136 Mr. Boots on the Analysis of Discontinuous Functions. 
Now let @(2) = f(a), then /[6(2)] —z = 0, and efm1—) ? = 1, 
therefore 
f | ey eee 
v= F[f(2)] (2). 
Suppose that f(w) =, then 
du 
— F(w) dx’ 
wherefore 
du aye = , ’ 
He) da ( AN cfr! ei b(x)]p'(x). (9) 
dx 
The transformation of the original integral v, by which the first member of this 
equation was obtained, was virtually @ = f'(a’), and that by which the second 
member was obtained was a = ¢(a’), and in both cases a’ must be real. This 
condition requires that a’ and f[¢(a’)] should be both real ; and as the limits of 
a’ determine those of 2, it is necessary that # and /[(#)] should be real, 
Apparently this is the only condition to which ¢(«) is subject. 
From (9) we have 
=} di d ae 
(zs) r(u) Ge — (— Ge)” etm e[ p(y] #'(2). 
Let = 
(fe) Oge= HCO, 
da 
then 
F[o(x)]o (v) = F[$(x)]¢'(2). 
Substituting, and dropping the suffix, 
r= — (— Fyre 42) 6) (10) 
uw being a root of the equation 
Fie) = 2, 
and ¢() an arbitrary function of x, rendering f[@(.)] real. 
3. In order to apply the above theorem, let us write for simplicity, 
x=2—f[9(2)], 
then 
