(^608) Series
and hence 0 is quite close to 1 even when n = 1. The above formula for n!
is a Stirling formula, and it is very useful.
20 We are familiar with the fact that
(1)
fl,
1
dx = lim rhx-n dx = lim x1-P lhJ =^1
xP h-»1-p 1 p
when p > 1 and
(2)
J
1 dx= lim f h- dx = lim log
x x h---1
Some questions and answers involving existence (or convergence) of Riemann-
Cauchy integrals are quite analogous to questions and answers involving infinite
series.
(3) Theorem Let f and g be Riemann integrable over each finite interval
a < x < h for which h > a and let
(4)
Then
(5)
0 < f (x) < g (x)
0 <
fam
f(x) dx < fag g(x) dx
when (5) is interpreted to mean that f f(x) dx < M whenever
is M, and that f g(x) dx = co if f f(x) dx = oo.
Proof of this theorem depends upon the fact that
fa
m
g(x) exists and
(6) 0 5
fah
f(x) dx <
fah
g(x) dx (h > a).
The functions in (6) are monotone increasing functions of h. In case f
'
g(x) dx
= M, the function fh g(x) dx has the leastupper bound M. The function
'f(x)
dx then has an upper bound M and hence must have a least upper
bound M1 for which M1 < M and
(7) f : f(x) dx= lim h f(z) dx= l.u.b.fhf(x) dx = M1 <= f'g(x) dx.
a h-+. h2a a a
In case fag f(z) dx = oo, the function fah f(x) dx does not havean upper bound,
so fa g(x) dx cannot have an upper bound and hence fa g(x) dx = co. This
proves the theorem. Supposing that p > 1 and that q is real, prove that(log x)Q
2 xP dx < oo