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.

Extension to a function with infinite values.

307

x = r cos c, = r sinll,
we have:
y2 - x2 -cos 2cp
(y d- x2)2 r2
therefore in a quadrant round the vertex x = 0, y 0, the integral:
- os 2 p d(Spj
0 r2
increases logarithmically beyond any limit. As another example:
Jm xq 2 c dxdy jj cos 2 (prdr dp
o 0
has a finite value, because although the function to be integrated is
discontinuous and indeterminate in the point x = 0, y = 0, it still
remains finite.
Again, the double integral:
- dxdy
JJ 2 + y
has a finite value for x > 0, y > 0, although the function becomes
infinite in the second order at the point x = 0 in the direction
of the axis of abscisse, but in this direction only. To demonstrate
this, let us calculate the value of the double integral for a rectangle
from x =0 to x = a, y = b to y = c, we have:
a c a
Arf dxlrS:4-i;2 /ta+C c n-1 a.
XJ x-J.+ J -X+ a + 2 1/c tan-2 tan
Now making c and b < c converge to zero in any way whatever, this
expression on the right side converges to zero.
The values resulting from the two successions of integrations are
not necessarily different, even when the double integral is unmeaning.
We have in the first example:
y fx ' -+ xx d 2 y = 0
(yJ2 + x2)2 d ({ - SI (2 + x2)2 x-~
0 0 0 0
while the double integral for the domain, that is here infinite, has
no existence.
When the function f becomes infinite along an entire curve in
the domain, let this be taken as the line p = const. Then the product:
f/(p, )q) aCP aq(p ) a OP -
20*