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Cauchy's Integral Formula

Module by: Doug Daniels, Steven J. Cox. E-mail the authors

Summary: This module gives a defintion of Cauchy's Integral Formula and describes its use and usefulness.

The Residue Theorem

After Cauchy's Theorem eqn6 and Cauchy's Theorem eqn7 perhaps the most useful consequence of Cauchy's Theorem is the

Lemma 1: The Curve Replacement Lemma

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 fzdz=fzdz z C 1 f z z C 2 f z .

Proof

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 fzdz=fzdz+fzdz+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.

Figure 1: The Curve Replacement Lemma
Curve Replacement Figure
Curve Replacement Figure (curvelemma2.png)

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 z(z λ 1 )(z λ 2 )d z 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 z(z λ 1 )(z λ 2 )d z =z(z λ 1 )(z λ 2 )d z +z(z λ 1 )(z λ 2 )d z 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 z(z λ 1 )(z λ 2 )d z = λ 1 λ 1 λ 2 1z λ 1 d z + λ 2 λ 2 λ 1 1z λ 2 d z =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, z(z λ 1 )(z λ 2 )d z =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

z(z λ 1 )(z λ 2 )d z =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)

Figure 2: Concentrating on the poles.
Figure 2 (notdone.png)

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

qzd z = j =1h k =1 m j q j , k z λ j kd z =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 fzza 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 fzzadz=fzzadz 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 r0 r 0 , in which case za z a and fzfa f z f a . Computing the integral of 1za 1 z a along the way we are led to the hope that fzzadz=fa2πi z C f z z a f a 2

In support of this conclusion we note that fzzadz=fzza+fazafazadz=fa1zadz+fzfazadz 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 r0 r 0 , the integrand of the second term approaches dd a fa 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

formula 1: Cauchy's Integral Formula

If ff is differentiable on and in the closed curve CC then

fa=12πifzzadz 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

dd a fa=12πifzza2dz 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,
d n da n fa=n!2πifzza1+nd z 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 12πifz(z λ 1 )z λ 2 2d z 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 12πifz(z λ 1 )z λ 2 2d z +12πifz(z λ 1 )z λ 2 2d z 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 12πifz(z λ 1 )z λ 2 2d z =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 12πifz(z λ 1 )z λ 2 2d z =dd a faa 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

Theorem 1: The Residue Theorem

If gg is a polynomial with roots λ j j=1h λ j j 1 h of degree d j j=1h 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 fzgzd z =2πi j =1hres λ j z C f z g z 2 j 1 h res λ j where res λ j =limit   z λ j 1( d j 1)!d d j 1 z λ j d j fzgzd z d j 1 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.

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