Riemann Zeta Function integral

Question

I was reading about the Riemann Zeta Function when they mentioned the contour integral $$\int_{+\infty}^{+\infty}\frac{(-x)^{s-1}}{e^x - 1} dx$$ where the path of integration "begins at $+\infty$, moves left down the positive real axis, circles the origin once in the positive (counterclockwise) direction, and returns up the positive real axis to $+\infty$." They specified that $(-x)^{s-1} = e^{(s-1)\log(-x)}$ so that $\log(-x)$ is real when x is a negative real number. Furthermore, $\log(-x)$ is undefined when x is a positive real number as a result of this choice of $\log$.

They proceeded to evaluate the integral by splitting it into $$\int_{+\infty}^{\delta}\frac{(-x)^{s-1}}{e^x - 1}dx + \int_{|x|=\delta}\frac{(-x)^{s-1}}{e^x - 1} dx + \int_{\delta}^{+\infty}\frac{(-x)^{s-1}}{e^x - 1} dx$$ I do not understand the justification for the path of the second integral; if $(-x)^{s-1}$ is undefined on the non-negative real axis, how can the path of integration cross the real axis at $x=\delta$, when $\delta$ is implicitly non-negative?

EDIT: I am aware that $\log(-x)$ is defined on the entire plane except for some ray from the origin. What I am specifically confused about is how the contour is allowed to cross the ray on which $\log(-x)$ is undefined.

EDIT: They describe the first and third integrals of the sum as being taken "slightly above" and "slightly below" the positive real axis as the function is undefined on the positive real axis.

Answer 1

Here is a picture that matches the description of the path of integration:

Note that we are taking the positive real axis as the branch cut for $\log(-x)$, and we have taken the branch of $\log(-x)$ whose argument goes from $-\pi$ to $\pi$.

The red and blue lines should be infinitesimally above and below the positive real axis.

Answer 2

There is more to what they specified than you've reproduced here. $\log(-x)$ is not necessarily undefined when the values are allowed to be complex. In fact, if $x = re^{i\theta}$, then $-x = re^{i(\theta + \pi)}$. Therefore we can define $\log(-x) = \log(r) + i(\theta + \pi)$. The problem is, $-x = re^{i(\theta + (2n+1)\pi)}$ as well, for any integer $n$. So which value of $\theta$ do we choose? No matter what the choice, at some place as you circle the origin, $\theta$ must suddenly jump in value to get back to the other side of its interval of definition. So we have to choose to place this discontinuity somewhere.

Since they need $\log(-x)$ defined on both the positive and negative real axis, and since they choose to circle $0$ on the upper side (counterclockwise from positive to negative circles above $0$), they probably chose to put the discontinuity on the negative imaginary axis. I.e., they defined $\log(-re^{i\theta}) = \log(r) + i\theta$ where $-\frac \pi 2 < \theta < \frac{3\pi} 2$.

So $(-x)^{s-1}$ is also defined everywhere except on the negative imaginary axis, and they circled the origin as they did exactly to avoid crossing that axis, even at its tip ($0$).