607] 
A MEMOIR ON PREPOTENTIALS. 
407 
To prove it, write w = A tan 6, then the integral is in the first place converted into 
IT 
2 /*2 _ 
I cos s ~ 3 6d0, which, putting cos Q—fx and therefore sin# = vl — x, becomes 
A 8 "Jo 
= —^ J a ^~ 1 (1 — ¿e)* (s-2)_1 dx, 
which has the value in question. 
Hence, replacing A by its value, we have 
T* ri( 5 -2) 
BfdS 
4m*r(k)f-'Sf 
or 
T f(s - 1) J {(a - x) 2 + ... + (c - s) 2 p~ 2) (s - 2) T i(s - 1) ((a 2 +... + c 2 )^ s ~ 2 > f s 
that is, 
BfdS 
4 7T^ S f S 1 Bf 
or 
{{a — x) 2 + ... + (c — ,z) 2 }4 (s-2) (s — 2) r^- (s — 2) [(a 2 + ... + c 2 )* (s 2) f s 2 
- 27r * 8 / g lg / L 
l(a 2 + ... + c 2 )^ s - 2 > ° r /*- 2 { ’ 
viz. this is the formula for the sphere with s — 1 instead of s. 
Annex VII. Example of Theorem D. Art. Nos. 133 and 134. 
133. The example relates to the (s + l)-dimensional sphere x 2 + ... + z 2 + w 2 =f 2 . 
Instead of at once assuming for V a form satisfying the proper conditions as to 
continuity, we assume a form with indeterminate coefficients, and make it satisfy the 
conditions in question. Write 
(a 2 + ... +c 2 + e 2 ) is * J 
= A (a 2 + ... + c 2 + e 2 ) + B for a 2 + ... + c 2 + e 2 < f*. 
In order that the two values may be equal at the surface, we must have 
M 
= a/ 2 + £ : 
dV 
in order that the derived functions , &c. may be equal, we must have 
-(s-l)aM . - 
— yTH = *<=•> 
viz. these are all satisfied if only —^jfJ~ = 2-d* 
We have thus the values of A and B; or the exterior potential being as above 
M 
~(a 2 + ... + c 2 + e 2 )4 s '-i ’