4.3 Properties of integrals 229
and, after observing that x log 2 = log 2x and hence f(x) = 2z, put (4) in the
form
limf(1 - s) -.f(0)= log
(6) &-I s 2.
But it follows from the definition of derivatives that the left member of (6) is
f'(0). From (5) we find that f'(0) = log 2. Therefore, (6) and (4) are correct.
The conclusion to be drawn from this story is that the function F defined by
i
(7) F(s) = fo (1+x)8dx
is a continuous function and, unless we can find another scapegoat, we must
blame the well-known perversity of inanimate matter for the strange fact that
F(s) is expressed in terms of exponentials when s p& 1 and is expressed as a loga-
rithm when s = 1.
4 This problem, like very many of the fundamental problems of science,
requires much more looking and thinking than calculating. Look at Figure
4.391, which shows the graph of a step function f defined over the interval
a 5 x < b, and observe that f(x)? 0. Remember that, in the problems at the
end of Section 4.2, we discovered (or almost discovered) that
(1) fabf(x) dx =ISI,
where ISI is the area of the set S of points (x,y) for which a S x <= b and 0 < y 5
f(x). The next step is the most difficult one. We should realize that, at the
present time, our ideas about areas of nonrectangular point sets are at best some-
what vague and nebulous and are at worst nonexistent or even erroneous. The
rest of our work is much easier. We look at Figure 4.392, which shows the graph
off over the interval a 5 x 5 b for the special case in which f(x) = x2, a = 0,
and b = 1. As above, let S be the set of points (x,y) for which a :s': x 5 b and
0 5 y 5 f(x). Our next step is to look again at Figure 4.392 and express the
_In
a b x
Figure 4.391 Figure 4.392
cheerful opinion that the set S ought to have an area which we can denote by
ISI and that the formula (1), which holds whenever f is a nonnegative step func-
tion, ought to hold whenever f is a nonnegative integrable function. Our final
step is to seek what a physicist could call experimental verification of this cheerful