PROFESSOR KELLAND ON GENERAL DIFFERENTIATION. 



605 



Theorem. — If (p(d + a) be an integral positive or negative function of 6 + a; 

 then will 



We shall be able to obtain an equation in other cases by means of the last equa- 

 lity but one in our process. 

 To prove this theorem : 



Let /:dd<p{6 + a) (z - ey be denoted by P. 



Assume 6=zy where ;j: is constant ; 



dQ=zdy 

 and 'P=f^ zd-yipiy z + a){z-6y . 



Let (f>{jz + a) — 1k{'yz + a)"' 



P=lAz/J dy(yz + ay(z-ey 

 = 2A2''+i/„'rf7(72+ a)" (l Y 



= 2 A zf^'-fo' d y (y z + tt.)'^ (l-yy 



= lAzf*^/c dy I 7'" a"' +my^^ . z^^ a 



'.{m-V) 

 1.2 



»/j«— 2 ^m~2 „3 



a^+ .... (1-7)" 



Now 



fjdyr'il-yy 



/'p + 1 lm + \ 



/m+p + 2 



by Euler's and Legendre's theorems. 



P = 2A^''*ViO + l 



lm + 1 .a" 

 lm+p + 2 



m Im z^^'- . a 

 /m+p + 1 



m(m—l) /m—1. a'"-^ ct^ 



"I ^ 71 — — + ... 



1.2 



/m+p 



] 



and since 



/m = /m + 1 



(m — V)m/m — l= /m + 1 



we get 



But 



P = 2A«f+7jo + l/w + l| 



/m+p + 2 /m + p + 1 



/m+p 1-2 /m+p—1 1.2.3 



+ ... 



dg-iP^i) ^ ' sin (»? + » + l)7r 



\{m+p + l)if /m+p + 2 



