After Cauchy's
Theorem eqn6 and Cauchy's Theorem eqn7 perhaps the most
useful consequence of Cauchy's
Theorem is the
Suppose that
C
2
C
2
is a closed curve that lies inside the region encircled by
the closed curve
C
1
C
1
. If f
is differentiable in the annular region outside
C
2
C
2
and inside
C
1
C
1
then
∫
C
1
fzdz=∫
C
2
fzdz
z
C
1
f
z
z
C
2
f
z
.
With reference to Figure 1 we
introduce two vertical segments and define the closed
curves
C
3
=abcda
C
3
a
b
c
d
a
(where the
bc
bc arc is clockwise and the
da da arc is counter-clockwise)
and
C
4
=adcba
C
4
a
d
c
b
a
(where the
ad
ad
arc is counter-clockwise and the
cb
cb
arc is clockwise). By merely following the arrows we learn that
∫
C
1
fzdz=∫
C
2
fzdz+∫
C
3
fzdz+∫
C
4
fzdz
z
C
1
f
z
z
C
2
f
z
z
C
3
f
z
z
C
4
f
z
As Cauchy's Theorem implies that
the integrals over
C
3
C
3
and
C
4
C
4
each vanish, we have our result.
This Lemma says that in order to integrate a function it
suffices to integrate it over regions where it is singular,
i.e. nondifferentiable.
Let us apply this reasoning to the integral
∫Czz−
λ
1
z−
λ
2
dz
z
C
z
z
λ
1
z
λ
2
where CC encircles both
λ
1
λ
1
and
λ
2
λ
2
as depicted in Figure 2. We find that
∫Czz−
λ
1
z−
λ
2
dz=∫
C
1
zz−
λ
1
z−
λ
2
dz+∫
C
2
zz−
λ
1
z−
λ
2
dz
z
C
z
z
λ
1
z
λ
2
z
C
1
z
z
λ
1
z
λ
2
z
C
2
z
z
λ
1
z
λ
2
Developing the integrand in partial fractions we find
∫
C
1
zz−
λ
1
z−
λ
2
dz=
λ
1
λ
1
−
λ
2
∫
C
1
1z−
λ
1
dz+
λ
2
λ
2
−
λ
1
∫
C
1
1z−
λ
2
dz=2πⅈ
λ
1
λ
1
−
λ
2
z
C
1
z
z
λ
1
z
λ
2
λ
1
λ
1
λ
2
z
C
1
1
z
λ
1
λ
2
λ
2
λ
1
z
C
1
1
z
λ
2
2
λ
1
λ
1
λ
2
Similarly,
∫
C
2
zz−
λ
1
z−
λ
2
dz=2πⅈ
λ
2
λ
2
−
λ
1
z
C
2
z
z
λ
1
z
λ
2
2
λ
2
λ
2
λ
1
Putting things back together we find
∫Czz−
λ
1
z−
λ
2
dz=2πⅈ
λ
1
λ
1
−
λ
2
+
λ
2
λ
2
−
λ
1
=2πⅈ
z
C
z
z
λ
1
z
λ
2
2
λ
1
λ
1
λ
2
λ
2
λ
2
λ
1
2
(1)
We may view Equation 1 as a special
instance of integrating a rational function around a curve
that encircles all of the zeros of its denominator. In
particular, recalling that cauchy's theorem eqn5 and Cauchy's theorem
eqn6, we find
∫Cqzdz=∑j=1h∑k=1
m
j
∫
C
j
q
j
,
k
z−
λ
j
kdz=2πⅈ∑j=1h
q
j
,
1
z
C
q
z
j
1
h
k
1
m
j
z
C
j
q
j
,
k
z
λ
j
k
2
j
1
h
q
j
,
1
(2)
To take a slightly more complicated example let us integrate
fzz−a
f
z
z
a
over some closed curve CC inside
of which ff is differentiable and
aa resides. Our Curve Replacement
Lemma now permits us to claim that
∫Cfzz−adz=∫Carfzz−adz
z
C
f
z
z
a
z
C
a
r
f
z
z
a
It appears that one can go no further without specifying
ff. The alert reader however
recognizes that in the integral over
Car
C
a
r
is independent of rr and so
proceeds to let
r→0
r
0
, in which case
z→a
z
a
and
fz→fa
f
z
f
a
. Computing the integral of
1z−a
1
z
a
along the way we are led to the hope that
∫Cfzz−adz=fa2πⅈ
z
C
f
z
z
a
f
a
2
In support of this conclusion we note that
∫Carfzz−adz=∫Carfzz−a+faz−a−faz−adz=fa∫Car1z−adz+∫Carfz−faz−adz
z
C
a
r
f
z
z
a
z
C
a
r
f
z
z
a
f
a
z
a
f
a
z
a
f
a
z
C
a
r
1
z
a
z
C
a
r
f
z
f
a
z
a
Now the first term is
fa2πⅈ
f
a
2
regardless of rr while, as
r→0
r
0
, the integrand of the second term approaches
ddafa
a
f
a
and the region of integration approaches the point
aa. Regarding this second term,
as the integrand remains bounded as the perimeter of
Car
C
a
r
approaches zero the value of the integral must itself be zero.
This result if typically known as
If ff is differentiable on
and in the closed curve CC
then
fa=12πⅈ∫Cfzz−adz
f
a
1
2
z
C
f
z
z
a
(3)
for each a lying inside
CC.
The consequences of such a formula run far and deep. We shall
delve into only one or two. First, we note that, as a does
not lie on CC, the right hand
side is a perfectly smooth function of
aa. Hence, differentiating each
side, we find
ddafa=12πⅈ∫Cfzz−a2dz
a
f
a
1
2
z
C
f
z
z
a
2
(4)
for each a lying inside
CC.
Applying this reasoning
nn times
we arrive at a formula for the n-th derivative of
ff at
aa,
dndanfa=n!2πⅈ∫Cfzz−a1+ndz
a
n
f
a
n
2
z
C
f
z
z
a
1
n
(5)
for each a lying inside
CC. The
upshot is that once
ff is shown
to be differentiable it must in fact be infinitely
differentiable. As a simple extension let us consider
12πⅈ∫Cfzz−
λ
1
z−
λ
2
2dz
1
2
z
C
f
z
z
λ
1
z
λ
2
2
where
ff is still assumed
differentiable on and in
CC and
that
CC encircles both
λ
1
λ
1
and
λ2
λ2
. By the curve replacement
lemma this integral is the sum
12πⅈ∫
C
1
fzz−
λ
1
z−
λ
2
2dz+12πⅈ∫
C
2
fzz−
λ
1
z−
λ
2
2dz
1
2
z
C
1
f
z
z
λ
1
z
λ
2
2
1
2
z
C
2
f
z
z
λ
1
z
λ
2
2
where
λ
j
λ
j
now lies in only
C
j
C
j
. As
fzz−
λ
2
f
z
z
λ
2
is well behaved in
C
1
C
1
we may use
Equation 3 to
conclude that
12πⅈ∫
C
1
fzz−
λ
1
z−
λ
2
2dz=f
λ
1
λ
1
−
λ
2
2
1
2
z
C
1
f
z
z
λ
1
z
λ
2
2
f
λ
1
λ
1
λ
2
2
Similarly, as
fzz−
λ
1
f
z
z
λ
1
is well behaved in
C
2
C
2
we may use
Equation 4 to conclude that
12πⅈ∫
C
2
fzz−
λ
1
z−
λ
2
2dz=ddafaa−
λ
1
|a=λ2
1
2
z
C
2
f
z
z
λ
1
z
λ
2
2
a
λ2
a
f
a
a
λ
1
These calculations can be read as a concrete instance of
If gg is a polynomial with roots
{
λ
j
|j=1…h}
λ
j
j
1
…
h
of degree
{
d
j
|j=1…h}
d
j
j
1
…
h
and CC is a closed curve
encircling each of the
λ
j
λ
j
and ff is differentiable on
and in CC then
∫Cfzgzdz=2πⅈ∑j=1hres
λ
j
z
C
f
z
g
z
2
j
1
h
res
λ
j
where
res
λ
j
=limz→
λ
j
1
d
j
−1!d
d
j
−1dz
d
j
−1z−
λ
j
d
j
fzgz
res
λ
j
z
λ
j
1
d
j
1
z
d
j
1
z
λ
j
d
j
f
z
g
z
is called the residue of
fg
f
g
at
λ
j
λ
j
.
One of the most important instances of
this theorem is the formula for the Inverse Laplace Transform.
