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I'd like to check my work. I'm trying to integrate $f(z)=\text{Log}(z)$ on the right half of the unit circle from $-i$ to $i$. $\text{Log} = \ln(r)+i\theta$ according to my book. So, $$\begin{align} \int_C f(z)\,dz&=\int_{z_1}^{z_2}f(z)\,dz\\\\& = \int_{-i}^i \text{Log}\,dz \\\\ &= \bigg[z\text{Log}-z\bigg]_{-i}^i \\\\ &= (i\text{Log}(i)-i)-\left(-i\text{Log}(-i)+i\right) \\\\ &= i\left(\ln(1)+\frac{\pi}{2}i\right)-i \ + i\left(\ln(1)-\frac{\pi}{2}i \right) \ -i \\\\&= -2i \\\\ \therefore \int_{-i}^i \text{Log}\,dz=-2i \end{align}$$

This may be incorrect, I'm not really sure how to incorporate the branch cut here.

Thanks!

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  • $\begingroup$ the circuit is a circle not the segment [-i,i]. $\endgroup$ – zwim Jun 15 '17 at 5:13
  • $\begingroup$ So, I'm required to use the formula $\int_C f(z)dz=\int_a^b f[z(\theta)]z'(\theta)d\theta$? Where $a$ and $b$ are $\frac{\pi}{2}$ and $\frac{-\pi}{2}$? What would my choice for $z'(\theta)$ be in this case? $\endgroup$ – Kosta Jun 15 '17 at 5:14
  • $\begingroup$ Have a look at these answers : math.stackexchange.com/questions/1602614/… $\endgroup$ – zwim Jun 15 '17 at 5:34
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I thought it might be instructive to present two alternative approaches to evaluate the integral of interest. To that end we proceed.


Let $\text{Log}(z)=\log(|z|)+i\text{Arg}(z)$ where $-\pi < \text{Arg}(z)\le \pi$. In addition, let $C$ denote the semicircular contour in the right-half plane of radisu $1$ and that begins at $z=-i$ and ends at $z=i$.


METHODOLOGY $1$:

On $C$, $z=e^{i\theta}$, $-\pi/2\le \theta \le \pi$. Hence, the integral $\int_C \text{Log}(z)\,dz$ is given by

$$\begin{align} \int_C \text{Log}(z)\,dz&=\int_{-\pi/2}^{\pi/2} \text{Log}(e^{i\theta})\,ie^{i\theta}\,d\theta\\\\ &=-\int_{-\pi/2}^{\pi/2} \theta e^{i\theta}\,d\theta\\\\ &=-2i\int_0^{\pi/2} \theta\sin(\theta)\,d\theta\\\\ &=-2i \end{align}$$


METHODOLOGY $2$:

Another approach is to exploit Cauchy's Integral Theorem. Inasmuch as $\text{Log}(z)$ is analytic for $z\in \mathbb{C}\setminus (-\infty,0]$. Then, the integral over $C$ can be deformed to connect $-i$ to $i$ along any rectifiable path that does not intersect the branch cut. Therefore, we write

$$\begin{align} \int_C \text{Log}(z)\,dz&= \lim_{\epsilon\to 0^+}\left(\int_{-1}^{-\epsilon }\text{Log}(iy)\,i\,dy+\int_{-\pi/2}^{\pi/2}\text{Log}(\epsilon e^{i\theta})\,ie^{i\theta}\,d\theta+\int_{\epsilon }^1\text{Log}(iy)\,i\,dy\right)\\\\ &=i \lim_{\epsilon\to 0^+}\int_\epsilon^1 \left(\text{Log}(-iy)+\text{Log}(iy)\right)\,dy\\\\ &=2i\int_0^1 \text{Log}(y)\,dy\\\\ &=-2i \end{align}$$

as expected!

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  • $\begingroup$ This is very helpful, thank you Mark! $\endgroup$ – Kosta Jun 16 '17 at 16:21
  • $\begingroup$ You're welcome. Pleased to hear that this is helpful! $\endgroup$ – Mark Viola Jun 16 '17 at 16:25
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An alternate way to check, without using contours, is to use $$\int \ln(x) \, dx = x \, \ln(x) - x$$ which provides \begin{align} \int_{-i}^{i} \ln(x) \, dx &= \left( i \, \ln(e^{\pi \, i/2}) - i \right) - \left( -i \, \ln(e^{-\pi \, i/2}) + i \right) \\ &= \frac{\pi \, i^2}{2} - \frac{\pi \, i^2}{2} - 2 i \\ &= -2i. \end{align} With this it can be stated that the real line integral result is the same as the contour integral result.

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looks good to me .Fundamental theorem of calculus wins again

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