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.

~ 181. 182.

Examples of integration.

333

It is easy to prove the theorem: that the complex integral of a
sum of functions is equal to the sum of the integrals formed for the
several summands; by its help we obtain from this the formula for
the integration of any integer rational function:
J(ao+alz+a2z2+.. a.z~n)dz=aO-+ -alz -[-3 a2Z3- n + C
2. The integral of a fractional rational function.
Every fractional rational function can be broken up into an
integer function and into partial fractions each having a constant
numerator and its denominator an integer power; for, the identities
developed in ~~ 111, 113 hold also for the complex variable. Accordingly
the integration of a fractional function requires us only to investigate:
- and more generally f
J za ^J f (-z - a)n
in place of these, writing z instead of - a, adopting therefore the
point a as zero or origin, we can deal with the simpler integrals:
J and
Considering first the case of n a positive integer greater than 1;
it is evident that the function - loses the character of an analytic
function at the singular point zero; in all the rest of the plane,
the point infinity included, it is regular. Surround the singular point
with a circle of radius r. Along the circumference of this circle, since
=- r(cos cp + i sin p), d =- r(- sin p -(- i cos q)dgp,
the value of the definite integral is:
27t
fra =_ 1 -sJin+tcosp d -p-: i (cosn-1(p-isinn-1 p)dg == 0,;n - n-i cos-t+m -_ isi-mg d -1
0
since n > 1. Thus the result of integrating round the boundary of
the origin vanishes, so that its boundary curve has not to be taken
into account. Along every path in the plane, even such as pass
through the point infinity, since for it
Lirm 'v -- 0,
we have the integral:
fd. - f+1
J-n -+l + C
and it vanishes along every closed path. But the point zero or origin
is a non-essential singular point of the integral.