130 Mr. W. H. L. Russell [Feb. 28, 



The integral (68.) calls for some particular remarks. Since by the 

 ordinary rules : 



• a , • on . • /o . i\n 2sin6>-sin(2r+3)6>+ sin(2r + l)6> 



sm0 + sm36> + . . . sm (2r+l)0= ^— - — ^ — ' x v 



2(1 — cos 20) 



whence also : 



2 sin hx — sin (2w, + 3)6a? + sir. (2n , + l)6a? 

 (l + a; a 0(l-cos2^) 



sin 36a? 



poo 



cfe, sin a; 



lo 



_ 2 f °° ^ sin ax sin 6a? , ^ f 00 da? sin cia? sin 

 ~ Jo _ ~~T+^ 7 Jo T+ti* 



+ . . . +2 



* 00 cfe sin ax sin (2n + l)6a? 



o l + ^ 2r 



Now 2 sin <xa? sin 6a? = cos (6 — &)a? — cos (b-\-a)x 



2 sin cm? sin 3 6.n= cos (36 — a)x— cos (36 + a)a? 



P" ^ cos mx = c ^_ Cxm+ Coe -c 2m ; _ > + 0,6-^; 

 Jo l + » 2 ' 



Where — c 1} — c 2 , — c 3 . . — c s are all the roots of the equation 

 x 2r + l = which have a negative modulus, and Oi, C 2 , Os . . . 0* are cer- 

 tain constants whose values have been assigned by Poisson : hence 



f w 7 . 2 sin 6a? — sin(2« + 3)6a? + sin(2n + l)6a? n >. , ~ . 



dx.smax. v J — -A l_=C 1 0(c 1 ) + C 2 0(c 2 ) 



Jo (l + a? jr )(l — cos 26a?) 



p -c6 -e(2n+3)6 



+ . . . +a0(c,)i,when0( 6 ) = ( fi «-e-<«) . 1 _ £ _ 2c6 . 

 Now let (n) increase without limit, then 



sin(2w- + 3) 6a? = sin 2n(l + — |6a? = sin 2nx. 



\ 2n) 



sin (2^ + l)6a? = sin 2w(l + — ^6a? = sin 2nx. 



\ 2n) 



Hence the integral in the second member vanishes, and 



— . - — —=yji- — — / — +^2- r-r — / — + . . . (s terms). 



j l + a; 2r sm6aj e 2Cl6 -l e***— 1 V J 



It is scarcely necessary to observe, that this process implies that (a) 



I w COS VYIX dx 



must be less than (6), otherwise the integral — '- — as here used 



Jo l + x 2r 



will be discontinuous. 



(70 ) I au sm 



Jo e cos * + l 2r(2r-2) ... 2 ' 2" 



<Z0sm 2r fl_(2r-l)(2r-3 . . 1 tt 



e cos, 



< 7a > f (TT 



c 2 sin 2 0)(e cos * + l) 2/1+? 



