I am having trouble computing an integral along a line segment on complex plane. Namely:
Let $\operatorname{Log}{z}$ denote the principal branch of logarithm on $\mathbb{C}-\mathbb{R_{\leq0}}$ and let $\gamma$ be a line segment from $-i$ to $1$. Compute the integral: $$\int_{\gamma}\overline{\operatorname{Log}(z)}dz$$
Attempt: The function $\overline{\operatorname{Log}(z)}$ is not holomorphic so we are not sure whether an antiderivative exists, so what seemed resonable for me was to parametrize $\gamma$ in polar coordinates and try direct computation by $\int_{\gamma}\log(|z|) + i\int_{\gamma}Arg(z)$ but the parametrization is rather ugly which makes computation rather too difficult taking into account that this was an exam task. Is there any neat way to compute this?
The key is: from the definition $\log(z)=\log|z|+i\arg z$ where $\arg z\in [-\pi,\pi)$, it is clear that $\overline{\log z}=\log \bar z$ on $\mathbb C\setminus \mathbb R_{\le 0}$.
Hence, by parametrizing the path as $z=\gamma(t)=-i+e^{i\pi/4}t$, $t\in[0,\sqrt 2]$, $$\int_{\gamma}\overline{\operatorname{Log}(z)}dz=\int_0^{\sqrt 2}\overline{\operatorname{Log}(-i+e^{i\pi/4}t)}e^{i\pi/4}dt=e^{i\pi/4}\int_0^{\sqrt 2}\operatorname{Log}(i+e^{-i\pi/4}t)dt$$
At this point, we have two approaches:
Direct calculation $$\begin{align} e^{i\pi/4}\int_0^{\sqrt 2}\operatorname{Log}(i+e^{-i\pi/4}t)dt &=e^{i\pi/4}\cdot\frac{(i+e^{-i\pi/4}t)(\operatorname{Log}(i+e^{-i\pi/4}t)-1)}{e^{-i\pi/4}}\bigg\vert^{t=\sqrt 2}_{t=0} \\ &= -1+\left(\frac{\pi}{2}-1\right)i \end{align} $$
Complex analysis approach $$\begin{align} e^{i\pi/4}\int_0^{\sqrt 2}\operatorname{Log}(i+e^{-i\pi/4}t)dt &=i\int_0^{\sqrt 2}\operatorname{Log}(i+e^{-i\pi/4}t)e^{-i\pi/4}dt\\ &= i\int_{\bar\gamma}\operatorname{Log}(z)dz \quad(1)\\ &= -i\int_{\text{quarter arc}}\operatorname{Log}(z)dz \quad(2)\\ &= -i\int^{\pi/2}_0(\ln|e^{i\theta}|+i\theta)ie^{i\theta}d\theta \\ &= i\int^{\pi/2}_0\theta e^{i\theta}d\theta \\ &=-1+\left(\frac{\pi}{2}-1\right)i \end{align} $$
$(1)$: $\bar\gamma$ is the straight line from $+i$ to $1$.
$(2)$: $\text{quarter arc}$ is the quarter arc from $+i$ to $1$. The deformation of integration path is justified by Cauchy's integral theorem.
Although method 1 looks simpler, indeed the algebra is more tedious. As a complex analysis enthusiaist, I view method 2 as a more elegant solution.