Elasticity_ Barber

18.5 Line integrals 281
Example
As a special case, we shall determine the value of the contour integral
∮
dζ
ζ n (ζ − a) ,
S
where the integration path S encloses the points ζ = 0, a. The integrand has
a simple pole at ζ = a and a pole of order n at ζ = 0. For the simple pole, we
have
Res(a) = 1
a n ,
from (18.33). For the pole at ζ =0, we have
so
f 1 (ζ) =
1
(ζ − a)
and
f [n−1]
1 (ζ) = (−1)(n−1) (n − 1)!
(ζ − a) n ,
f [n−1]
1 (0) = (−1)(n−1) (n − 1)!
(−a) n and Res(0) = (−1)(n−1)
(−a) n = − 1
a n ,
(18.41)
from (18.39). Using (18.40), we then have
∮
(
dζ
1
ζ n (ζ − a) = 2πı a n − 1 )
a n = 0 ; n ≥ 1 (18.42)
S
since if n=0 we have only the simple pole at ζ =a.
= 2πı ; n = 0 , (18.43)
18.5.2 The Cauchy integral theorem
Changing the symbols in equation (18.32), we have
f(ζ 0 ) = 1 ∮
f(s)ds
2πı S (s − ζ 0 ) , (18.44)
where the path of the integral, S, encloses the point s = ζ 0 . If S is chosen to
coincide with the boundary of a simply connected region Ω within which f
is holomorphic, this provides an expression for f at a general point in Ω in
terms of its boundary values. We shall refer to this as the interior problem,
since the region Ω is interior to the contour. Notice that equation (18.32) and
hence (18.44) is based on the assumption that the contour S is traversed in the
anticlockwise direction. Changing the direction would lead to a sign change
in the result. An equivalent statement of this requirement is that the contour
is traversed in a direction such that the enclosed simply connected region Ω
always lies on the left of the path S.