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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.

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

enter image description here

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.

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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 ($0$).

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  • $\begingroup$ Sorry if I was vague. I know that log(-x) is not necessarily undefined; however, in the text that I was reading they defined the branch of log in the integral specifically so that log(-x) is undefined when x is a positive real number. Furthermore, even if they did choose some other angle, the contour in the second integral is a circle about the origin, so some point of discontinuity occurs on the contour, which is precisely what I am confused about. I will edit the question to make this more clear. $\endgroup$ – arbitrary username Nov 24 '15 at 3:42
  • $\begingroup$ They are a little sloppy with the notation, but if you read carefully the text you posted here, you see that they mean to only integrate on the upper half of the circle, passing from the positive real axis to the negative real axis. It doesn't even work to go all the way around, as that would leave them back on the positive axis, not the negative axis where the curve continues. And if as you claim that defined $\log(-x)$ to be undefined on the positive real axis, then the entire 3rd integral is undefined. Check again. The discontinuity is the negative imaginary axis. $\endgroup$ – Paul Sinclair Nov 24 '15 at 3:48
  • $\begingroup$ The curve is supposed to return up the positive real axis. They require the first and third integrals to be taken "slightly above" and "slightly below" the positive real axis, which I take to mean they are integrating over the lines parameterized by f(t) = t ± i\delta, where \delta can be arbitrarily small. I understand your confusion about the contour, as that is precisely what I am confused about. $\endgroup$ – arbitrary username Nov 24 '15 at 3:56
  • $\begingroup$ Now you are giving critical additional information: the integrals are not actually on the real line. Then yes, they can be undefined on the line itself. The contour around the origin is also off the line (note again, they describe it as only going halfway around, not all the way). so it never reaches the undefined place (though why it is even needed if they are not integrating along the real line, I don't know). $\endgroup$ – Paul Sinclair Nov 24 '15 at 4:22

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