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.

338 The definite integral of a unique analytic function.

Bk. IV. ch. I.

When we require this integral along a path to infinity along
which the abscissa x increases arbitrarily negatively and the ordinate y
positively, we can proceed by describing the path of integration first
up to - a along the axis of abscissa, then the path b parallel to
ordinates, and ultimately causing a and b to increase arbitrarily.
Conceiving ex. gr. the parabola y = x2 as path of integration, let us
consider the point
z = - a -+- ia2
upon it. The value of the integral extended to this point is:
Je3 dz = edx + i e-a+iydy
0 0 0
= (e- 1) + e-~(eia2 - 1)= - 1 + e-a+"a2
When a converges to infinity, the value of the right side passes over
into -- 1. But it is only along a determinate path of integration
that a determinate value of the integral is generated.
In order to study in this example the behaviour of the essential
singular point in integration generally, we transfer it to the origin
or point zero by the substitution z = and consider the integral
_ e~ dz'
zr'2 s -j +'l + '4I2 1 - ____ + r
~'~ = yt2 =-fig + z" g + By'L +'" + 2z' 1+ 'LI
taken in a negative circuit round the point '-== 0.
We have here a series advancing by powers of the variable -;
we shall therefore first prove. in general the theorem (see ~ 131):
When F(z) is a series which advances by powers of a function
f(S), -its integral within a domain wherein the series converges and'
f(s) is an analytic function, is obtained by integrating its several
terms.
With a view to this proof, we have to demonstrate that the series:
F(z) -= a + a,{ ft()} + a2{f(z)}2 +' + a {/f (z)} +.
converges uniformly within its convergency. It is obvious that if
the series converge for a determinate value of Z, it is absolutely
convergent for every value of s for which absf(z) < absf(Z). For,
inasmuch as the series converges for f(Z), its terms must decrease
in amount in such a way that, calling An = abs [a,,, = — abs [f(Z)],
we shall have:
An+l nB+l < AnR", and that, LimAn A
shall at most = 1.