Complex Integrals, Antiderivatives, Logarithms

In summary, the question is about whether the line integral of a function with a multiple-valued antiderivative is still path independent and how to prove it.
  • #1
Local
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I've been teaching myself a little bit of Complex Variables this semester, and I had a question concerning complex integrals.

If I understand correctly, then if a function [itex] f [/itex] has an antiderivative [itex] F [/itex], then the line integral [itex] \int_C f(z) dz [/itex] is path independent and always evaluates to [itex] F(z_1) - F(z_0) [/itex], where [itex] z_1 [/itex] and [itex] z_0 [/itex] are the final and initial points of the contour [itex] C [/itex].

Now let [itex] C [/itex] be the unit circle, oriented counterclockwise. Let's also say that 1 is the initial and final point of [itex] C [/itex]. Then by direct computation, we know that [tex] \int_C \frac{1}{z} dz = 2 \pi i .[/tex] Now, I think I understand that [itex] \log z [/itex] is not quite an antiderivative of [itex] \frac{1}{z} [/itex], because there is no way to define a single-valued branch of the complex logarithm over a region containing the unit circle in such a way that it will be differentiable at every point of the circle. So we can't quite say that [tex] \int_C \frac{1}{z} dz = \log(1) - \log(1). [/tex] But on the other hand, it kind of looks like the formula still holds, if we understand that the final value of [itex] \log(1) [/itex] is [itex] 2\pi i [/itex] more than the initial value, because we have gone around full circle.

At this point, I'm not too sure what my question is. I guess my question is: If a function [itex] f [/itex] has an antiderivative [itex] F [/itex] that is multiple-valued, like in my example above, can we still say that [tex] \int_C f(z) dz = F(z_1) - F(z_0), [/tex] as long as we choose the "appropriate" value of [itex] F(z_1) [/itex] and [itex] F(z_0) [/itex]? And if so, how could one go about proving this?

Thanks for reading all of that.
 
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  • #2
You will see the beautiful solution of this when studying the Riemann surface of a function.
 
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Local said:
I've been teaching myself a little bit of Complex Variables this semester, and I had a question concerning complex integrals.

If I understand correctly, then if a function [itex] f [/itex] has an antiderivative [itex] F [/itex], then the line integral [itex] \int_C f(z) dz [/itex] is path independent and always evaluates to [itex] F(z_1) - F(z_0) [/itex], where [itex] z_1 [/itex] and [itex] z_0 [/itex] are the final and initial points of the contour [itex] C [/itex].

Now let [itex] C [/itex] be the unit circle, oriented counterclockwise. Let's also say that 1 is the initial and final point of [itex] C [/itex]. Then by direct computation, we know that [tex] \int_C \frac{1}{z} dz = 2 \pi i .[/tex] Now, I think I understand that [itex] \log z [/itex] is not quite an antiderivative of [itex] \frac{1}{z} [/itex], because there is no way to define a single-valued branch of the complex logarithm over a region containing the unit circle in such a way that it will be differentiable at every point of the circle. So we can't quite say that [tex] \int_C \frac{1}{z} dz = \log(1) - \log(1). [/tex] But on the other hand, it kind of looks like the formula still holds, if we understand that the final value of [itex] \log(1) [/itex] is [itex] 2\pi i [/itex] more than the initial value, because we have gone around full circle.

At this point, I'm not too sure what my question is. I guess my question is: If a function [itex] f [/itex] has an antiderivative [itex] F [/itex] that is multiple-valued, like in my example above, can we still say that [tex] \int_C f(z) dz = F(z_1) - F(z_0), [/tex] as long as we choose the "appropriate" value of [itex] F(z_1) [/itex] and [itex] F(z_0) [/itex]? And if so, how could one go about proving this?

Thanks for reading all of that.

An actual antiderivative exists within a branch cut. EDIT: the antiderivative can be translated by
integer multiples of ## 2\pi i ## to get a different antiderivative in a different branch cut.
 
Last edited:

Related to Complex Integrals, Antiderivatives, Logarithms

1. What is the difference between a complex integral and a regular integral?

A complex integral is an integral that is evaluated over a complex domain, while a regular integral is evaluated over a real domain. This means that the limits of integration, as well as the function being integrated, can contain complex numbers in a complex integral.

2. How do you find the antiderivative of a complex function?

The process for finding the antiderivative of a complex function is the same as finding the antiderivative of a real function. Use the rules of integration, such as the power rule and substitution, to find the antiderivative. Just be sure to keep in mind the properties of complex numbers, such as the fact that the square root of a negative number is a complex number.

3. Can complex numbers be used in logarithmic functions?

Yes, complex numbers can be used in logarithmic functions. However, the base of the logarithm must also be a complex number, and the result will be a complex number as well.

4. What is the significance of the Cauchy-Riemann equations in complex integrals?

The Cauchy-Riemann equations are a set of necessary conditions for a complex function to be differentiable at a point. In complex integrals, these equations are used to determine whether a function has an antiderivative, and if so, what form it takes.

5. Are there any special techniques for evaluating complex integrals?

Yes, there are several special techniques, such as contour integration and the residue theorem, that are used to evaluate complex integrals. These techniques involve using complex analysis and complex variables to simplify the integral and make it easier to evaluate.

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