352 



First we investigate the number of poles of the 



Bessel functions, and using the definition of C 



integrands for e = 0, hence, the number of positive we find 

 real roots of 



a = ? 



K (5) 





 Ki(C) 



, ? > 0. 



(47) 



6 5 = 



i £ K (5 )Ki(C ) 

 0-^0 



(56) 



{2Ki(S )K (5 )-S Ki2(5 )+5 K 2(5 )} 



where now (43) a = r^/u^r ^ = 2 (p^ - p )/pu2. 



Hence by (49) we find that 



From the well known expansions of Kq{£;) and 

 Ki(?) it follows that the right hand side of (47) 

 is zero for C = and tends to infinity for 5 -s- ■». 

 We prove that this function increases monotonically 

 with C, hence, we have to consider 



Im(S +6 5) > 0, 







(57) 



K (5) K (S) 

 d_ J. _o ^ 2 _o 



dC ^ Ki(5) Ki{S) 



S + S 



K M5) 







(48) 



Instead of proving that the right hand side of (48) 

 is positive we will show that 



or the pole of the integrand of the first integral 

 in (46) is slightly above the real axis for e small 

 and E > . In the same way the pole of the inte- 

 grand of the second integral in (46) is slightly 

 below the real axis. 



Now we want to give a different representation 

 of (46) . We distinguish between two cases x > 

 and X < 0. In the case of x > we rotate the 

 direction of integration of Ij and I2 as follows. 



2 K (S) Ki(5) - 5 Ki2{5) + C K 2(5) > 

 o 



(49) 



(0,-i") , 



Ij -> (0,+i=>) , I 



and in the case of x < 



Ij ^ (0,-i<») , I2 ^ (0,+i") 



(58) 



(59) 



This is easily shown to be true for g ->• <» . Hence, 



when the derivative of (49) is negative the 



function itself has to be positive. This derivative. 



From the foregoing it follows, that for x > 0, a 

 -2 K (C) [Ki(C)-l- 5 K (£)]/5 - Ki'(5) + K 2(C), (50) pole has to be added to Ij as well as to Ij. The 

 ° ° question arises: are there still other poles in 



is negative, since K^ (S) for < C < ". This means 

 that the right hand side of (47) increases mono- 

 tonically, hence, there is one and only one root 

 5 = 5 of (47) in < 5 < ■»■ 



We will estimate the value of £ 



show that 



(£+1) K^(5) - £ Ki(5) > 0. 



Therefore we 



(51) 



From well-known expansions for K and Kj , this 

 inequality holds for £ -^ ". The derivative of the 

 left hand side of (51) being 



(S+1) [Ko(5) - Ki(£) 



(52) 



is clearly negative, and hence (51) holds in < 



5 < ". From Kj (C) > Kq(£) and (51) it follows 



that the root 5 of (47) satisfies 

 



a < C < 

 



a + (h a^ 



a) 



(53) 



Second we have to determine at which side of the 

 real axis this root is situated when e is small 

 but not zero. Consider the denominator of the 

 first integral of (46) , hence a root of 



(C-i I) K (5) - a Ki (?) = 0. 

 o 



(54) 



The zero in the neighborhood of the real axis of 

 (54) is assumed as 



5=5^+6 5, 



(55) 



the complex half plane Re £ > which are passed 

 by rotating the lines of integration? We now 

 shall give a proof that this does not happen. This 

 proof was kindly given to us by our colleague Prof. 

 Dr. B. L. J. Braaksma. 

 Consider the function 



F(5) ^L E. K^(5) - a Ki(a 



- [CKj- (?) 



+ {a+l)Ki(?)] , (60) 



which is real for real values of £. Suppose Sj 

 with Re si > and Im s^ 7^ is a zero of F(C), 

 then also S2 = sj (complex conjugated value) is 

 such a zero. The functions Ki(s.C), j = 1,2, 

 satisfy 



[S 



dC 



2 "^51 



(s^ 52 ^. 1)] Ki(s.C) 

 1 3 



, j = 1,2. 

 (61) 



Multiplying (61) by Kj (s £) with k = 2 for j = 1 

 and k = 1 for j = 2 , we find by subtracting the 

 results 



(Si - S2)5Ki{siS)Ki(s2S) = TT C[Ki(S2C) ^ Ki{siC) 



d_ 

 d? 



d_ 

 d? 



Ki(si5)^l(S2?)] (62) 



Hence 



2 2 

 (sj - S2) 



/? Ki(si£)Ki{S2C)d£ = £[(Ki(S2C)f^i(si5)^- 



d? 



a? 



Kl(si?)^K,<V)l| 



(63) 



where ? satisfies (47) or (54) with e = 0. Sub- 

 stituting (55) into (54), expanding the modified 



It is easily seen that the right hand side 

 vanishes for C ^ <» and because sj and S2 are zeros 



