POINTS OF THE INTEGRAL CALCULUS. [131] 



If V be integrable per se, so are also V x and V y ; and moreover, V is identically equivalent 



with 



fv x dx + (v p - v; + ...) P + (v q - v; + ...) q +(v r - v; + ...) r + ... 



If V be integrable per se, so also is 



v y w+v v w+ V q W" + V r W" + 



where W is any function whatsoever of w, y, p, q,..., or of any of them. For common 

 integration by parts shows that this is integrable or not with f(V y — V'+ ...)Wdx. 

 This is an easy way of making complicated integrable functions from simple ones.) 



I have put the preceding remarks in parentheses, because, though immediately arising 

 out of what goes before, they are not wanted in what follows, namely, in proving the 

 sufficiency of the criterion. For this purpose observe, first, that if the criterion exist, or if 

 V - VJ + ... = 0, we can re-establish the last six equations (and thence the first) in form, 

 without any assumption as to X, Y, &c. being all partial differential coefficients of one 

 function. For we have a right to assume S = V t , R = V s — S' = V s — V t ', &c, and pro- 

 ceeding thus, we obtain F= V p ~ P' = V p - V q ' + .... But, by the criterion, V y = V p ' - ... = Y', 

 whence the second equation follows: lastly, assume X so that V=X+ Yp + ... and we 

 have, as before shewn, V x = X'. 



From these forms alone, and the universal theorem (W), it follows that in every case 

 M n = N„„ or that X+ Yp + ... is a complete differential coefficient. Take any letter, say 

 r. If the partial tests of integrability be all true whenever one of the letters m, n, is r, they 

 are all true when one of them is q, the letter preceding r. To prove this, observe that 



( V q ) r = Q r ' + Q q + P T = R q +Q q + R p by hypothesis, 

 (V,) q = R q ' + R p +Q q . 



This proves nothing, being only an identity. Now take 



(r,) r - p; + p q +Y r = r; + p q + R y , 



(F,), = R p ' + R y + Q p , whence P q = Q p . 

 Again, (r y ) r =F/+F,= i?/+F„ 



(r r ),-Bj+Q,i wh ence Q y = Y q . 

 Lastly, ( V,) r = X r ' + X q = Rj + X g , 



( V r ) x = RJ + Q x ; whence Q x = X q , 



whence the theorem, if true for any case in which r enters, is true for any case in which q 

 enters. Everything then depends upon proving the tests in which s enters, since no one of 

 the capitals is a function of t. 



Now V ta = S„ F>0 + ,S, + 



which is an identity. But V tr = S r , and V rt = + R s + 0, whence S r = R t ; and so on. 

 Hence it is proved, from first principles, that the criterion is sufficient as well as necessary. 



If the function X+ Yp + Pq+ ... be integrable per se, it appears that Xdx+ Ydy + Pdp+ ... 

 is integrable per se, independently of the connexion between y, p, q, &c. If then 



Ada + Rdb + Cdc + Ede + ... 



41—2 



