PSI - Issue 3

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

541

16

1 

1  0

1  0

1 2

1 2

1

1 2

1 2

1 2

1 2

  

  

   





  2 











 ln 1











  1

1

1

1

ln ln 2 1  t dt t

ln 2

,

 d t

t

t

dt





 















   

 

1



where    x Euler Phi function. Thus, it is found

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

1 2

z z

  

  

  

      

 

  1

.

2

ln 2

L z











 

Let’s take here

1   z x . Then

1 

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

1 2

dt

x x

  

  

  

      





  1

(31)

ln

2

ln 2

 x t











 

2

1

t



1



and it leads from the obtained equality that for

0  n

1 2

  

  

   

 

  1

ln 2

when

1 0   x . It can be seen there is no singularity here.

0 L x



 

   

 

After the differentiation of (31) one gets

  

  

1 1 3 1 2 1 2 2 1 3         x F x 1, ; ;  

1 3 5 1 2, ; ; 1 2 2 1      x F x  

x x

x x

  

2

2





 









2

1

1

.

1 L x

x

  x



 



1 2

2 1





   L x 1





Hence finally for

, if

1 0   x .

x





0 n 

2

1 0    x . The equality (29) should be integrated along some curve

Let’s consider the case when

1   z with arbitrary point z

connecting point

1 

z

z ds

1

1 2

1

1 2

1





  2 



  2 











1

1

1   

1

.

s

s

ds

  d









s







1

1

1







1 1

1 1

u u

 

 

 

Integral in left hand part with the change of variable

, where

, is calculated by the

s

 u z z

 

analogous way

 z

1

1 2

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

z z

  

  2 







1

1

2

.

s

s

ds





 

1



1     z ds

 ln 1       ln  



, the integral in right hand part takes the form

With the fact that

z

s

1 

1 

1 

z ds

1

1 2

1

1 2

1

1 2



  2 



  2 



  2 











 ln 1

   d







 z d

1

1

1  

1

1  

1

ln

.

  d



























s





1

1

1

1









Let’s change variable

2 1    t of the first integral in the right hand part. Then