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