712 
PROFESSOR CAYLEY ON PREPOTENTIALS. 
60. It is now easy to see in general how the foregoing transformed value 
x q ~Hl — x'f s ~ l dx> where q is negative and fractional, gives at once the value of the 
Jx 
e 2 
term in e~ 2q . Observe that in the integral x is always between 1 and X (= e8 a 
positive quantity less than 1) ; the function to be integrated never becomes infinite. 
Imagine for a moment an integral f x a dx, where a is positive pr negative. We may con- 
Jx 
r*l fX _ jl+a 
ventionally write this =1 x a dx— i x a dx, understanding the first symbol to mean , 
Jo Jo l+« 
X 1 + “ jl + a Ql J 
and the second to mean ; they of course properly mean — — — 
and 
X 1+ “_o ] 
+ «' J A * l+« l+« 
but the terms in 0 1+ “, whether zero or infinite, destroy each other, the original form 
, in fact, showing that no such terms can appear in the result. 
In accordance with the convention we write 
f x q ~\l — xf s ~'dx={ x q ~'(l — x) is ~'dx— f x q ~ l (l— x) ¥ ~'dx ; 
Jx Jo Jo 
and it follows that the term in e~ 2q is 
\e~ 2q f x q ~\l —x) ¥ ~ l dx, 
this last expression (wherein q, it will be remembered, is a negative fraction) being 
understood according to the convention ; and so understanding it the value of the 
term is 
i/,-2? 
“ 2 r wwr 
where the T of the negative q is to be interpreted in accordance with the equation 
T(q+l)=qTq; viz. we have r^ + l), = g( g + i) F(g~l-2), &c., so as to make the 
argument of the T positive. Observe that under this convention we have 
r^IYl— a )— or the term is -e~ 2q — - — 
i^i(i q )— sin??r , oi tneterm is 2 e . sin q7[ r(is+?)r(l- ? ) 
61. An example in which -|s— 1 is integral will make the process clearer, and will 
serve instead of a general proof. Suppose q= — y, ys — 1 = 4, the expression 
( x^ (1— x) 4 dx = ( (x~*— 4#“^+ 6^— 4x&-t-x?r) dx 
^0 Jo 
is used to denote the value 
_7_1A T 42. _ 7l 7_ 
1 3 T^13 5T 27 
— 7.2401 -7 5 
7/ 1 2. j 6_ 1 j 1_\ 7/ 44 i 1 6_\ _ 
— / ^ — 1 — 3 -f is'- 5 ~r 2 7b — H — 27 — 5 i - 13b — 5.13.27’ “5 . 13.27' 
