Question about Contour Integration

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In summary, the conversation discusses a question about complex analysis, specifically regarding the calculation of a contour integral over the unit circle using different methods. The first method, using Cauchy's Integral Formula, results in a value of 2*pi*i. The second method, using the path z(t)=e^(it), also gives 2*pi*i as the answer. The conversation then delves into a possible substitution using u=e^(it), but it is ultimately explained that this substitution does not change the value of the integral due to the multi-valued nature of the antiderivative. The summary concludes that the result of 2*pi*i is still recovered regardless of the method used.
  • #1
Apogee
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This question deals specifically with complex analysis.

Let C be the unit circle in the complex plane (|z| = 1). If you calculate the contour integral of (1/z)dz over C using Cauchy's Integral Formula, you get 2*pi*i. If you calculate it using the path z(t)=e^(it), t in [0,2pi], you also receive 2*pi*i.

However, if I do the second method, but make the substitution u = e^(it), the limits of integration (specifically 0 and 2pi) both become 1. Hence, after substitution, the integral is 0. Why is this the case? I imagine that u-substitutions work differently with complex integrals, but I'm trying to understand what is wrong with making this substitution.
 
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  • #2
If you use the second method you end up with the integral ##\int_0^{2\pi}i\,dt## which gives ##2\pi i##. If you substitute ##u=e^{it}## in there you'll just get back to the original integral ##\int_C\frac{dz}z##, since ##u=z##. It's not clear why you believe that gives an integral of zero. If you show your working we may be able to show where it goes wrong.
 
  • #3
Apogee said:
However, if I do the second method, but make the substitution u = e^(it), the limits of integration (specifically 0 and 2pi) both become 1. Hence, after substitution, the integral is 0. Why is this the case?

I assume you mean the following situation:

##\displaystyle\biggr. \oint \frac{1}{z}dz= log(z)\biggr |_1^1=(0+2\pi i)-0=2\pi i##

And that is because the antiderivative is a multi-valued function. So in Complex Analysis, it is not always the case that the integral is zero when the limits "appear" to be the same. One must consider the (analytically-continuous) path along a multivalued antiderivative (if antiderivatives are used) which in general does not return to the same spot over the antiderivative when the path is closed. Now, if you plot the real or imaginary component of the log(z) function and plot an analytically-continuous path along it starting from z=1, wrap around the origin ##2\pi## and go back to 1, you'll see that you end up at a point on the function equal to ##0+2\pi i## or if you just plot the imaginary part of log(z), you'll see that you end up at the point ##2\pi##. So that if you understand this, you can explain why, if the winding number is 2, then we have:

##\displaystyle\biggr. \oint \frac{1}{z}dz= log(z)\biggr |_1^1=4\pi i##

But this is tough to understand I know when seeing for the first time. Better for now to just use the substitution ##z(t)=e^{it}## which is of course an analytically continuous path, then the integral becomes

##\displaystyle \oint \frac{1}{z}dz=\int_0^{2\pi} i dt=2\pi i##
 
  • #4
Thank you very much! I understand what the issue is now. logz is multivalued. So, though the limits of integration are the same, each results in two different values of the antiderivative. So, the result 2*pi*I is still recovered. :)
 

Related to Question about Contour Integration

1. What is contour integration?

Contour integration is a mathematical technique used to evaluate integrals along a path in the complex plane. It involves breaking up a complex function into smaller parts and integrating along a specific contour or path.

2. Why is contour integration important?

Contour integration is important in many areas of mathematics and science, including complex analysis, engineering, physics, and statistics. It allows us to solve difficult integrals and understand complex functions in a more manageable way.

3. How is contour integration different from regular integration?

Contour integration involves integrating along a specific path in the complex plane, whereas regular integration involves integrating over a specific interval on the real number line. The techniques used in contour integration are also different, as it involves methods such as the Cauchy integral theorem and residue calculus.

4. What are some applications of contour integration?

Contour integration has many practical applications, including solving differential equations, calculating probabilities in statistics, and understanding the behavior of electric and magnetic fields in physics. It is also used in image processing and signal analysis, among other areas.

5. What are some common challenges in contour integration?

One of the main challenges in contour integration is choosing the right contour or path to integrate along, as this can greatly affect the final result. Another challenge is dealing with singularities or poles in the complex plane, which require special techniques such as residue calculus. Additionally, complex functions can be difficult to visualize and understand, making it challenging to apply contour integration in certain situations.

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