PSI - Issue 3

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

539

14

This expression is symmetrical with respect to the  and , t so this formula is true as

,   t and as

  t too.

Appendix B. The limit values of the integrals (25) Let’s find the limit values of the integrals (25) when

1 0   x . It simpler to provide all   , .   n P x Let’s use the equality

1 0    x and

reasoning for the integrals of a more common structure with Jacobi polynomials that leads from the formulas ( 10.8.19 and 10.8.20, Beitman et al (1974))             1 , , 1 1 1 1 1 sin                     n n t t x x P t dt P x x t

(28)

           n n  1     1



1

x





  

  1;1   x

where

1, 1       ,

2

1,       F n n

;1 ;   

,



2

    

and   , ; ; F a b c x Gauss hypergeometric function. Let’s integrate this integral by parts

dt

        

        

ln 1

u

du





 x t

 x t

1 

     t 1

   t P t dt ,   

        dv t 1 1

   t P t dt ,   



ln 1 1



n

n

 x t

 1 1   

1



  1  

1 1, 1 1     n P   



  t

1

v

t

t



2

n

1

1 1

1 1   n x t 2

  1  

1

  1  

1













  t





 

1, 1 1    

1, 1

 

 

ln

1

1

|

1

1

t

t

P

t

t

P

t dt

 















1



1

1

n

n





2

 x t

n

1



  1 1 1  

1





        n 1  







1

x

x







1

x





  

  x

1, 1

1

 

 

 

 

2

,      

1; ;   

.

P

F n n

 











1

n



2 sin 1  n n n It is seen from the result that the initial integral has final limits when  

2       

2

1 0    x and

1 0   x .

 1 , 1 2 2 1 ! 3 n

   n

!

n



1 , 1

 

 

 

Taking into account that     n T x   n

  x

а

,

U x

P x

P



2 2







n

n

1

2

  n

2

2

one gets    

1 1

x







 

  

  x U x 2 1   n 1

, ; ;

.

0 L x

  F n n



2 2

n

n

  1    n L x n   0

  L x n

1 0    x and   0

when

when

This implies that

1 0.   x

Let’s consider the integral     1 ln 1 1 1    t

1 

     t 1





1

t

,   

d



  t P t dt ,   

  P x dx





n

n

dx

 x t

 t x

1 

1





     x 1





        n 



1

1

x

1

x





  

 

,   P x

  

2

1,       F n n

;1 ;   

.











n

sin n Taking into consideration equality (28), one should pass here from Jacobi’s polynomials to the Chebyshov’s polynomials. It gives     1 2 3 1 2 1,1 ; ; . 2 2 1                n T x x L x nF n n x 1 2        