46 



Mr. I. Todhunter on Jacobi's Theorem, 



between the limits and I, and for the negative part the integral between 

 the limits i and 1. Or we may put the integral in the form 



Jo (i+ivy Jo (i+?v)' • 



I do not know which of these two ways Ivory adopted. It is true that in 

 each of them the positive part decreases as I increases ; but in each of them 

 the negative part is finite when I is finite, and is infinitesimal when I is in- 

 finite, and so does not always increase with I as Ivory supposes. 



9. However, although Ivory's reasoning is unsound, the result which he 

 wishes to establish is correct, as I shall now show. 



Consider the integral I — ® V>Tl^ > We have to show that 

 Jo (l+l x ) 



as I increases from unity to infinity, the integral vanishes and changes sign 

 once, and only once. 



It is obvious that the integral must vanish once, because it is positive 

 when 1=1, and negative when 1= oc : it is indeed infinitesimal in the latter 

 case, but the sign is certainly negative. 



Put g for Ix ; thus the integral becomes 



/(z 2 -s 2 )(i-z 2 yz. 



We have to show that the equation w=0 can have only one root between 

 I— 1 and 1= oc . If the equation could have more than one root it must 



have three roots at least; and then the equation ^=0 must have two 



1 dl 



roots at least. Now 



glf 1 * 1 -?* ^/,, say, 

 dl Jo 



and 



dv_ l\\-l 2 ) 

 dl (i+O 3 * 



Thus v is positive when 1=1, and while I increases ^ is always negative ; 



etc 



thus v continually diminishes algebraically, and so cannot change sign more 



than once. Hence ~ cannot vanish more than once ; it will vanish once 

 dl 



because v is positive when 1= 1, and has a finite negative value when l=<x. , 

 namely, 



n 



f 2 sin 2 0(1 - 2 sin 2 0)^0, that is-S 

 Jo 8 



10. There is also another way in which the result may be established. 

 For from (4), by the known method of integration, we have 



