$\left|x^{s-1} \right|=x^{\left| s-1 \right|}$? http://planetmath.org/analyticcontinuationofriemannzetausingintegral
The above link is a page of planetmath about analytic continuation of Riemann zeta function. The fourth equation in the page contains the following :
$$
\sum_{n=1}^\infty \int_0^\infty \left|e^{-nx} x^{s-1} \right| dx = 
\sum_{n=1}^\infty \int_0^\infty e^{-nx} x^{\left| s-1 \right|} dx
$$
where $s$ is a complex number such that $\Re(s) > 1$. Is it right? 
I think it should be the next inequality rather than the equality.
$$
\sum_{n=1}^\infty \int_0^\infty \left|e^{-nx} x^{s-1} \right| dx \le
\sum_{n=1}^\infty \int_0^\infty e^{-nx} x^{\left| s-1 \right|} dx
$$

If $x$ is positive real and $s=\sigma+it$ with $\sigma,t\in\mathbb R$ then $x^{s-1}=x^{\sigma-1}\cdot x^{it}$. The factor $x^{it}=e^{it\ln x}$ has modulus $1$, so after taking absolute values we have  $$\left |x^{s-1}\right|=x^{\sigma-1}$$instead.
