686] 
ON A FUNCTIONAL EQUATION. 
299 
which determine \ R, S, R', S' and then 
RS' 
RR' 
Z = * los WS’ v = * V ogX+hg W r ) ’ £ = 2 lo g x - 
There is some difficulty as to the definite integral, on account of the denominator 
factor sin ££, which becomes = 0 for the series of values t = r — > but this is a point 
which I do not enter into. 
I will in the first instance verify the result. Writing x x in place of x, and 
taking Vi to denote the corresponding values of tj, tj, it will be shown that 
fi = f, Vi = V + 2£ see post, (1). 
Hence in the difference <f>x — <f)X x we have the integral 
f sin %t {sin 7jt — sin (rj + 2£) t\ dt 
J sin £"£ sinh 7rt 
(where and in all that follows the limits are oo, 0 as before); here, since 
sin 7]t — sin (r] + 2£) t = — 2 sin ty cos (7) + £) t, 
the factor sin ty divides out, and the numerator is 
= — 2 sin %t cos (tj + £) t, 
which is 
= sin (tj + f - £) t - sin (tj + £+ £) t. 
Hence the integral in question is 
sin (tj + £ — £) t dt f sin(?7 + £+ %)tdt 
-S ! 
sinh irt 
-J 
Now we have in general 
— h ~ 
sinh 7rt 
sin at dt 
sinh 7rt ’ 
1 + exp. a 
(this is, in fact, Poisson’s formula 
1 1 a f sin (2n log /3 + log k) t. dt 
~ 1 + kS m ~ 2 
pirt . p Tit 
in the second Memoir on the distribution of Electricity, &c., Mem. de VInst., 1811, 
p. 223); and hence the value is 
1 1 
or since 
we have 
1 + exp. (tj + K ~ £) + 1 + ex P- (v + £ + I) ’ 
. . RR' f. , , RS' 
V + £ = log ^ 2 log ~S8' 5 * ^ ^ R'S ’ 
J?' 2 R' 
y + %- £ = log x + 1 log S T 2 = log A -g , 
77 + f+| : = logA + |log ^ 2 - = log X, 
S 
38—2