360 



is unknown. We sxibstitute (A, 9) into (A, 7) and 

 separate functions which are analytic in the upper 

 half plane and respectively in the lower one. There- 

 fore we write : 



i(2TT)"^/^{C-iX)"-^ it.+i)''^^^ h'^(5) = 



h+(C) + h-{C) , (A, 11) 



where : 



-i h"^ (iA) 1 



h (C) = 



(2Tr)^/2 (iX+i)l/2 5_ix 



(A, 12) 



is analytic in the lower half of the complex £;-plane, 

 and: 



h+,n i r H''(iA) _ H"'(g) J^_ 



*^' ~ ,1/2 ^., . 1/2 ,^ .1/2^ S-iX ^^'^^' 



(2ti) ' (iX+i) (5+i) 



f(x) = 



-i^x (C-i 



1/2 



(27,) 



1/2 



H (0 



r2_ 1:2. 



[C^ + CiS + h (C) (5^- Cp]dC . 



(A, 19) 



Since the integrand is analytic in the lower half 

 of the 5-plane, the right hand side of (A, 19) 

 vanishes for x > 0, as follows from the calculus 

 of residues. In order to obtain an expression for 

 f (x) for X 1- we investigate the integrand in 

 (A, 19) for ?->•+'». The function f (x) is continuous 

 at X = 0, since~the integrand is 0(£;~3/2)_ Hence 

 f(0~) = 0. For real values of 5 we have, by (A,6) : 



+ +1/2 



H-(5) = (H(C)) "-^ exp 





(f;)] fr^M^'2o' 



is analytic in the upper half plane. Using (A, 9) 

 and (A, 11) we can write (A, 7) as: 



where the integral is a Cauchy principal-value. 

 Hence for real £ we have 



[<5r ""(S) (5+i)"-^''^ h''(5)]+ h"^(5)] (C^ - C^) = 

 c o 



-h (£) a' 



5^) - (C 







i)-^'^^ H (C) f(C) 



(A, 14) 



The function 6r (C) is analytic in the upper 



half plane by virtue of (A, 10). Hence the left 



hand side of (A, 14) is analytic in the upper half 



plane. Since f (x) vanishes for x > 0, f (C) is 



analytic in the lower half plane and hence the 



right hand side of (A, 14) is also analytic there. 



Hence, both sideg of (A, 14) represent an entire 



function. The H~(5) tend to 1 and the h (5) are 



0(1/0 for C -»• <». We assume 6r (?) Z'^^^ and f(C) 



5"'-/^ to be bounded for 5 -s- ■». Then the entire 



function must be a first order polynomial C + Ci 



C, where the values of the constants C and Cj will 



be given later. We can now solve for the unknowns 



6r and f: 

 c 



.r^p=i^±iI^ [^^^^-h^Q], (A,15) 



c __+ . 



f(5) = 



(C-i) 



H (C) 



-1/2 



H (5) 



52-52 



[C^ + Ci C + h (C) (C -5^)] (A, 16 



-1/2 

 The value of Ci is chosen so that f (r) is 0(5 ) 



for 5 ->• to. Hence, by (A,12) : 



[H (S)] 



1 + H(S) , 



(A, 21) 



where H(C) is a continuous function, which is 0(1/5) 

 for 5 -* + " by (A, 4). Therefore, if the factor 

 H~(£) in the denominator in the integrand in (A, 19) 

 is omitted, the value of the integral changes by a 

 term which is (x) for x t 0. The other factors 

 in the integrand in (A, 19) are an exponential and 

 a rational function of C. Using the calculus of 

 residues and Tauberian theorems we can obtain an 

 asymptotic expression for the value of this integral 

 for X t 0. We do not go into details and give the 

 result only: 



f(x) ; 2Tr -^/^ B Ixl ^'^^ , x t (A, 22) 



where : 



B = (X2 + £2) (^ ^ j^) 1/2 jj+(.^) 



(a, 23) 



In a similar way we investigate 6r (x) for x ■(• 0. 

 Substitution of (A, 17) and A, 18) into'^(A,15) and 

 then into (A, 9) gives: 



(S) 



H+{iX) (£+1 



1/2 



2p2 



X^S 



(271)1/2 (iA + i)l/2 H'*'(5) 



(52-c2) {5-iX) 



(A, 24) 



Ci = i (27t) ^^^ (iX + i)"^/^ H"'(iX) . 



(A, 17) 



We choose C so that f(C) is an order smaller, i.e., 

 0(£"3/2), fgr 5 -^ »; hence: 



(277)""^/^ X(iX+i) ■'■■^^ H"^(iX) 



(A, 18) 



The meaning of this choice for f (x) will be discussed 

 later. 



We obtain f(x) with the inverse Fourier transform: 



-5/2 

 This expression is 0(5 ) for C ->■ "> and hence 



6r (x) and its first derivative are continuous at 



X = 0. An expression for x 4- is obtained in the 



same way as for f (x) : 



<5r (x) = 5r (0) + 5r' (0) 

 ceo 



3t7 



1/2 



3/2 , „ 



B X x 4- 0. 



(A, 25) 



At this point we return to our choice (A, 18) for 

 Cq. If (A, 18) does not hold, then it can be shown 

 that: 



