Following Riemann paper about analytic continuation of Zeta Function:


I can't understand the contour integral step:

"If one now considers the contour integral from +infinity to +infinity taken in a positive sense around a domain which includes the value 0 but no other point of discontinuity of the integrand in its interior, then this is easily seen to be equal to:"

$$\int_\gamma\frac{(-x)^{s-1}}{e^x-1}dx=(e^{-\pi s i}-e^{\pi s i})\int_0^\infty\frac{{x}^{s-1}}{e^x-1}dx$$ Can someone explain how can I get this result ?


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    $\begingroup$ or some basic information about writing math at this site see e.g. here, here, here and here. $\endgroup$ Jun 3, 2014 at 16:02

2 Answers 2


Please allow me to elaborate a little on Riemann's and Jeb's elegant exposition.

So imagine the contour $\gamma$ as coming just "above" the real axis from $+\infty$ to $0$ and then swinging counterclockwise around the origin, and then going back just "below" the real axis back to $+\infty$.

As Riemann suggests, we can write $(-x)^{s-1} = e^{(s-1)\log(-x)}$. But what should $\log(-x)$ be equal to for positive $x$?

As Jeb suggested, $-1 = e^{i\pi} = e^{-i\pi}$, so we can write $\log(-x)$ as $\log(e^{i\pi}x) = i\pi+\log{x}$, or as $\log(e^{-i\pi}x) = -i\pi+\log{x}$. However, we must make choices that preserve the continuity of $\log(-x)$ as $x$ moves along the contour in the complex plane. Also, as Riemann specifies, we want $\log(-x)$ to be real when $x$ is negative.

For the piece of the contour from $+\infty$ to $0$, we must choose $$\log(-x) = -i\pi+\log{x}.$$

On the way back from $0$ to $+\infty$, in order to preserve continuity we must have $$\log(-x) = i\pi+\log{x}.$$

Thus, the integral for the part of the contour from $+\infty$ to $0$ is equal to $$\int_{+\infty}^0 \frac{e^{-(s-1)i\pi}e^{(s-1)\log{x}}}{e^x-1}\,dx = e^{-\pi si} \int_0^{+\infty} \frac{e^{(s-1)\log{x}}}{e^x-1}\,dx.$$ And the integral for the part of the contour from $0$ to $+\infty$ is equal to $$\int_0^{+\infty} \frac{e^{(s-1)i\pi}e^{(s-1)\log{x}}}{e^x-1}\,dx = -e^{\pi si} \int_0^{+\infty} \frac{e^{(s-1)\log{x}}}{e^x-1}\,dx.$$

Now add those two pieces together to get Riemann's result.

  • $\begingroup$ Thanks for your answer, it was really helpful $\endgroup$
    – emiliano
    Jun 2, 2014 at 14:00
  • $\begingroup$ Hey John, i just forget it ... can you recommend me some bibliography related to this kind of complex function analysis? $\endgroup$
    – emiliano
    Jun 2, 2014 at 15:12
  • $\begingroup$ Personally I like Ahlfors' book, but it is overpriced. Many recommend Frietag's book for its number theory applications at the end of the book. Or look here for recommendations. Often books on analytic number theory will teach you enough complex analysis as you go along. Try Murty's Problems in Analytic Number Theory. $\endgroup$
    – John M
    Jun 2, 2014 at 15:57

Use the fact that $-1 = e^{\pi i}$ and calculate the residue at $0$. Then the result pops out via the residue theorem.


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