378 Contour Integration in the Problem of Di fraction. 



while when r<r\ or #<#', we have only to interchange r 

 and ?•', or 6 and 0', in the above expression. 



Fio. 2. 



TH£ ?n-P£/!S/£ 



for* fC'J //V TH£ 7n-PL/iH£ 



To deduce the solution of! the diffraction problem, where 

 the boundary is given by the planes = and 6 — a, we have 

 to balance the expression just obtained by the addition of 

 solutions o£ the differential equation 





which remain finite in the space considered, O<0<:a, and 

 cause the boundary conditions to be satisfied. 

 It will be seen that 



if r 



,,= Uc.L' 



cos m 



(V_ + 0') -cos mQrr-a + 6') 



sm m 



e 



>m »<« 



cos ?H 



sm ?»a J 



|A sin m (a— 6)~\ J TO ( b J )~U m (kr)dm 

 sin wi7r 



satisfies all these conditions when r>r / , and a>6>6', while 

 when r</, or 0< 6\ we shall have only to interchange r and 

 i J , or and 0', in the above. 



To prove this we note first that the different elements of 

 the integral satisfy the differential equation, and that the 

 boundary conditions are satisfied. Further, that no new 

 sources are introduced by the additional terms, since the in- 

 tegrals they give converge for all the values of r and 6 in 

 the space considered, unlike that for the first term, which 

 has a singularity at the point (/, 6'). 



