Calculus(89)——Complex Indefinite Integral

2023-06-27 18:14 作者:Mark-McCutcheon 0人读过 | 我要投稿

With the Cauchy-Goursat theorem propared,we can study the complex indefinite integral.

Before we start to study the indefinite integral,I want to introduce the residue briefly.

We have discussed the singular point of analytic functions(generalized analytic functions),it means that the function is not analytic on it but there are always analytic points in any of its neighborhoods.Now,if the function is analytic in a deleted neighborhood of the singular point,we call this singular point "isolated singular point".

Now let's consider an isolated singular point $a$ and a function $f(z)$ is analytic in its deleted neighborhood

$%5C%5C0%3C%5Cvert%20z-a%20%5Cvert%20%3CR$

then we call the value of the integral

$%5C%5C%5Cmathop%7B%5Cmathrm%20%7BRes%7D%7D_%7Bz%3Da%7D%20f(z)%3D%5Cfrac%7B1%7D%7B2%5Cpi%20i%7D%20%5Cint_%7B%5CGamma%20%7Df(z)%20%5C%20dz%20%5C%20%2C%20%5C%20%5CGamma%20%3A%5Cvert%20z-a%20%5Cvert%20%3D%5Crho%20%2C0%3C%5Crho%20%3CR$

the residue of the function at the point $a$ .Actually with the Cauchy integral theorem we know that the value has nothing to do with $%5Crho%20$ (why?),the residue only have relation with the function and the isolated singular point.Now,consider a complex-connective domain $D$ with boundary

$%5C%5C%5C%20C%3DC_0%20%5C%20or%20%5C%20C%3DC_0%2BC_1%5E-%2B%E2%80%A6%2BC_n%5E-%20$

Function $f(z)$ is analytic on this domain except for points

$%5C%5Ca_1%2Ca_2%2C%E2%80%A6%2Ca_n%5Cin%20D$

and is continuous on the closed domain

$%5C%5C%5Cbar%7BD%7D%20%3DD%5Ccup%20C$

except for points

$%5C%5Ca_1%2Ca_2%2C%E2%80%A6%2Ca_n$

Then we have

$%5C%5C%5Cint_%7BC%7Df(z)%20%5C%20dz%3D2%5Cpi%20i%5Csum_%7Bi%3D1%7D%5En%20%5Cmathop%7B%5Cmathrm%7BRes%7D%7D_%7Bz%3Da_i%7Df(z)$

This formula is called "Cauchy Residue Theorem".To prove it,we should create circles

$%5C%5C%5CGamma%20_1%2C%5CGamma%20_2%2C%E2%80%A6%2C%5CGamma%20_n$ which are small enough and the points

$%5C%5Ca_1%2Ca_2%2C%E2%80%A6%2Ca_n$

are the circles' centers.Then we just need to use the Cauchy integral theorem along curve

$C_0%2B%5CGamma%20_1%5E-%2B%5CGamma%20_2%5E-%2B%E2%80%A6%2B%5CGamma%20_n%5E-%5C%5Cor%5C%5CC_0%2BC_1%5E-%2B%E2%80%A6%2BC_n%5E-%2B%5CGamma%20_1%5E-%2B%5CGamma%20_2%5E-%2B%E2%80%A6%2B%5CGamma%20_n%5E-$

This theorem tells us that when we need to calculate a integral,we can only concern the isolated singular points in the domain.

Another question si that why we need to multiply the integrall by a coefficient $%5Cfrac%7B1%7D%7B2%5Cpi%20i%7D%20$ ?The answer is that if we do that,we can tackle the problems easierly with the latter theories.

Residue theorem is a powerful theorem to handle many real integrals in mathematical analysis.

Complex Indefinite Integral

Since the value of integral of analytic functions has nothing to do with the integral path,as long as we choose the start and finish point we can determine the integral value.Let function $f(z)$ is analytic on domain $D$ .we can choose a fixed point $z_0$ and a moving point $z$ on $D$ as the start and the finish.As the moving point is moving,the integral value will change.Then there is a new function on $D$ which is determined by $f(z)%2Cz_0$ and the variation is $z$ :

$%5C%5CF(z)%3D%5Cint_%7Bz_0%7D%5E%7Bz%7D%20f(%5Czeta%20)%20%5C%20d%5Czeta%20$

To avert misleading,we should make the integral variation different to the integral bounds,here I use letter $%5Czeta%20$ .

Actually we have a theorem about the $F(z)$ above like real analysis:

If $f(z)$ is continuous on single-connective domain $D$ and its integral vlue has nothing to do with the integral path on $D$ ,then $F(z)$ is analytic on $D$ and

$%5C%5CF'(z)%3Df(z)%2Cz%5Cin%20D$

Apparently,as long as the function is analytic,it meets the condition in this generalized theorem.

proof

Because that the integral value is only determined by the start and finish point and that

$%5C%5C%5Cint_%7BC%7Df(z)%20%5C%20dz%3D%5Cint_%7BC_1%7Df(z)%20%5C%20dz%2B%5Cint_%7BC_2%7Df(z)%20%5C%20dz%2C%5Cmathrm%7BC%20%5C%20is%20%5C%20created%20%5C%20by%20%5C%20combining%20%5C%20C_1%20%5C%20and%20%5C%20C_2%7D$

we have

$%5C%5C%5Cint_%7BA%7D%5E%7BC%7D%20f(z)%20%5C%20dz%3D%5Cint_%7BA%7D%5E%7BB%7D%20f(z)%20%5C%20dz%2B%5Cint_%7BB%7D%5E%7BC%7D%20f(z)%20%5C%20dz$

hence,

$%5Cfrac%7BF(z%2B%5CDelta%20z)-F(z)%7D%7B%5CDelta%20z%7D%20%5C%5C%3D%5Cfrac%7B1%7D%7B%5CDelta%20z%7D%20(%5Cint_%7Bz_0%7D%5E%7Bz%2B%5CDelta%20z%7Df(%5Czeta%20)%20%5C%20d%5Czeta%20-%5Cint_%7Bz_0%7D%5E%7Bz%7Df(%5Czeta)%20%5C%20d%5Czeta)%5C%5C%3D%5Cfrac%7B1%7D%7B%5CDelta%20z%7D%20%5Cint_%7Bz%7D%5E%7Bz%2B%5CDelta%20z%7D%20f(%5Czeta)%20%5C%20d%5Czeta$

For every $z%5Cin%20D$ ,we consider the consequence making $%5CDelta%20z$ goes to zero.Now $%5CDelta%20z$ is variation but $z$ is regarded as a constant.Then value of $f(z)$ is a constant,too.hence

$%5C%5C%5Cfrac%7B1%7D%7B%5CDelta%20z%7D%20%5Cint_%7Bz%7D%5E%7Bz%2B%5CDelta%20z%7D%20f(z)%20%5C%20d%5Czeta%20%3D%5Cfrac%7B1%7D%7B%5CDelta%20z%7D%20f(z)%5Cint_%7Bz%7D%5E%7Bz%2B%5CDelta%20z%7D%5C%20d%5Czeta$

Back to the definition,

$%5C%5C%5Cint_%7Ba%7D%5E%7Bb%7D%20dz%3D%5Clim%20%5Csum_%7Ba%7D%5Eb%20(z_k-z_%7Bk-1%7D)%3D%5Clim%20(b-a)%3Db-a$

$%5C%5C%5Cfrac%7B1%7D%7B%5CDelta%20z%7D%20f(z)%5Cint_%7Bz%7D%5E%7Bz%2B%5CDelta%20z%7D%5C%20d%5Czeta%3Df(z)$

Hence

$%5Cvert%20%5Cfrac%7BF(z%2B%5CDelta%20z)-F(z)%7D%7B%5CDelta%20z%7D-f(z)%20%20%5Cvert%20%5C%5C%3D%5Cvert%20%5Cfrac%7B1%7D%7B%5CDelta%20z%7D%20%20%5Cint_%7Bz%7D%5E%7Bz%2B%5CDelta%20z%7D(f(%5Czeta%20)-f(z))%20%5C%20dz%20%5Cvert%20$

$%5Czeta%20$ is the complex along the integral path whose start and finish are $z$ and $z%2B%5CDelta%20z$ .As long as $%5Cvert%5CDelta%20z%5Cvert$ is small enoughly, $z$ and $z%2B%5CDelta%20z$ will be close enoughly and we can choose a integral path that the points on it are all close enoughly with $z$ ,namely

$%5C%5C%5Cvert%5Czeta%20-z%5Cvert%3C%5Cdelta%20$

Because $f(z)$ is continuous,

$%5C%5C%5Cvert%20f(%5Czeta%20)-f(z)%5Cvert%3C%5Cvarepsilon%20$

Another aspect,baecause $D$ is an open set,so it must be analytic in a neighborhood of $z$ ,It allows us directly choose the line segment whose ends are $z$ and $z%2B%5CDelta%20z$ as the integral path,as long as $%5Cvert%5CDelta%20z%5Cvert$ is small enoughly.Now let us use the integral approximation formula:

$%5Cvert%20%5Cfrac%7B1%7D%7B%5CDelta%20z%7D%20%20%5Cint_%7Bz%7D%5E%7Bz%2B%5CDelta%20z%7D(f(%5Czeta%20)-f(z))%20%5C%20dz%20%5Cvert%20%5C%5C%5Cleq%20%5Cfrac%7B1%7D%7B%5Cvert%20%5CDelta%20z%20%5Cvert%20%7D%20%5Cvarepsilon%20%5Cvert%20%5CDelta%20z%20%5Cvert%20%3D%5Cvarepsilon%20$