( 257 ) 



Putting under the sign of the integnil a factor si?i "^ ta7i %• outside 



1 



the brackets and, by regarding that factor as - — {l-\-'cos^(f>tafi^r), 



cos r 



writing the integral as sum of two integrals to the former of 



which the same division in two is applied, etc., we find, if we write 



I sin^ r dr =z Sh : 







(n-l)/(r) . 



n— 1^ _- — g^f^ 7j— 2y. QQs r Sn—2 — siil »— 4/> cos r aS„_4 . . 

 K-l 



.... — sin ^r cos r S^ -\- ji {1 — cos r) 



(for n even) 

 =. — sin «— 2r cos r Sn—2 — sin n—4j. (.qs r /S„_4 .... 

 .... — sin r cos r S^ -\- 2 r 



(for n odd) 



(n-l)(n-3).... 



(n— 2)(/j — 4) 







— I sin n—ir dr = (w — 1) Sn—2 i sin »— V c?r, 







(for n even) 



^^ (n-l)(n-3).... 



— I sin "— V dr =. (n — 1) S,i—2 | sin ^—^r dr. 



(71—2) (/I— 4). 







(for n odd) 

 If we write ê„ for 2 . jr . 2 . jr . 2 . . . ., to n factors, we have 



kn = » and =z On—}. 



" (n-2) (n-4) . . . .' K 



Therefore : 



r 



ƒ (?') sin n— V = ^„ I sin "— ir c?/* , 







and the potential (y) becomes : 



r 

 COS y /* , 



kn— isin"~^r dr. 



sin ""~'r^ 



II. We find the field {(f) by taking difference of field (^) multi- 



1 1 



plied by è and field (y) by 7-^ — = 7 — , i. e. 



