Two questions regarding $\mathrm {Li}$ from "Edwards" I would appreciate help understanding a relation in Edwards's "Riemann's Zeta Function."
On page 30 he has:
$$\int_{C^{+}} \frac{t^{\beta - 1}}{\log t}dt = \int_{0}^{x^{\beta}}\frac{du}{\log u}= \mathrm {Li} (x^{\beta}) - i\pi$$ 
He states for $\beta$ positive and real, change variables $u = t^{\beta}$ which implies $\log t = \log u/\beta$ and $dt/t = du/u \beta$. 
Here $C^{+}$ is a path which is a line segment from $0$ to $1 - \epsilon$ and passes over the singularity at $u = 1$ in a semi-circle in the upper half-plane and continues in a line segment from $1 + \epsilon$ to $1$.
I would appreciate help with two aspects:
-- Since most of the discussions of the logarithmic integral I have seen take the integral from $2$ rather than from $0$, how do you treat what looks like a $- \mathrm {Li}(0)$ term?
-- How do you actually get the $- i \pi$ term. I would guess it's from integrating around the half-circle above $u = 1$ in a clockwise direction. But I have tried parametrization with $u = r e^{i \theta}$. Maybe this is something I should know from complex analysis.
Thanks very much.

Here is a try for part two, the $- \pi i$ term:
Change variable from $u$ to $e^{t}$. Then $\log u = t$ and $du = e^{t}dt$.
Considering the integral around a half circle, now about and above $t = 0$, in the clockwise direction, the residue of the integrand is $-i\pi\left(e^{t}\right)_{z=0} = - i \pi$.
