70 Mr Mathews, Reduction of Generating Functions 



then since 



r = a -1 + a~ 2 + a~ s + ... 



a — 1 



the coefficient of a -1 in the expansion of <£ (a) | (a— 1) is 



Co "T Cj + C 2 T • • • } 



that is, 0</>. Hence 



the integral being taken over a closed simple contour at every 

 point of which the above expansion of <£ (a) \ (a — 1 ) is legitimate. 

 But the integral is also equal to 



2wiXR 



when 1R is the sum of the residues of <f>(a) \ (a— 1) within the 

 contour. Calculating this sum by the method of partial fractions, 

 the value of X2<£ is determined. 

 For example, if 



<#>(«) = ' 



(i-«-)(i-; 



we must suppose | ax \ < 1, | y \ < | a \, as well as | a \ > 1. We can 

 take as the path of integration the circle which has its radius 

 equal to ] a \ and centre at the origin ; the poles within it are 

 1 and y, so that 



SR ( 1 — , r\ = ,-, ^ ; + 



^<a-l)(l-a.)(l-*)) t 1 — X 1 -*) (^Dd-^) 



{\-x){l-xy) 

 on reduction. 



The same result may be obtained in this case by taking the 

 residue with respect to the one external pole, for which a = x~ x , 

 and changing sign ; the value of 0<£ then comes out immediately. 



As another example, let it be required to find 



n 



w 

 a 



a— )fl-<w> (i : D(i~ 



Here 



(j> aw 



(a — 1) (a — 1)(1 — ax) (1 — ay) (a — z) (a— w) 



