522 PROFESSOR KELLAND ON THE VIBRATIONS OF 



Now / 2 cos v dv=v 2 sin v + 2 v cos r 2 sin 



r dr f cosAr 



_ -- 



Substituting all these results in equation (1), it gives 



/ > - oc Co, (b'rr 8 , sin A r , . cos X; r sin kr\ 



-^irg I drlr 2 (t>r + i~ -- rtyr +r $'r -^ -- <f)'r ^ 



This is evidently the same as equation (a). 

 Finally let us sum this expression for r. 

 If we integrate, the result is 



sn 



to be taken between the limits r~o, r=cc. For the latter limit the expression 

 generally vanishes. Hence 



g_ 47r Pf ^ 4y4 A:^ 4 ) 



^ (1.2.3.4 1.2.3.4.5 j /r r= 



For one value of fr this process fails, viz. when/>=r 2 . In this case it is 

 evident that the above expression does not vanish when r=cc. The reason why 

 this is so, is that r 2 fr is a finite quantity. Now, if we return to the original 

 expression for <?, we shall find that it is the representation of the difference 

 between two terms, which in the operation have assumed different forms. To 

 obtain the correct result, I would suggest that the terms retain both the same 

 form, which they will do, if we write cos a $ x instead of 1, and, finally, put a=0. 

 We shall then have, 



2 /J. 4>' r * 9N / * , S \ 



e* = 2 m (9 r + L- - ojr) (cos a o x cos k o x) 

 which, when the law of force is that of the inverse square of the distance becomes, 



a_ y m J" sin ar 3 sin a r 3 cos a r 

 r 3 \ a r a 3 r 3 a 2 r 2 



/sin kr 3 sin kr 3 cos k r' 

 v fc y tc "i K> f 



If we now substitute integrals in place of sums, we obtain 



fs\n ar 3 sin ar 3 cos a r 



/<*> /<* 

 dr( 



sin k r 3 sin A r 3 cos k r\ 



