The antiderivative of $\frac{1}{x-ib}$

Question

I know if my domain $D_{\mathbb{R}}\subset\mathbb{R}$ and $b$ is a real constant, then:

$$\int_{D_\mathbb{R}}\dfrac{1}{x-ib}\mathrm{dx}=\ln(|x-ib|)$$

But if my domain $D_{\mathbb{C}}\subset\mathbb{C}$:

$$\int_{D_\mathbb{C}}\frac{1}{x-ib}\mathrm{dx}=\log(x-ib)=\ln|x-ib|+i\mathrm{arg}(z)$$

This feels a bit counterintuitive to me, but is it correct?

Answer 1

In the complex case, even for real domain, do not use $|\cdot|$. That is incorrect.
Correct is: [some branch of] $\log(x-ib)$.

Example
Computed numerically, we have to $10$ decimals $$ \int_0^1 \frac{1}{x-i}\;dx \approx 0.3465735903 + i\; 0.7853981634 $$ But $$ \log\big(|x-i|\big)\Big|_{x=0}^{x=1} = \log\big(|1-i|\big) - \log\big(|-i|\big) = \log(\sqrt2) - \log(1) \approx 0.3465735903 $$ incorrect.

On the other hand, $$ \log(1-i) - \log(-i) = \big(\log\big(\sqrt2\big) - \frac{i\pi}{4}\big) - \big(\frac{-i\pi}{2}\big) = \log\big(\sqrt{2}\big) + \frac{i\pi}{4} \approx 0.3465735903 + i\;0.7853981635 $$ correct
I used a branch of $\log$ that is continuous on the line segment from $1-i$ to $-i$.

NOTE. When you ask Maple or Mathematica for $\int \frac{dx}{x}$ they give you $\log (x)$ and not $\log\big(|x|\big)$. This is (one of) the reasons. The "answer" $\log\big(|x|\big)$ is just wrong in many cases.

Answer 2

For the first one, since $D\subset\mathbb R$, we know that $D$ is an interval such as $(a,b)$ with $a<b$ since $D$ is connected.

This means the first integral becomes $$ \int_{D_\mathbb{R}}\dfrac{1}{x-ib}\mathrm{dx}=\ln(|x-ib|)\Bigg|_{x=a}^b. $$

For the integral over $\mathbb C$: Look up the definition of the complex logarithm function first (on Wikipedia or any standard textbook). You'll find that there is no unique logarithm function.