---
abstract: 'We solve connection problem between fundamental solutions at singular points $0$ and $1$ for the generalized hypergeometric function, using analytic continuation of the integral representation. All connection coefficients are products of the sine and the cosecant.'
address:
- 'YM: School of Mathematics and Physics, Kanazawa University, Kanazawa, Ishikawa 920-1192, Japan'
- 'HN: School of Mathematics and Physics, Kanazawa University, Kanazawa, Ishikawa 920-1192, Japan'
author:
- 'Y. Matsuhira and H. Nagoya'
---

[Connection problem for the generalized hypergeometric function]{}

Introduction
============

Let $n \in {{\mathbb Z}}$ ($n>0$), $\alpha_1, \dots, \alpha_{n+1}, \beta_1, \dots, \beta_n \in {{\mathbb C}}$. The generalized hypergeometric series is $$\begin{aligned}
 _{n+1} F _n \left( \begin{matrix}
\alpha_1 ,  \dots ,  \alpha_{n+1} \\
\beta_1 ,  \dots ,   \beta_n \\
\end{matrix} ; z \right) =\sum_{k=0}^\infty \frac{(\alpha_1)_k \cdots (\alpha_{n+1})_k} {(\beta_1)_k \cdots (\beta_n)_k k!}z^k, \end{aligned}$$ where $(a)_k = a(a+1) \cdots (a+k-1)$. The generalized hypergeometric series converges on $|z|<1$ and satisfies the Fuchsian differential equation with three singular points $0$, $1$, $\infty$: $$\left\{ \frac{d}{dz}  \prod_{k=1}^{n}\left(\frac{d}{dz} + \beta_k -1\right) 
 -z \prod_{k=1}^{n+1}\left(\frac{d}{dz}+\alpha_k\right) \right\}F=0.$$ We call this differential equation the generalized hypergeometric equation and a solution to the equation a generalized hypergeometric function.

The Riemann scheme, which is the table of the characteristic exponents, of the generalized hypergeometric equation is $$\begin{aligned}
\left\{
 \begin{array}{ccc}
      z=0 & z=1 &  z=\infty \\
      0 & 0 & \alpha_1 \\
      1-\beta_1 & 1 & \alpha_2 \\
      \vdots & \vdots  & \vdots \\
      1-\beta_{n-1} & n-1 &  \alpha_n \\
      1-\beta_n & \sum_{i=1}^{n}\beta_i-\sum_{i=1}^{n+1}\alpha_i & \alpha_{n+1}
    \end{array}
\right\}. \end{aligned}$$ Assume $\alpha_i-\beta_j\notin{{\mathbb Z}}$ and $\beta_i-\beta_j\notin{{\mathbb Z}}$ ($i\neq j$). Then the generalized hypergeometric equation admit a fundamental system of solutions at $z=0$ given by $$\begin{aligned}
f_i^{(0)}(z)=(-z)^{1-\beta_i} {}_{n+1} F _n \left( \begin{matrix}
\alpha_1-\beta_i+1 ,  \alpha_2-\beta_i+1, \dots ,  \alpha_{n+1}-\beta_i+1 \\
\beta_1-\beta_i+1 ,  \dots ,  \widehat{\beta_i-\beta_i+1} ,  \dots ,  \beta_{n+1}-\beta_i+1 \\
\end{matrix} ; z \right)\quad (i=1,\ldots,n+1), \end{aligned}$$ where the symbol $\widehat{A}$ means omitting $A$ and $\beta_{n+1}=1$, and a fundamental system of solutions at $z=\infty$ given by $$\begin{aligned}
f_i^{(\infty)}(z)=(-z)^{-\alpha_i}  {}_{n+1} F _n \left( \begin{matrix}
\alpha_i-\beta_1+1 ,  \alpha_i-\beta_2+1, \dots ,  \alpha_i-\beta_{n+1}+1 \\
\alpha_i-\alpha_1+1 ,  \dots ,  \widehat{\alpha_i-\alpha_i+1} ,  \dots ,  \alpha_i-\alpha_{n+1}+1 \\
\end{matrix} ; \frac{1}{z} \right)\quad (i=1,\ldots,n+1). \end{aligned}$$ Fundamental systems of solutions at $z=1$ consist of one non-holomorphic solution with the characteristic exponent $\sum_{i=1}^{n}\beta_i-\sum_{i=1}^{n+1}\alpha_i$ and $n$ holomorphic solutions.

Connection problem between fundamental systems of solutions of the generalized hypergeometric equation has been solved by various authors by several methods [@Kawabata; @1], [@Kawabata; @2], [@Kawabata; @3] [@Mimachi; @intersection; @numbers; @for; @n+1Fn], [@Norlund], [@Okubo; @Takano; @Yoshida], [@S], [@Winkler]. The choice of the fundamental systems of solutions $$X_0=\left\{ f_1^{(0)}(z),\ldots, f_{n+1}^{(0)}(z)\right\},\quad 
X_\infty=\left\{ f_1^{(\infty)}(z),\ldots, f_{n+1}^{(\infty)}(z)\right\}$$ at $z=0$ and $z=\infty$ is canonical. All connection coefficients associated with $X_0$ and $X_\infty$ are products of the Gamma function and the inverse of the Gamma function. We note that if we multiply $f_i^{(0)}(z)$ and $f_i^{(\infty)}(z)$ by suitable scalars, then connection coefficients become products of the sine and the cosecant.

On the other hand, there is no canonical choice of fundamental systems of solutions at $z=1$. A connection matrix depends on choice of the fundamental system of solutions and connection coefficients are not necessarily products of the Gamma function and the inverse of the Gamma function. If we take the fundamental system of solutions at $z=1$ by $(z-1)^i+O((z-1)^n)$ ($i=0,1,\ldots,n-1$), $(z-1)^a(1+O(z-1))$ with $a=\sum_{i=1}^{n}\beta_i-\sum_{i=1}^{n+1}\alpha_i$, then the connection coefficients with $X_0$ or $X_\infty$ involve values of the generalized hypergeometric series at $z=1$. Kawabata reported in [@Kawabata; @3] that there is a fundamental system of solutions at $z=1$ such that non-diagonal elements of the connection matrix with $X_0$ or $X_\infty$ are products of the sine and the cosecant and diagonal elements of that are one minus products of the sine and the cosecant.

To the author’s knowledge, connection problem between fundamental systems of solutions at singular points $0$ and $1$ for the generalized hypergeometric equation has not been solved [*completely*]{}. The aim of the present note is to do that. Namely, we give connection matrices whose elements are products of the sine and the cosecant. In order to obtain the connection formula expressing a fundamental system of solution at $z=1$ in terms of the integral representations corresponding to $X_0$, we calculate analytic continuation of the connection formula expressing the integral representations corresponding to $X_\infty$ in terms of the integral representations corresponding to $X_0$ [@Mimachi; @intersection; @numbers; @for; @n+1Fn]. We deform the domains of integration loaded with the integrand of the generalized hypergeometric function. The technique dealing with deformation for the loaded cycles associated with Selberg type integrals is elaborated in [@Mimachi; @Iwahori-Hecke].

Unfortunately, by using analytic continuation of the connection formula expressing the integral representations corresponding to $X_0$ in terms of the integral representations corresponding to $X_\infty$ [@Mimachi; @intersection; @numbers; @for; @n+1Fn], we have not found a way to obtain the connection formula expressing the integral representations corresponding to $X_0$ in terms of the fundamental system of solution at $z=1$. Instead, we present directly the inverse matrix of the connection matrix expressing a fundamental system of solution at $z=1$ in terms of the integral representations corresponding to $X_0$, and give a proof by the residue calculus.

There had been two methods deriving connection formulas of rigid Fuchsian systems by hypergeometric integrals. One is to use the Cauchy’s integral theorem for obtaining linear relations among loaded cycles [@Aomoto; @1], [@Aomoto; @2]. Such method was used for obtaining the connection formulas of the Simpson’s Even four [@Haraoka; @Mimachi], ${}_3F_2$ [@Mimachi; @3F2], sixth order rigid Fuchsian systems derived from conformal field theory [@BHS]. However, in the case of a general $n$-multiple hypergeometric integrals, there are too many linear relations among loaded cycles to obtain connection formulas between fundamental systems of solutions. Another method is to compute intersection numbers of loaded cycles [@Mimachi; @intersection; @numbers; @for; @n+1Fn], which avoids to solve too many linear relations among loaded cycles. We believe that to compute analytic continuation of known connection formulas is also useful for obtaining another connection formulas. Especially because the connection coefficients for multiplicity-free case, such as the connection coefficients for elements of $X_0$ and $X_\infty$, is solved in [@Oshima].

The plan of the paper is as follows. In Section 2, we prepare notations and recall the connection matrix of the generalized hypergeometric function between $X_0$ and $X_\infty$ in [@Mimachi; @intersection; @numbers; @for; @n+1Fn]. In Section 3, we introduce domains of integration for a fundamental system of solutions at $z=1$, and prove the connection formulas between the fundamental system of solutions at $z=1$ and $z=0$. At the end, we remark on periodicity of the connection matrices.

Preliminary
===========

We consider a multi-valued function $$\begin{aligned}
u(t)=\prod_{i=1}^n t_i^{\lambda_i} \prod_{i=1}^{n} (t_{i}-t_{i-1})^{\mu_i}
(t_n-z)^{\mu_{n+1}}\end{aligned}$$ for $t=(t_1,\ldots,  t_n)\in{{\mathbb C}}^n$ with parameters $\lambda_i,\mu_i\in{{\mathbb C}}$ defined on $$\begin{aligned}
T_z={{\mathbb C}}^n-\bigcup_{i=1}^n \{t_i=0\} \cup \bigcup_{i=1}^{n+1} \{t_{i-1}-t_i=0\}, \end{aligned}$$ where $$\begin{aligned}
t_0=1,\quad t_{n+1}=z. \end{aligned}$$

The function $u(t)$ is the integrand of an integral representation of the generalized hypergeometric series ${}_{n+1}F_n(z)$. Namely, we have $$\begin{aligned}
\label{eq_integral_representation_for_GHS}
{}_{n+1}F_n\left( \begin{matrix}
\alpha_1 ,  \dots ,  \alpha_{n+1} \\
\beta_1 ,  \dots ,   \beta_n \\
\end{matrix} ; z \right)=&\prod_{1\le i\le n}B(\alpha_i,\beta_i-\alpha_i)
 \int_{D^{(0)}_{n+1}} u(t)dt,\end{aligned}$$ where $B(\alpha,\beta)$ is the Beta function and $D^{(0)}_{n+1}=\{t\in T_z\mid 1<t_1<t_2\cdots<t_n\}$ with $$\label{eq_transformation_lambda_alpha_beta}
\lambda_i=\alpha_{i+1}-\beta_i\quad (1 \leq i \leq n), \quad 
\mu_i=\beta_i-\alpha_i-1\quad (1 \leq i \leq n+1).$$ Here, we suppose $$\begin{aligned}
\mathrm{Re}(\alpha_i)>0,\quad \mathrm{Re}(\beta_i-\alpha_i)>0 \quad (1\le i\le n)\end{aligned}$$ for the convergence of the integral and fix the arguments as $$\begin{aligned}
\mathrm{arg}(t_i)=\mathrm{arg}(t_i-t_{i-1})=0\quad (1\le i\le n),
\quad |\mathrm{arg}(t_n-z)|<\frac{\pi}{2}. \end{aligned}$$

The formula implies that the domain $D^{(0)}_{n+1}$ of integration describes the asymptotic behaviour of the holomorphic solution at $z=0$. There are domains of integration yielding the asymptotic behaviours of the non-holomorphic solutions at $z=0$ with the characteristic exponents $1-\beta_i$ ($1\le i\le n$), and the non-holomorphic solutions at $z=\infty$ with the characteristic exponents $1-\beta_i$ ($1\le i\le n+1$) [@Mimachi; @intersection; @numbers; @for; @n+1Fn]. In order to define integrals for domains of integration, we should fix branches of $u(t)$.

For the convenience, suppose $z \in {{\mathbb R}}-\{0,1\}$. We first fix branches of $u(t)$ for real $z$, and then we consider analytic continuation of $u(t)$ for general $z$. For a simply connected domain $D$ in the real part $T_{{\mathbb R}}$ of $T_z$, let us define $u_D(t)$ as $$\begin{aligned}
u_D(t)=\prod_{i=1}^n (\epsilon_it_i)^{\lambda_i} 
\prod_{i=1}^{n+1} (\eta_i(t_{i-1}-t_i))^{\mu_i}, \end{aligned}$$ where $\epsilon_i,\eta_i\in\{1,-1\}$ such that $\epsilon_it_i>0$ and $\eta_i(t_{i-1}-t_i)>0$ on $D$. We fix the arguments of all $\epsilon_i t_i$ and $\eta_i(t_{i-1}-t_i)$ as $0$. The function $u_D(t)$ is a branch of $u(t)$ multiplied by a scalar.

In what follows, for convenience we use $$\begin{aligned}
&e(A) =\exp (\pi \sqrt{-1} A), \quad
s(A) =\sin (\pi A) \quad (A \in {{\mathbb C}}), 
\\
&\alpha_{i,j}=\sum_{s=i}^j\alpha_s,\quad 
\beta_{i,j}=\sum_{s=i}^j\beta_s\quad (i<j), \end{aligned}$$ and for the exponents $\lambda_i, \mu_i$ of $u(t)$ $$\begin{aligned}
\lambda_{i, j} &=
\begin{cases}
\lambda_i + \cdots +\lambda_j & (i \leq j) , \\
0 & (i=j+1) , \\
-(\lambda_{j+1}+ \cdots +\lambda_{i-1}) & (i \geq j+2) , 
\end{cases} \\
\mu_{i, j} &=
\begin{cases}
\mu_i + \cdots +\mu_j & (i \leq j) , \\
0 & (i=j+1) , \\
-(\mu_{j+1}+ \cdots +\mu_{i-1}) & (i \geq j+2) ,
\end{cases} \\
e_{i, j} &= e(\lambda_{i, j}), \\
\tilde{e}_{i, j} &= e(\mu_{i, j}).   \end{aligned}$$ We note that for all $i$ we have $$\begin{aligned}
\lambda_{i}=\lambda_{i, i}, 
\quad \mu_{i}=\mu_{i, i}.\end{aligned}$$

Connection problem between fundamental solutions at $z=0$ and $z=\infty$
------------------------------------------------------------------------

In this subsection, we recall the results in [@Mimachi; @intersection; @numbers; @for; @n+1Fn]. We fix $z \in {{\mathbb C}}$ such that $z < 0$. Set $$\begin{aligned}
D_i^{(0)}=&\{t\in T_{{\mathbb R}}\mid z<t_n< \cdots <t_i <0,\ 1<t_1< \cdots < t_{i-1} \} \quad (1 \leq i \leq  n+1),
\\
D_i^{(\infty)}=&\{t\in T_{{\mathbb R}}\mid t_i< \cdots <t_n <z,\ 0<t_{i-1}< \cdots < t_1< 1 \} \quad (1 \leq i \leq  n+1).\end{aligned}$$ The domains of integration in the case of $n=2$ are pictured as follows.

(-3,0)–(3.5,0)node\[anchor=west\][$t_2=0$]{}; (-3,-1.5)–(3.5,-1.5)node\[anchor=west\][$t_2=z$]{}; (0,3.5)–(0,-3)node\[anchor=north\][$t_1=0$]{}; (2,3.5)–(2,-3)node\[anchor=north\][$t_1=1$]{}; (-3,-3)–(3.5,3.5)node\[anchor=west\][$t_2=t_1$]{}; node at (2.5,3) [$D_1^{(0)}$]{}; node at (2.5,-0.75) [$D_2^{(0)}$]{}; node at (1.2,0.5) [$D_3^{(\infty)}$]{}; node at (1,-2) [$D_2^{(\infty)}$]{}; node at (-0.5,-1) [$D_3^{(0)}$]{}; node at (-2.5,-2) [$D_1^{(\infty)}$]{};

The orientation of the domains $D_i^{(0)}$ or $D_i^{(\infty)}$ of integration is fixed to be natural one induced from $T_{{\mathbb R}}$. In [@Mimachi; @intersection; @numbers; @for; @n+1Fn], it was shown that the domains $D_i^{(0)}$ of integration give a fundamental system of solutions at $z=0$, and $D_{i}^{(\infty)}$ gives a fundamental system of solutions at $z=\infty$.

\[[@Mimachi; @intersection; @numbers; @for; @n+1Fn], Proposition 2.1\] \[\[M2\], prop 2.1\] (1) For a fixed $i$ such that $1 \leq i \leq n+1$, if ${\rm Re}(\alpha_i-\beta_s+1)>0$ and ${\rm Re}(\beta_s-\alpha_s)>0$ for $1 \leq s \leq n+1$ with $s \neq i$ and $|z|>1$, then we have $$\begin{aligned}
&\int _{D_i^{(\infty)}} u_{D_i^{(\infty)}}(t)dt_1 \cdots dt_n \label{sol_infty} 
=\prod_{1 \leq s \leq n+1, s \neq i} B(\alpha_i-\beta_s+1, \beta_s-\alpha_s)f_i^{(\infty)}(z),\end{aligned}$$ where $$\begin{aligned}
f_i^{(\infty)}(z)&=(-z)^{-\alpha_i}  {}_{n+1} F _n \left( \begin{matrix}
\alpha_i-\beta_1+1 ,  \alpha_i-\beta_2+1, \dots ,  \alpha_i-\beta_{n+1}+1 \\
\alpha_i-\alpha_1+1 ,  \dots ,  \widehat{\alpha_i-\alpha_i+1} ,  \dots ,  \alpha_i-\alpha_{n+1}+1 \\
\end{matrix} ; \frac{1}{z} \right). \end{aligned}$$

\(2) For a fixed $i$ such that $1 \leq i \leq n+1$, if ${\rm Re}(\alpha_s-\beta_i+1)>0$, ${\rm Re}(\beta_s-\alpha_s)>0$ for $1 \leq s \leq n+1$ with $s \neq i$ and $|z|<1$, then we have $$\begin{aligned}
&\int _{D_i^{(0)}} u_{D_i^{(0)}}(t)dt_1 \cdots dt_n \label{sol_0} 
=\prod_{1 \leq s \leq n+1, s \neq i} B(\alpha_s-\beta_i+1, \beta_s-\alpha_s)f_i^{(0)}(z),\end{aligned}$$ where $$\begin{aligned}
f_i^{(0)}(z)&=(-z)^{1-\beta_i} {}_{n+1} F _n \left( \begin{matrix}
\alpha_1-\beta_i+1 ,  \alpha_2-\beta_i+1, \dots ,  \alpha_{n+1}-\beta_i+1 \\
\beta_1-\beta_i+1 ,  \dots ,  \widehat{\beta_i-\beta_i+1} ,  \dots ,  \beta_{n+1}-\beta_i+1 \\
\end{matrix} ; z \right). \end{aligned}$$

Let $F_D(z)=\int_D u_D(t)dt$.

\[\[M2\], prop 2.5\] For $i$ and $j$ such that $1\leq i,j\leq n+1$, suppose that $$\begin{aligned}
&\mathrm{Re}(\alpha_i-\beta_j+1)>0,\quad \mathrm{Re}(\beta_j-\alpha_j)>0
\quad (i\neq j), 
\\
&\alpha_i-\beta_j\notin{{\mathbb Z}},\quad \beta_i-\beta_j\notin{{\mathbb Z}}\quad (i\neq j). \end{aligned}$$ Then we have $$\label{eq_connection_formula_i0}
F_{D_i^{(\infty)}}(z)=\sum_{1 \leq j \leq n+1} \frac{s(\beta_i-\alpha_i)}{s(\beta_j-\alpha_i)} \prod_{1 \leq s \leq n+1,\atop s \neq j} \frac{s(\alpha_s-\beta_j)}{s(\beta_s-\beta_j)} \times F_{D_j^{(0)}}(z)$$ for $1\leq i \leq n+1$.

Connection problem between fundamental system of solutions at $z=0$ and $z=1$
=============================================================================

Asymptotic behaviour
--------------------

We fix $z \in {{\mathbb C}}$ such that $0<z < 1$. Set $$\begin{aligned}
\tilde{D}_i^{(0)}&=\{t\in T_{{\mathbb R}}\mid 0<t_i< \cdots <t_n <z,\ 1<t_1< \cdots < t_{i-1} \} \quad (1 \leq i \leq  n+1), \label{domain0_0<z<1} 
\\
\tilde{D}_i^{(1)}&=\{t\in T_{{\mathbb R}}| t_i< \cdots <t_n <0,\ 0<t_{i-1}< \cdots < t_1< 1 \} \quad (1 \leq i \leq  n), \label{holo_domain1_0<z<1}
 \\
\tilde{D}_{n+1}^{(1)}&=\{t\in T_{{\mathbb R}}| z<t_n< \cdots < t_1< 1 \}. \label{nonholo_domain1_0<z<1}\end{aligned}$$ The domains of integration in the case of $n=2$ are pictured as follows.

(-3,0)–(3.5,0)node\[anchor=west\][$t_2=z$]{}; (-3,-1.5)–(3.5,-1.5)node\[anchor=west\][$t_2=0$]{}; (-1.5,3.5)–(-1.5,-3)node\[anchor=north\][$t_1=0$]{}; (2,3.5)–(2,-3)node\[anchor=north\][$t_1=1$]{}; (-3,-3)–(3.5,3.5)node\[anchor=west\][$t_2=t_1$]{}; node at (2.5,3) [$\tilde{D}_3^{(0)}$]{}; node at (2.5,-0.75) [$\tilde{D}_2^{(0)}$]{}; node at (1.25,0.5) [$\tilde{D}_3^{(1)}$]{}; node at (-1,-0.5) [$\tilde{D}_1^{(0)}$]{}; node at (0.25,-2) [$\tilde{D}_2^{(1)}$]{}; node at (-2.5,-2) [$\tilde{D}_{1}^{(1)}$]{};

The orientation of the domains $\tilde{D}_i^{(0)}$ or $\tilde{D}_i^{(1)}$ of integration is fixed to be natural one induced from $T_{{\mathbb R}}$. In [@Mimachi; @intersection; @numbers; @for; @n+1Fn], it was shown that the domains $\tilde{D}_i^{(0)}$ of integration give a fundamental system of solutions at $z=0$, and $\tilde{D}_{n+1}^{(1)}$ gives the non-holomorhic solution at $z=1$.

\[\[M2\], prop 3.1\] (1) For a fixed $i$ such that $1 \leq i \leq n+1$, if ${\rm Re}(\alpha_s-\beta_i+1)>0$ and ${\rm Re}(\beta_s-\alpha_s)>0$ for $1 \leq s \leq n+1$ with $s \neq i$ and $|z|<1$, then we have $$\begin{aligned}
&\int _{\tilde{D}_i^{(0)}} u_{\tilde{D}_i^{(0)}}(t)dt_1 \cdots dt_n
=\prod_{1 \leq s \leq n+1,\atop s \neq i} B(\alpha_s-\beta_i+1, \beta_s-\alpha_s)f_i^{(0)}(z),\end{aligned}$$ where $$\begin{aligned}
f_i^{(0)}(z)&=z^{1-\beta_i} {}_{n+1} F _n \left( \begin{matrix}
\alpha_1-\beta_i+1 ,  \alpha_2-\beta_i+1, \dots ,  \alpha_{n+1}-\beta_i+1 \\
\beta_1-\beta_i+1 ,  \dots ,  \widehat{\beta_i-\beta_i+1} ,  \dots ,  \beta_{n+1}-\beta_i+1 \\
\end{matrix} ; z \right).  \end{aligned}$$

\(2) If ${\rm Re}(\beta_{1,s}-\alpha_{1,s})>0$ for $1 \leq s \leq n$ and ${\rm Re}(\beta_s-\alpha_s)>0$ for $1 \leq s \leq n+1$, and $|1-z|<1$, then we have $$\begin{aligned}
&\int _{\tilde{D}_{n+1}^{(1)}} u_{\tilde{D}_{n+1}^{(1)}}(t)dt_1 \cdots dt_n \label{nonholosol_1}
=\prod_{s=1}^{n} B(\beta_{1,s}-\alpha_{1,s}, \beta_{s+1}-\alpha_{s+1})f_{n+1}^{(1)}(z), \end{aligned}$$ where $$\begin{aligned}
f_{n+1}^{(1)}(z)=&(1-z)^{\beta_{1,n}-\alpha_{1,n+1}}  \sum_{i_1, \dots, i_n \geq 0} 
\prod_{s=1}^{n} \frac{(\beta_s-\alpha_{s+1})}{i_s!}\prod_{s=1}^{n}\frac{(\sum_{k=1}^{s} (\beta_k-\alpha_k))_{i_1+\cdots+i_s}}
{(\sum_{k=1}^{s+1}(\beta_k-\alpha_k))_{i1+\cdots+i_s}}(1-z)^{i_1+\cdots+i_n}.\end{aligned}$$

\[holo\_sols\_at\_1\] For a fixed $i$ such that $1 \leq i \leq n$, if $$\begin{aligned}
&\mathrm{Re}(\alpha_i-\beta_s+1)>0\quad (1\leq s\leq n+1,\ s\neq i),
\\
&\mathrm{Re}(\beta_s-\alpha_s)>0\quad (1\leq s\leq n,\ s\neq i),
\\
&\mathrm{Re}(\alpha_{n+1}-\beta_i+1)>0, \end{aligned}$$ and $|1-z|<1$, then we have $$\begin{aligned}
\label{eq_holosol_1}
\int _{\tilde{D}_{i}^{(1)}} u_{\tilde{D}_{i}^{(1)}}(t)dt_1 \cdots dt_n =&\prod_{s=1,s\neq i}^{n} B(\alpha_i-\beta_s+1, \beta_s-\alpha_s) 
  B(\alpha_i, \alpha_{n+1}-\beta_i+1) f_{i}^{(1)}(z), \end{aligned}$$ where $$\begin{aligned}
f_{i}^{(1)}(z)=&\sum_{m_1, m_2 \geq 0}\frac{(\alpha_{n+1})_{m_1}
(\alpha_i-\beta_i+1)_{m_2}
 (\alpha_1)_{m_1+m_2}}{m_1!(\alpha_i+\alpha_{n+1}-\beta_i+1)_{m_1+m_2}} \sum_{m_3=0}^{m_2}
 \frac{(-1)^{m_3}}{ m_3 ! (m_2-m_3)!}
 \prod_{s=1,s\neq i}^{n}\frac{(\alpha_i-\beta_s+1)_{m_3}}
{(\alpha_i-\alpha_s+1)_{m_3}}
 (1-z)^{m_1}.\end{aligned}$$

We change the integration variables as $$\begin{aligned}
t_s&=u_1u_2 \cdots u_s \ (1 \leq s \leq i-1) , \\
t_s&=u_s^{-1}u_{s+1}^{-1} \cdots u_n^{-1}(u_n-1) \ (i \leq s \leq n).\end{aligned}$$ Its Jacobian is $$\begin{aligned}
\frac{\partial(t_1, \dots , t_n)}{\partial(u_1, \dots, u_n)}=u_1^{i-2}u_2^{i-3}\cdots u_{i-2}^{1}u_{i-1}^{0}u_{i}^{-2}u_{i+1}^{-3}\cdots u_{n-1}^{i-n-1}u_n^{i-n-2}(1-u_n)^{n-i}. \end{aligned}$$ Hence, we have $$\begin{aligned}
&\int _{\tilde{D}_{i}^{(1)}} u_{\tilde{D}_{i}^{(1)}}(t)dt_1 \cdots dt_n
\\
=&\int_{(0, 1)^n} \prod_{s=1}^{i-1}u_s^{\lambda_{s, i-1}+\mu_{s+1, i-1}+i-s-1}\prod_{s=i}^{n}u_s^{-\lambda_{i, s}-\mu_{i, s+1}+i-s-2}\prod_{s=1}^{i-1}(1-u_s)^{\mu_s}  \prod_{s=i}^{n-1}(1-u_s)^{\mu_{s+1}} (1-u_n)^{\lambda_{i, n}+\mu_{i+1, n}+n-i}
 \\
& \times \left( 1-u_n(1-u_1\cdots u_{n-1}) \right)^{\mu_j} \left( 1-(1-z)u_n \right)^{\mu_{n+1}}du_1\cdots du_n
\\
=&\sum_{m_1, m_2 \geq 0} \sum_{m_3=0}^{m_2} \frac{(-\mu_{n+1})_{m_1}(-\mu_i)_{m_2}}{m_1 ! m_3 ! (m_2-m_3)!}(1-z)^{m_1}  \prod_{s=1,s\neq i}^{n} \int_{0}^{1} u_s^{\lambda_{s, i-1}+\mu_{s+1, i-1}+i-s+m_3-1}(1-u_s)^{\mu_s}du_s
 \\
&\times \int_{0}^{1} u_n^{-\lambda_{i, n}-\mu_{i, n+1}+i-n+m_1+m_2-2}(1-u_n)^{\lambda_{i, n}+\mu_{i+1, n}+n-i}du_n\end{aligned}$$ by the binomial theorem $$\begin{aligned}
(1-(1-z)u_n)^{\mu_{n+1}}&= \sum_{m_1 \geq 0} \frac{(-\mu_{n+1})_{m_1}}{m_1 !} u_n^{m_1}(1-z)^{m_1} ,
 \\
(1-u_n(1-u_1\cdots u_{n-1}) )^{\mu_i}
&=\sum_{m_2 \geq 0} \sum_{m_3=0}^{m_2} \frac{(-\mu_i)_{m_2}(-1)^{m_3}}{m_3 !(m_2-m_3)!}u_1^{m_3}\cdots u_{n-1}^{m_3}u_n^{m_2}.  \end{aligned}$$ Using the formula $$\begin{aligned}
B(\alpha+m,\beta)=\frac{(\alpha)_m}{(\alpha+\beta)_m}B(\alpha,\beta), \end{aligned}$$ we obtain $$\begin{aligned}
 &\int _{\tilde{D}_{i}^{(1)}} u_{\tilde{D}_{i}^{(1)}}(t)dt_1 \cdots dt_n
\\
=&\sum_{n_1, n_2 \geq 0} \sum_{n_3=0}^{n_2} \frac{(-\mu_{n+1})_{n_1}(-\mu_i)_{n_2}}{n_1 ! n_3 ! (n_2-n_3)!}(1-z)^{n_1}  \prod_{s=1,s\neq i}^{n} B(\lambda_{s, i-1}+\mu_{s+1, i-1}+i-s+n_3, \mu_s +1) \\
&\times B(-\lambda_{i, n}-\mu_{i, n+1}+i-n+n_1+n_2-1, \lambda_{i, n}+\mu_{i+1, n}+n-i+1) \\
=&\prod_{s=1,s\neq i}^{n} B(\lambda_{s, i-1}+\mu_{s+1, i-1}+i-s, \mu_s +1)  B(-\lambda_{i, n}-\mu_{i, n+1}+i-n-1, \lambda_{i, n}+\mu_{i+1, n}+n-i+1) \\
&\times \sum_{n_1, n_2 \geq 0} \sum_{n_3=0}^{n_2} \frac{(-\mu_{n+1})_{n_1}(-\mu_i)_{n_2}(-\lambda_{i, n}-\mu_{i, n+1}-n-1)_{n_1+n_2}}{n_1 !n_3 ! (n_2-n_3)!(-\mu_j-\mu_{n+1})_{n_1+n_2}} 
 \prod_{s=1,s\neq i}^{n}\frac{(\lambda_{s, i-1}+\mu_{s+1, i-1}+i-s)_{n_3}}{(\lambda_{s, i-1}+\mu_{s, i-1}+i-s+1)_{n_3}}
 (1-z)^{n_1}.\end{aligned}$$ Finally the transformation yields .

Main theorem
------------

In this subsection, we solve connection problem between fundamental systems of solutions at singular points $0$ and $1$ for the generalized hypergeometric equation.

\[thm\_connection\_formula\_10\] Assume for $1 \leq i,  j \leq n+1$ $$\begin{aligned}
&\mathrm{Re}(\alpha_i-\beta_j+1)>0\quad (j\neq i),
\\
&\mathrm{Re}(\beta_j-\alpha_j)>0, 
\\
&\alpha_i-\beta_j \notin {{\mathbb Z}}, \quad \beta_i-\beta_j \notin {{\mathbb Z}}\quad (j\neq i). \end{aligned}$$ Then we have $$\label{eq_connection_formula_10}
F_{\tilde{D}_i^{(1)}}(z)=\sum_{1\le j\le n+1}\frac{s(\beta_i-\alpha_i)s(\alpha_{n+1})}
{s(\beta_j-\alpha_i)s(\beta_j-\alpha_{n+1})}
\prod_{1\le k\le n+1,\atop k\neq j}\frac{s(\alpha_k-\beta_j)}{s(\beta_k-\beta_j)}
\times 
F_{\tilde{D}_j^{(0)}}(z)$$ for $1\le i\le n$ and $$\label{eq_connection_formula_10_n+1}
F_{\tilde{D}_{n+1}^{(1)}}(z)=
\sum_{j=1}^{n+1} \prod_{1 \leq k \leq n+1, \atop k \neq j} 
\frac{s(\alpha_k-\beta_j)}{s(\beta_k-\beta_j)} \times 
F_{\tilde{D}_j^{(0)}}(z).$$

By induction. If $n=1$, the connection formulas and are for the Gauss hypergeometric function.

Suppose the identity is true for $k<n$. Our strategy to prove the theorem is to calculate analytic continuation of the connection formula along the following path $\gamma$ on the plane of the variable $z$.

(-3,0)–(3,0); (-1.5,0) to \[out=270, in=180\] (0,-1); (0,-1) to \[out=0, in=180\] (1.75,0); (0, 0)node\[anchor=south\][$0$]{} circle \[radius=2pt\]; (2.5, 0)node\[anchor=south\][$1$]{} circle \[radius=2pt\]; (-1.5, 0)node\[anchor=south\][$z$]{} circle \[radius=2pt\];

First we compute analytic continuation of the integral representations along the path $\gamma$. In what follows, we abbreviate $F_D(z)$ as $D$, and $\{t\in T_{{\mathbb R}}\mid \cdots\}$ as $\{\cdots\}$.

\[lem\_analytic\_continuation\_of\_domain\] By analytic continuation along the path $\gamma$, we have $$\begin{aligned}
 \gamma^{*}\left(D_j^{(0)}\right)=&(-1)^{n-j+1}e_{j, n}\tilde{e}_{j+1, n+1}\tilde{D}_j^{(0)}\quad (1 \leq j \leq n+1),
 \\
\label{eq_analytic_continuation_infinity} 
\gamma^{*}\left(D_j^{(\infty)}\right)=&\tilde{D}_{j}^{(1)}
 +\sum_{k=j+1}^{n}e_{k, n} \biggl\{ \begin{array}{c} 
t_j< \cdots <t_{k-1} <0, 
\\
0<t_k< \cdots < t_n<z,\ 0<t_{j-1}< \cdots <t_1<1
\end{array} \biggr\}
\\
&+e_{j, n} \{ 0<t_j< \cdots <t_n <z,\ 0<t_j< \cdots < t_1<1\}\nonumber
\\
&+e_{j, n} \tilde{e}_{j} \{ 0<t_{j-1}<t_j< \cdots <t_n <z, \ 
0<t_{j-1}< \cdots < t_1< 1\}\quad (1 \leq j \leq n), 
\nonumber
\\
\gamma^{*}\left(D_{n+1}^{(\infty)}\right)=&\tilde{D}_{n+1}^{(1)}
+\tilde{e}_{n+1}\{ t_n<z,\ 0<t_n<\cdots <t_1<1\}, 
\label{eq_analytic_continuation_infinity_n+1}\end{aligned}$$ where $\gamma^{*}(D)$ stands for the result of analytic continuation along the path $\gamma$ for $D$.

By the definition of $D_j^{(0)}$, analytic continuation along the path $\gamma$ increases the arguments of $t_k$ ($j \leq k \leq n$), $t_{k-1}-t_k$ ($j+1 \leq k \leq n+1)$ by $\pi$. Hence, we have $$\begin{aligned}
\gamma^{*}(D_j^{(0)})=e_{j, n}\tilde{e}_{j+1, n+1} \{ \overleftarrow{0< t_j< \cdots <t_n <z},\ 1<t_1< \cdots < t_{j-1}\}. \end{aligned}$$ Here, $ \overleftarrow{0< t_j< \cdots <t_n <z}$ means that the orientation is inverse. Hence we obtain $$\begin{aligned}
\gamma^{*}(D_j^{(0)})&=e_{j, n}\tilde{e}_{j+1, n+1} \{ \overleftarrow{0< t_j< \cdots <t_n <z},\
 1<t_1< \cdots < t_{j-1}\} \\
&=(-1)^{n-j+1}e_{j, n}\tilde{e}_{j+1, n+1} \{ 0< t_j< \cdots <t_n <z,\ 1<t_1< \cdots < t_{j-1} \}. \end{aligned}$$

Next we compute analytic continuation of $D_j^{(\infty)}$. By the definition, after analytic continuation along the path $\gamma$, the line $\{ t_n <z\}$ divides into two lines $\{ t_n <0\}$ and $\{0<t_n<z\}$ on the plane of the variable $t_n$. Since the argument of $t_n$ increases $\pi$ on $\{0<t_n<z\}$, the line $\{0<t_n<z\}$ is multiplied by $e_n$. Similarly, $\{t_{n-1}<t_n,\ 0<t_n<z\}$ divides into two lines $\{t_{n-1}<0\}$ and $\{0<t_{n-1}<t_n\}$ on the plane of the variable $t_{n-1}$. Since the argument of $t_{n-1}$ increases $\pi$ on $\{0<t_{n-1}<t_n\}$, the line $\{0<t_{n-1}<t_n\}$ is multiplied by $e_{n-1}$. By repeating this procedure from $t_{n-2}$ plane to $t_j$ plane, we have $$\begin{aligned}
\gamma^{*}(D_j^{(\infty)})=&\tilde{D}_{j}^{(1)}
+\sum_{k=j+1}^{n}e_{k, n} \biggl\{
    \begin{array}{c}
      t_j< \cdots <t_{k-1} <0,  \\
      0<t_k< \cdots < t_n<z,\ 0<t_{j-1}< \cdots <t_1<1
    \end{array}
  \biggr\} \\
&+e_{j, n} \left\{ 0<t_{j}< \cdots <t_n <z,\  
0<t_{j-1}< \cdots < t_1<1\right\}.\end{aligned}$$ The last domain is actually divided as $$\begin{aligned}
 \{ 0<t_j< \cdots <t_n <z,\ 0<t_{j-1}< \cdots < t_1<1 \} 
=&\tilde{e}_{j} \{ 0<t_{j-1}<t_j< \cdots <t_n <z,\ 0<t_{j-1}<t_{j-2}< \cdots < t_1<1 \} \\
&+ \{ 0<t_j< \cdots <t_n <z,\ t_j<t_{j-1}< \cdots < t_1<1 \},  \end{aligned}$$ because the line $\{0<t_{j-1}<t_{j-2}\}$ splits into two lines $\{0<t_{j-1}<t_j\}$ and $\{0<t_j<t_{j-1}<t_{j-2}\}$ on the plane of the variable $t_{j-1}$, and the argument of $(t_{j-1}-t_j)$ increases $\pi$ on $\{0<t_{j-1}<t_j\}$. Therefore, we obtain . In the same way, we obtain .

We back to the proof of the theorem. The connection formula in Proposition \[\[M2\], prop 2.5\] reads as $$\begin{aligned}
D_j^{(\infty)}=\sum_{ k =1}^{n+1} (-1)^{n+k-j} \frac{s(\mu_j)}{s(\lambda_{j, k-1}+\mu_{j, k})} \prod_{1 \leq l \leq n+1,\atop l \neq k} \frac{s(\lambda_{l, k-1}+\mu_{l, k})}{s(\lambda_{l, k-1}+\mu_{l+1, k})}  D_k^{(0)}\end{aligned}$$ for the parameters $\lambda_i$, $\mu_i$ by the relations . Due to Lemma \[lem\_analytic\_continuation\_of\_domain\], analytic continuation of this for $j=1,\ldots,n$ along the path $\gamma$ is equal to $$\begin{aligned}
\label{eq_annalytic_continuation_connection_formula_1}
\tilde{D}_{j}^{(1)}
=&\sum_{ k=1}^{ n+1} (-1)^{n+k-j} \frac{s(\mu_j)}{s(\lambda_{j, k-1}+\mu_{j, k})} \prod_{1 \leq l \leq n+1,\atop l \neq k} 
\frac{s(\lambda_{l, k-1}+\mu_{l, k})}{s(\lambda_{l, k-1}+\mu_{l+1, k})} (-1)^{n-k+1}e_{k, n}\tilde{e}_{k+1, n+1}\tilde{D}_k^{(0)}{\nonumber}\\
&-\sum_{k=j+1}^{n}e_{k, n} \{ -\infty<t_j< \cdots <t_{k-1} <0,\ 0<t_k< \cdots < t_n<z,\ 0<t_{j-1}< \cdots <t_1<1\}\nonumber
\\
&-e_{j, n} \{ 0<t_j< \cdots <t_n <z,\ 0<t_j< \cdots < t_1<1\}\\
&-e_{j, n} \tilde{e}_{j} \{ 0<t_{j-1}<t_j< \cdots <t_n <z,\ 
 0<t_{j-1}< \cdots < t_1< 1\}. \nonumber\end{aligned}$$ In the integral associated with $\{ t_j< \cdots <t_{k-1} <0,\ 0<t_k< \cdots < t_n<z,\ 0<t_{j-1}< \cdots <t_1<1\}$ in the second term of , the integral with respect to the variables $t_1,\ldots, t_{k-1}$ corresponds to the integral defined for $\tilde{D}_{j}^{(1)}$ of ${}_kF_{k-1}(t_k)$ because of $0<t_k<1$. Hence, we can apply the induction hypothesis to $\{ t_j< \cdots <t_{k-1} <0,\ 0<t_k< \cdots < t_n<z,\ 0<t_{j-1}< \cdots <t_1<1\}$ and obtain a sum of integrals associated with $\{ 0<t_k< \cdots < t_n<z,\ 0<t_m<\cdots<t_k,\ 1<t_1<\cdots<t_{m-1}\}$ ($m=1,\ldots,k$), which is nothing but $\tilde{D}_m^{(0)}$ of ${}_{n+1}F_n(z)$. In the same way, we can regard the third term $\{ 0<t_j< \cdots <t_n <z,\ 0<t_j< \cdots < t_1<1\}$ in as $\tilde{D}_j^{(1)}$ of ${}_jF_{j-1}(t_j)$. Hence, we can apply the induction hypothesis to $\{ 0<t_j< \cdots <t_n <z,\ 0<t_j< \cdots < t_1<1\}$ and obtain a sum of integrals associated with $\{ 0<t_j< \cdots < t_n<z,\ 0<t_k<\cdots<t_j,\ 1<t_1<\cdots<t_{k-1}\}$ ($k=1,\ldots,j$), which is nothing but $\tilde{D}_k^{(0)}$ of ${}_{n+1}F_n(z)$. We can also regard the fourth term $\{ 0<t_{j-1}<t_j< \cdots <t_n <z,\ 
 0<t_{j-1}< \cdots < t_1< 1\}$ in as $\tilde{D}_{j-1}^{(1)}$ of ${}_{j-1}F_{j-2}(t_{j-1})$. Hence, we can apply the induction hypothesis to $\{ 0<t_{j-1}<t_j< \cdots <t_n <z,\ 
 0<t_{j-1}< \cdots < t_1< 1\}$ and obtain a sum of integrals associated with $\{ 0<t_{j-1}< \cdots < t_n<z,\ 0<t_k<\cdots<t_{j-1},\ 
1<t_1<\cdots<t_{k-1}\}$ ($k=1,\ldots,j-1$), which is nothing but $\tilde{D}_k^{(0)}$ of ${}_{n+1}F_n(z)$.

Since the identities and read as $$\begin{aligned}
\tilde{D}_j^{(1)}=&(-1)^{j-1} \sum_{k=1}^{n}
\frac{s(\mu_j)s(\mu_{n+1})}{s(\lambda_{j, k-1}+\mu_{j, k})
s(\lambda_{n+1, k-1}+\mu_{n+2, k})} 
\prod_{l=1, l\neq k}^{n}\frac{s(\lambda_{l, k-1}+\mu_{l, k})}
{s(\lambda_{l, k-1}+\mu_{l+1, k})}   \tilde{D}_{k}^{(0)} 
\\
&+(-1)^{j-1}\frac{s(\mu_j)}{s(\lambda_{j, n}+\mu_{j, n+1})} 
\prod_{l=1}^{n}\frac{s(\lambda_{l, n}+\mu_{l, n+1})}
{s(\lambda_{l, n}+\mu_{l+1, n+1})} \tilde{D}_{n+1}^{(0)} \end{aligned}$$ for the parameters $\lambda_i$, $\mu_i$ by the relations , for $j=1,\ldots, n$ we obtain $$\begin{aligned}
\label{eq_connection_formula_proof_j=1-n}
\tilde{D}_{j}^{(1)} 
=&\sum_{ k=1}^{  n+1} (-1)^{-j+1} e_{k, n}\tilde{e}_{k+1, n+1} \frac{s(\mu_j)}{s(\lambda_{j, k-1}+\mu_{j, k})} \prod_{1 \leq l \leq n+1, l \neq k} \frac{s(\lambda_{l, k-1}+\mu_{l, k})}
{s(\lambda_{l, k-1}+\mu_{l+1, k})}  \tilde{D}_k^{(0)} 
\\
&-\sum_{k=j+1}^{n} e_{k, n} \left\{ (-1)^{j-1} \sum_{m=1}^{k-1}
\frac{s(\mu_j)s(\mu_{k})}{s(\lambda_{j, m-1}+\mu_{j, m})
s(\lambda_{k, m-1}+\mu_{k+1, m})} 
\prod_{l=1, l\neq m}^{k-1}\frac{s(\lambda_{l, m-1}+\mu_{l, m})}{s(\lambda_{l, m-1}+\mu_{l+1, m})}   \tilde{D}_{m}^{(0)}
\right. {\nonumber}\\
&\left.+(-1)^{j-1}\frac{s(\mu_j)}{s(\lambda_{j, k-1}+\mu_{j, k})} 
\prod_{l=1}^{k-1}\frac{s(\lambda_{l, k-1}+\mu_{l, k})}
{s(\lambda_{l, k-1}+\mu_{l+1, k})} \tilde{D}_{k}^{(0)} 
\right\}{\nonumber}\\
&-e_{j, n}\sum_{k=1}^{j}(-1)^{j-1}
 \prod_{l=1, l \neq k}^{j} \frac{s(\lambda_{l, k-1}+\mu_{l, k})}{s(\lambda_{l, k-1}+\mu_{l+1, k})}\tilde{D}_{k}^{(0)}
-e_{j, n}\tilde{e}_{j}\sum_{k=1}^{j-1}(-1)^{j} \prod_{l=1, l \neq k}^{j-1} \frac{s(\lambda_{l, k-1}+\mu_{l, k})}
{s(\lambda_{l, k-1}+\mu_{l+1, k})}\tilde{D}_{k}^{(0)} {\nonumber}\\
=&\sum_{k=1}^{j-1} \left(A_k-\sum_{m=j+1}^{n}B_{k, m}-E_k-F_k \right) \tilde{D}_{k}^{(0)}+\sum_{k=j}^{n} \left( A_k-\sum_{m=k+1}^{n+1}B_{k, m}-C_k \right) \tilde{D}_{k}^{(0)},{\nonumber}\end{aligned}$$ where $$\begin{aligned}
A_k&=(-1)^{j-1}e_{k, n}\tilde{e}_{k+1, n+1}\frac{s(\mu_j)}{s(\lambda_{j, k-1}+\mu_{j, k})}\prod_{l=1, l \neq k}^{n+1} \frac{s(\lambda_{l, k-1}+\mu_{l, k})}{s(\lambda_{l, k-1}+\mu_{l+1, k})}\quad (1\leq k\leq n+1) , \\
B_{k, m}&=e_{m, n} (-1)^{j-1} \frac{s(\mu_j)s(\mu_{m})}{s(\lambda_{j, k-1}+\mu_{j, k})
s(\lambda_{m,k-1}+\mu_{m+1,k})}
\prod_{l=1, l \neq k}^{m-1} \frac{s(\lambda_{l, k-1}+\mu_{l, k})}{s(\lambda_{l, k-1}+\mu_{l+1, k})} \quad (1\leq m\leq n,\ 1\leq k\leq m), 
 \\
C_k&=e_{k, n}(-1)^{j-1}\frac{s(\mu_j)}{s(\lambda_{j, k-1}+\mu_{j, k})}
 \prod_{l=1}^{k-1}\frac{s(\lambda_{l, k-1}+\mu_{l, k})}{s(\lambda_{l, k-1}+\mu_{l+1, k})}
\quad (j\leq k \leq n+1) 
 , \\
E_k&=e_{j, n}(-1)^{j-1}\prod_{l=1, l \neq k}^{j} \frac{s(\lambda_{l, k-1}+\mu_{l, k})}{s(\lambda_{l, k-1}+\mu_{l+1, k})}
\quad (1\leq k\leq j-1) , \\
F_k&=e_{j, n}\tilde{e}_{j}(-1)^{j} \prod_{l=1, l \neq k}^{j-1} \frac{s(\lambda_{l, k-1}+\mu_{l, k})}{s(\lambda_{l, k-1}+\mu_{l+1, k})}
\quad (1\leq k\leq j-1).\end{aligned}$$

First, we compute the first term in the right hand side of as follows. For $k=1,\ldots, j-1$ we have $$\begin{aligned}
A_k-\sum_{m=j+1}^{n}B_{k, m}-E_k-F_k=&(-1)^{j-1}
\prod_{l=1, l \neq k}^{j-1} \frac{s(\lambda_{l, k-1}+\mu_{l, k})}{s(\lambda_{l, k-1}+\mu_{l+1, k})}
\\
&\times
\left(
e_{k, n}\tilde{e}_{k+1, n+1}\frac{s(\mu_j)}{s(\lambda_{j, k-1}+\mu_{j, k})}\prod_{l=j}^{n+1} \frac{s(\lambda_{l, k-1}+\mu_{l, k})}{s(\lambda_{l, k-1}+\mu_{l+1, k})}
\right.
\\
&-\sum_{m=j+1}^ne_{m, n}  \frac{s(\mu_j)s(\mu_{m})}{s(\lambda_{j, k-1}+\mu_{j, k})
s(\lambda_{m,k-1}+\mu_{m,k})}
\prod_{l=j}^{m-1} \frac{s(\lambda_{l, k-1}+\mu_{l, k})}{s(\lambda_{l, k-1}+\mu_{l+1, k})}
\\
&\left.-e_{j, n} \frac{s(\lambda_{j, k-1}+\mu_{j, k})}
{s(\lambda_{j, k-1}+\mu_{j+1, k})}+e_{j, n}\tilde{e}_{j}
\right). \end{aligned}$$ The sum of the last term and the second last term above is computed as $$\begin{aligned}
&-e_{j, n} \frac{s(\lambda_{k, j-1}+\mu_{k+1, j-1})}{s(\lambda_{k, j-1}+\mu_{k+1, j})}+e_{j, n}\tilde{e}_{j} 
\\
&=\frac{1}{s(\lambda_{k, j-1}+\mu_{k+1, j})} \frac{-e_{j, n}(e_{k, j-1}\tilde{e}_{k+1, j-1}-e_{k, j-1}^{-1}\tilde{e}_{k+1, j-1}^{-1})+e_{j, n}\tilde{e}_{j}(e_{k, j-1}\tilde{e}_{k+1, j}-e_{k, j-1}^{-1}\tilde{e}_{k+1, j}^{-1})}{2\sqrt{-1}}
\\
&=e_{k, n}\tilde{e}_{k+1, j}\frac{s(\mu_j)}
{s(\lambda_{k, j-1}+\mu_{k+1, j})}.\end{aligned}$$

Suppose for $i\geq j$ that it holds $$\begin{aligned}
&-\sum_{m=j+1}^ie_{m,n}
\frac{s(\mu_j)s(\mu_{m})}{s(\lambda_{j, k-1}+\mu_{j, k})
s(\lambda_{m,k-1}+\mu_{m+1,k})}
\prod_{l=j}^{m-1} \frac{s(\lambda_{l, k-1}+\mu_{l, k})}{s(\lambda_{l, k-1}+\mu_{l+1, k})}+e_{k, n}\tilde{e}_{k+1, j}\frac{s(\mu_j)}
{s(\lambda_{k, j-1}+\mu_{k+1, j})}
\\
&=e_{k, n}\tilde{e}_{k+1, i}\frac{s(\mu_j)}
{s(\lambda_{k,j-1}+\mu_{k+1,j-1})}\prod_{l=j}^{i}
\frac{ s(\lambda_{k, l-1}+\mu_{k+1, l-1})}
 {s(\lambda_{k, l-1}+\mu_{k+1, l})}.\end{aligned}$$ Then we have $$\begin{aligned}
&-\sum_{m=j+1}^{i+1}e_{m,n}
\frac{s(\mu_j)s(\mu_{m})}{s(\lambda_{j, k-1}+\mu_{j, k})
s(\lambda_{m,k-1}+\mu_{m+1,k})}
\prod_{l=j}^{m-1} \frac{s(\lambda_{l, k-1}+\mu_{l, k})}{s(\lambda_{l, k-1}+\mu_{l+1, k})}+e_{k, n}\tilde{e}_{k+1, j}\frac{s(\mu_j)}
{s(\lambda_{k, j-1}+\mu_{k+1, j})}
\\
&=\frac{s(\mu_j)}{s(\lambda_{k, j-1}+\mu_{k+1, j-1})}
\frac{\prod_{l=j}^{i} s(\lambda_{k, l-1}+\mu_{k+1, l-1})}
 {\prod_{l=j}^{i+1}s(\lambda_{k, l-1}+\mu_{k+1, l})}
\\
&\times  \frac{-e_{i+1,n}\left(\tilde{e}_{i+1}-\tilde{e}_{i+1}^{-1}\right)
+e_{k, n}\tilde{e}_{k+1, i}\left(e_{k,i}\tilde{e}_{k+1,i+1}
-e_{k,i}^{-1}\tilde{e}_{k+1,i+1}^{-1}\right)
}
{2\sqrt{-1}} 
 \\
&=e_{k, n}\tilde{e}_{k+1, i+1}\frac{s(\mu_j)}
{s(\lambda_{k,j-1}+\mu_{k+1,j-1})}\prod_{l=j}^{i+1}
\frac{ s(\lambda_{k, l-1}+\mu_{k+1, l-1})}
 {s(\lambda_{k, l-1}+\mu_{k+1, l})}.
 \end{aligned}$$ Hence we obtain $$\begin{aligned}
-\sum_{m=j+1}^{n}B_{k, m}-E_k-F_k 
=e_{k, n}\tilde{e}_{k+1, n}\frac{s(\mu_j)}
{s(\lambda_{k,j-1}+\mu_{k+1,j-1})}\prod_{l=j}^{n}
\frac{ s(\lambda_{k, l-1}+\mu_{k+1, l-1})}
 {s(\lambda_{k, l-1}+\mu_{k+1, l})}.\end{aligned}$$ Therefore, we have $$\begin{aligned}
A_k-\sum_{m=j+1}^{n}B_{k, m}-E_k-F_k 
=&\frac{(-1)^{j-1}e_{k,n}\tilde{e}_{k+1,n}s(\mu_j)}
{s(\lambda_{k,j-1}+\mu_{k+1,j-1})}
 \frac{\prod_{l=1, l \neq k}^{n}s(\lambda_{l, k-1}+\mu_{l, k})}
 {\prod_{l=1, l \neq k}^{n+1}s(\lambda_{l, k-1}+\mu_{l+1, k})}
 \\
 &\times 
  \frac{-\tilde{e}_{n+1}(e_{k,n}\tilde{e}_{k+1,n}-e_{k,n}^{-1}\tilde{e}_{k+1,n}^{-1})
 +e_{k,n}\tilde{e}_{k+1,n+1}-e_{k,n}^{-1}\tilde{e}_{k+1,n+1}^{-1}}{2\sqrt{-1}}
 \\
 =&\frac{(-1)^{j-1}s(\mu_j)s(\mu_{n+1})}
{s(\lambda_{k,j-1}+\mu_{k+1,j-1})}
 \frac{\prod_{l=1, l \neq k}^{n}s(\lambda_{l, k-1}+\mu_{l, k})}
 {\prod_{l=1, l \neq k}^{n+1}s(\lambda_{l, k-1}+\mu_{l+1, k})}.\end{aligned}$$ Similarly, we can compute the second term in the right hand side of . This completes the proof for the connection formula .

The connection formula for $j=n+1$, namely, the connection formula can be proved in the same way above. By analytic continuation of the connection formula in Proposition \[\[M2\], prop 2.5\] and Lemma \[lem\_analytic\_continuation\_of\_domain\], we get $$\begin{aligned}
\tilde{D}_{n+1}^{(1)}=&
\sum_{ k=1}^{ n+1} (-1)^{n} \frac{s(\mu_{n+1})}{s(\lambda_{n+1, k-1}+\mu_{n+1, k})}
 \prod_{1 \leq l \leq n+1,\atop l \neq k} 
\frac{s(\lambda_{l, k-1}+\mu_{l, k})}{s(\lambda_{l, k-1}+\mu_{l+1, k})} e_{k, n}\tilde{e}_{k+1, n+1}\tilde{D}_k^{(0)}{\nonumber}\\
&-\tilde{e}_{n+1}\{ t_n<z,\ 0<t_n<\cdots <t_1<1\}. \end{aligned}$$ Then, in the integral associated with $\{ t_n<z,\ 0<t_n<\cdots <t_1<1\}$, the integral with respect to the variables $t_1,\ldots, t_{n-1}$ corresponds to the integral defined for $\tilde{D}_n^{(1)}$ of ${}_nF_{n-1}(t_n)$ because $0<t_n<1$. Hence we can apply the induction hypothesis to $\{ t_n<z,\ 0<t_n<\cdots <t_1<1\}$ and obtain a sum of integrals associated with $\{0<t_n<z,\ 0<t_k<\cdots<t_n,\ 1<t_1<\cdots<t_{k-1} \}$ ($k=1,\ldots,n$), which is nothing but $\tilde{D}_k^{(0)}$ of ${}_{n+1}F_n(z)$.

Therefore, we obtain $$\begin{aligned}
\tilde{D}_{n+1}^{(1)}
=&\sum_{k=1}^{n} \left\{ (-1)^ne_{k, n}\tilde{e}_{k+1, n+1} \frac{s(\mu_{n+1})}{s(\lambda_{n+1, k-1}+\mu_{n+1, k})} \prod_{l=1, l \neq k}^{n+1} \frac{s(\lambda_{l, k-1}+\mu_{l, k})}
{s(\lambda_{l,k-1}+\mu_{l+1, k})} \right.
\\
&\left.+(-1)^n\tilde{e}_{n+1} \prod_{l=1, l \neq k}^{n} \frac{s(\lambda_{l, k-1}+\mu_{l, k})}
{s(\lambda_{l, k-1}+\mu_{l+1, k})} \right\}\tilde{D}_k^{(0)} \\
&+(-1)^{n-1}\frac{s(\mu_{n+1})}{s(\lambda_{n+1, n}+\mu_{n+1, n+1})} \prod_{l=1}^{n} \frac{s(\lambda_{l, n}+\mu_{l, n+1})}{s(\lambda_{l, n}+\mu_{l+1, n+1})}\tilde{D}_{n+1}^{(0)}. \end{aligned}$$ Straightforward computation yields the connection formula .

The connection formulas and for $n=2$ was computed in [@Mimachi; @3F2]. The connection formula was obtained in [@Kawabata; @1], [@Mimachi; @intersection; @numbers; @for; @n+1Fn], [@Okubo; @Takano; @Yoshida].

From Theorem \[thm\_connection\_formula\_10\], we have the connection matrix $C^{(10)}$ given by $$\left(\tilde{D}_1^{(1)},\ldots, \tilde{D}_{n+1}^{(1)}\right)=
\left(\tilde{D}_1^{(0)},\ldots,\tilde{D}_{n+1}^{(0)}\right)C^{(10)}.$$ The elements of $C^{(10)}$ are products of the sine and the cosecant: $$\begin{aligned}
\left(C^{(10)}\right)_{i,j}=&\frac{s(\beta_j-\alpha_j)s(\alpha_{n+1})}{s(\beta_i-\alpha_j)s(\beta_i-\alpha_{n+1})} \prod_{k=1, k \neq i}^{n+1}\frac{s(\alpha_k-\beta_i)}{s(\beta_k-\beta_i)}\quad 
(1\leq i\leq n+1,\ 1\leq j\leq n),
\\
 \left(C^{(10)}\right)_{i,n+1}=&\prod_{k=1, k \neq i}^{n+1}
 \frac{s(\alpha_k-\beta_i)}{s(\beta_k-\beta_i)}\quad 
(1\leq i\leq n+1). \end{aligned}$$ We expect that there exists the inverse matrix $C^{(01)}$ of $C^{(10)}$, which implies the set $\left\{\tilde{D}_1^{(1)},\ldots, \tilde{D}_{n+1}^{(1)}\right\}$ of the domains of integration gives a fundamental system of solutions at $z=1$, and the elements of $C^{(01)}$ are products of the sine and the cosecant. Recall that $C^{(10)}$ are derived from analytic continuation of the connection formulas in Proposition \[\[M2\], prop 2.5\], which relates the fundamental system of solutions at $z=\infty$ to that at $z=0$. However, we have not found a way to use the induction hypothesis to the analytic continuation along the path $\gamma$ of the connection formula relating the fundamental system of solutions at $z=0$ to that at $z=\infty$. Instead, we present the inverse matrix $C^{(01)}$ below and give a proof by direct calculation.

Let a square matrix $C^{(01)}$ of order $n+1$ be defined by $$\begin{aligned}
\left(C^{(01)}\right)_{i,j}=&-
\frac{s(\alpha_i)s(\beta_{1, n} -\alpha_{1, n+1}-\beta_j +\alpha_i)s(\beta_j-\alpha_j)}{s(\alpha_{n+1})s(\beta_{1, n}-\alpha_{1, n+1})s(\beta_j-\alpha_i)}
\prod_{k=1, k \neq i}^{n} \frac{s(\beta_k - \alpha_i)}{s(\alpha_k-\alpha_i)}\end{aligned}$$ for $1\le i\le n$, $1\le j\le n+1$, and $$\left(C^{(01)}\right)_{n+1,i}=-\frac{s(\beta_i-\alpha_i)}{s(\beta_{1, n}-\alpha_{1, n+1})}$$ for $1\le i\le n+1$.

\[thm\_connection\_matrix\_01\] Assume for $1\leq i,j \leq n+1$, $\alpha_i-\beta_j,\alpha_i-\alpha_j,\beta_i-\beta_j,\beta_{1, n} -\alpha_{1, n+1}\not\in{{\mathbb Z}}$. Then the matrix $C^{(01)}$ is the inverse of the connection matrix $C^{(10)}$.

We compute the elements $\left(C^{(01)} C^{(10)}\right)_{i, j}$ of the product directly. We consider separately the following cases: (i) $1\leq i,j\leq n$, (ii) $1\leq i\leq n$, $j=n+1$, (iii) $i=n+1$, $1\leq j\leq n$, (iv) $i=j=n+1$.

In the case of (i), we have $$\begin{aligned}
\left(C^{(01)} C^{(10)}\right)_{i, j}=&-\sum_{k=1}^{n+1}\frac{s(\alpha_i)s(\beta_{1, n}-\alpha_{1, n+1}-\beta_k+\alpha_i)s(\beta_k-\alpha_k)}{s(\alpha_{n+1})s(\beta_{1, n}-\alpha_{1, n+1})s(\beta_k-\alpha_i)}\\
& \times \frac{s(\beta_j-\alpha_j)s(\alpha_{n+1})}{s(\beta_k-\alpha_j)s(\beta_k-\alpha_{n+1})}\prod_{l=1, l \neq i}^{n} \frac{s(\beta_l-\alpha_i)}{s(\alpha_l-\alpha_i)} \prod_{l=1, l \neq k}^{n+1}\frac{s(\alpha_l-\beta_k)}{s(\beta_l-\beta_k)} \\
=&-\frac{s(\alpha_i)s(\beta_j-\alpha_j)}{s(\beta_{1, n}-\alpha_{1, n+1})}\prod_{l=1, l \neq i}^{n} \frac{s(\beta_l-\alpha_i)}{s(\alpha_l-\alpha_i)} \\
& \times \left( \sum_{k=1}^{n+1} \frac{s(\beta_{1, n}-\alpha_{1, n+1}-\beta_k+\alpha_i)s(\beta_k-\alpha_k)}{s(\beta_k-\alpha_i)s(\beta_k-\alpha_j)s(\beta_k-\alpha_{n+1})}\prod_{l=1, l \neq k}^{n+1}\frac{s(\beta_k-\alpha_l)}{s(\beta_k-\beta_l)} \right). \end{aligned}$$ Put $a_i=e^{\pi \sqrt{-1} \alpha_i}, b_i=e^{\pi \sqrt{-1} \beta_i}$. Then by definition of the sine, we get $$\begin{aligned}
\label{eq_1_ab}
&\sum_{k=1}^{n+1}\frac{s(\beta_{1, n}-\alpha_{1, n+1}-\beta_k+\alpha_i)s(\beta_k-\alpha_k)}{s(\beta_k-\alpha_i)s(\beta_k-\alpha_j)s(\beta_k-\alpha_{n+1})}\prod_{l=1, l \neq k}^{n+1}\frac{s(\beta_k-\alpha_l)}{s(\beta_k-\beta_l)} 
\\
&=- 2\sqrt{-1}\frac{ a_j a_{n+1} }{a_1^2 \cdots a_{n+1}^2}\sum_{k=1}^{n+1}  \frac{(b_1^2 \cdots b_n^2 a_i^2 - a_1^2 \cdots a_{n+1}^2 b_k^2)}{(b_k^2-a_i^2)(b_k^2-a_j^2)(b_k^2-a_{n+1}^2)} \frac{\prod_{l=1}^{n+1} (b_k^2-a_l^2)}{\prod_{l=1, l \neq k}^{n+1} (b_k^2-b_l^2)}. {\nonumber}\end{aligned}$$ It is sufficient to prove that the right hand side above is equal to zero as a rational function of $a_i$ and $b_i$.

Let the rational function $f_1(x)$ be defined by $$\begin{aligned}
f_1(x)=& -2\sqrt{-1} \frac{ a_i a_{n+1} }{a_1^2 \cdots a_{n+1}^2} \frac{(b_1^2 \cdots b_n^2 a_i^2 - a_1^2 \cdots a_{n+1}^2 x)}{(x-a_i^2)(x-a_j^2)(x-a_{n+1}^2)} \prod_{l=1}^{n+1} \frac{(x-a_l^2)}{ (x-b_l^2)} . \end{aligned}$$ In the case of $j\neq i$, $f_1(x)$ has only simple poles $x=b_k^2$ ($k=1,\ldots, n+1$). The residues are given as $$\begin{aligned}
\underset{x=b_k^2}{\rm Res}f_1(x)dx&=-2\sqrt{-1} \frac{ a_i a_{n+1} }{a_1^2 \cdots a_{n+1}^2} \frac{(b_1^2 \cdots b_n^2 a_i^2 - a_1^2 \cdots a_{n+1}^2 b_k^2)}{(b_k^2-a_i^2)(b_k^2-a_j^2)(b_k^2-a_{n+1}^2)} \frac{\prod_{l=1}^{n+1} (b_k^2-a_l^2)}{\prod_{l=1, l \neq k}^{n+1} (b_k^2-b_l^2)}.\end{aligned}$$ Since the sum of all residues of a rational function on $\mathbb{P}$ is equal to zero, the right hand side of is zero.

In the case of $i=j$, $f_1(x)$ has only simple poles $x=a_i^2$, and $x=b_k^2$ ($k=1,\ldots, n+1$). The sum of the residues at $x=b_k^2$ ($k=1,\ldots, n+1$) is equal to the right hand side of . The residues at $x=a_i^2$ is computed as $$\begin{aligned}
\underset{x=a_i^2}{\rm Res}f_1(x)dx=& -2\sqrt{-1} \frac{ a_i a_{n+1} }{a_1^2 \cdots a_{n+1}^2} \frac{(b_1^2 \cdots b_n^2 a_i^2 - a_1^2 \cdots a_{n+1}^2 a_i^2)}{(a_i^2-a_{n+1}^2)} \frac{\prod_{l=1, l \neq i}^{n+1} (a_i^2-a_l^2)}{\prod_{l=1}^{n+1} (a_i^2-b_l^2)}  \\
=& 2\sqrt{-1} \frac{ a_i^2 b_i }{a_1 \cdots a_{n+1} b_1 \cdots b_n} \frac{(b_1^2 \cdots b_n^2 - a_1^2 \cdots a_{n+1}^2)}{(b_i^2-a_i^2)(a_i^2-1)} \prod_{l=1, l \neq i}^{n} \frac{b_l (a_i^2-a_l^2)}{a_l (a_i^2-b_l^2)}  \\
=& \frac{s(\beta_{1, n}-\alpha_{1, n+1})}{s(\beta_i -\alpha_i)s(\alpha_i)}\prod_{l=1, l \neq i}^{n} \frac{s(\alpha_i -\alpha_l)}{s(\alpha_i - \beta_l)}.\end{aligned}$$ Hence, we have $$\begin{aligned}
\text{R.H.S. of \eqref{eq_1_ab}}
=&\sum_{k=1}^{n+1} \underset{x=b_k^2}{\rm Res} f_1(x)dx \\
=& -\underset{x=a_i^2}{\rm Res}f_1(x)dx \\
=& -\frac{s(\beta_{1, n}-\alpha_{1, n+1})}{s(\beta_i -\alpha_i)s(\alpha_i)}\prod_{l=1, l \neq i}^{n} \frac{s(\alpha_i -\alpha_l)}{s(\alpha_i - \beta_l)}. \end{aligned}$$ Therefore, we obtain $\left(C^{(01)} C^{(10)}\right)_{i, i}=1$ for $1\le i, j\le n$.

In the case of (ii), we have $$\begin{aligned}
\left(C^{(01)} C^{(10)}\right)_{i, n+1}=&-\sum_{k=1}^{n+1} \frac{s(\alpha_i) s(\beta_{1, n}-\alpha_{1, n+1}-\beta_{k}+\alpha_{i})s(\beta_k-\alpha_k)}{s(\alpha_{n+1})s(\beta_{1, n}-\alpha_{1, n+1})s(\beta_k -\alpha_i)}
 \prod_{l=1, l \neq i}^{n} \frac{s(\beta_l-\alpha_i)}{s(\alpha_l-\alpha_i)}\prod_{l=1, l \neq k}^{n+1}\frac{s(\alpha_l-\beta_k)}{s(\beta_l-\beta_k)} \\
=&-\frac{s(\alpha_i)}{s(\alpha_{n+1})s(\beta_{1, n}-\alpha_{1, n+1})}\prod_{l=1, l \neq i}^{n} \frac{s(\beta_l-\alpha_i)}{s(\alpha_l-\alpha_i)} \\
&\times \sum_{k=1}^{n+1} \frac{s(\beta_{1, n}-\alpha_{1, n+1}-\beta_{k}+\alpha_{i})s(\beta_k-\alpha_k)}{s(\beta_k -\alpha_i)} \prod_{l=1, l \neq k}^{n+1}\frac{s(\beta_k-\alpha_l)}{s(\beta_k-\beta_l)}.  \end{aligned}$$ Rewriting the parameters $\alpha_i$, $\beta_i$ to $a_i$, $b_i$, we get $$\begin{aligned}
&\sum_{k=1}^{n+1} \frac{s(\beta_{1, n}-\alpha_{1, n+1}-\beta_k +\alpha_i)s(\beta_k -\alpha_k)}{s(\beta_k -\alpha_i)} \prod_{l=1, l \neq k}^{n+1} \frac{s(\beta_k -\alpha_l)}{s(\beta_k -\beta_l)} \\
&=\sum_{k=1}^{n+1} \frac{1}{2 \sqrt{-1}} \frac{1}{a_1 \cdots a_{n+1} b_k^2} \frac{(b_1^2 \cdots b_n^2 a_i^2 - a_1^2 \cdots a_{n+1}^2 b_k^2)}{(b_k^2-a_i^2)} \frac{\prod_{l=1}^{n+1} (b_k^2-a_l^2)}{\prod_{l=1, l \neq k}^{n+1} (b_k^2 -b_l^2)}. \end{aligned}$$ Let the rational function $f_2(x)$ be defined by $$\begin{aligned}
f_2(x)=\frac{1}{2 \sqrt{-1}} \frac{1}{a_1 \cdots a_{n+1} x} \frac{(b_1^2 \cdots b_n^2 a_i^2 - a_1^2 \cdots a_{n+1}^2 x)}{(x-a_i^2)} \prod_{l=1}^{n+1} \frac{(x-a_l^2)}{(x -b_l^2)}.\end{aligned}$$ Then $f_2(x)$ has only simple poles $x=0$, $x=\infty$, and $x=b_k^2$ ($k=1,\ldots, n+1$). The residues are given as $$\begin{aligned}
 \underset{x=0}{\rm Res}f_2(x)dx 
=& - \frac{1}{2 \sqrt{-1}},
\\
\underset{x=\infty}{\rm Res}f_2(x)dx=&\frac{1}{2 \sqrt{-1}}, 
\\
\underset{x=b_k^2}{\rm Res}f_2(x)dx=&\frac{1}{2 \sqrt{-1}} \frac{1}{a_1 \cdots a_{n+1} b_k^2} \frac{(b_1^2 \cdots b_n^2 a_i^2 - a_1^2 \cdots a_{n+1}^2 b_k^2)}{(b_k^2-a_i^2)} \frac{\prod_{l=1}^{n+1} (b_k^2-a_l^2)}{\prod_{l=1, l \neq k}^{n+1} (b_k^2 -b_l^2)}. \end{aligned}$$ Hence, we have $$\begin{aligned}
\sum_{k=1}^{n+1}\underset{x=b_k^2}{\rm Res}f_2(x)dx 
=&-\underset{x=0}{\rm Res}f_2(x)dx-\underset{x=\infty}{\rm Res}f_2(x)dx=0.\end{aligned}$$ Therefore, we obtain $\left(C^{(01)} C^{(10)}\right)_{i, n+1}=0$ for $1\leq i\leq n$.

In the case of (iii), we have $$\begin{aligned}
\left(C^{(01)} C^{(10)}\right)_{n+1, j}=& -\frac{s(\beta_j -\alpha_j)s(\alpha_{n+1})}{s(\beta_{1, n}-\alpha_{1, n+1})}\left( \sum_{k=1}^{n+1} \frac{s(\beta_k- \alpha_k)}{s(\beta_k -\alpha_j)s(\beta_k -\alpha_{n+1})} \prod_{l=1, l \neq k}^{n+1} \frac{s(\beta_k-\alpha_l)}{s(\beta_k- \beta_l)} \right). \end{aligned}$$ Rewriting the parameters $\alpha_i$, $\beta_i$ to $a_i$, $b_i$, we get $$\begin{aligned}
&\sum_{k=1}^{n+1} \frac{s(\beta_k- \alpha_k)}{s(\beta_k -\alpha_j)s(\beta_k -\alpha_{n+1})} \prod_{l=1, l \neq k}^{n+1} \frac{s(\beta_k-\alpha_l)}{s(\beta_k- \beta_l)}  \\
&= \sum_{k=1}^{n+1} 2 \sqrt{-1} \frac{b_1 \cdots b_n a_j a_{n+1}}{a_1 \cdots a_{n+1}} \frac{1}{(b_k^2-a_j^2)(b_k^2-a_{n+1}^2)} \frac{\prod_{l=1}^{n+1} (b_k^2-a_l^2)}{\prod_{l=1, l \neq k}^{n+1} (b_k^2-b_l^2)}. \end{aligned}$$ Let the rational function $f_3(x)$ be defined by $$\begin{aligned}
f_3(x)= 2 \sqrt{-1} \frac{b_1 \cdots b_n a_j a_{n+1}}{a_1 \cdots a_{n+1}} \frac{1}{(x-a_j^2)(x-a_{n+1}^2)} \prod_{l=1}^{n+1} \frac{ (x-a_l^2)}{ (x-b_l^2)}. \end{aligned}$$ Then $f_3(x)$ has only simple poles $x=b_k^2$ ($k=1,\ldots, n+1$). The residues are given as $$\begin{aligned}
\underset{x=b_k^2}{\rm Res}f_3(x)dx
=2 \sqrt{-1} \frac{b_1 \cdots b_n a_j a_{n+1}}{a_1 \cdots a_{n+1}} \frac{1}{(b_k^2-a_j^2)(b_k^2-a_{n+1}^2)} \frac{\prod_{l=1}^{n+1} (b_k^2-a_l^2)}{\prod_{l=1, l \neq k}^{n+1} (b_k^2-b_l^2)}. \end{aligned}$$ Hence we obtain $\left(C^{(01)} C^{(10)}\right)_{n+1, j}=0$ for $1\leq j\leq n$.

In the case of (iv), we have $$\begin{aligned}
\left(C^{(01)} C^{(10)}\right)_{n+1, n+1}=-\frac{1}{s(\beta_{1, n}-\alpha_{1, n+1})} \sum_{k=1}^{n+1} s(\beta_k -\alpha_k) \prod_{l=1, l \neq k}^{n+1} \frac{s(\beta_k -\alpha_l)}{s(\beta_k -\beta_l)}. \end{aligned}$$ Rewriting the parameters $\alpha_i$, $\beta_i$ to $a_i$, $b_i$, we get $$\begin{aligned}
&\sum_{k=1}^{n+1} s(\beta_k -\alpha_k) \prod_{l=1, l \neq k}^{n+1} \frac{s(\beta_k -\alpha_l)}{s(\beta_k -\beta_l)} 
= -\sum_{k=1}^{n+1} \frac{b_1 \cdots b_n}{2 \sqrt{-1} a_1 \cdots a_{n+1} b_k^2} \frac{\prod_{l=1}^{n+1} (b_k^2 -a_l^2)}{\prod_{l=1, l \neq k}^{n+1} (b_k^2 -b_l^2)}. \end{aligned}$$ Let the rational function $f_4(x)$ be defined by $$\begin{aligned}
f_4(x)=-\frac{b_1 \cdots b_n}{2 \sqrt{-1} a_1 \cdots a_{n+1}x} \prod_{l=1}^{n+1} \frac{x-a_l^2}{x-b_l^2}. \end{aligned}$$ Then $f_4(x)$ has only simple poles $x=0$, $x=\infty$, and $x=b_k^2$ ($k=1,\ldots, n+1$). The residues are given as $$\begin{aligned}
\underset{x=0}{\rm Res}f_4(x)dx 
=& - \frac{a_1 \cdots a_{n+1}}{2 \sqrt{-1} b_1 \cdots b_n},
\\
\underset{x=\infty}{\rm Res}f_4(x)dx
=& \frac{b_1 \cdots  b_n}{2 \sqrt{-1} a_1 \cdots a_{n+1}},
\\
\underset{x=b_k^2}{\rm Res}f_4(x)dx =& -\sum_{k=1}^{n+1} \frac{b_1 \cdots b_n}{2 \sqrt{-1} a_1 \cdots a_{n+1} b_k^2} \frac{\prod_{l=1}^{n+1} (b_k^2 -a_l^2)}{\prod_{l=1, l \neq k}^{n+1} (b_k^2 -b_l^2)}. \end{aligned}$$ Hence, we have $$\begin{aligned}
\sum_{k=1}^{n+1}\underset{x=b_k^2}{\rm Res}f_4(x)dx 
=& -\underset{x=0}{\rm Res}f_4(x)dx -\underset{x=\infty}{\rm Res}f_4(x)dx
 \\
=& -s(\beta_{1, n}-\alpha_{1, n+1}). \end{aligned}$$ Therefore, we obtain $\left(C^{(01)} C^{(10)}\right)_{n+1,n+1}=1$.

Assume for $1 \leq i,  j \leq n+1$ $$\begin{aligned}
&\mathrm{Re}(\alpha_i-\beta_j+1)>0\quad (j\neq i),
\\
&\mathrm{Re}(\beta_j-\alpha_j)>0, 
\\
&\alpha_i-\alpha_j\notin {{\mathbb Z}}\quad (j\neq i),\quad 
\alpha_i-\beta_j \notin {{\mathbb Z}}, \quad \beta_i-\beta_j \notin {{\mathbb Z}}\quad  (j\neq i). \end{aligned}$$ Then we have $$\begin{aligned}
F_{\tilde{D}_{i}^{(0)}}(z)=&-\sum_{j=1}^{n}\frac{s(\alpha_j)s(\beta_{1, n} -\alpha_{1, n+1}-\beta_i +\alpha_j)s(\beta_i-\alpha_i)}{s(\alpha_{n+1})s(\beta_{1, n}-\alpha_{1, n+1})s(\beta_i-\alpha_j)}
\prod_{l=1, l \neq j}^{n} \frac{s(\beta_l - \alpha_j)}{s(\alpha_l-\alpha_j)}F_{\tilde{D}_{j}^{(1)}}(z)
 \\
&-\frac{s(\beta_i-\alpha_i)}{s(\beta_{1, n}-\alpha_{1, n+1})}F_{\tilde{D}_{n+1}^{(1)} }(z).\end{aligned}$$ Moreover, $\left\{ F_{\tilde{D}_1^{(1)}}(z),\ldots,F_{\tilde{D}_{n+1}^{(1)}}\right\}$ forms a fundamental system of solutions to the generalized hypergeometric equation.

We note that together with Proposition \[\[M2\], prop 2.1\] and Theorem \[thm\_connection\_formula\_10\], we have the connection matrix $C^{(1\infty)}$ expressing the fundamental system of solutions $\left\{ F_{\tilde {D}^{(1)}_1}(z),\ldots, F_{\tilde{D}^{(1)}_1}(z)\right\}$ at $z=1$ in terms of the fundamental system of solutions $\left\{ F_{D^{(\infty)}_1}(z),\ldots, F_{D^{(\infty)}_1}(z)\right\}$ at $\infty$ as a product of the connection matrices. It is observed that the elements of the connection matrix $C^{(1\infty)}$ for $n\le 3$ are products of the sine and cosecant. However, there are elements of the inverse of $C^{(1\infty)}$ for $n\le 3$ which are not products of the sine and cosecant. If we want to have connection matrices between the fundamental system of solutions of singular points $z=1$ and $z=\infty$ whose elements are products of the sine and cosecant, then we should perform the change of variable $z=1/w$ to the connection formulas and . Due to Theorem \[thm\_connection\_matrix\_01\], all elements of both the connection matrix and its inverse are products of the sine and cosecant.

Periodicity
-----------

In this subsection, we point out that the connection matrices multiplied by diagonal matrices in the previous sections are invariant under integer shifts of the characteristic exponents $\alpha_i$, $\beta_i$ of the generalized hypergeometric equation. This is true for the connection matrix with the fundamental systems $X_0$ and $X_\infty$ of solutions at $z=0$ and $z=\infty$, and it was used for the construction of the monodromy-invariant general solutions of Fuchsian systems of rank $2$ in [@Iorgov; @Lisovyy; @Teschner] and rank $N\geq 2$ in [@GIL2] with help of the $W_N$ conformal field theory with central charge $c=N-1$. When $N=2$, $W_N$-algebra is the Virasoro algebra. As a result, for $N=2$ the tau function of the sixth Painlevé equation is obtained in terms of the Virasoro conformal blocks with central charge $c=1$ [@Iorgov; @Lisovyy; @Teschner], and for $N\ge 2$ the isomonodromic tau function of the so-called Fuji-Suzuki-Tsuda system [@FS], [@Suzuki], [@Tsuda1] is obtained in terms of semi-degenerate conformal blocks of $W_N$-algebra with central charge $c=N-1$, which are equivalent to the Nekrasov partition functions [@Nekrasov] by AGT correspondence [@AGT], [@Wyllard]. The series representation of the tau function of the Fuji-Suzuki-Tsuda system is also derived by expanding the Fredholm determinant [@GIL3]. The periodicity of the connection matrix of $q$-hypergeometric series was used in [@Jimbo; @Nagoya; @Sakai] in order to obtain a fundamental solution to the connection-preserving deformation of the $2$ by $2$ $q$-difference linear system associated with the $q$-difference Painlevé VI equation.

We take $$\begin{aligned}
D_i^{(0)}=&\{t\in T_{{\mathbb R}}\mid z<t_n< \cdots <t_i <0,\ 1<t_1< \cdots < t_{i-1} \} \quad (1 \leq i \leq  n+1),
\\
D_i^{(\infty)}=&\{t\in T_{{\mathbb R}}\mid t_i< \cdots <t_n <z,\ 0<t_{i-1}< \cdots < t_1< 1 \} \quad (1 \leq i \leq  n+1).\end{aligned}$$ as domains of integration giving the fundamental systems of solutions at $z=0$ and $z=\infty$, and $$\begin{aligned}
\tilde{D}_i^{(1)}&=\{t\in T_{{\mathbb R}}| t_i< \cdots <t_n <0,\ 0<t_{i-1}< \cdots < t_1< 1 \} \quad (1 \leq i \leq  n), 
 \\
\tilde{D}_{n+1}^{(1)}&=\{t\in T_{{\mathbb R}}| z<t_n< \cdots < t_1< 1 \}\end{aligned}$$ as domains of integration giving the fundamental system of solutions at $z=1$.

From Proposition \[\[M2\], prop 2.5\], we have the connection matrix $C^{(\infty 0)}$ given by $$\left(D_1^{(\infty)},\ldots, D_{n+1}^{(\infty)}\right)=
\left(D_1^{(0)},\ldots,D_{n+1}^{(0)}\right)C^{(\infty 0)}.$$ The elements of $C^{(\infty 0)}$ read as $$\begin{aligned}
\left( C^{(\infty 0)}\right)_{i,j}=
\frac{s(\beta_j-\alpha_j)}{s(\beta_i-\alpha_j)}
\prod_{k=1, k \neq i}^{n+1}
 \frac{s(\alpha_k-\beta_i)}{s(\beta_k-\beta_i)} \end{aligned}$$ for $1\leq i,j\leq n+1$.

Let the diagonal matrices $N_0$, $N_1$, $N_\infty$ of order $n+1$ be defined by $$\begin{aligned}
 N_0=&{\rm diag} (e(\alpha_1),\ldots,e(\alpha_{n+1})),
 \\
 N_1=&e(\beta_{1,n}-\alpha_{1,n+1})
 {\rm diag}(e(\beta_1),\ldots,e(\beta_{n+1})),
 \\
 N_\infty=&N_1. 
 \end{aligned}$$

The connection matrices associated with the fundamental systems of solutions $$\begin{aligned}
 \left(D_1^{(0)},\ldots,D_{n+1}^{(0)}\right)N_0,
 \quad
 \left(\tilde{D}_1^{(1)},\ldots, \tilde{D}_{n+1}^{(1)}\right)N_1,
 \quad 
 \left(D_1^{(\infty)},\ldots, D_{n+1}^{(\infty)}\right)N_\infty
 \end{aligned}$$ are invariant under integer shifts of the characteristic exponents of the generalized hypergeometric equation.

It is sufficient to consider the shifts $\alpha_i\mapsto \alpha_i\pm 1$ and $\beta_i\mapsto \beta_i\pm 1$ for $1\leq i\leq n+1$. The connection matrices which express the fundamental system of solutions at $z=1$, $z=\infty$ in terms of the fundamental system of solutions at $z=\infty$ is $$\begin{aligned}
\widehat{C}^{(10)}=N_0^{-1} {\rm diag}(-e(\beta_1),\ldots,-e(\beta_{n+1}))C^{(10)}N_1, 
\quad 
\widehat{C}^{(\infty 0)}=N_0^{-1}C^{(\infty 0)}N_\infty, \end{aligned}$$ respectively, because of Lemma \[lem\_analytic\_continuation\_of\_domain\] and the relations of the parameters . Since the matrices $C^{(10)}$, $C^{(\infty 0)}$ are explicitly written, it is immediate to check that the shift $\alpha_i\mapsto \alpha_i\pm 1$ or $\beta_i\mapsto \beta_i\pm 1$ preserves the connection matrices $\widehat{C}^{(1 0)}$, $\widehat{C}^{(\infty 0)}$.

**Acknowledgments.** This work is partially supported by JSPS KAKENHI Grant Number JP18K03326.

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