Evaluating $\int_{-\infty}^0\frac{1-e^{(1+7i)z}}{e^{-z}-1}dz$ as the generalized harmonic number $H_{1+7i}$

Question

After I was interested in the entire function that appears in the Proof of this Wikipedia's article for Faulhaber formula, I was asking to Wolfram Alpha online calculator, and I've obtained examples as $$\int_{-\infty}^0\frac{1-e^{(1+7i)z}}{e^{-z}-1}dz=H_{1+7i},$$

where $H_{z}$ is a generalized harmonic number. See it with such example

integrate (1-e^((1+7i)z))/(e^(-z)-1)dz, from z=-infty to z=0.

I presume that the corresponding generalization should be in the literature, but

Question. Can you explain us previous example? Why $$\int_{-\infty}^0\frac{1-e^{(1+7i)z}}{e^{-z}-1}dz=H_{1+7i}?$$ And if it is possible what is $H_{1+7i}$ (how one defines such harmonic number, from an understandable way). Thank you for reading my question.

Answer 1

Considering the more general case $$I=\int \frac{1-e^{k z}}{e^{-z}-1}\,dz$$ Change variable $z=\log(x)$ and get $$I=\int \frac{x^k-1}{x-1}\,dx$$

This makes $$J=\int_{-\infty}^0 \frac{1-e^{k z}}{e^{-z}-1}\,dz=\int_0^1\frac{x^k-1}{x-1}\,dx=\int_0^1 \sum_{n=0}^{k-1}x^n\,dx =\sum_{n=0}^{k-1}\int_0^1 x^n\,dx= \sum_{n=0}^{k-1}\frac 1{n+1}=H_k$$ This also applies in the complex domain provided $\Re(k)>-1$ .