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.

~ 171. 172.

Infinite domain of integration.

309

When it is to be formed for the inside of one branch of a
hyperbola whose equation is: x > 0, xy = -, the double integral can
be defined as:
a b aZ
j f/x y) d T = Lim J dx f(x, y) cdy Lim dy f (x, y) d x.
(a =oo, 6 = ) k k (a=oo, b-c) k k
X=- Y - By= - x ---
b x a y
The value of the double integral becomes singular, when it
depends on the way in which a and b become infinite.
When for arbitrarily increasing value of x and y, the function to be
integrated f(x, y) = f(r cos p, r sin p) has ultimately neither maxima
nor minima, so that, within its domain of integration for every value
of p as r increases infinitely, f converges to zero; then, if its order
of vanishing (nullitude) be higher than the seconrd, since from a
A
certain value rl, f(r cos p,q r sin p) is constantly < a where a > 2,
the double integral is finite and completely independent of the method
of the transition to the infinite region. For, the part of the integral:
f(x, y)cT
relative to a domain for which r > rl, is less than:
A fl
and this expression vanishes for arbitrarily increasing values of r.
But the double integral has no finite value whatever, when the
function f(x, y), having neither maxima nor minima as the values of
x and y increase, becomes infinitely small in an order lower than the
second or continues finite.
In case the function undergoes incessant oscillations as the values
of x and y increase infinitely, the double integral exists, as did the
simple integral, without involving any limit as to the order in which
the function becomes infinitely small.
When the double integral has no existence, there may yet be a
singular value for a determinate succession of integrations. One
example of this was given in last ~.; another important example is
the following: The function f(x, y) = cos (xy) is not integrable in
the infinite strip from x = 0 to x = b, y = 0 to y -= o, for we have:
h b b b h b
C - r / \7 \ / - I sin hx
fdyf cos (x y) dcx fdx cos (xy) dy == sif j dx fn - dx,
0 0 0 0 0 0
and if we first make Ih increase arbitrarily in the function to be
integrated, this becomes quite indeterminate; but the integral: