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Article

On the Domain of Analytical Continuation of the Ratios of Generalized Hypergeometric Functions 3F2

1
Faculty of Mathematics and Computer Science, Vasyl Stefanyk Carpathian National University, 57 Shevchenko Str., 76018 Ivano-Frankivsk, Ukraine
2
Department of Management and Administration, West Ukrainian National University, 11 Lvivska Str., 46009 Ternopil, Ukraine
3
Institute of Applied Mathematics and Fundamental Sciences, Lviv Polytechnic National University, 12 Stepan Bandera Str., 79013 Lviv, Ukraine
*
Author to whom correspondence should be addressed.
Axioms 2025, 14(12), 871; https://doi.org/10.3390/axioms14120871 (registering DOI)
Submission received: 4 November 2025 / Revised: 20 November 2025 / Accepted: 25 November 2025 / Published: 27 November 2025

Abstract

The paper considers the problem of the analytical extension of the ratios of generalized hypergeometric functions F 2 3 A new domain of analytic continuation for these ratios under certain conditions to parameters is established. In this case, the domain of analytic extension of the special function is the domain of convergence of its branched continued fraction expansion. This paper also provides an example of applying the obtained results to dilogarithm function.

1. Introduction

Special functions, including the generalized hypergeometric function, find diverse applications in almost all fields of science and engineering (see, for example, [1,2,3]). This paper continues the study of the ratios of generalized hypergeometric functions F 2 3 through their branched continued fraction expansions, starting in [4,5].
Branched continued fractions [6], as well as their confluent case, continued fractions [7], play a particularly important role here due to their elegant structure and good approximating properties, such as wide domains of convergence, faster convergence rates under certain conditions compared to series, and numerical stability, which allows them to be an effective tool for approximating the special functions [8,9,10,11]. A description of the resulting branched continued fraction structures can be found in [12].
Recall that the function F 2 3 is defined as follows ([13], p. 8):
F 2 3 ( α 1 , α 2 , α 3 ; β 1 , β 2 ; z ) = k = 0 ( α 1 ) k ( α 2 ) k ( α 3 ) k ( β 1 ) k ( β 2 ) k z k k ! ,
where α 1 , α 2 , α 3 , β 1 , β 2 C ,   β 1 , β 2 Z ,   Z = { 0 , 1 , 2 , } ,   ( · ) k is the Pochhammer symbol, z C .
The main goal of this work is to establish a new domain of analytical extension of the certain ratios of generalized hypergeometric functions F 2 3 (see (3)), which is simultaneously the domain of convergence of their branched continued fraction expansions (4). In [4], it is established that
H ε = z C : z 1 ε 4 , + , 0 < ε < 1 ,
is the domain of the analytic continuation of these ratios under certain conditions to the real parameters of the function (1). Another domain
H ϱ = z C : z 1 8 ϱ , +
is established in [2], where ϱ is a positive number that depend only on the coefficients of the branched continued fraction expansions. Obviously, this will be a wider domain provided that 2 ϱ ( 1 ε ) < 1 . In addition, it is also proved here that the union of domains
H μ = z C : | z | < 1 + cos ( arg ( z ) ) μ
and
H ν = z C : | z | < 1 8 ν
is the domain of the analytic extension under certain conditions to the complex parameters of (1), where μ and ν are positive numbers that depend only on the coefficients of the branched continued fraction expansions. The problem of analytical continuation of generalized hypergeometric function F 2 3 using other methods was studied, particularly in [14,15,16,17,18]. A description of the various uses of this function can be found in [2].

2. Auxiliary Results

In this section, we present results designed to provide a greater understanding of the research object and help to prove the main result.
Let ( i j ) 0 = ( i 0 , j 0 ) ,
I = { ( 1 , 1 ) , ( 1 , 2 ) , ( 2 , 1 ) , ( 2 , 2 ) } ,
and
I k ( i j ) 0 = { ( i j ) k : ( i j ) k = ( i 1 , j 1 , i 2 , j 2 , , i k , j k ) , 1 + δ i k 1 1 i k 2 , j k { 1 , 2 } , | i k j k | | i k 1 j k 1 | } , k 1 ,
where δ p q is the Kronecker symbol. Then for ( i j ) 0 = ( 1 , 1 ) we have
I 1 ( 1 , 1 ) = { ( 2 , 1 ) } , I 2 ( 1 , 1 ) = { ( 2 , 1 , 1 , 1 ) , ( 2 , 1 , 2 , 2 ) } , I 3 ( 1 , 1 ) = { ( 2 , 1 , 1 , 1 , 2 , 1 ) , ( 2 , 1 , 2 , 2 , 1 , 2 ) , ( 2 , 1 , 2 , 2 , 2 , 1 ) } , I 4 ( 1 , 1 ) = { ( 2 , 1 , 1 , 1 , 2 , 1 , 1 , 1 ) , ( 2 , 1 , 1 , 1 , 2 , 1 , 2 , 2 ) , ( 2 , 1 , 2 , 2 , 1 , 2 , 2 , 2 ) , ( 2 , 1 , 2 , 2 , 2 , 1 , 1 , 1 ) , ( 2 , 1 , 2 , 2 , 2 , 1 , 2 , 2 ) } , etc .
For ( i j ) 0 = ( 1 , 2 ) we get
I 1 ( 1 , 2 ) = { ( 2 , 2 ) } , I 2 ( 1 , 2 ) = { ( 2 , 2 , 1 , 2 ) , ( 2 , 2 , 2 , 1 ) } , I 3 ( 1 , 2 ) = { ( 2 , 2 , 1 , 2 , 2 , 2 ) , ( 2 , 2 , 2 , 1 , 1 , 1 ) , ( 2 , 2 , 2 , 1 , 2 , 2 ) } , I 4 ( 1 , 2 ) = { ( 2 , 2 , 1 , 2 , 2 , 2 , 1 , 2 ) , ( 2 , 2 , 1 , 2 , 2 , 2 , 2 , 1 ) , ( 2 , 2 , 2 , 1 , 1 , 1 , 2 , 1 ) , ( 2 , 2 , 2 , 1 , 2 , 2 , 1 , 2 ) , ( 2 , 2 , 2 , 1 , 2 , 1 , 2 , 1 ) } , etc .
Now, for ( i j ) 0 = ( 2 , 1 ) we obtain
I 1 ( 2 , 1 ) = { ( 1 , 1 ) , ( 2 , 2 ) } , I 2 ( 2 , 1 ) = { ( 1 , 1 , 2 , 1 ) , ( 2 , 2 , 1 , 2 ) , ( 2 , 2 , 2 , 1 ) } , I 3 ( 2 , 1 ) = { ( 1 , 1 , 2 , 1 , 1 , 1 ) , ( 1 , 1 , 2 , 1 , 2 , 2 ) , ( 2 , 2 , 1 , 2 , 2 , 2 ) , ( 2 , 2 , 2 , 1 , 1 , 1 ) , ( 2 , 2 , 2 , 1 , 2 , 2 ) } , I 4 ( 2 , 1 ) = { ( 1 , 1 , 2 , 1 , 1 , 1 , 2 , 1 ) , ( 1 , 1 , 2 , 1 , 2 , 2 , 1 , 2 ) , ( 1 , 1 , 2 , 1 , 2 , 2 , 2 , 1 ) , ( 2 , 2 , 1 , 2 , 2 , 2 , 1 , 2 ) , ( 2 , 2 , 1 , 2 , 2 , 2 , 2 , 1 ) , ( 2 , 2 , 2 , 1 , 1 , 1 , 2 , 1 ) , ( 2 , 2 , 2 , 1 , 2 , 2 , 1 , 2 ) , ( 2 , 2 , 2 , 1 , 2 , 2 , 2 , 1 ) } , etc .
Finally, for ( i j ) 0 = ( 2 , 2 ) we have
I 1 ( 2 , 2 ) = { ( 1 , 2 ) , ( 2 , 1 ) } , I 2 ( 2 , 2 ) = { ( 1 , 2 , 2 , 2 ) , ( 2 , 1 , 1 , 1 ) , ( 2 , 1 , 2 , 2 ) } , I 3 ( 2 , 2 ) = { ( 1 , 2 , 2 , 2 , 1 , 2 ) , ( 1 , 2 , 2 , 2 , 2 , 1 ) , ( 2 , 1 , 1 , 1 , 2 , 1 ) , ( 2 , 1 , 2 , 2 , 1 , 2 ) , ( 2 , 1 , 2 , 2 , 2 , 1 ) } , I 4 ( 2 , 2 ) = { ( 1 , 2 , 2 , 2 , 1 , 2 , 2 , 2 ) , ( 1 , 2 , 2 , 2 , 2 , 1 , 1 , 1 ) , ( 1 , 2 , 2 , 2 , 2 , 1 , 2 , 2 ) , ( 2 , 1 , 1 , 1 , 2 , 1 , 1 , 1 ) , ( 2 , 1 , 1 , 1 , 2 , 1 , 2 , 1 ) , ( 2 , 1 , 2 , 1 , 1 , 2 , 2 , 2 ) , ( 2 , 1 , 2 , 2 , 2 , 1 , 1 , 1 ) , ( 2 , 1 , 2 , 2 , 2 , 1 , 2 , 2 ) } , etc .
Remark 1.
If | I k ( 1 , 1 ) | denotes the cardinality of the set I k ( 1 , 1 ) , then the sequence { | I k ( 1 , 1 ) | } is a sequence of Fibonacci numbers, starting from the third number ([4], Proposition 1). It is obvious that the sequence of the set { I k ( 1 , 2 ) } has the same property. Furthermore, it is easy to show that the sequence { | I k ( 2 , 1 ) | } (or a similar { | I k ( 2 , 2 ) | } ) is a Fibonacci sequence starting with the fourth number.
In ([4], Subsection 2.2) the following is proven:
Theorem 1.
For each ( i j ) 0 I , the ratio
F 2 3 ( α 1 , α 2 , α 3 ; β 1 , β 2 ; z ) F 2 3 ( α 1 + δ i 0 1 δ j 0 1 + δ i 0 2 δ j 0 2 , α 2 + δ i 0 1 δ j 0 2 + δ i 0 2 δ j 0 1 , α 3 + δ i 0 2 ; β 1 + δ i 0 1 δ j 0 1 + δ i 0 2 , β 2 + δ i 0 1 δ j 0 2 + δ i 0 2 ; z )
has a formal branched continued fraction
1 i 1 = 1 + δ i 0 1 | i 1 j 1 | | i 0 j 0 | , j 1 { 1 , 2 } 2 u ( i j ) 1 ( i j ) 0 z 1 i 2 = 1 + δ i 1 1 | i 2 j 2 | | i 1 j 1 | , j 2 { 1 , 2 } 2 u ( i j ) 2 ( i j ) 0 z 1   i k = 1 + δ i k 1 1 | i k j k | | i k 1 j k 1 | , j k { 1 , 2 } 2 u ( i j ) k ( i j ) 0 z 1   ,
where for ( i j ) k I k ( i j ) 0 ,   k 1 ,   ( i j ) 0 I ,
u ( i j ) k ( i j ) 0 = β 1 α 3 + r = 0 k 2 δ i r 1 δ j r 1 α 1 + r = 0 k 2 ( δ i r 1 δ j r 1 + δ i r 2 δ j r 2 ) α 2 + r = 0 k 2 ( δ i r 1 δ j r 2 + δ i r 2 δ j r 1 ) β 1 + 1 + r = 0 k 2 ( δ i r 1 δ j r 1 + δ i r 2 ) β 1 + r = 0 k 2 ( δ i r 1 δ j r 1 + δ i r 2 ) β 2 + r = 0 k 2 ( δ i r 1 δ j r 2 + δ i r 2 ) ,
if i k 1 = 2 ,   j k 1 = i k = j k = 1 ,
u ( i j ) k ( i j ) 0 = β 2 α 3 + r = 0 k 2 δ i r 1 δ j r 2 α 1 + r = 0 k 2 ( δ i r 1 δ j r 1 + δ i r 2 δ j r 2 ) α 2 + r = 0 k 2 ( δ i r 1 δ j r 2 + δ i r 2 δ j r 1 ) β 2 + 1 + r = 0 k 2 ( δ i r 1 δ j r 2 + δ i r 2 ) β 1 + r = 0 k 2 ( δ i r 1 δ j r 1 + δ i r 2 ) β 2 + r = 0 k 2 ( δ i r 1 δ j r 2 + δ i r 2 ) ,
if i k 1 = j k 1 = j k = 2 ,   i k = 1 ,
u ( i j ) k ( i j ) 0 = β 1 α 1 + r = 0 k 2 δ i r 2 δ j r 1 α 2 + r = 0 k 2 ( δ i r 1 δ j r 2 + δ i r 2 δ j r 1 ) α 3 + r = 0 k 2 δ i r 2 β 1 + 1 + r = 0 k 2 ( δ i r 1 δ j r 1 + δ i r 2 ) β 1 + r = 0 k 2 ( δ i r 1 δ j r 1 + δ i r 2 ) β 2 + r = 0 k 2 ( δ i r 1 δ j r 2 + δ i r 2 ) ,
if i k 1 = j k 1 = j k = 1 ,   i k = 2 ,
u ( i j ) k ( i j ) 0 = β 1 α 1 + r = 0 k 2 δ i r 2 δ j r 1 α 3 + 1 + r = 0 k 2 δ i r 2 α 2 + r = 0 k 2 ( δ i r 1 δ j r 2 + δ i r 2 δ j r 1 ) β 1 + 1 + r = 0 k 2 ( δ i r 1 δ j r 1 + δ i r 2 ) β 2 + 1 + r = 0 k 2 ( δ i r 1 δ j r 2 + δ i r 2 ) β 1 + r = 0 k 2 ( δ i r 1 δ j r 1 + δ i r 2 ) ,
if i k 1 = j k 1 = i k = 2 ,   j k = 1 ,
u ( i j ) k ( i j ) 0 = β 2 α 2 + r = 0 k 2 δ i r 2 δ j r 2 α 1 + r = 0 k 2 ( δ i r 1 δ j r 1 + δ i r 2 δ j r 2 ) α 3 + r = 0 k 2 δ i r 2 β 2 + 1 + r = 0 k 2 ( δ i r 1 δ j r 2 + δ i r 2 ) β 1 + r = 0 k 2 ( δ i r 1 δ j r 1 + δ i r 2 ) β 2 + r = 0 k 2 ( δ i r 1 δ j r 2 + δ i r 2 ) ,
if i k 1 = 1 ,   j k 1 = i k = j k = 2 ,
u ( i j ) k ( i j ) 0 = β 2 α 2 + r = 0 k 2 δ i r 2 δ j r 2 α 3 + 1 + r = 0 k 2 δ i r 2 α 1 + r = 0 k 2 ( δ i r 1 δ j r 1 + δ i r 2 δ j r 2 ) β 1 + 1 + r = 0 k 2 ( δ i r 1 δ j r 1 + δ i r 2 ) β 2 + 1 + r = 0 k 2 ( δ i r 1 δ j r 2 + δ i r 2 ) β 2 + r = 0 k 2 ( δ i r 1 δ j r 2 + δ i r 2 ) ,
if j k 1 = 1 ,   i k 1 = i k = j k = 2 .
Remark 2.
Although Formulas (5)–(10) look cumbersome, they are quite convenient for calculating coefficients of the branched continued fraction expansions (4). A detailed description of the process of obtaining these formulas can be found in ([2], Section 6).
Example 1.
For ( i j ) 0 = ( 1 , 1 ) the ratio
F 2 3 ( α 1 , α 2 , α 3 ; β 1 , β 2 ; z ) F 2 3 ( α 1 + 1 , α 2 , α 3 ; β 1 + 1 , β 2 ; z )
has the following formal expansion
1 u 2 , 1 1 , 1 z 1 u 2 , 1 , 1 , 1 1 , 1 z 1 u 2 , 1 , 1 , 1 , 2 , 1 1 , 1 z 1 u 2 , 1 , 1 , 1 , 2 , 1 , 1 , 1 1 , 1 z 1   u 2 , 1 , 1 , 1 , 2 , 1 , 2 , 2 1 , 1 z 1   u 2 , 1 , 2 , 2 1 , 1 z 1 u 2 , 1 , 2 , 2 , 1 , 2 1 , 1 z 1 u 2 , 1 , 2 , 2 , 1 , 2 , 2 , 2 1 , 1 z 1   u 2 , 1 , 2 , 2 , 2 , 1 1 , 1 z 1 u 2 , 1 , 2 , 2 , 2 , 1 , 1 , 1 1 , 1 z 1   u 2 , 1 , 2 , 2 , 2 , 1 , 2 , 2 1 , 1 z 1   .
Remark 3.
For ( i j ) 0 = ( 1 , 1 ) , the ratio (11) was considered in [5], where a description of the process of computing the coefficients of (12) through the parameters of (1), without the expansion itself, are given.
Similarly, we have the following example.
Example 2.
For ( i j ) 0 = ( 1 , 2 ) the ratio
F 2 3 ( α 1 , α 2 , α 3 ; β 1 , β 2 ; z ) F 2 3 ( α 1 , α 2 + 1 , α 3 ; β 1 , β 2 + 1 ; z )
has the following formal branched continued fraction
1 u 2 , 2 1 , 2 z 1 u 2 , 2 , 1 , 2 1 , 2 z 1 u 2 , 2 , 1 , 2 , 2 , 2 1 , 2 z 1 u 2 , 2 , 1 , 2 , 2 , 2 , 1 , 2 1 , 2 z 1   u 2 , 2 , 1 , 2 , 2 , 2 , 2 , 1 1 , 2 z 1   u 2 , 2 , 2 , 1 1 , 2 z 1 u 2 , 2 , 2 , 1 , 1 , 1 1 , 2 z 1 u 2 , 2 , 2 , 1 , 1 , 1 , 2 , 1 1 , 2 z 1   u 2 , 2 , 2 , 1 , 2 , 2 1 , 2 z 1 u 2 , 2 , 2 , 1 , 2 , 2 , 1 , 2 1 , 2 z 1   u 2 , 2 , 2 , 1 , 2 , 2 , 2 , 1 1 , 2 z 1   .
Example 3.
For ( i j ) 0 = ( 2 , 1 ) the ratio
F 2 3 ( α 1 , α 2 , α 3 ; β 1 , β 2 ; z ) F 2 3 ( α 1 , α 2 + 1 , α 3 + 1 ; β 1 + 1 , β 2 + 1 ; z )
has the following formal branched continued fraction
1 u 1 , 1 2 , 1 z 1 u 1 , 1 , 2 , 1 2 , 1 z 1 u 1 , 1 , 2 , 1 , 1 , 1 2 , 1 z 1 u 1 , 1 , 2 , 1 , 1 , 1 , 2 , 1 2 , 1 z 1   u 1 , 1 , 2 , 1 , 2 , 2 2 , 1 z 1 u 1 , 1 , 2 , 1 , 2 , 2 , 1 , 2 2 , 1 z 1   u 1 , 1 , 2 , 1 , 2 , 2 , 2 , 1 2 , 1 z 1   u 2 , 2 2 , 1 z 1 u 2 , 2 , 1 , 2 2 , 1 z 1 u 2 , 2 , 1 , 2 , 2 , 2 2 , 1 z 1 u 2 , 2 , 1 , 2 , 2 , 2 , 1 , 2 2 , 1 z 1   u 2 , 2 , 1 , 2 , 2 , 2 , 2 , 1 2 , 1 z 1   u 2 , 2 , 2 , 1 2 , 1 z 1 u 2 , 2 , 2 , 1 , 1 , 1 2 , 1 z 1 u 2 , 2 , 2 , 1 , 1 , 1 , 2 , 1 2 , 1 z 1   u 2 , 2 , 2 , 1 , 2 , 2 2 , 1 z 1 u 2 , 2 , 2 , 1 , 2 , 2 , 1 , 2 2 , 1 z 1   u 2 , 2 , 2 , 1 , 2 , 2 , 2 , 1 2 , 1 z 1   .
Finally, let us consider one more example.
Example 4.
For ( i j ) 0 = ( 2 , 2 ) the ratio
F 2 3 ( α 1 , α 2 , α 3 ; β 1 , β 2 ; z ) F 2 3 ( α 1 + 1 , α 2 , α 3 + 1 ; β 1 + 1 , β 2 + 1 ; z )
has the following formal branched continued fraction
1 u 1 , 2 2 , 2 z 1 u 1 , 2 , 2 , 2 2 , 2 z 1 u 1 , 2 , 2 , 2 , 1 , 2 2 , 2 z 1 u 1 , 2 , 2 , 2 , 1 , 2 , 2 , 2 2 , 2 z 1   u 1 , 2 , 2 , 2 , 2 , 1 2 , 2 z 1 u 1 , 2 , 2 , 2 , 2 , 1 , 1 , 1 2 , 2 z 1   u 1 , 2 , 2 , 2 , 2 , 1 , 2 , 2 2 , 2 z 1   u 2 , 1 2 , 2 z 1 u 2 , 1 , 1 , 1 2 , 2 z 1 u 2 , 1 , 1 , 1 , 2 , 1 2 , 2 z 1 u 2 , 1 , 1 , 1 , 2 , 1 , 1 , 1 2 , 2 z 1   u 2 , 1 , 1 , 1 , 2 , 1 , 2 , 2 2 , 2 z 1   u 2 , 1 , 2 , 2 2 , 2 z 1 u 2 , 1 , 2 , 2 , 1 , 2 2 , 2 z 1 u 2 , 1 , 2 , 2 , 1 , 2 , 2 , 2 2 , 2 z 1   u 2 , 1 , 2 , 2 , 2 , 1 2 , 2 z 1 u 2 , 1 , 2 , 2 , 2 , 1 , 1 , 1 2 , 2 z 1   u 2 , 1 , 2 , 2 , 2 , 1 , 2 , 2 2 , 2 z 1   .
The following result is proved in ([2], Theorem 2):
Theorem 2.
Let ( i j ) 0 be an arbitrary pair in I and let g ( i j ) k ( i j ) 0 ,   ( i j ) k I k ( i j ) 0 ,   k 1 , be the real numbers satisfying the inequalities
0 < g ( i j ) k ( i j ) 0 1 , ( i j ) k I k ( i j ) 0 , k 1 .
Then the branched continued fraction
1 + i 1 = 1 + δ i 0 1 | i 1 j 1 | | i 0 j 0 | , j 1 { 1 , 2 } 2 g ( i j ) 1 ( i j ) 0 z ( i j ) 1 ( i j ) 0 1 + i 2 = 1 + δ i 1 1 | i 2 j 2 | | i 1 j 1 | , j 2 { 1 , 2 } 2 g ( i j ) 2 ( i j ) 0 ( 1 g ( i j ) 1 ( i j ) 0 ) z ( i j ) 2 ( i j ) 0 1 +   + i k = 1 + δ i k 1 1 | i k j k | | i k 1 j k 1 | , j k { 1 , 2 } 2 g ( i j ) k ( i j ) 0 ( 1 g ( i j ) k 1 ( i j ) 0 ) z ( i j ) k ( i j ) 0 1 +  
converges absolutely and uniformly for
| z ( i j ) k ( i j ) 0 | 1 i k 1 , ( i j ) k I k ( i j ) 0 , k 1 .

3. Domain of Analytical Extension

In this section, we prove our main result and give an example of its application to the dilogarithm function.
The following is true:
Theorem 3.
Let ( i j ) 0 be an arbitrary pair in I and let (1) be a generalized hypergeometric function with parameters satisfying the inequalities
u ( i j ) k ( i j ) 0 0 , i k = 1 + δ i k 1 1 | i k j k | | i k 1 j k 1 | , j k { 1 , 2 } 2 u ( i j ) k ( i j ) 0 κ , ( i j ) k I k ( i j ) 0 , k 2 ,
where u ( i j ) k ( i j ) 0 ,   ( i j ) k I k ( i j ) 0 ,   k 2 , are defined by (5)–(10) herewith β 1 , β 2 Z , κ is a positive number. Then,
(A)
The branched continued fraction (4) converges uniformly on every compact subset of the domain
H κ = z C : z 1 4 κ , +
to the function f ( z ) , holomorphic in this domain;
(B)
The function f ( z ) is an analytic continuation of the ratio (3) in the domain (14).
Remark 4.
Conditions (13) are satisfied when
a k 0 , k { 1 , 2 , 3 } , a n d b k a k , b k a 3 , b k 0 , k { 1 , 2 } .
Proof Theorem 3.
To prove (A), we will use the convergence continuation theorem ([19], Theorem 3), which provides an extension of the domain of convergence from the already known small one to a wider one. To do this, we need to show that the approximants of the branched continued fraction (4) form a sequence of holomorphic functions in the domain (14), uniformly bounded on each compact subset of this domain.
First, we introduce the notation of the so-called tails of the expansion (4), which will allow us to write the approximants in a convenient form. Let ( i j ) 0 be an arbitrary pair in I . We set
W ( i j ) n ( i j ) 0 n ( z ) = 1 , ( i j ) n I n ( i j ) 0 , n 1 ,
and
W ( i j ) k ( i j ) 0 n ( z ) = 1 i k + 1 = 1 + δ i k 1 | i k + 1 j k + 1 | | i k j k | , j k + 1 { 1 , 2 } 2 u ( i j ) k + 1 ( i j ) 0 z 1 i k + 2 = 1 + δ i k + 1 1 | i k + 2 j k + 2 | | i k + 1 j k + 1 | , j k + 2 { 1 , 2 } 2 u ( i j ) k + 2 ( i j ) 0 z 1   i n = 1 + δ i n 1 1 | i n j n | | i n 1 j n 1 | , j n { 1 , 2 } 2 u ( i j ) n ( i j ) 0 z 1 ,
where ( i j ) k I k ( i j ) 0 ,   1 k n 1 ,   n 2 . Then it is obvious that the following recurrence relation holds
W ( i j ) k ( i j ) 0 n ( z ) = 1 i k + 1 = 1 + δ i k 1 | i k + 1 j k + 1 | | i k j k | , j k + 1 { 1 , 2 } 2 u ( i j ) k + 1 ( i j ) 0 z W ( i j ) k + 1 ( i j ) 0 n ( z ) , ( i j ) k I k ( i j ) 0 , 1 k n 1 , n 2 .
Furthermore, if f n ( i j ) 0 ( z ) denotes nth approximant of branched continued fraction (4), then
f n ( i j ) 0 ( z ) = 1 + i 1 = 1 δ i 0 1 | i 1 j 1 | | i 0 j 0 | , j 1 { 1 , 2 } 2 u ( i j ) 1 ( i j ) 0 z 1 i 2 = 1 + δ i 1 1 | i 2 j 2 | | i 1 j 1 | , j 2 { 1 , 2 } 2 u ( i j ) 2 ( i j ) 0 z 1   1 i n = 1 + δ i n 1 1 | i n j n | | i n 1 j n 1 | , j n { 1 , 2 } 2 u ( i j ) n ( i j ) 0 z 1 = 1 i 1 = 1 + δ i 0 1 | i 1 j 1 | | i 0 j 0 | , j 1 { 1 , 2 } 2 u ( i j ) 1 ( i j ) 0 z W ( i j ) 1 ( i j ) 0 n ( z ) ,
where u ( i j ) k ( i j ) 0 ,   ( i j ) k I k ( i j ) 0 ,   k 1 , are defined by (5)–(10).
In what follows, we will show that { f n ( i j ) 0 ( z ) } is a sequence of functions holomorphic in the domain (14). Since f n ( i j ) 0 ( z ) ,   n 1 , are rational functions, it suffices to prove that W ( i j ) 1 ( i j ) 0 n ( z ) 0 for all indices and for all z H κ .
We write the domain (14) in the form
H κ = α ( π / 2 , π / 2 ) H κ , α ,
where
H κ , α = z C : | z | + Re ( z e 2 i α ) < cos 2 α 2 κ .
Let n be an arbitrary natural number and α be an arbitrary number from the interval ( π / 2 , π / 2 ) . By induction on k ,   1 k n , for ( i j ) k I ( i j ) 0 , we prove that
Re ( W ( i j ) k ( i j ) 0 n ( z ) e i α ) > cos α 2 > 0 .
From (15), it is obvious that for k = n and ( i j ) n I n ( i j ) 0 , the inequalities (18) are valid. By the induction hypothesis that (18) hold for k = r + 1 and ( i j ) r + 1 I r + 1 ( i j ) 0 such that r + 1 n , we prove the inequalities (18) for k = r and ( i j ) r I r ( i j ) 0 . The use of (16) for ( i j ) r I ( i j ) 0 leads to
W ( i j ) r ( i j ) 0 n ( z ) e α = e α i r + 1 = 1 + δ i r 1 | i r + 1 j r + 1 | | i r j r | , j r + 1 { 1 , 2 } 2 u ( i j ) r + 1 ( i j ) 0 z e 2 i α W ( i j ) r + 1 ( i j ) 0 n ( z ) e α .
Now, for an arbitrary i r + 1 such that 1 + δ i r 1 i r + 1 2 ,   | i r + 1 j r + 1 | | i r j r | ,   j r + 1 { 1 , 2 } , it follows from (13) and (17) that
u ( i j ) r + 1 ( i j ) 0 ( | z | + Re ( z e 2 i α ) ) < cos 2 α 2 .
From this inequality, it is easy to show that
Im ( u ( i j ) r + 1 ( i j ) 0 z e 2 i α ) 2 4 4 Re ( u ( i j ) r + 1 ( i j ) 0 z e 2 i α ) .
Next, using ([20], Corollary 2), (13), (17), and the induction hypothesis, we have
Re ( W ( i j ) k ( i j ) 0 n ( z ) e i α ) cos α i r + 1 = 1 + δ i r 1 | i r + 1 j r + 1 | | i r j r | , j r + 1 { 1 , 2 } 2 u ( i j ) r + 1 ( i j ) 0 ( | z | + Re ( z e 2 i α ) ) Re ( W ( i j ) r + 1 ( i j ) 0 n ( z ) e α ) > cos α cos α 2 κ i r + 1 = 1 + δ i r 1 | i r + 1 j r + 1 | | i r j r | , j r + 1 { 1 , 2 } 2 u ( i j ) r + 1 ( i j ) 0 cos α 2 > 0 .
Thus,
W ( i j ) 1 ( i j ) 0 n ( z ) 0 , ( i j ) 1 I 1 ( i j ) 0 , n 1 , a n d z H κ , α .
Furthermore, the approximants f n ( i j ) 0 ( z ) ,   n 1 , of expansion (4) are functions holomorphic in the domain (17), and, consequently, in (14) by virtue of arbitrariness α .
In what follows, we will show that { f n ( i j ) 0 ( z ) } is a sequence of functions uniformly bounded on every compact subset of the domain (14).
Let K be an arbitrary compact subset of H κ . Then there exists an open disk
H τ = { z C : | z | < τ }
such that K H τ . Let us cover K by domains of the form
H κ , α , τ = H κ , α H τ .
From this cover we choose a finite subcover
H κ , α 1 , τ , H κ , α 2 , τ , , H κ , α k , τ .
Then, using (18), for the arbitraries r { 1 , 2 , , k } , z H κ , α r , τ , and f n ( i j ) 0 ( z ) ,   n 1 , we obtain
| f n ( i j ) 0 ( z ) | 1 + i 1 = 1 + δ i 0 1 | i 1 j 1 | | i 0 j 0 | , j 1 { 1 , 2 } 2 u ( i j ) 1 ( i j ) 0 | z | | W ( i j ) 1 ( i j ) 0 n ( z ) e i α | 1 + i 1 = 1 + δ i 0 1 | i 1 j 1 | | i 0 j 0 | , j 1 { 1 , 2 } 2 u ( i j ) 1 ( i j ) 0 τ Re ( W ( i j ) 1 ( i j ) 0 n ( z ) e i α ) < 1 + i 1 = 1 + δ i 0 1 | i 1 j 1 | | i 0 j 0 | , j 1 { 1 , 2 } 2 2 u ( i j ) 1 ( i j ) 0 τ cos α = C ( H κ , α r , τ ) .
We set
C ( K ) = max 1 r k C ( H κ , α r , τ ) .
Then for arbitrary z K we have
| f n ( i j ) 0 ( z ) | C ( K ) , n 1 ,
i.e., the sequence { f n ( i j ) 0 ( z ) } is uniformly bounded on every compact subset of the domain (14).
Let
λ = i 1 = 1 + δ i 0 1 | i 1 j 1 | | i 0 j 0 | , j 1 { 1 , 2 } 2 | u ( i j ) 1 ( i j ) 0 | , μ = min 1 4 λ , 1 4 κ
and
H η = { z R : 0 < z < η } , 0 < η < μ .
Then for arbitraries ( i j ) k I k ( i j ) 0 ,   k 1 , and z H η we obtain
| u ( i j ) k ( i j ) 0 z | < η max { λ , κ } 1 4 < 1 i k 1 ,
i.e., the elements of (4) satisfy the conditions of Theorem 2, with
g ( i j ) k ( i j ) 0 = 1 2 , ( i j ) k I k ( i j ) 0 , k 1 , z ( i j ) 1 ( i j ) 0 = 2 u ( i j ) 1 ( i j ) 0 z , ( i j ) 1 I 1 ( i j ) 0 , z ( i j ) k ( i j ) 0 = 4 u ( i j ) k ( i j ) 0 z , ( i j ) k I k ( i j ) 0 , k 2 .
According to Theorem 2, branched continued fraction (4) converges in the domain H η . Evidently H η H κ for each 0 < η < μ , in particular H μ / 2 H κ , Finally, by ([19], Theorem 3), the convergence of (4) is uniform on compact subsets of the domain (14).
The proof of (B) is similar to the proof of ([4], Theorem 2), hence it is omitted. □
From Theorem 3 we have the following consequence.
Corollary 1.
Let ( i j ) 0 = ( 1 , 1 ) and F 2 3 ( 1 , α 2 , α 3 ; β 1 , β 2 ; z ) be a generalized hypergeometric function with parameters satisfying the inequalities
u ( i j ) k ( i j ) 0 > 0 , i k = 1 + δ i k 1 1 | i k j k | | i k 1 j k 1 | , j k { 1 , 2 } 2 u ( i j ) k ( i j ) 0 κ , ( i j ) k I k ( i j ) 0 , k 1 ,
where u ( i j ) k ( i j ) 0 ,   ( i j ) k I k ( i j ) 0 ,   k 1 are defined by (5)–(10), α 1 = 0 ,   β 1 is replaced by β 1 1 and β 1 Z { 1 } ,   β 2 Z , κ is a positive number. Then, the branched continued fraction
1 1 i 1 = 1 + δ i 0 1 | i 1 j 1 | | i 0 j 0 | , j 1 { 1 , 2 } 2 u ( i j ) 1 ( i j ) 0 z 1 + i 2 = 1 + δ i 1 1 | i 2 j 2 | | i 1 j 1 | , j 2 { 1 , 2 } 2 u ( i j ) 2 ( i j ) 0 z 1   i k = 1 + δ i k 1 1 | i k j k | | i k 1 j k 1 | , j k { 1 , 2 } 2 u ( i j ) k ( i j ) 0 z 1  
converges uniformly on every compact subset of (14) to the function f ( i j ) 0 ( z ) , holomorphic in H κ ; in addition, f ( i j ) 0 ( z ) is an analytic continuation of the function F 2 3 ( 1 , α 2 , α 3 ; β 1 , β 2 ; z ) in the domain (14).
Remark 5.
The similar consequences are valid when
(a)
( i j ) 0 = ( 1 , 2 ) and α 2 = 0 ,   β 2 is replaced by β 2 1 ;
(b)
( i j ) 0 = ( 2 , 1 ) and α 2 = 0 (or α 3 = 0 ), α 3 (or α 2 ), β 1 ,   β 2 are replaced by α 3 1 (or α 2 1 ), β 1 1 ,   β 2 1 , respectively;
(c)
( i j ) 0 = ( 2 , 2 ) and α 1 = 0 (or α 3 = 0 ), α 1 (or α 3 ), β 1 ,   β 2 are replaced by α 1 1 (or α 3 1 ), β 1 1 ,   β 2 1 , respectively.
Example 5.
Consider the dilogarithm function (see, [21])
Li 2 ( z ) = z F 2 3 ( 1 , 1 , 1 ; 2 , 2 ; z ) = k = 1 z k k 2 .
It follows from Corollary 1 that the expansion
z 1 i 1 = 1 + δ i 0 1 | i 1 j 1 | | i 0 j 0 | , j 1 { 1 , 2 } 2 u ( i j ) 1 ( i j ) 0 z 1 + i 2 = 1 + δ i 1 1 | i 2 j 2 | | i 1 j 1 | , j 2 { 1 , 2 } 2 u ( i j ) 2 ( i j ) 0 z 1   i k = 1 + δ i k 1 1 | i k j k | | i k 1 j k 1 | , j k { 1 , 2 } 2 u ( i j ) k ( i j ) 0 z 1  
is an analytic continuation of the function Li 2 ( z ) in (14), where u ( i j ) k ( i j ) 0 ,   ( i j ) k I k ( i j ) 0 ,   k 1 , are defined by (5)–(10), ( i j ) 0 = ( 1 , 1 ) ,   α 1 = 0 , and β 1 is replaced by β 1 1 , κ is defined by (19).
Indeed, we will show that the coefficients of the branched continued fraction (20) satisfy conditions (19). From (7) we have
u 2 , 1 1 , 1 = ( 2 1 ) 1 · 1 ( 2 + 1 ) 2 · 2 = 1 12 .
Now, from (5) and (10) we get
u 2 , 1 , 1 , 1 1 , 1 = ( 2 1 + δ 1 1 δ 1 1 ) ( 1 + δ 1 1 δ 1 1 + δ 1 2 δ 1 2 ) ( 1 + δ 1 1 δ 1 2 + δ 1 2 δ 1 1 ) ( 2 + 1 + δ 1 1 δ 1 1 + δ 1 2 ) ( 2 + δ 1 1 δ 1 1 + δ 1 2 ) ( 2 + δ 1 1 δ 1 2 + δ 1 2 ) = 1 6 , u 2 , 1 , 2 , 2 1 , 1 = ( 2 1 + δ 1 2 δ 1 2 ) ( 1 + 1 + δ 1 2 ) ( 1 + δ 1 1 δ 1 1 + δ 1 2 δ 1 2 ) ( 2 + 1 + δ 1 1 δ 1 1 + δ 1 2 ) ) ( 2 + 1 + δ 1 1 δ 1 2 + δ 1 2 ) ) ( 2 + δ 1 1 δ 1 2 + δ 1 2 ) = 1 6 ,
respectively. Next, from (7) we obtain
u 2 , 1 , 1 , 1 , 2 , 1 1 , 1 = ( 2 1 + δ 1 2 δ 1 1 + δ 2 2 δ 1 1 ) ( 1 + δ 1 1 δ 1 2 + δ 1 2 δ 1 1 + δ 2 1 δ 1 2 + δ 2 2 δ 1 1 ) ( 1 + δ 1 2 + δ 2 2 ) ( 2 + 1 + δ 1 1 δ 1 1 + δ 1 2 + δ 2 1 δ 1 1 + δ 2 2 ) ( 2 + δ 1 1 δ 1 1 + δ 1 2 + δ 2 1 δ 2 1 + δ 2 2 ) ( 2 + δ 1 1 δ 1 2 + δ 1 2 + δ 2 1 δ 1 2 + δ 2 2 ) = 2 15
and from (6) and (8) we have
u 2 , 1 , 2 , 2 , 1 , 2 1 , 1 = ( 2 1 + δ 1 1 δ 1 2 + δ 2 1 δ 1 2 ) ( 1 + δ 1 1 δ 1 1 + δ 1 2 δ 1 2 + δ 2 1 δ 1 1 + δ 2 2 δ 1 2 ) ( 1 + δ 1 1 δ 1 2 + δ 1 2 δ 1 1 + δ 2 1 δ 1 2 + δ 2 2 δ 1 1 ) ( 2 + 1 + δ 1 1 δ 1 2 + δ 1 2 + δ 2 1 δ 1 2 + δ 2 2 ) ( 2 + δ 1 1 δ 1 1 + δ 1 2 + δ 2 1 δ 1 1 + δ 2 2 ) ( 2 + δ 1 1 δ 1 2 + δ 1 2 + δ 2 1 δ 1 2 + δ 2 2 ) = 1 12 , u 2 , 1 , 2 , 2 , 2 , 1 1 , 1 = ( 2 1 + δ 1 2 δ 1 1 + δ 2 2 δ 1 1 ) ( 1 + 1 + δ 1 2 + δ 2 2 ) ( 1 + δ 1 1 δ 1 2 + δ 1 2 δ 1 1 + δ 2 1 δ 1 2 + δ 2 2 δ 1 1 ) ( 2 + 1 + δ 1 1 δ 1 1 + δ 1 2 + δ 2 1 δ 1 1 + δ 2 2 ) ( 2 + 1 + δ 1 1 δ 1 2 + δ 1 2 + δ 2 1 δ 1 2 + δ 2 2 ) ( 2 + δ 1 1 δ 1 1 + δ 1 2 + δ 2 1 δ 1 1 + δ 2 2 ) = 3 20 ,
respectively.
In the next step, using (5), (10), (9), (5), and (10), we compute
u 2 , 1 , 1 , 1 , 2 , 1 , 1 , 1 1 , 1 , u 2 , 1 , 1 , 1 , 2 , 1 , 2 , 2 1 , 1 , u 2 , 1 , 2 , 2 , 1 , 2 , 2 , 2 1 , 1 , u 2 , 1 , 2 , 2 , 2 , 1 , 1 , 1 1 , 1 , a n d u 2 , 1 , 2 , 2 , 2 , 1 , 2 , 2 1 , 1 ,
respectively. It is easy to see that these and all other coefficients of the branched continued fraction (20) will be positive numbers. The validity of the second inequality in (19) follows from the fact that for each ( i j ) k I k 1 , 1 ) ,   k 1 , there is a finite limit
lim k + u ( i j ) k 1 , 1 .

4. Conclusions

The paper establishes a new domain of analytical extension of ratios (3), which is a plane with a section along the real axis from 1 / ( 4 κ ) to + , where κ is a positive number that depends only on the coefficients of the branched continued fraction expansions (4). Provided that κ < 2 ϱ , this domain will be wider than (2). Theorem 3, ([4], Theorem 2), and ([2], Theorem 3) use three different methods to prove the convergence of the expansions (4) in the corresponding domains. The methodology of proving the result of this paper can be used to establish the domains of analytical continuation of other ratios of hypergeometric functions that have representations in the form of branched continued fractions.
Further study of branched continued fraction (4) is possible in the following directions. First of all, this is the study of the convergence regions and the rate of convergence of these expansions for both real and complex coefficients. In this direction, the results of [22,23,24,25,26] are interesting and very promising. Collections of results on the convergence of branched continued fractions can be found in [27,28]. Some convergence problems related to continued fractions and branched continued fractions are described in [29,30]. Also of no less importance is the study of computational stability. There are also some interesting results and ideas for proving them [11,31,32].
We can also study other special functions using branched continued fractions, such as those discussed in [33,34]. Finally, we can also try to apply quantum calculus to branched continued fraction expansions for some special functions (some interesting results in this direction related to polynomials can be found in [35,36]).

Author Contributions

Conceptualization, R.D.; writing—original draft, R.D. and S.H.; writing—review & editing, R.D., M.D. and S.H.; project administration, R.D. and M.D. All authors have read and agreed to the published version of the manuscript.

Funding

This research received no external funding.

Data Availability Statement

Data are contained within the article.

Acknowledgments

This research was supported by the Ministry of Education and Science of Ukraine, project registration number 0123U101791.

Conflicts of Interest

The authors declare no conflicts of interest.

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Dmytryshyn, R.; Dmytryshyn, M.; Hladun, S. On the Domain of Analytical Continuation of the Ratios of Generalized Hypergeometric Functions 3F2. Axioms 2025, 14, 871. https://doi.org/10.3390/axioms14120871

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Dmytryshyn R, Dmytryshyn M, Hladun S. On the Domain of Analytical Continuation of the Ratios of Generalized Hypergeometric Functions 3F2. Axioms. 2025; 14(12):871. https://doi.org/10.3390/axioms14120871

Chicago/Turabian Style

Dmytryshyn, Roman, Marta Dmytryshyn, and Sofiia Hladun. 2025. "On the Domain of Analytical Continuation of the Ratios of Generalized Hypergeometric Functions 3F2" Axioms 14, no. 12: 871. https://doi.org/10.3390/axioms14120871

APA Style

Dmytryshyn, R., Dmytryshyn, M., & Hladun, S. (2025). On the Domain of Analytical Continuation of the Ratios of Generalized Hypergeometric Functions 3F2. Axioms, 14(12), 871. https://doi.org/10.3390/axioms14120871

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