CONFORM A L TRANSFORMATION 119 



Green's function is replaced by —log r. A similar result holds for the other 

 point in question, namely (0, tt). 



Integrating (7-1) from 6 = to 6 '= w and using equations (A2-1, 3) of 

 Appendix II gives 



ao(v) = nh(v, v) - h{v, v)] + 003") 



00 



= 4/32 X! «"' «"'"" + 0(|8') (7-2) 



n^l.3.5,- ■ • 



where i; > 0. \\'e consider only positive values of v since g{v, d) and the 

 a„(t>)'s are even functions of v. Thus (4-5) yields 



R'^' = -ik2^' Z «"'('^' + kY' + 0(i3') (7-3) 



n-1,3.5.- • 



This is an approximation to the exact value given by the double series in 

 (6-2). Comparison of (6-5) and (7-3) when ^ and k approach zero gives, 

 incidentally, 



±m-'t(n- l/ir =ltfn-'. 



m=-l n"»l ^ m—1 



From (4-2), (7-1) and the expansions (A2-2) of log (chv ± cos 6) it follows 

 that 



fl„(iO = 4/3m-ie-"l''l + 0(/32), m = 1, 3, 5, • • • 



(7-4) 

 aM = 0(/32) , m - 0, 2, 4, 6 • • • 



Equations (4-5), (A2-4), the relation 7„ = m^ — k-, and (A2-8) give us the 

 answer we seek: 



^^3) ^ ^(^2) _ .j^z^^ Yl y-' in-' J{m, m, k, y„ , 0, 0) + 0(^') 



m=1.3.5.--- 



= -ik2^ Z 7»'^"' + 0(^') (7-5) 



m-l.3.5,--- 



It is not necessary to go to i?^ ' because it differs from Rg^ by only 0(/3^). 



When // lies in the plane of the bend the reflection from a small angle 

 corner with no truncation may be obtained by much the same procedure. 

 For brevity we shall not write down the order of magnitude of the remainder 

 terms. From (5-9), (A2-1), and (A2-3) 



b,{v) = ao(r) - a2(r)/2 (7-6) 



= ^[h - h - (h - h)/2] 



