Integral of $\text{Log}(z)$ 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!
 A: 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!
A: 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.
A: looks good to me .Fundamental theorem of calculus wins again
