I lost in this proof of Riemann's paper: On the Number of Prime Numbers less than a Given Quantity.
If one now considers the integral $$ \int \frac{(-x)^{s-1}}{e^x-1} dx $$ from $\infty$ to $\infty$ 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
$$ (e^{-\pi si}-e^{\pi si}) \int_0^{\infty} \frac{x^{s-1}}{e^x-1}dx $$.
That contour of integration is equivalent to the contour consisting of a circle $C_0$ of radius $\epsilon$ around the origin and two half-lines $C_+=(\infty\rightarrow \epsilon)$ and $C_-=(\epsilon\rightarrow\infty)$.
The value of the integrand at the point $z=-\epsilon$ on the circle is real and negative.
Turning by $\pi$ counterclockwise around the origin produces a phase factor $e^{i\pi s}$, since on the corresponding demi-circle $x$ is parameterized as $x=\epsilon e^{i\varphi}$, $\varphi\in[\pi,2\pi]$. The integral over $C_-$ will therefore be given by $$-e^{i\pi s} \int_{\epsilon}^{\infty}\frac{x^{s-1}dx}{e^x-1}.\tag{1}$$
Similarly, turning by $\pi$ clockwise, the integral over $C_+$ gives $$e^{-i\pi s} \int_{\epsilon}^{\infty}\frac{x^{s-1}dx}{e^x-1}.\tag{2}$$
Now let $\epsilon\rightarrow 0$, then the integral over $C_0$ vanishes whereas (1) and (2) give the quoted result.