lambda S(\xi))\,d\xi, \\ \textit{let\/}\;S_{\xi\xi}''(\xi^0)\,{=}\,\biggl(\!{\binom{S_{ss}''(\xi^0)}{S_{st}''(\xi^0)}} {\binom {S_{st}''(\xi^0)}{S_{tt}''(\xi^0)}}\!\biggr)\textit{ be the Hesse matrix, and let} \; \det{S_{\xi\xi}''(\xi^0)}\ne0. \end{gathered} \end{equation*}$$

Then, as $\lambda\to+\infty$,

$$\begin{equation} F(\lambda)=\frac{2\pi}{\lambda}\exp\bigl(\lambda S(\xi^0)\bigr) \bigl(\det{S_{\xi\xi}''(\xi^0)}\bigr)^{-{1}/{2}} \bigl(\varphi(\xi^0)+O(\lambda^{-1})\bigr). \end{equation} \tag{ 38 }$$

Proof.  This assertion is proved in the Fedoryuk's monograph [7], p. 259, Proposition 1.1. $\square$

Lemma 6.  For the linear form (21) we have the equation

$$\begin{equation} \begin{aligned} \, \gamma_1&\equiv \lim_{n\to\infty}\biggl(\frac{1}{n}\ln|\Lambda_1(n)\sqrt{2}+\Lambda_2(n)|\biggr) \\ &=140\ln(3-2\sqrt{2}) -125\ln{2}+56-\Delta+\ln{|f(s_1, t_1)|}, \\ \gamma_1&\approx 112.931654,\quad\textit{where}\quad \Delta=\lim_{n\to\infty}\frac{1}{n}\ln{\Delta_n}\approx 33.573984. \end{aligned} \end{equation} \tag{ 39 }$$

Proof.  Consider the function

$$\begin{equation*} g(\delta,\tau)\equiv f\biggl(\frac{1}{\delta}, \frac{1}{\tau}\biggr) =-\frac{x^{-56}\delta^{32}\tau^{38}}{(1-\delta)^{19}(\delta-\tau)^{20}(\tau-{1}/{x})^{56}}. \end{equation*}$$

Let $L_1$ be the circle $|\tau-{1}/{x}|={1}/{t_1}-{1}/{x}$ and $L_2$ the circle $|\delta-{1}/{x}|={1}/{s_1}-{1}/{x}$. We note that, by (3) and (36), we have ${1}/{x}<{1}/{t_1}<{1}/{s_1}< 1$. Obviously, $\max_{(\delta, \tau)\in L_2\times L_1}\ln{|g(\delta, \tau)|}$ is attained only at the point $({1}/{s_1},{1}/{\tau_1})$. We denote by $L_1^*$ the image of the circle $L_1$ under the map $t={1}/{\tau}$ and by $L_2^*$ the image of the circle $L_2$ under the map $s={1}/{\delta}$. Then it follows from the definition of the function $g(\delta, \tau)$ that $\max_{(s, t)\in L_2^*\times L_1^*}\ln{|f(s, t)|}$ is attained only at the point $(s_1, t_1)$. For the integral $J_n$, by (35), we have

$$\begin{equation} J_n=A_n\ln{2}+B_n, \end{equation} \tag{ 40 }$$

where $A_n,B_n \in\mathbb{Q}\cdot \sqrt{2}\oplus\mathbb{Q}$ (see Corollary 1).

We claim that

$$\begin{equation} A_n=\frac{1}{2(2\pi i)^2}\int_{L_2^*} ds\int_{L_1^*}G(s, t)\, dt, \end{equation} \tag{ 41 }$$

where the circles $L_1^*$ and $L_2^*$ are traversed in the positive direction. Equations of the form (41) are standard and occur in many papers, for example, in [3] and [8].

In the proof of Lemma 1, for the integral in (10) we have, by (12) and (14)

$$\begin{equation*} \int_0^1R_{l_1}(z)\, dz=A(l_1)\ln{2}+B(l_1), { } \end{equation*}$$

where $A(l_1),B(l_1)\in\mathbb{Q}\cdot \sqrt{2}\oplus\mathbb{Q}$, $A(l_1)=k_1/2=-(\operatorname{res}_{z=z_0}R_{l_1}(z))/2$ and $z_0=3+2\sqrt{2}$, and then it follows from (7), (9) and (40) that $A_n=(\operatorname{res}_{z=z_0}R(z))/2$. Applying (6), we obtain

$$\begin{equation*} A_n=\frac{1}{2}\cdot\frac{1}{2\pi i}\int_{l}R(z)\, dz=\frac{1}{2}\cdot\frac{1}{2\pi i} \int_{L_2^*}\operatorname{res}_{t=x}G(s, t)\, ds=\frac{1}{2}\cdot\frac{1}{(2\pi i)^2}\int_{L_2^*}ds\int_{L_1^*}G(s, t)\, dt, \end{equation*}$$

where $l$ stands for the contour going around the point $z_0$ in the positive direction and mapping onto the circle $L_2^*$ under the map $s={z^2}/{(z^2-1)}$. Thus, the formula (41) is proved.

Let $\gamma_1$ be a small arc of the circle $L_1^*$ with centre at the point $t_1$, let $\gamma_2$ be a small arc of the circle $L_2^*$ with centre at the point $s_1$, let $\gamma=\gamma_2\times\gamma_1$, let $\Gamma=(L_2^*\times L_1^*)\setminus\gamma$; let $\xi^0=(s_1, t_1)$, and let $\max_{\xi\in\Gamma}|f(\xi)|=F<|f(\xi^0)|$. We can define some branch $\ln{f(\xi)}=\ln{|f(\xi)|}+ih(\xi)$ on $\gamma$ holomorphic at the point $\xi^0$.

It follows from (35) and (36) that $f(\xi^0)=-|f(\xi^0)|$, and therefore one can choose $h(\xi^0)=\pi$. Further, the function $\varphi(\xi)$ is holomorphic at the point $\xi^0$ since $\operatorname{Re}s> 1$, $s\in\gamma_2$. Obviously, $\varphi(\xi^0)\ne 0$. By (41) we have

$$\begin{equation} A_n=\frac{1}{2(2\pi i)^2}\biggl(\int_{\gamma}\varphi(\xi)\exp\bigl(n(\ln{|f(\xi)|} +ih(\xi))\bigr)\,d\xi+\int_{\Gamma}\varphi(\xi)(f(\xi))^{n}\,d\xi\biggr). \end{equation} \tag{ 42 }$$

We apply Lemma 5 to the first integral in (42) with $\lambda=n$ and $S(\xi)=\ln{f(\xi)}$. The conditions of Lemma 5 are satisfied, since the remaining non-degeneracy condition for the saddle point $\xi^0$ can readily be verified: $\det{S_{\xi\xi}''(\xi^0)}\approx4.38471\times10^{11}\ne 0$. Then by (38) we obtain

$$\begin{equation*} \begin{aligned} \, &\int_{\gamma}\varphi(\xi)\exp(n(\ln{|f(\xi)|}+ih(\xi)))\,d\xi \\ &\qquad=\frac{2\pi}{n}(-1)^{n}\exp\bigl(n\ln{|f(\xi^0)|}\bigr) \bigl(\det{S_{\xi\xi}''(\xi^0)}\bigr)^{-{1}/{2}}\biggl(\varphi(\xi^0) +O\biggl(\frac{1}{n}\biggr)\biggr). \end{aligned} \end{equation*}$$

We estimate the second integral in (42) trivially:

$$\begin{equation*} \biggl|\int_{\Gamma}\bigl(\varphi(\xi)(f(\xi))^{n}\,d\xi\bigr)\biggr|<CF^n \end{equation*}$$

for some positive constant $C$. Then $\lim_{n\to\infty}({1}/{n})\ln{|A_n|}=\ln{|f(s_1, t_1)|}$.

To complete the proof of the lemma, it remains to evaluate the limit

$$\begin{equation*} \lim_{n\to\infty}\frac1{n}\ln\biggl(\frac{q_{56n}}{\Delta_n}\biggr). \end{equation*}$$

We note that $\lim_{n\to\infty}({1}/{n})\ln{(q_{56n})}=56$ and evaluate $\Delta=\lim_{n\to\infty}({1}/{n})\ln{\Delta_n}$. Let us write out the inequalities (19) for the family of parameters (4):

$$\begin{equation} \begin{aligned} \, &[50\omega]+[19\omega]+[38\omega]+[32\omega]-[25\omega]-[38\omega]-[20\omega]-[56\omega]>0, \\ &[14\omega]+[19\omega]+[38\omega]+[50\omega]-[7\omega]-[38\omega]-[20\omega]-[56\omega]>0, \\ &[38\omega]+[38\omega]-[20\omega]-[56\omega]>0, \\ &[50\omega]+[19\omega]+[50\omega]-[25\omega]-[38\omega]-[56\omega]>0, \\ &[14\omega]+[19\omega]+[32\omega]-[7\omega]-[38\omega]-[20\omega]>0. \end{aligned} \end{equation} \tag{ 43 }$$

The set $E$ of numbers $\omega$ satisfying at least one of the inequalities (43) has the form

$$\begin{equation*} \begin{aligned} \, E=&\biggl[\frac{1}{50},\frac{1}{28}\biggr)\cup\biggl[\frac{1}{19},\frac{3}{56}\biggr)\cup \biggl[\frac{3}{50},\frac{5}{56}\biggr)\cup\biggl[\frac{3}{32},\frac{3}{28}\biggr)\cup \biggl[\frac{3}{25},\frac{1}{7}\biggr)\cup\biggl[\frac{3}{19},\frac{9}{56}\biggr) \\ &\qquad\cup \biggl[\frac{9}{50},\frac{11}{56}\biggr)\cup\biggl[\frac{4}{19},\frac{3}{14}\biggr)\cup \biggl[\frac{7}{32},\frac{1}{4}\biggr)\cup\biggl[\frac{13}{50},\frac{15}{56}\biggr)\cup \biggl[\frac{7}{25},\frac{17}{56}\biggr)\cup\biggl[\frac{6}{19},\frac{9}{28}\biggr) \\ &\qquad\cup \biggl[\frac{17}{50},\frac{7}{20}\biggr)\cup\biggl[\frac{7}{19},\frac{2}{5}\biggr)\cup \biggl[\frac{21}{50},\frac{3}{7}\biggr)\cup\biggl[\frac{7}{16},\frac{9}{20}\biggr)\cup \biggl[\frac{23}{50},\frac{13}{28}\biggr)\cup\biggl[\frac{9}{19},\frac{27}{56}\biggr) \\ &\qquad\cup \biggl[\frac{10}{19},\frac{31}{56}\biggr)\cup\biggl[\frac{11}{19},\frac{33}{56}\biggr)\cup \biggl[\frac{19}{32},\frac{17}{28}\biggr)\cup\biggl[\frac{31}{50},\frac{5}{8}\biggr)\cup \biggl[\frac{12}{19},\frac{13}{20}\biggr)\cup\biggl[\frac{21}{32},\frac{37}{56}\biggr) \\ &\qquad\cup \biggl[\frac{13}{19},\frac{5}{7}\biggr)\cup\biggl[\frac{14}{19},\frac{3}{4}\biggr)\cup \biggl[\frac{19}{25},\frac{43}{56}\biggr)\cup\biggl[\frac{39}{50},\frac{45}{56}\biggr)\cup \biggl[\frac{13}{16},\frac{23}{28}\biggr)\cup\biggl[\frac{16}{19},\frac{17}{20}\biggr) \\ &\qquad\cup \biggl[\frac{43}{50},\frac{7}{8}\biggr)\cup\biggl[\frac{17}{19},\frac{51}{56}\biggr)\cup \biggl[\frac{23}{25},\frac{13}{14}\biggr)\cup\biggl[\frac{15}{16},\frac{19}{20}\biggr)\cup \biggl[\frac{24}{25},\frac{27}{28}\biggr)\cup\biggl[\frac{31}{32},\frac{55}{56}\biggr). \end{aligned} \end{equation*}$$

Let $\psi(x)={\Gamma'(x)}/{\Gamma(x)}$, where $\Gamma(x)$ stands for the gamma function. Then, in the standard way (see Lemma 6 in [9]), we obtain

$$\begin{equation} \begin{aligned} \, \Delta=&\lim_{n\to\infty}\biggl(\frac{1}{n}\ln{\Delta_n}\biggr) =\biggl(\psi\biggl(\frac{1}{28} \biggr)-\psi\biggl(\frac{1}{50}\biggr)\biggr) +\biggl(\psi\biggl(\frac{3}{56}\biggr) -\psi\biggl(\frac{1}{19}\biggr)\biggr) \\ &\qquad+\dots+\biggl(\psi\biggl(\frac{55}{56}\biggr)-\psi\biggl(\frac{31}{32}\biggr)\biggr) \approx 33.573984. \end{aligned} \end{equation} \tag{ 44 }$$

This completes the proof of the lemma. $\square$

Lemma 7.  For the linear form (21) let

$$\begin{equation} \gamma_3=140\ln{(3+2\sqrt{2})}+125\ln{2}-56+\Delta-\ln{|f(s_2,t_2)|}\approx 360.410552, \end{equation} \tag{ 45 }$$

where the function $f(s,t)$ is defined in (35), the point $(s_2,t_2)$ is of the form (37), and $\Delta=\lim_{n\to\infty}({1}/{n})\ln{\Delta_n}$ is evaluated in (44).

Futher, let $\varepsilon_1,\varepsilon_2> 0$. Then there is an $N\in\mathbb{N}$, $N=N(\varepsilon_1, \varepsilon_2)$, such that the following inequalities hold for all $n\geq N$ and for at least one of the values $m\in\{n, n+1\}$:

$$\begin{equation} e^{-(\gamma_3+\varepsilon_1)m}<|L_m|<e^{-(\gamma_3-\varepsilon_2)m}. \end{equation} \tag{ 46 }$$

Proof.  Let $l_1$ be a ray, in the complex plane $s$, which issues from zero and passes through the point $s_2$ and let $l_2$ be an analogous ray, in the plane $t$, which passes through the point $t_2$. By (35), we have

$$\begin{equation*} \begin{aligned} \, \int_0^{-\infty}ds\int_{-i\infty}^{i\infty}G(s,t)\,dt &=\int_0^{-\infty}ds\int_0^{i\infty}G(s,t)\,dt -\int_0^{-\infty}ds\int_0^{-i\infty}G(s,t)\,dt \\ &\equiv J^{(1)}-J^{(2)}. \end{aligned} \end{equation*}$$

Let $J^{(1)}=A+Bi$. Then $J^{(2)}=A-Bi$, that is, $J_n=({1}/{(2\pi i)})(J^{(1)}-J^{(2)})={B}/{\pi}=-({1}/{\pi})\operatorname{Im}J^{(2)}$. Let us move the integration with respect to $s$ in the integral $J^{(2)}$: $(0,-\infty)\to l_1$, and let us similarly move the integration with respect to the variable $t$: $(0, -i\infty)\to l_2$. Since there are no singular points of the integrand in the domains between the indicated rays, it follows that $J^{(2)}=\int_{l_1}ds\int_{l_2}G(s,t)\, dt$. Hence,

$$\begin{equation} J_n=-\frac{1}{\pi}\operatorname{Im}\biggl(\int_{l_1}ds\int_{l_2}G(s,t)\, dt\biggr). \end{equation} \tag{ 47 }$$

As in Lemma 6, we write $\xi=(s,t)$ and $\xi^0=(s_2,t_2)$.

Simple computer calculations show that $\max_{\xi\in\,l_1\times l_2}|f(\xi)|$ is attained only at the point $\xi^0$. Let $\gamma_1$ be a small segment of $l_1$ containing $s_2$ as an interior point, let $\gamma_2$ be a similar segment of $l_2$ containing $t_2$ as an interior point, and let $\gamma=\gamma_1\times\gamma_2$, $\Gamma=(l_1\times l_2)\setminus\gamma$ and $\max_{\xi\in\Gamma}|f(\xi)|=F<|f(\xi^0)|$.

As in Lemma 6, one can apply Lemma 5 to the integral $\int_{\gamma}G(\xi)\,d\xi$, since

$$\begin{equation*} \det{S_{\xi\xi}''(\xi^0)}\approx 2.69736\times10^4-1.13489i\times10^4\ne0. \end{equation*}$$

In our situation, the equation (38) becomes

$$\begin{equation*} \begin{aligned} \, &\int_{\gamma}\varphi(\xi)\exp{\bigl(n(\ln{|f(\xi)|}+ih(\xi))\bigr)\,d\xi} \\ &\qquad=\frac{2\pi}{n}\exp{\bigl(n(\ln{|f(\xi^0)|}+ih(\xi^0))\bigr)} \bigl(\det{S_{\xi\xi}''(\xi^0)}\bigr)^{-{1}/{2}}\biggl(\varphi(\xi^0) +O\biggl(\frac{1}{n}\biggr)\biggr). \end{aligned} \end{equation*}$$

We have $h(\xi^{(0)})\equiv\omega\approx 1.9062$. Let $\varphi(\xi^0)(\det{S_{\xi\xi}''(\xi^0)})^{-{1}/{2}}=re^{i\alpha}$, $r> 0$. Then (see (47))

$$\begin{equation*} -\frac{1}{\pi}\operatorname{Im} \int_{\gamma}G(\xi)\,d\xi =-\frac{2r}{n}\exp{\bigl(n\ln{|f(\xi^0)|}\bigr)}\sin{(n\omega+\alpha)} +o\biggl(\frac{1}{n}\exp\bigl(n\ln{|f(\xi^0)|}\bigr)\biggr). \end{equation*}$$

Obviously,

$$\begin{equation*} \int_{\Gamma}G(\xi)\,d\xi=o\biggl(\frac{1}{n}\exp\bigl(n\ln{|f(\xi^0)|}\bigr)\biggr). \end{equation*}$$

Let us write $\omega_0={(\pi-\omega)}/{2}$. We claim that for every $n\in\mathbb{N}$

$$\begin{equation*} \max{\bigl(|\sin{(n\omega+\alpha)}|,\,|\sin{((n+1)\omega+\alpha))}|\bigr)}\geq \sin{(\omega_0)}. \end{equation*}$$

Indeed, let $n\omega+\alpha=2\pi n_1+\alpha_1$, $(n+1)\omega+\alpha=2\pi n_2+\alpha_2$, where $n_1,n_2\in \mathbb{Z}$, $\alpha_1,\alpha_2\in[0,2\pi)$. If $|\sin{(n\omega+\alpha)}|<\sin{(\omega_0)}$, then

$$\begin{equation*} \alpha_1\in A\equiv[0, \omega_0)\cup(\pi-\omega_0, \pi+\omega_0)\cup(2\pi-\omega_0,2\pi). \end{equation*}$$

In this case, $\alpha_2\notin A$, that is, $|\sin{((n+1)\omega+\alpha)}|\geq \sin{(\omega_0)}$. Let $m\in\{n, n+1\}$ be the value for which $|\sin{(m\omega+\alpha)}|\geq \sin{(\omega_0)}$. We choose $C_1=r\sin{(\omega_0)}$ and $C_2=4r$. Then for $n\geq N_1$, we obtain from (17) that the integral $J_m$ has the bounds

$$\begin{equation*} \frac{C_1}{m}\,e^{m\ln{|f(s_2,t_2)|}}<|J_m|<\frac{C_2}{m}\,e^{m\ln{|f(s_2,t_2)|}}. { } \end{equation*}$$

Since $\lim_{n\to\infty}({1}/{n})\ln({q_{56n}}/{\Delta_n})=56-\Delta$, it follows that for all $m\geq N_2$ we have

$$\begin{equation*} e^{(56-\Delta-{\varepsilon_1}/{2})m}<\frac{q_{56m}}{\Delta_m}<e^{(56-\Delta+{\varepsilon_2}/{2})m}. \end{equation*}$$

Let ${C_1}/{m}\,{>}\,e^{-({\varepsilon_1}/{2})m}$, ${C_2}/{m}\,{<}\,e^{({\varepsilon_2}/{2})m}$ and $N\,{=}\,\max{(N_1, N_2, N_3)}$ for all $n\,{\geq}\,N_3$. Then, by (21), for $n\geq N$ we have

$$\begin{equation*} \begin{aligned} \, |L_m|&<\exp{\bigl(m\bigl(-140\ln{(3+2\sqrt{2})}-125\ln{2}+56-\Delta+\varepsilon_2+\ln{|f(s_2, t_2)|}\bigr)\bigr)} \\ &=e^{-(\gamma_3-\varepsilon_2)m}, \end{aligned} \end{equation*}$$

and similarly $|L_m|>e^{-(\gamma_3+\varepsilon_1)m}$. This completes the proof of the lemma. $\square$

For the integral (2) we consider the function

$$\begin{equation} g(x,\alpha)=\frac{1}{2\pi i}\int_0^{a(\alpha)}ds\int_{l_t}\frac{s^{k}t^{h-k}\, dt}{\sqrt{{s}/{(s-1)}}(1-s)^{k+l-j+1}(s-t)(t-x)}, \end{equation} \tag{ 48 }$$

where $a(\alpha)={\alpha}/{(\alpha-1)}$, $\alpha\in(0,1)$, $l_t$ is the circle in the complex plane $t$ whose diameter is the segment $[{x}/{2},2x]$ on the real line, and the integration with respect to $l_t$ is carried out in the negative direction.

In this section, we write

$$\begin{equation*} D_N(f(x,\alpha))=\frac{1}{N!}\,\frac{\partial^{N}f(x,\alpha)}{\partial x^N},\qquad D_N(f(x))=\frac{1}{N!}f^{(N)}(x),\quad N\in\mathbb{Z^+}. \end{equation*}$$

In the following lemma we establish a relationship between the function $g(x, \alpha)$ and the integral (2).

Lemma 8.  The integral (2) and the function (48) satisfy the relation

$$\begin{equation} J=\lim_{\alpha\to 1-0}{T(g(x, \alpha))}, \end{equation} \tag{ 49 }$$

where $T$ is the operator $D_{k+m-h}x^{j}D_{h+j-k}$.

Proof.  By (48), we obtain in the standard way (see, for example, [3], pp. 167–169) that

$$\begin{equation} D_{h+j-k}(g(x,\alpha))\,{=}\,\frac{1}{2\pi i}\int_0^{a(\alpha)}\!ds\int_{l_t}\frac{s^{k}t^{h-k}\, dt}{\sqrt{{s}/{(s\,{-}\,1)}}(1\,{-}\,s)^{k+l-j+1}(s\,{-}\,t)(t\,{-}\,x)^{h+j-k+1}}. \end{equation} \tag{ 50 }$$

In (50) we make the change of variables $s_1=s$, $t_1={xs}/{t}$. For all $s< 0$ we obtain the integration with respect to the variable $t_1$ in the negative direction over the circle $L_{t_1}$ whose diameter is the segment $[2s,{s}/{2}]$. Further, $dt\,ds=-({xs_1}/{t_1^2})\, dt_1\, ds_1$, $s-t=({s_1}/{t_1})(t_1-x)$ and $t-x=({x}/{t_1})(s_1-t_1)$. Therefore,

$$\begin{equation*} \begin{aligned} \, &\frac{s^{k}t^{h-k}\, ds\,dt}{\sqrt{{s}/{(s-1)}}(1-s)^{k+l-j+1}(s-t)(t-x)^{h+j-k+1}} \\ &\quad=-\frac{s_1^k({x^{h-k}s_1^{h-k}}/{t_1^{h-k}})({xs_1}/{t_1^2})\, ds_1 \, dt_1}{\sqrt{{s_1}/{(s_1-1)}}(1-s_1)^{k+l-j+1}({s_1}/{t_1})(t_1-x)}\, \frac{1}{{(x(s_1-t_1))^{h+j-k+1}}/{t_1^{h+j-k+1}}} \\ &\quad=-x^{-j}\frac{s_1^h t_1^j \, ds_1 \, dt_1}{\sqrt{{s_1}/{(s_1-1)}}(1-s_1)^{k+l-j+1}(s_1-t_1)^{h+j-k+1}(t_1-x)}. \end{aligned} \end{equation*}$$

The formula (50) becomes

$$\begin{equation} \begin{aligned} \, &D_{h+j-k}(g(x, \alpha)) \\ &\qquad=\frac{x^{-j}}{2\pi i}\int_0^{a(\alpha)}ds_1\int_{L_{t_1}}\frac{s_1^{h}t_1^{j}\, dt_1}{\sqrt{{s_1}/{(s_1-1)}}(1-s_1)^{k+l-j+1}(s_1-t_1)^{h+j-k+1}(t_1-x)}, \end{aligned} \end{equation} \tag{ 51 }$$

where the integration over $L_{t_1}$ is carried out in the positive direction.

Let us multiply both the sides of the identity (51) by $x^j$ and apply the operator $D_{k+m-h}$ to both sides. One can now replace the integration over $L_{t_1}$ by integration over the line $(-i\infty, i\infty)$ since both integrals are equal to $2\pi i\cdot \operatorname{res}_{t_1=s_1}G(s_1,t_1)$. Finally, passing to the limit as $\alpha\to 1-0$, we obtain the identity (49). $\square$

Our next task is to evaluate the function $g(x, \alpha)$.

Lemma 9.  The function (48) satisfies the relation

$$\begin{equation} g(x,\alpha)=\frac{2(-1)^{k+l-j}x^{h-k}}{x-1}\int_0^{\sqrt{\alpha}}\frac{z^{2k}(z^2-1)^{l-j}\, dz}{z^2-{x}/{(x-1)}}. \end{equation} \tag{ 52 }$$

Proof.  We evaluate the inner integral in (48) using Cauchy's residue theorem:

$$\begin{equation*} g(x, \alpha)=-x^{h-k}\int_0^{a(\alpha)}\frac{s^{k}\, ds}{\sqrt{{s}/{(s-1)}}(1-s)^{k+l-j+1}(s-x)}. \end{equation*}$$

Let us make the change $u={s}/{(s-1)}$ in the integral thus obtained. Setting $s={u}/{(u-1)}$, $ds=-{du}/{(1-u)^2}$, $1-s={1}/{(1-u)}$ and $u\in[0, \alpha]$, we obtain

$$\begin{equation*} \begin{aligned} \, g(x,\alpha)&=x^{h-k}\int_0^{\alpha}\frac{({u^k}/{(u-1)^k})({du}/{(1-u)^2})} {\sqrt{u}({1}/{(1-u)^{k+l-j+1}})({u}/{(u-1)}-x)} \\ &=(-1)^{k+1}x^{h-k}\int_0^{\alpha}\frac{u^k(1-u)^{l-j}\, du}{\sqrt{u}\,(u+x(1-u))} \\ &=\frac{(-1)^{k+1}x^{h-k}}{x-1}\int_0^{\alpha}\frac{u^k(1-u)^{l-j}\, du}{\sqrt{u}\,({x}/{(x-1)}-u)}. \end{aligned} \end{equation*}$$

Finally, let us make the change $u=z^2$, $z\in[0, \sqrt{\alpha}\,]$. We have

$$\begin{equation*} g(x,\alpha)=\frac{2(-1)^{k+l-j}x^{h-k}}{x-1}\int_0^{\sqrt{\alpha}}\frac{z^{2k}(z^2-1)^{l-j}\, dz}{z^2-{x}/{(x-1)}}. \end{equation*}$$

This completes the proof of the lemma. $\square$

The following lemma gives a representation of the integral (2) in a form convenient for the evaluation of the constant $\gamma_2$. We restrict ourselves to the case $l\leq j$, which holds for the family (4) of parameters.

Lemma 10.  Let $l\leq j$. Then the integral (2) satisfies the relation

$$\begin{equation} \begin{aligned} \, &J=2(-1)^{k+l-j}T\biggl(\sum_{\nu=0}^{k+l-j-1}\frac{1}{2k+2l-2j-2\nu-1}\, \frac{x^{m-q+\nu}}{(x-1)^{\nu+1}} \\ &\qquad-\frac{x^{h-{1}/{2}}}{(x-1)^{k+l-j+{1}/{2}}}\ln{(\sqrt{x}+\sqrt{x-1})}\biggr). \end{aligned} \end{equation} \tag{ 53 }$$

Proof.  We write $A=z^2-1$, $B=1/(x-1)$, $l-j=-d$, $d\in\mathbb Z^+$. Then

$$\begin{equation*} \begin{aligned} \, \frac{(z^2-1)^{l-j}}{z^2-{x}/{(x-1)}}&=\frac{A^{-d}-B^{-d}}{A-B}+\frac{B^{-d}}{A-B} =-\frac{A^d-B^d}{(A-B)A^{d}B^{d}}+\frac{B^{-d}}{A-B} \\ &=-\sum _{\nu=0}^{d-1}A^{-\nu-1}B^{\nu-d}+\frac{B^{-d}}{A-B} \\ &=-\sum _{\nu=0}^{d-1}(z^2-1)^{-\nu-1}(x-1)^{d-\nu}+\frac{(x-1)^d}{z^2-{x}/{(x-1)}}. \end{aligned} \end{equation*}$$

Further,

$$\begin{equation*} \begin{aligned} \, \frac{z^{2k}}{z^2-{x}/{(x-1)}}&=\frac{z^{2k}-({x}/{(x-1)})^k}{z^2-{x}/{(x-1)}} +\frac{({x}/{(x-1)})^k}{z^2-{x}/{(x-1)}} \\ &=\sum_{\mu=0}^{k-1}z^{2k-2\mu-2}\biggl(\frac{x}{x-1}\biggr)^{\mu} +\frac{({x}/{(x-1)})^k}{z^2-{x}/{(x-1)}}. \end{aligned} \end{equation*}$$

Therefore,

$$\begin{equation*} \begin{aligned} \, \frac{z^{2k}(z^2-1)^{l-j}}{z^2-{x}/{(x-1)}}=&-\sum_{\nu=0}^{d-1} \frac{z^{2k}} {(z^2-1)^{\nu+1}}(x-1)^{d-\nu} +(x-1)^{d}\biggl(\sum _{\mu=0}^{k-1}z^{2k-2\mu-2} \biggl(\frac{x}{x-1}\biggr)^{\mu} \\ &\qquad+\biggl(\frac{x}{x-1}\biggr)^{k} \frac{1}{z^2-{x}/{(x-1)}}\biggr). \end{aligned} \end{equation*}$$

Hence, it follows from (52) that

$$\begin{equation*} \begin{aligned} \, g(x, \alpha)&=2(-1)^{k+l-j}\biggl(-x^{h-k}\sum_{\nu=0}^{d-1}(x-1)^{d-\nu-1} \int_0^{\sqrt{\alpha}}\frac{z^{2k}\, dz} {(z^2-1)^{\nu+1}}\biggr) \\ &\qquad+2(-1)^{k+l-j}\biggl((x-1)^{d-1}x^{h-k}\sum_{\mu=0}^{k-1}\biggl(\frac{x}{x-1} \biggr)^{\mu}\int_0^{\sqrt{\alpha}}z^{2k-2\mu-2}\, dz \\ &\qquad\qquad+\frac{x^h}{(x-1)^{k+l-j+1}}\int_0^{\sqrt{\alpha}} \frac{dz}{z^2-{x}/{(x-1)}}\biggr). \end{aligned} \end{equation*}$$

By (49), we have

$$\begin{equation*} \begin{aligned} \, J=&\lim_{\alpha\to 1-0}2(-1)^{k+l-j+1}T\biggl(\sum_{\nu=0}^{d-1}x^{h-k}(x-1)^{d-\nu-1} \int_0^{\sqrt{\alpha}}\frac{z^{2k}\, dz}{(z^2-1)^{\nu+1}}\biggr) \\ &\quad+\lim_{\alpha\to 1-0}2(-1)^{k+l-j}T\biggl(\sum_{\mu=0}^{d-1}(x-1)^{d-\mu-1}x^{h-k+\mu} \frac{(\sqrt{\alpha})^{2k-2\mu-1}}{2k-2\mu-1}\biggr) \\ &\quad+\lim_{\alpha\to 1-0}2(-1)^{k+l-j}T\biggl(\sum_{\mu=d}^{k-1}\frac{x^{h-k+\mu}}{(x-1)^{\mu-d+1}}\, \frac{(\sqrt{\alpha})^{2k-2\mu-1}}{2k-2\mu-1}\biggr) \\ &\quad+2(-1)^{k+l-j}T\biggl(\frac{x^h}{(x-1)^{k+l-j+1}} \int_0^1\frac{dz}{z^2-{x}/{(x-1)}}\biggr) \equiv S_1+S_2+S_3+S_4. \end{aligned} \end{equation*}$$

We claim that $S_1=S_2= 0$. Consider the sum $S_1$. For all $\nu=0,1,\dots,d-1$, we obtain from Leibniz' formula that

$$\begin{equation*} \begin{aligned} \, &x^{j}D_{h+j-k}\bigl(x^{h-k}(x-1)^{d-\nu-1}\bigr) \\ &\qquad=\sum _{\substack{r_1,r_2\in\mathbb{Z^+}\!,\, r_2<d-\nu \\r_1+r_2=h+j-k}} {\binom{h-k}{r_1}}{\binom{d-\nu-1}{r_2}}x^{h+j-k-r_1}(x-1)^{d-\nu-1-r_2} \\ &\qquad\equiv P_{\nu}(x)\in\mathbb{Z}[x],\qquad \operatorname{deg}P_{\nu}(x)\leq d-\nu-1\leq j-l-1<k+m-h. \end{aligned} \end{equation*}$$

Therefore, $D_{k+m-h}(P_{\nu}(x))= 0$. Hence, $S_1= 0$. Similarly, $S_2= 0$. Further, in the sum $S_3$ we set $\mu=d+\nu$, $\nu=0,1,\dots,k+l-j-1$. Then

$$\begin{equation*} S_3=2(-1)^{k+l-j}T\biggl(\sum_{\nu=0}^{k+l-j-1}\frac{x^{m-q+\nu}}{(x-1)^{\nu+1}}\, \frac{1}{2k+2l-2j-2\nu-1}\biggr). \end{equation*}$$

Finally,

$$\begin{equation*} \int_0^1\frac{dz}{z^2-{x}/{(x-1)}}=-\frac{(x-1)^{{1}/{2}}}{x^{{1}/{2}}} \ln{(\sqrt{x}+\sqrt{x-1})}, \end{equation*}$$

that is,

$$\begin{equation*} S_4=2(-1)^{k+l-j+1}T\biggl(\frac{x^{h-{1}/{2}}}{(x-1)^{k+l-j+{1}/{2}}} \ln{(\sqrt{x}+\sqrt{x-1})}\biggr). \end{equation*}$$

We have $J=S_3+S_4$, which coincides with the assertion of the lemma. $\square$

To evaluate the integral (2) by applying the formula (53), two more lemmas will be useful.

Lemma 11.  Let $M\in\mathbb{N}$, $a,b\in\mathbb{R}$. Then

$$\begin{equation} D_M\biggl(\frac{x^a}{(x-1)^b}\biggr) =(-1)^M\sum_{r=0}^{M}{\binom{a}{r}}{\binom {b-a+M-1}{M-r}}\frac{x^{a-r}}{(x-1)^{b+M}}. \end{equation} \tag{ 54 }$$

Proof.  We proceed by induction on $M$. When $M= 1$ we have

$$\begin{equation*} \begin{aligned} \, D_1\biggl(\frac{x^a}{(x-1)^b}\biggr)&=ax^{a-1}(x-1)^{-b}-bx^{a}(x-1)^{-b-1} \\ &=-\bigl(ax^{a-1}(x-1)^{-b-1}+(b-a)x^{a}(x-1)^{-b-1}\bigr), \end{aligned} \end{equation*}$$

which coincides with (54) when $M= 1$.

Let us make the induction step $M\to M+1$. Using the induction assumption, we obtain

$$\begin{equation*} \begin{aligned} \, &D_{M+1}\biggl(\frac{x^a}{(x-1)^b}\biggr) =\frac{1}{M+1}D_1 \biggl(D_M\biggl(\frac{x^a}{(x-1)^b}\biggr)\biggr) \\ &\qquad=\frac{(-1)^M}{M+1}D_1 \biggl(\sum_{r=0}^{M}{\binom{a}{r}}{\binom {b-a+M-1}{M-r}}\frac{x^{a-r}}{(x-1)^{b+M}}\biggr) \\ &\qquad=\frac{(-1)^{M+1}}{M+1}\sum_{r=0}^{M}{\binom{a}{r}}{\binom{b-a+M-1}{M-r}} \bigl((a-r)x^{a-r-1}(x-1)^{-b-M-1} \\ &\qquad\qquad+(b-a+M+r)x^{a-r}(x-1)^{-b-M-1}\bigr) \\ &\qquad\equiv(-1)^{M+1}\sum_{\rho=0}^{M+1}C_{\rho}\frac{x^{a-\rho}}{(x-1)^{b+M+1}}. \end{aligned} \end{equation*}$$

We write

$$\begin{equation*} d_{\rho}={\binom{a}{\rho}}{\binom{b-a+M}{M+1-\rho}}. { } \end{equation*}$$

We need to prove that $C_{\rho}=d_{\rho}$ for all $\rho=0,1,\dots,M+1$.

The following equation holds:

$$\begin{equation*} C_0={\binom{b-a+M-1}{M}}\frac{b-a+M}{M+1}={\binom{b-a+M}{M+1}}=d_0. \end{equation*}$$

Further,

$$\begin{equation*} C_{M+1}={\binom{a}{M}}\frac{a-M}{M+1}={\binom{a}{M+1}}=d_{M+1}. { } \end{equation*}$$

Finally, when $\rho\in[1, M]$ we need to show that

$$\begin{equation*} \begin{aligned} \, &\frac{1}{M+1}\biggl({\binom{a}{\rho-1}} {\binom{b-a+M-1}{M-(\rho-1)}}(a-(\rho-1)) \\ &\qquad+{\binom{a}{\rho}} {\binom{b-a+M-1}{M-\rho}}(b-a+M+\rho)\biggr)=d_{\rho}. \end{aligned} \end{equation*}$$

Consider two cases.

1. Let $b-a+M\ne 0$. Here

$$\begin{equation*} \begin{gathered} \, {\binom{a}{\rho-1}}(a-(\rho-1))={\binom{a}{\rho}}\rho, \\ \begin{split} {\binom {b-a+M-1}{M-(\rho-1)}}&={\binom{b-a+M}{M+1-\rho}}\frac{b-a+\rho-1}{b-a+M}, \\ {\binom {b-a+M-1}{M-\rho}}&={\binom{b-a+M}{M+1-\rho}}\frac{M+1-\rho}{b-a+M}. \end{split} \end{gathered} \end{equation*}$$

Therefore,

$$\begin{equation*} \begin{aligned} \, C_{\rho}&=\frac{1}{M+1}{\binom{a}{\rho}}{\binom {b-a+M}{M+1-\rho}}\frac{\rho(b-a+\rho-1)+(M+1-\rho)(b-a+M+\rho)}{b-a+M} \\ &={\binom{a}{\rho}}{\binom{b-a+M}{M+1-\rho}}=d_{\rho}. \end{aligned} \end{equation*}$$

2. Let $b-a+M= 0$. Here

$$\begin{equation*} \begin{gathered} \, {\binom{b-a+M-1}{M-(\rho-1)}}=(-1)^{M-\rho+1}, \qquad {\binom {b-a+M-1}{M-\rho}}=(-1)^{M-\rho}, \\ C_{\rho}=\frac{1}{M+1}\biggl({\binom{a}{\rho}}\rho(-1)^{M-\rho+1}+{\binom {a}{\rho}}(-1)^{M-\rho}\rho\biggr)=0={\binom{a}{\rho}} {\binom{b-a+M}{M+1-\rho}}=d_{\rho}. \end{gathered} \end{equation*}$$

This completes the proof of the lemma. $\square$

Lemma 12.  The following equation holds for every $N\in\mathbb{N}$ and for arbitrary analytic functions $u=u(x)$ and $\vartheta=\vartheta(x)$:

$$\begin{equation} D_N(u\vartheta)=\vartheta D_N(u)+\sum_{\lambda=0}^{N-1}\frac{\lambda!\, (N-1-\lambda)!}{N!}D_{N-1-\lambda} (D_{\lambda}(u)\vartheta'). \end{equation} \tag{ 55 }$$

Proof.  By the definition of the operator $D_N$, it suffices to show that

$$\begin{equation} (u\vartheta)^{(N)}=\vartheta u^{(N)} +\sum_{\lambda=0}^{N-1}(u^{(\lambda)}\vartheta')^{(N-1-\lambda)}. \end{equation} \tag{ 56 }$$

In turn, to prove the formula (56), it suffices to prove that the coefficients of $u^{(r)}\,\vartheta^{(N-r)}$, $r=0,1,\dots,N-1$, on both the sides of the equation (56) coincide, that is,

$$\begin{equation} {\binom{N}{r}}=\sum_{\lambda=0}^{r}{\binom{N-1-\lambda}{r-\lambda}},\qquad r=0, 1, \dots, N-1. \end{equation} \tag{ 57 }$$

We shall prove this by induction on $r$. When $r= 0$ we have

$$\begin{equation*} {\binom{N}{0}}={\binom{N-1}{0}}=1. \end{equation*}$$

To carry out the induction step $(r-1)\to r$, note that

$$\begin{equation*} {\binom{N}{r}}={\binom{N-1}{r}}+{\binom{N-1}{r-1}}. \end{equation*}$$

By the induction assumption,

$$\begin{equation*} {\binom{N-1}{r-1}}=\sum_{\lambda=0}^{r-1}{\binom {N-2-\lambda}{r-1-\lambda}} =\sum_{\lambda=1}^{r}{\binom{N-1-\lambda}{r-\lambda}}, \end{equation*}$$

and then the equation (57) holds. $\square$

In what follows, it is useful to note the asymptotics of generalized binomial coefficients:

$$\begin{equation*} {\binom {b}{r}}=\frac{b(b-1)\dotsb(b-r+1)}{r!},\qquad b\in\mathbb{R},\quad r\in\mathbb{N}. \end{equation*}$$

For $x\in\mathbb{R}$ we introduce the function

$$\begin{equation} x^* = \begin{cases} x\ln{x} &\text{for }x>0, \\ 0 &\text{for }x=0, \\ x\ln{(-x)} &\text{for }x<0. \end{cases} \end{equation} \tag{ 58 }$$

Obviously, the function $x^*$ is odd.

Lemma 13.  Let $n\in\mathbb{N}$, $n\to+\infty$, $b=b_0n+O(1)$, $r=r_0n+O(1)$, $b_0,r_0\in\mathbb{R}$, $r\in\mathbb{Z^+}$ and ${\binom{b}{r}}\ne 0$. Then the following equation holds:

$$\begin{equation} \lim_{n\to\infty}\frac{1}{n}\ln{\biggl|{\binom{b}{r}}\biggr|}=b_0^*-r_0^*-(b_0-r_0)^*. \end{equation} \tag{ 59 }$$

Proof.  It is clear that $r_0\geq 0$. If $r_0= 0$, then

$$\begin{equation*} \lim_{n\to\infty}\frac{1}{n}\ln{\biggl|{\binom{b}{r}}\biggr|}=0, \end{equation*}$$

and the formula (59) holds. Let $r_0> 0$ everywhere below. We consider several cases.

1) Let $b_0>r_0$. The following formula holds:

$$\begin{equation*} {\binom{b}{r}}=\frac{\Gamma(b+1)}{\Gamma(b-r+1)\Gamma(r+1)}. \end{equation*}$$

It follows from Stirling's formula that $\ln{\Gamma(x+1)}\,{=}\,x\ln{x}-x+O(\ln{x})$ as $x\,{\to}\,+\infty$. Taking into account that $b$, $b-r$, $r\to +\infty$ as $n\to+\infty$, we obtain

$$\begin{equation*} \begin{aligned} \, \ln{\Gamma(b+1)}&=(b_0n+O(1))\ln{(b_0 n+O(1))}-b_0n+O(\ln{n}) \\ &=b_0n\ln{(b_0n)}-b_0n+O(\ln{n}). \end{aligned} \end{equation*}$$

Similar formulae hold for $\Gamma(b-r+1)$ and $\Gamma(r+1)$. Therefore,

$$\begin{equation*} \begin{gathered} \, \begin{split} \ln{{\binom{b}{r}}}&=b_0n\ln{(b_0n)}-b_0n-(b_0-r_0)\ln{((b_0-r_0)n)} \\ &\qquad+(b_0-r_0)n-r_0n\ln{(r_0n)}+r_0n+O(\ln{n}) \\ &=n\bigl(b_0\ln{b_0}-(b_0-r_0)\ln{(b_0-r_0)}-r_0\ln{r_0}\bigr)+O(\ln{n}), \end{split} \\ \lim_{n\to+\infty}\frac{1}{n}\ln{{\binom{b}{r}}}=b_0^*-(b_0-r_0)^*-r_0^*, \end{gathered} \end{equation*}$$

and the formula (59) is proved in the case under consideration.

2) Let $b_0=r_0$. Here

$$\begin{equation*} {\binom{b}{r}}=C_1\frac{\Gamma(b+1)}{\Gamma(r+1)} \end{equation*}$$

for some constant $C_1\ne 0$, and the formula (59) is verified trivially.

3) Let $0<b_0<r_0$. In this case,

$$\begin{equation*} {\binom{b}{r}}=C_2\frac{\Gamma(b+1)\Gamma(r-b)}{\Gamma(r+1)}, \end{equation*}$$

where, as above, the constant $C_2$ is non-zero,

$$\begin{equation*} \lim_{n\to+\infty}\frac{1}{n}\ln{\biggl|{\binom {b}{r}}\biggr|} =b_0^*+(r_0-b_0)^*-r_0^*=b_0^*-(b_0-r_0)^*-r_0^*, \end{equation*}$$

and we have used the fact that the function (58) is odd.

4) Let $b_0= 0$. As in cases 2) and 3), the following equation holds:

$$\begin{equation*} {\binom{b}{r}}=C_3\frac{\Gamma(r-b)}{\Gamma(r+1)},\qquad C_3\ne 0, { } \end{equation*}$$

that is,

$$\begin{equation*} \lim_{n\to\infty}\frac{1}{n}\ln{\biggl|{\binom {b}{r}}\biggr|}=r_0^*-r_0^*=0=b_0^*-(b_0-r_0)^*-r_0^*. \end{equation*}$$

The last case remains.

5) Let $b_0< 0$. Here

$$\begin{equation*} \begin{gathered} \, {\binom {b}{r}}=(-1)^{r}\frac{(-b)(-b+1)\dotsb(-b+r-1)}{r!} =(-1)^{r} \frac{\Gamma(-b+r)}{\Gamma(-b)\Gamma(r+1)}, \\ \lim_{n\to\infty}\frac{1}{n}\ln{\biggl|{\binom {b}{r}}\biggr|} =(r_0-b_0)^*-(-b_0)^*-r_0^*=b_0^*-(b_0-r_0)^*-r_0^*. \end{gathered} \end{equation*}$$

This completes the proof of the lemma. $\square$

Let us apply the results thus obtained to the linear form (21). We first evaluate the integral $J_n$ in (35) using the formula (53):

$$\begin{equation} \begin{aligned} \, J_n&=2(-1)^n T\biggl(\sum_{\nu=0}^{19n-1}\frac{1}{38n-2\nu-1}\, \frac{x^{-12n+\nu}}{(x-1)^{\nu+1}} \\ &\qquad-\frac{x^{7n-1/2}}{(x-1)^{19n+1/2}}\ln{(\sqrt{x}+\sqrt{x-1})}\biggr), \end{aligned} \end{equation} \tag{ 60 }$$

where $T$ stands for the operator $D_{56n}x^{38n}D_{20n}$.

We write $A=(3+2\sqrt{2})^{-140n}2^{-125n}$ and

$$\begin{equation} \Sigma_{1,\nu}=A T\biggl(\frac{1}{x^{12n-\nu}(x-1)^{\nu+1}}\biggr),\qquad \nu=0, 1, \dots, 12n-1, \end{equation} \tag{ 61 }$$

$$\begin{equation} \Sigma_{2,\nu}=A T\biggl(\frac{x^{\nu}}{(x-1)^{12n+\nu+1}}\biggr),\qquad \nu=0, 1, \dots, 7n-1, \end{equation} \tag{ 62 }$$

$$\begin{equation} \Sigma_{3}=A T\biggl(\frac{x^{7n-1/2}}{(x-1)^{19n+1/2}} \ln{(\sqrt{x}+\sqrt{x-1})}\biggr). \end{equation} \tag{ 63 }$$

Then it follows from (21) and (60)–(63) that

$$\begin{equation} L_n=\frac{q_{56n}}{\Delta_n}\,2(-1)^n \biggl(\sum_{\nu=0}^{12n-1}\frac{1}{38n-2\nu-1}\Sigma_{1,\nu} +\sum_{\nu=0}^{7n-1}\frac{1}{14n-2\nu-1}\Sigma_{2,\nu}-\Sigma_{3}\biggr). \end{equation} \tag{ 64 }$$

Let us combine the evaluation of the operators $D_M$ by the ordinary Leibniz formula and by the formula (54). In the latter case, we denote the operator $D_M$ by $D_{M}'$.

Let us begin with the evaluation of $\Sigma_{1,\nu}$. We have

$$\begin{equation*} \begin{aligned} \, D_{20n}\biggl(\frac{1}{x^{12n-\nu}(x-1)^{\nu+1}}\biggr) &=\sum_{r_1=0}^{20n}{\binom {12n-\nu+r_1-1}{r_1}}{\binom{20n-r_1+\nu}{20n-r_1}} \\ &\qquad\times\frac{1}{x^{12n-\nu+r_1}(x-1)^{\nu+20n-r_1+1}}. \end{aligned} \end{equation*}$$

By Lemma 11,

$$\begin{equation*} \begin{aligned} \, &D_{56n}'\biggl(x^{38n}\frac{1}{x^{12n-\nu+r_1}(x-1)^{\nu+20n-r_1+1}}\biggr) =D_{56n}'\biggl(\frac{x^{26n+\nu-r_1}}{(x-1)^{20n+\nu-r_1+1}}\biggr) \\ &\qquad=\sum_{r_2=0}^{56n} {\binom{26n+\nu-r_1}{r_2}}{\binom {50n}{56n-r_2}}\frac{x^{26n+\nu-r_1-r_2}}{(x-1)^{76n+\nu-r_1+1}}. \end{aligned} \end{equation*}$$

Non-zero summands occur in the last sum only when $r_2\in[6n, 26n+\nu-r_1]$. Further, it follows from (3) that

$$\begin{equation*} \begin{aligned} \, \frac{x^{26n+\nu-r_1-r_2}}{(x-1)^{76n+\nu-r_1+1}} &=\biggl(\frac{3+2\sqrt{2}}{4\sqrt{2}} \biggr)^{26n+\nu-r_1-r_2}((3+2\sqrt{2})4\sqrt{2})^{76n+\nu-r_1+1} \\ &=(3+2\sqrt{2})^{102n+2\nu-2r_1-r_2+1}(4\sqrt{2})^{50n+r_2+1} \\ &=A^{-1}(3-2\sqrt{2})^{38n-2\nu+2r_1+r_2-1}(4\sqrt{2})^{r_2+1}. \end{aligned} \end{equation*}$$

Therefore, it follows from (61) for $\nu=0,1,\dots,12n-1$ that

$$\begin{equation} \begin{aligned} \, \Sigma_{1,\nu}&=\sum_{r_1=0}^{20n}\sum_{r_2=6n}^{26n+\nu-r_1} \binom{12n\,{-}\,\nu\,{+}\,r_1\,{-}\,1}{r_1} \binom {20n\,{-}\,r_1\,{+}\,\nu}{20n\,{-}\,r_1}\binom{26n\,{+}\,\nu\,{-}\,r_1}{r_2} \binom{50n}{56n\,{-}\,r_2} \\ &\qquad\times(3-2\sqrt{2})^{38n-2\nu+2r_1+r_2-1}(4\sqrt{2})^{r_2+1}. \end{aligned} \end{equation} \tag{ 65 }$$

Let us now evaluate $\Sigma_{2,\nu}$ for $\nu=0,1,\dots,7n-1$. We write

$$\begin{equation*} D_{20n}=\frac{\Lambda!\,(20n-\Lambda)!}{(20n)!}D_{20n-\Lambda}'D_{\Lambda}, \end{equation*}$$

where $\Lambda= \Lambda_0n+O(1)$, $\Lambda\in\mathbb{Z^+}$, will be chosen below to optimize the bound for $\gamma_2$.

We have

$$\begin{equation*} \begin{aligned} \, &D_{\Lambda}\biggl(\frac{x^{\nu}}{(x-1)^{12n+\nu+1}}\biggr) \\ &\qquad=\sum_{r_1=0}^{\min{(\Lambda,\nu)}}(-1)^{\Lambda-r_1}{\binom {\nu}{r_1}} {\binom {12n+\nu+\Lambda-r_1}{\Lambda-r_1}}\frac{x^{\nu-r_1}}{(x-1)^{12n+\nu+\Lambda-r_1+1}}, \\ &D_{20n-\Lambda}'\biggl(\frac{x^{\nu-r_1}}{(x-1)^{12n+\nu+\Lambda-r_1+1}}\biggr) \\ &\qquad=(-1)^{\Lambda}\sum_{r_2=0}^{\nu-r_1}{\binom {\nu-r_1}{r_2}} {\binom{32n}{20n-\Lambda-r_2}} \frac{x^{\nu-r_1-r_2}}{(x-1)^{32n+\nu-r_1+1}}, \\ &D_{56n}'\biggl(\frac{x^{38n+\nu-r_1-r_2}}{(x-1)^{32n+\nu-r_1+1}}\biggr) \\ &\qquad=\sum_{\rho=0}^{38n+\nu-r_1-r_2}{\binom {38n+\nu-r_1-r_2}{\rho}} {\binom {50n+r_2}{56n-\rho}}\frac{x^{38n+\nu-r_1-r_2-\rho}}{(x-1)^{88n+\nu-r_1+1}}. \end{aligned} \end{equation*}$$

As in the evaluation of $\Sigma_{1,\nu}$, we have

$$\begin{equation*} \frac{x^{38n+\nu-r_1-r_2-\rho}}{(x-1)^{88n+\nu-r_1+1}} =A^{-1}(3-2\sqrt{2})^{14n-2\nu+2r_1+r_2+\rho-1}(4\sqrt{2})^{r_2+\rho+1}. \end{equation*}$$

Therefore, it follows from (62) that

$$\begin{equation} \begin{aligned} \, \Sigma_{2,\nu}&=\frac{\Lambda!\,(20n-\Lambda)!}{(20n)!} \sum_{r_1=0}^{\min{(\Lambda,\nu)}}\sum_{r_2=0}^{\nu-r_1} \sum_{\rho=\max{(0,\, 6n-r_2)}}^{38n+\nu-r_1-r_2}(-1)^{r_1}{\binom{\nu}{r_1}} {\binom {12n+\nu+\Lambda-r_1}{\Lambda-r_1}} \\ &\qquad\times{\binom{\nu-r_1}{r_2}}{\binom{32n}{20n-\Lambda-r_2}} {\binom{38n+\nu-r_1-r_2}{\rho}}{\binom {50n+r_2}{56n-\rho}} \\ &\qquad\times(3-2\sqrt{2})^{14n-2\nu+2r_1+r_2+\rho-1}(4\sqrt{2})^{r_2+\rho+1}. \end{aligned} \end{equation} \tag{ 66 }$$

Finally, let us calculate the value of $\Sigma_{3}$ using Lemma 12. We write

$$\begin{equation*} u=\frac{x^{7n-1/2}}{(x-1)^{19n+1/2}},\qquad v=\ln{(\sqrt{x}+\sqrt{x-1})},\qquad v'=\frac{1}{2\sqrt{x}\,\sqrt{x-1}}. \end{equation*}$$

Then

$$\begin{equation*} D_{20n}(uv)=vD_{20n}(u)+\frac{1}{2}\sum_{\lambda=0}^{20n-1} \frac{\lambda!\,(20n-1-\lambda)!}{(20n)!} D_{20n-1-\lambda}\biggl(D_{\lambda}(u)\frac{1}{\sqrt{x}\,\sqrt{x-1}}\biggr). \end{equation*}$$

Similarly,

$$\begin{equation} \begin{gathered} \, \begin{split} &T(uv)=vT(u) \\ &\qquad+\frac{1}{2}\sum_{\lambda_1=0}^{56n-1}\frac{\lambda_1!\,(56n-1-\lambda_1)!} {(56n)!}D_{56n-1-\lambda_1}\biggl(D_{\lambda_1}(x^{38n}D_{20n}(u)) \frac{1}{\sqrt{x}\,\sqrt{x-1}}\biggr) \\ &\qquad+\frac{1}{2}\sum_{\lambda=0}^{20n-1}\frac{\lambda!\,(20n-1-\lambda)!}{(20n)!} D_{56n}\biggl(x^{38n}D_{20n-1-\lambda}\biggl(D_{\lambda}(u)\frac{1}{\sqrt{x}\, \sqrt{x-1}}\biggr)\biggr), \end{split} \\ \Sigma_{3}=AT(uv)\equiv AT(u)v+\frac{1}{2}\sum_{\lambda_1=0}^{56n-1}S_3(\lambda_1) +\frac{1}{2}\sum_{\lambda=0}^{20n-1}S_{3}^*(\lambda). \end{gathered} \end{equation} \tag{ 67 }$$

We note that $v=({1}/{2})\ln{\bigl(x+x-1+2\sqrt{x(x-1)}\bigr)=({1}/{2})\ln{\sqrt{2}}} =({1}/{4})\ln{2}$.

Hence, the coefficient at $\ln{2}$ in the linear form (21) is

$$\begin{equation} \Lambda_1(n)\sqrt{2}+\Lambda_2(n)=\frac{q_{56n}}{4\Delta_n} AT\biggl(\frac{x^{7n-1/2}}{(x-1)^{19n+1/2}}\biggr). \end{equation} \tag{ 68 }$$

Remark 2.  The formula (68) enables one to find the constant $\gamma_1$ (see Lemma 6) using the terms of some positive series rather than by applying the saddle-point method. The former method is very simple, and therefore it is popular. It has been used many times in recent years; see, for example, [3], [6] and [9]. Therefore, we restrict ourselves to rather brief comments.

Since ${1}/{x}< 1$, it follows that

$$\begin{equation*} \frac{x^{7n-1/2}}{(x-1)^{19n+1/2}} =\frac{1}{x^{12n+1}}\biggl(1-\frac{1}{x}\biggr)^{-19n-1/2} =x^{-12n-1}\sum_{s=0}^{\infty}{\binom {19n+s-1/2}{s}}x^{-s} \end{equation*}$$

is a convergent series. Therefore, differentiating termwise, we obtain the positive series

$$\begin{equation*} T\biggl(\frac{x^{7n-1/2}}{(x-1)^{19n+1/2}}\biggr) =\sum_{s=6n}^{\infty}{\binom{19n+s-1/2}{s}}{\binom {s+32n}{20n}} {\binom{s+50n}{56n}}\biggl(\frac{1}{x}\biggr)^{s+50n+1}. \end{equation*}$$

The desired asymptotic behaviour is given by the maximal term of the series. Let us find this term by solving the equation

$$\begin{equation*} \frac{s'+19}{s'}\,\frac{s'+32}{s'+12}\,\frac{s'+50}{s'-6}\,\frac{1}{x}=1, \qquad s'>6. \end{equation*}$$

We have $s'\approx 3160.03$. Correspondingly, the index of the maximal term of the series is $s=s'n+O(1)$. Therefore, as in the papers indicated above,

$$\begin{equation*} \begin{aligned} \, &\lim_{n\to\infty}\frac{1}{n}\ln{T\biggl(\frac{x^{7n-1/2}}{(x-1)^{19n+1/2}}\biggr)} =(19+s')^*-19^*-s'^*+(s'+32)^*-20^* \\ &\qquad-(s'+12)^*+(s'+50)^*-56^*-(s'-6)^*-(s'+50)\ln{x}=\ln{|f(s_1,t_1)|} \end{aligned} \end{equation*}$$

by Lemma 13, and, using the equation (68) we obtain the result coinciding with that of Lemma 6. We note that the reasoning used in Remark 2 does not involve the evaluation of the constant $\gamma_2$.

Let us return to the proof of Theorem 1. The first term in the sum (67) is evaluated in the same way as $\Sigma_{2,\nu}$, where $\nu=7n-{1}/{2}$. The half-integer value of $\nu$ enables us to extend the bounds of variation of $r_1$, $r_2$, $\rho$. Instead of (66), we obtain

$$\begin{equation} \begin{aligned} \, AvT(u)&=\frac{\Lambda!\,(20n-\Lambda)!}{(20n)!}\sum_{r_1=0}^{\Lambda} \sum_{r_2=0}^{20n-\Lambda}\sum_{\rho=\max{(0,\,6n-r_2)}}^{56n}(-1)^{r_1}{\binom {7n-1/2}{r_1}} \\ &\qquad\times{\binom{19n+\Lambda-r_1-1/2}{\Lambda-r_1}} {\binom{7n-r_1-1/2}{r_2}}{\binom{32n}{20n-\Lambda-r_2}} \\ &\qquad\times {\binom{45n-r_1-r_2-1/2}{\rho}}{\binom {50n+r_2}{56n-\rho}} \\ &\qquad\times (3-2\sqrt{2})^{2r_1+r_2+\rho}(4\sqrt{2})^{r_2+\rho+1}\frac{1}{4}\ln{2}. \end{aligned} \end{equation} \tag{ 69 }$$

Let us now evaluate the summands $S_3(\lambda_1)$ in the sum in (67). As above, let

$$\begin{equation*} D_{20n}=\frac{\Lambda!\,(20n-\Lambda)!}{(20n)!}D_{20n-\Lambda}'D_{\Lambda}. \end{equation*}$$

We obtain in succession

$$\begin{equation*} \begin{gathered} \, D_{\Lambda}(u)=\sum_{r_1=0}^{\Lambda}{\binom{7n-1/2}{r_1}} {\binom{19n+\Lambda-r_1-1/2}{\Lambda-r_1}}(-1)^{\Lambda-r_1} \frac{x^{7n-r_1-1/2}}{(x-1)^{19n+\Lambda-r_1+1/2}}, \\ \begin{split} D_{20n-\Lambda}'\biggl(\frac{x^{7n-r_1-1/2}}{(x-1)^{19n+\Lambda-r_1+1/2}}\biggr) &=(-1)^{\Lambda}\sum_{r_2=0}^{20n-\Lambda}{\binom {7n-r_1-1/2}{r_2}}{\binom{32n}{20n-\Lambda-r_2}} \\ &\qquad\times\frac{x^{7n-r_1-r_2-1/2}}{(x-1)^{39n-r_1+1/2}}, \end{split} \\ \begin{split} D_{\lambda_1}'\biggl(\frac{x^{45n-r_1-r_2-1/2}}{(x-1)^{39n-r_1+1/2}}\biggr) &=(-1)^{\lambda_1}\sum_{\rho_1=0}^{\lambda_1}{\binom {45n-r_1-r_2-1/2}{\rho_1}} {\binom{-6n+r_2+\lambda_1}{\lambda_1-\rho_1}} \\ &\qquad\times\frac{x^{45n-r_1-r_2-\rho_1-1/2}}{(x-1)^{39n+\lambda_1-r_1+1/2}}, \end{split} \\ \begin{split} &D_{56n-\lambda_1-1}'\biggl(\frac{x^{45n-r_1-r_2-\rho_1-1}}{(x-1)^{39n+\lambda_1-r_1+1}} \biggr)=(-1)^{\lambda_1+1} \\ &\quad\times\sum_{\rho_2=0}^{56n-\lambda_1-1}{\binom{45n-r_1-r_2-\rho_1-1}{\rho_2}} {\binom{50n+r_2+\rho_1}{56n-\lambda_1-\rho_2-1}} \frac{x^{45n-r_1-r_2-\rho_1-\rho_2-1}}{(x-1)^{95n-r_1}}, \end{split} \\ \frac{x^{45n-r_1-r_2-\rho_1-\rho_2-1}}{(x-1)^{95n-r_1}} =A^{-1}(3-2\sqrt{2})^{2r_1+r_2+\rho_1+\rho_2+1}(4\sqrt{2})^{r_2+\rho_1+\rho_2+1}. \end{gathered} \end{equation*}$$

Therefore,

$$\begin{equation} \begin{aligned} \, S_3(\lambda_1)&=\frac{\Lambda!\,(20n-\Lambda)!}{(20n)!}\, \frac{\lambda_1!\,(56n-1-\lambda_1)!}{(56n)!} \sum_{r_1=0}^{\Lambda}\sum_{r_2=0}^{20n-\Lambda}\sum_{\rho_1=0}^{\lambda_1} \sum_{\rho_2=0}^{56n-\lambda_1-1}(-1)^{r_1+1} \\ &\qquad\times{\binom {7n-1/2}{r_1}}{\binom{19n+\Lambda-r_1-1/2}{\Lambda-r_1}}{\binom{7n-r_1-1/2}{r_2}} \\ &\qquad\times{\binom {32n}{20n-\Lambda-r_2}}{\binom{45n-r_1-r_2-1/2}{\rho_1}} {\binom{-6n+r_2+\lambda_1}{\lambda_1-\rho_1}} \\ &\qquad\times{\binom{45n-r_1-r_2-\rho_1-1} {\rho_2}}{\binom {50n+r_2+\rho_1}{56n-\lambda_1-\rho_2-1}} \\ &\qquad\times (3-2\sqrt{2})^{2r_1+r_2+\rho_1+\rho_2+1}(4\sqrt{2})^{r_2+\rho_1+\rho_2+1}. \end{aligned} \end{equation} \tag{ 70 }$$

It remains to evaluate $S_3^*(\lambda)$. Let

$$\begin{equation*} D_{\lambda}=\frac{(\lambda-\Lambda)!\,\Lambda!}{\lambda!}D_{\lambda-\Lambda}' D_{\Lambda}. \end{equation*}$$

We obtain in the standard way that

$$\begin{equation*} \begin{aligned} \, D_{\Lambda}\biggl(\frac{x^{7n-1/2}}{(x-1)^{19n+1/2}}\biggr) &=\sum_{r_1=0}^{\Lambda}{\binom {7n-1/2}{r_1}}(-1)^{\Lambda-r_1} \\ &\quad\times{\binom {19n+\Lambda-r_1-1/2}{\Lambda-r_1}} \frac{x^{7n-r_1-1/2}}{(x-1)^{19n+\Lambda-r_1+1/2}}, \\ D_{\lambda-\Lambda}'\biggl(\frac{x^{7n-r_1-1/2}}{(x-1)^{19n+\Lambda-r_1+1/2}}\biggr) &=\sum_{r_2=0}^{\lambda-\Lambda}(-1)^{\lambda-\Lambda}{\binom {7n-r_1-1/2}{r_2}} {\binom {12n+\lambda}{\lambda-\Lambda-r_2}} \\ &\quad\times\frac{x^{7n-r_1-r_2-1/2}}{(x-1)^{19n+\lambda-r_1+1/2}}, \\ D_{20n-1-\lambda}'\biggl(\frac{x^{7n-r_1-r_2-1}}{(x-1)^{19n+\lambda-r_1+1}}\biggr) &=(-1)^{\lambda+1}\sum_{r_3=0}^{20n-1-\lambda}{\binom {7n-r_1-r_2-1}{r_3}} \\ &\quad\times{\binom{32n+r_2}{20n-1-\lambda-r_3}} \frac{x^{7n-r_1-r_2-r_3-1}}{(x-1)^{39n-r_1}}, \\ D_{56n}'\biggl(\frac{x^{45n-r_1-r_2-r_3-1}}{(x-1)^{39n-r_1}}\biggr) &= \sum_{\rho=\max{(0,\,6n-r_2-r_3)}}^{45n-r_1-r_2-r_3-1}{\binom {45n-r_1-r_2-r_3-1}{\rho}} \\ &\quad\times{\binom{50n+r_2+r_3}{56n-\rho}} \frac{x^{45n-r_1-r_2-r_3-\rho-1}}{(x-1)^{95n-r_1}}, \\ \frac{x^{45n-r_1-r_2-r_3-\rho-1}}{(x-1)^{95n-r_1}} &=A^{-1}(3-2\sqrt{2})^{2r_1+r_2+r_3+\rho+1}(4\sqrt{2})^{r_2+r_3+\rho+1}. \end{aligned} \end{equation*}$$

Thus, in the second case, we have

$$\begin{equation} \begin{aligned} \, S_3^*(\lambda)&=\frac{\lambda!\,(20n-1-\lambda)!}{(20n)!}\, \frac{(\lambda-\Lambda)!\,\Lambda!}{\lambda!}\sum_{r_1=0}^{\Lambda} \sum_{r_0=0}^{\lambda-\Lambda}\sum_{r_3=0}^{20n-1-\lambda} \sum_{\rho=\max{(0,\,6n-r_2-r_3)}}^{45n-r_1-r_2-r_3-1}(-1)^{r_1+1} \\ &\quad\times{\binom{7n-1/2}{r_1}}{\binom{19n+\Lambda-r_1-1/2}{\Lambda-r_1}}{\binom {7n-r_1-1/2}{r_2}} {\binom{12n+\lambda}{\lambda-\Lambda-r_2}} \\ &\quad\times{\binom{7n-r_1-r_2-1}{r_3}}{\binom{32n+r_2}{20n-1-\lambda-r_3}} {\binom {45n-r_1-r_2-r_3-1}{\rho}} \\ &\quad\times{\binom{50n+r_2+r_3}{56n-\rho}}(3-2\sqrt{2})^{2r_1+r_2+r_3+\rho+1} (4\sqrt{2})^{r_2+r_3+\rho+1}. \end{aligned} \end{equation} \tag{ 71 }$$

All the summands in (64) evaluated using the formulae (65), (66), (70) and (71) are of the form $(3-2\sqrt{2})^{N}R$ or $(3-2\sqrt{2})^{N}\sqrt{2}\,R$, and the summands in the formula (69) are of the form $(3-2\sqrt{2})^{N}R\ln{2}$ or $(3-2\sqrt{2})^{N}R\sqrt{2}\,\ln{2}$, where $R\in\mathbb{Q}$ and $N\in\mathbb{Z^+}$. We note that $(3+2\sqrt{2})^{N}=A_N+B_N\sqrt{2}$ and $(3-2\sqrt{2})^{N}=A_N-B_N\sqrt{2}$ for $N\in\mathbb{N}$, where $A_N,B_N\in\mathbb{N}$. Correspondingly, $\sqrt{2}(3-2\sqrt{2})^N=-2B_N+A_N\sqrt{2}$. Obviously, $\lim_{N\to\infty}({1}/{N})\ln{A_N}=\lim_{N\to\infty}({1}/{N})\ln{B_N}=3+2\sqrt{2}$.

The total number of summands in $L_n$ is estimated as $O(n^5)$. Thus, for $\gamma_2$ we have the bound $\lim_{n\to\infty}\sup({1}/{n})\ln{\Lambda}\leq 56-\Delta+\lim_{n\to\infty}({1}/{n})\ln{|S|}$, where $S$ is the summand of maximal modulus among all above sums after replacing $3-2\sqrt{2}$ by $3+2\sqrt{2}$. The asymptotic behaviour of the binomial coefficients is calculated using Lemma 13.

Computer calculations show that the corresponding maximal summand is attained in the sum (69) for the following values of parameters: $\Lambda=\Lambda'n+O(1)$, $r_1=r_1'n+O(1)$, $r_2=r_2'n+O(1)$ and $\rho=\rho'n+O(1)$, where

$$\begin{equation*} (\Lambda';\,r_1';\,r_2';\,\rho')\approx(1.202;\,0.2043;\,4.6937;\,37.8025). \end{equation*}$$

Hence by Lemma 13, it follows from (69) that

$$\begin{equation*} \begin{aligned} \, &\lim_{n\to\infty}\sup\frac{1}{n}\ln{\Lambda}\leq 56 -\Delta+\Lambda'^*+(20-\Lambda')^*-20^*+7^*-(7-r_1')^*-r_1'^* \\ &\qquad+ (19+\Lambda'-r_1')^*-19^*-(\Lambda'-r_1')^*+(7-r_1')^*- (7-r_1'-r_2')^*-r_2'^* \\ &\qquad-(20-\Lambda'-r_2')^*+32^*-(12+\Lambda'+r_2')^*+(45-r_1'-r_2')^* \\ &\qquad- (45-r_1'-r_2'-\rho')^*-\rho'^*+(50+r_2')^*-(56-\rho')^*-(\rho'+r_2'-6)^* \\ &\qquad+(2r_1'+r_2'+\rho') \ln{(3+2\sqrt{2})}+(r_2'+\rho')\ln{(4\sqrt{2})} \approx 241.793372\equiv\gamma_2. \end{aligned} \end{equation*}$$

Then, by Lemma 4, the inequality (1) holds for

$$\begin{equation*} \mu=\frac{2(\gamma_1+\gamma_3)}{(\gamma_3-\gamma_2)}\approx 7.98101. \end{equation*}$$

This completes the proof of Theorem 1.

4.1. The number $\ln{2}$ occupies a special position in the theory of Diophantine approximation. As Nesterenko said in the paper [9], ``$\ln{2}$ is a natural model for comparing the different methods developed for estimating the irrationality exponent for logarithms of rational numbers''. The best estimate for the irrationality measure of the number $\ln{2}$ is due to Marcovecchio [3]: $\mu\left(\ln{2}\right)\leq 3.57455390{\dots}\,$, and the estimate for the measure of quadratic irrationality is due to Polyansky [10]: $\mu_2(\ln{2})\leq 12.84161{\dots}\,$ . For obvious reasons, the estimate obtained in this paper is between these values.

4.2. In [1], to obtain an estimate of the form (1), the classical hypergeometric construction with integer parameters was used, while in [2] an analogous integral (but with half-integral parameters) was applied. In essence, the integral (2) is a linear combination of hypergeometric integrals with half-integral parameters (see the equations (7)–(9) in this paper). This linear combination is arranged in such a way that the coefficients of the corresponding linear form (see (35) and (21)) have a relatively small common denominator. We note that an integral of the form (2) was first applied in the paper [6] to obtain a new bound for the irrationality measure of the number ${\pi}/{\sqrt{3}}$.

4.3. The use of a combined differentiation with the help of Lemma 11, which enables us to reduce the value of the constant $\gamma_2$, also makes an important contribution.

4.4. The most difficult and laborious part of the work was to obtain a sufficiently small value of the constant $\gamma_2$. It is probable that the bound given in this paper is not definitive and that the methods developed in §3 can be improved.

4.5. The first author has obtained several new results using the integral (2). In particular, he managed to improve the bound for the approximation of the number $\pi$ by numbers in the field $\mathbb{Q}(\sqrt{3})$. These results are currently being prepared for publication.

The authors dedicate this paper to the centenary jubilee of Professor A. B. Shidlovskii, who was the teacher of the second author for many years.