An introduction to the study of the elements of the differential and integral calculus. From the German of the late Axel Harnack, With the permission of the author.

266 The definite integral as the limiting value of a sum. Bk, 111. oh. VI.
not hold, ex. gr. for any functions f for which, while 9p has always
the same sign, there is a linear set of points x at which the inequality:
abs [p(x, a, ]t.)f = abs f(xa + h) - f(x, a. ()f(x, a)] <6
abs [Lp(x, a, b)] == abs | -- h (' --- - a m J < 6
cannot be satisfied by an assignable value 7.*)
But the course of our investigation reveals to us a condition
that is sufficient. In fact if f(x, a) for every value of x is also
a continuous function of a, the above inequality can be written by
the help of the Theorem of the Mean Value in the form:
abs [f(x, a+ + ) f(x, a)] <.
Now when this condition is to be fulfilled generally for all values of x
within the interval of integration (with exception, possibly, of only a
discrete set of points), f- must be generally a uniformly continuous
function of a, and therefore also generally a continuous function
of both variables. We have therefore the theorem: When for
x
*) Suppose the integral (x, a)dx = x si (4tan-1 and therefore is
0
a continuous function of both variables, we have:
f(x, a) = sin (4 tan-1- - x cos (4 tan-
At x =, a = 0, /(0, 0) is to be =0; this convention has no influence upon
the integral. Moreover the equation holds for every value of x; for a = 0 we
have f(x, 0) = 0; for x = 0, f(o, a) = 0. If we differentiate the integral with
respect to a we obtain:
ajf(x) d x a cos (4 tan-1 ),
oa x2
0 1+2
and for a - 0 this value is equal to 4. But the value of:
x
0 o
because:
(f(x, a)) - i i(4 tn — t- a) 0
-- +- cos I4 tan-1 -0.
a==0
The theorem of the interchange therefore does not hold, although for every value
f(x, a)
of x both f(x, a) and - are continuous functions of a.
of bthf~z ~ a da