Integrate Complex Function w/o Cauchy's or Residuals

In summary, the given problem is to integrate the function \frac{1}{z^2+1} around a circle of radius 2 centered at the origin, without using Cauchy's Integral Formula or Residuals. This can be solved by using the identity \int \frac{du}{a^2 + u^2} = \frac{1}{a}arctan\frac{u}{a} +c and applying it to the closed path, resulting in an answer of 0. The original approach using a u-substitution with the bounds of 0 and 2pi was incorrect since they are for the parametrization of the circle, not the integral in terms of z.
  • #1
fauboca
158
0
Without using Cauchy's Integral Formula or Residuals, I am trying to integrate

[tex]\int_{\gamma}\frac{dz}{z^2+1}[/tex]

Around a circle of radius 2 centered at the origin oriented counterclockwise.

[tex]\frac{i}{2}\left[\int_0^{2\pi}\frac{1}{z+i}dz-\int_0^{2\pi}\frac{1}{z-i}dz\right][/tex]

[tex]\gamma(t)=2e^{it}, \quad \gamma'(t)=2ie^{it}[/tex]

The answer is zero. I am supposed to get each integral to be [itex]2\pi i[/itex] which is 0 when subtracted.

I know it is related to the fact that [itex]\int_{\gamma}\frac{1}{z}dz = 2\pi i[/itex].

And using u-sub isn't correct since any closed path would be zero when that isn't true. The integral of 1/z shows that not all closed paths will be zero.
 
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  • #2
Try using the identity:

[tex]\int \frac {du}{a^2 + u^2} = \frac{1}{a}arctan\frac{u}{a} +c[/tex]
 
  • #3
Totalderiv said:
Try using the identity:

[tex]\int \frac {du}{a^2 + u^2} = \frac{1}{a}arctan\frac{u}{a} +c[/tex]

That is nice and a lot easier than the route I was taking. Without the substitution of the integral, would it sill have the bounds of 0 and 2pi though? I know it is over gamma but the bounds were for 2e^{it}.
 
  • #4
The circle starts and ends at the same point. Your bounds are actually from 2 to 2. The integral identity applied to your problem requires bounds in z, not in terms of its parametrization.
 

1. What is the definition of a complex function?

A complex function is a function that maps complex numbers to complex numbers. It can be written in the form f(z) = u(x, y) + iv(x, y), where u and v are real-valued functions of the complex variables x and y, and i is the imaginary unit √(-1).

2. Why is Cauchy's theorem important in complex integration?

Cauchy's theorem states that the value of a contour integral around a closed path depends only on the values of the function inside the contour. This allows for the evaluation of complex integrals without having to consider the specific path of integration, making it a powerful tool in complex analysis.

3. How do you integrate a complex function without using Cauchy's or Residuals?

One approach is to use the Cauchy-Riemann equations, which provide a relationship between the real and imaginary parts of a complex function. By solving these equations, the integral can be expressed in terms of real integrals, which can then be evaluated using traditional integration techniques.

4. Can complex functions be integrated using residue theory?

Yes, residue theory is another powerful method for evaluating complex integrals. It involves finding the residues (complex numbers) of a function at its singular points, and then using these residues to evaluate the integral. This method is particularly useful for calculating integrals along paths that enclose singularities.

5. Are there any limitations to integrating complex functions without Cauchy's or Residuals?

While both the Cauchy-Riemann equations and residue theory provide alternative methods for integrating complex functions, they may not always be applicable. In some cases, the use of these methods can result in more complicated integrals than using Cauchy's or Residuals. Additionally, these methods may not be as efficient as using Cauchy's or Residuals for certain types of integrals.

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