27 T. H. Havelock. 
We may obtain this result without analysing the expressions for ¢ and 0¢/0z. 
We obtain from (3) the following complete expressions in real terms, at z = 0, 
: ” wu COS K& — ( — Kp) SIN KE _py 
§ = 2M |, 7 eee a es (8) 
od. M * uk, Cos Ka — {xe (1 — Ko) + p2} sin Kx Aebhe (9) 
dz do (ke = kg)? + 2 
To carry out the integration of the product over the surface, we use the follow- 
ing theorem : if 
i@ = | (A, cos Kx + B, sin xx) dk, 
0 
v@= { (A, cos xx + B, sin xa) dx, 
0 
where A,, Ay, B,, B, are functions of x, then 
[. iQ b@m— a (i Ny HET, 1B) Ge (10) 
This theorem is derived from the Fourier double integral 
Bw =| dk | 5 ®) eee = oe) de, (11) 
Tw JO —00 
and is subject to the same conditions. 
In the present case, comparing (8) and (11) we have 
2nKyMye 4 
ie ¢ (a) cos Ka da = (aaa 
‘0D ) wee —xf 
| ¢ (x) sin xa dx = — aS (12) 
- © ( — Ko)? + 
Hence we have 
© oo OY tai] 
| ) ae dx = 4k )M? | —__ dk, 
-a Oz 0 (K — Ky)? + p 
; LS Kee KS 
R= Lim trecgMPou | ———— dk 
u>0 0 (kK — Ko)” sil ye 
= 47? 0K 2M 2e~ 2nof, (13) 
281 
