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If I have an integral $\int_C \frac{1}{z} dz$ where $C$ is the left half of the circle from $i$ to $-i$, then I can choose the branch cut to be the positive axis, and have the antiderivative to be log(z). Then using the antiderivative, I can evaluate the integral:

$\int_C \frac{1}{z} dz = log(-i) - log(i)$

However, if I let $0 \leq arg(z) < 2\pi$, my integral will evaluate to $\frac{3\pi}{2} - \frac{\pi}{2} = \pi$. If I choose $-\pi \leq arg(z) < \pi$, then my integral will evaluate to $\frac{-\pi}{2} - \frac{\pi}{2} = -\pi$. I feel like I should get the same value. What am I doing wrong here.

CuriousAlpaca
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  • When you choose $0 \leq \arg(z) < 2\pi$, the branch cut is the positive real axis. Your contour $C$ passes through this, so this branch of $\log(z)$ is not a primitive for $\frac{1}{z}$ in the area you're integrating over. The correct answer is $-i\pi$, which you can see by parametrizing the curve. Notice that you have also left out the factor of $i$. – Thusle Gadelankz Oct 28 '20 at 20:22
  • Oh I forgot about the $i$. I see what you are saying. Since $C$ is the left half circle, isn't choosing the branch to be positive axis the correct choice? Because $C$ doesn't pass through the positive x-axis – CuriousAlpaca Oct 28 '20 at 20:35
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    Yes, of course. That was my fault, apologies. In that case, the contour passes through the negative real axis, making the principle branch of the logarithm an ineligible choice. With that orientarion of the semicircle, the correct answer is of course $i \pi$. – Thusle Gadelankz Oct 28 '20 at 20:40
  • I have now added a more thorough answer below. I hope that will clarify! – Thusle Gadelankz Oct 29 '20 at 05:50

1 Answers1

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Solving the integral by parametrization yields:

$$\int_{C} \frac{dz}{z} = \int_{\pi/2}^{3\pi/2} \frac{1}{e^{i \theta}} ie^{i \theta} \, d \theta=i \pi$$

You may arrive at the same result by choosing the branch of the logarithm given by

$$\log_0 (x) = \text{Log} |z| + i \arg_0(z),$$

where $\arg_0(z) \in [0, 2\pi)$, and using it as a primitive for $\frac{1}{z}$. As mentioned, we choose this branch since the contour passes through the negative real axis. This branch of the logarithm however, has its branch cut at the positive real axis. We get

$$\int_{C} \frac{dz}{z} = \log_0 (-i) - \log_0(i) =i (\frac{3\pi}{2}-\frac{\pi}{2})=i \pi$$

Using the principal branch of the logarithm instead will give a wrong result because it is not a primitive of the function we are integrating in an open set containing the contour over which we integrate.

Thusle Gadelankz
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