The dilogarithm is defined in $\mathbb{C}$ as
$$ Li_2(z) = -\int_0^1 \frac{\ln(1 - zt)}{t} dt $$
Because $1-zt \in \mathbb{C}$, then you can write $\ln(1 - zt) = \ln|1 - zt| + i·\arg(1 - zt)$
As pointed out by @Maxim in Dilogarithm: real and complex you can pick the principal determination of the argument $\arg$, which is the angle in $(-\pi, \pi]$ so
$\arg(1 - zt) = \begin{cases}2\arctan\left(\frac{\Im[1 - zt]}{|1 - zt| + \Re[1 - zt]}\right),\ \mbox{if $\Im \neq 0$ or $\Re > 0$}\\ \pi,\ \mbox{if $\Im = 0$ and $\Re < 0$}\\\mbox{undefined}, \ \mbox{if $\Re = 0$ and $\Im = 0$} \end{cases}$
and therefore
$$ Li_2(z) = -\int_0^1 \frac{\ln|1 - zt|}{t}dt - i\int_0^1 2\arctan\left(\frac{\Im[1 - zt]}{|1 - zt| + \Re[1 - zt]}\right)\frac{dt}{t}, \quad \mbox{for $z \in \mathbb{C} - (1, +\infty)$} \tag1$$
I want to pay attention to the special case $z\in(1, +\infty)$. For this one, $1 - zt < 0$ when $t>1/z$, so for this interval the complex logarithm is not defined within the principal analytic determination, i.e, $\arg \in (-\pi, \pi)$, that's why we can take just the principal determination as before. Then we can write the complex logarithm for this case as $\ln(1 - zt) = \ln|1 - zt| + i\pi$, so the dilogarithm is now
$$ Li_2(z) = -\int_0^1 \frac{\ln|1 - zt|}{t}dt - i\pi\int_{1/z}^1\frac{dt}{t} = -\int_0^1 \frac{\ln|1 - zt|}{t}dt - i\pi\ln(|z|), \quad \mbox{for $z\in(1, +\infty)$}\tag2 $$
According to Mathematica/Wolfram-Alpha website, $\Im[Li_2(r)] = -\pi\ln(r)$, where $r\in(1, +\infty)$. You can see that here, take a look to the plot.
Hence, it seems I'm right but is it really good to allow $\arg$ to be equal to $\pi$ even if this breaks the analyticity of the complex logarithm? The definition of $\arg$ is independent of the logarithm, but when you use it within the logarithm, it make me wonder if it's a good thing to do or maybe we should never take into account the region $z\in(1, +\infty)$. To what extend is analyticity something to worry about?