§2. Absolutely convergent series
113
If you replace

is equal to l/aqk, so that 'ljJ(x) = q m('ljJ) = qm( So
n(c 1 / n - 1)/c 1 / n ::; L(a, b) ::; n(c 1 / n - 1).
As n increases indefinitely, the left hand side tends to log c, and the third
does so too, since the denominator c 1 / n tends to 1. We have thus proved that
the function 1/ x is integrable and that
(11.4)
lb dx/x = 10gb -loga (O Example 2. The same method, a little less simply, can be applied to calculate
the integral of the function f(x) = X S between x = a > 0 and x = b > a,
where 8 E Z, 8 #- -1 (the case where 8 = -1 has just been treated and is quite
different). One uses the same subdivision of [a, bJ and the same definitions of

must now be those which f(x) takes at the end points 30 , i.e. (aqk)s = asqsk
and (aqk+l)s. Now
m( L (aqk+l - aqk)asqsk = (q - 1)a s + 1 L q(s+l)k =
O~k O~k [
]
qn(s+l) 1
(q _ 1)a s + 1 1 + qs+l + ... + q(s+l)(n-l) = (q _ l)a s +l
-
qs+l - 1
q -1 [(aqn)S+l _ as+1] = q - 1 (bS+1 _ as+1)
qS+l _ 1
qS+l - 1
since aqn = b. The area sought is, on the other hand, smaller than m('ljJ),
the number one obtains on replacing f(aqk) by f(aqk+l) = qS f(aqk) in the
preceding calculations, so m('ljJ) = qSm( As n increases indefinitely, q = (b/a)l/n tends to 1, so that the ratio
(qB+l_1)/(q_1) = (qs+l_ pH )/(q-1) tends, by definition, to the derivative
of the function x I-> x s + 1 at x = 1, i.e. t0 31 8 + 1 #- 0; its reciprocal thus
tends to 1/(8 + 1), so that
limm( Since m('ljJ) = qSm( 30 This choice assumes that f is increasing, i.e. s > O. We leave to the reader the
trouble of switching the letters r.p and 1/J when s < O.
31 We have seen this for sEN in nO 4, but the result persists for 8 E Z and even
for s E R, so the calculations and result apply in this last case too.