PSI - Issue 3

Nataly Vaysfeld et al. / Procedia Structural Integrity 3 (2017) 526–544 Author name / Structural Integrity Procedia 00 (2017) 000–000

540

15



  1 1  



n

n

1

n



1 2

2 1





        n x  1 2 

j

j

This implies that   L x

when

1 0    x

n





  3

1

2

j

0

j

2 ! j



 2 1 2    x  1 2



and  

n when

1 0.   x

1  L x





2

Let’s consider the special case when

0. n  One can use the approach proposed by G. Ya. Popov in Popov (1968).

One considers the integral

1

1

1 2

   L z z  ln 



 1;1 .

  2 



   



1  

1

,  

d z



 

1



1

1 2

, given on the complex variable z along the cut segment   1;1  . It has value

  2 



and function    g z z  



1

1

z



2        i

2         i

  exp

  exp

on the upper branch of cut, and value

on the lower branch. On the base of Cauchy

g z

g z

  

   C g

1

 C is arbitrary closed contour, covering segment   1;1  ,

formula one can write   g z

, where

d





2

   i

z

 C . We contract the contour to the segment 

 1;1  . As a result one gets

point z is situated outside contour

1 

1

1 2

1

1 2

1

d





 1;1 .

  2 



  2 











(29)

1

1   

1

1

,

z

z

z











 

z







1



1  z with arbitrary point z :

Let’s integrate the equality (29) along some curve connecting point

1 

z

z ds

1

1 2

1

1 2

1

1 

1 

  2 



  2 











(30)

1

1

1   

1

.

s

s

ds

  d









s







1



1 1

1 1

u u

 

 

 

where

is done at the left hand part of the equality

The variable change

s

 u z z



z

1 1 1

1 2

1

1 2

1 3 2 2

1 1 3 1 1, ; ; 1 2 2 1      z F  

z z

  

  

  



1 

0 

1

  2 













1

1

1

2 1, ; ;

2

,

s

s

 ds u

  u d u F 

u











2

2

z with regard to the integral expression (9.111, Gradshtein et al (1963)) for Gauss hypergeometric function. Let’s consider the right hand part of the equality (30). With the known expression     1 ln 1 ln ,          z ds z s one gets

1 

1 

z ds

1

1 2

1

1 2

1 

  2 



  2 











 ln 1



 

1

1

1  

1

.

  d

 d L z















 

1 s In the first integral on the right hand side the change of variable is provided   

1



1 2    t :