# Riemann Zeta function Analytic continuation integral

Following Riemann paper about analytic continuation of Zeta Function:

http://www.maths.tcd.ie/pub/HistMath/People/Riemann/Zeta/EZeta.pdf

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 ?

Cheers,

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

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.

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