113] THE INTEGRAL Jdx + n/(^To)^T6) (x + c). 55 
the arbitrary constant which should have formed the second side of the equation 
having been determined by observing that the algebraical equation gives for p = 6, 
k — 00 > a s yst em of values, which, when the signs are properly chosen, satisfy the 
transcendental equation. In fact, arranging the rational algebraical equation according 
to the powers of k, it becomes 
k 2 (p - e) 2 - 2k {p9(p + 9) + 2(a + b + c)p9 + (be + ca + ah) (p + 0) + 2 abc] 
+ p-6- — 2 (be + ca + ab)p9 — 4abe (p + 9) + b 2 c 2 + c 2 a 2 + a 2 b 2 — 2a?bc — 2b 2 ca — 2c 2 ab = 0; (*) 
which proves the property in question, and is besides a very convenient form of the 
algebraical integral. The ambiguous signs in the transcendental integral are not of 
course arbitrary (indeed it has just been assumed that for p = d, lip and II9 are to 
be taken with opposite signs), but the discussion of the proper values to be given 
to the ambiguous signs would be at all events tedious, and must be passed over for 
the present. 
It is proper to remark, that 9=p gives not only, as above supposed, k= oo, 
but another value of k, which, however, corresponds to the transcendental equation 
+ II& 4 2 lip = 0 5 
the value in question is obviously 
p 4 — 2 (be + ca + ab)p 2 — 8abep + b-c 2 + c 2 a 2 + arb 2 — 2ci 2 bc — 2b 2 ca — 2chib 
(p + a) (p + b) (p + c) 
Consider, in general, a cubic function a« 3 + 3bx 2 y + 3cxy 2 + dp 3 , or, as I now write 
it in the theory of invariants, (a, b, c, d) (oc, y'f, the Hessian of this function is 
(ac - b 2 , | (ad - be), bd - c 2 ) (x, y) 2 , 
and applying this formula to the function (p + a) (p + b) (p + c), it is easy to write 
the equation last preceding in the form 
4 k = p — (a+ b + c)- 
9 Hessian {(p + a) (p + b) (p + c)} 
(p + a) (p + b)(p + c) 
which is a formula for the duplication of the transcendent Tlx. 
Reverting now to the general transcendental equation 
±Uk±Up±U9 = 0, 
we have in like manner 
± n& + np + Tld’ = 0; 
and assuming a proper correspondence of the signs, the elimination of np gives 
U9'-U9 = 2Uk-,