The dilogarithm function. Question on an identity of it Upon reading a journal article about manipulating series using the dilogarithm function, I have a few questions. But before I ask them, let me give the information the article provides.
Consider the series $\displaystyle\sum \frac{1}{k^2} = 1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16} + ...$
The related power series is $f(z) = \displaystyle\sum \frac{1}{k^2}z^k = z + \frac{z^2}{4} + \frac{z^3}{9} + ...$
The derivative of this is $f'(z) = \displaystyle\sum \frac{1}{k}z^{k-1} = 1 + \frac{z}{2} + \frac{z^2}{3} + ... = -\frac{\ln(1-z)}{z}$
The series we have provided is just the derivative of the dilogarithm function, and so
$Li'_{2}(z) = -\displaystyle\frac{\ln(1-z)}{z}$.
Now there is an identity that says $Li_{2}( -\displaystyle\frac{1}{z}) + Li_{2}(-z) + \displaystyle\frac{1}{2}(\ln(z))^2 = C$ where $C$ is a constant. 
From here, taking $z = 1$ gives us $C = 2Li_{2}(-1) = 2(-1 + \displaystyle\frac{1}{4} - \displaystyle\frac{1}{9} + \displaystyle\frac{1}{16} - ...)$.
This can be related to $Li_{2}(1)$ by using the fact that the even terms of $\displaystyle\sum \frac{1}{k^2} = \sum \frac{1}{(2k)^2} = \frac{1}{4}Li_{2}(1)$. And so the odd terms must be $\displaystyle\frac{3}{4}Li_{2}(1)$ and so the alternating sum I gave earlier is the difference of the even terms and the odd terms which is $-\displaystyle\frac{1}{2}Li_{2}(1)$. This shows that $C = -Li_{2}(1)$.
Here are my questions now.
Where on earth does $Li_{2}( -\displaystyle\frac{1}{z}) + Li_{2}(-z) + \displaystyle\frac{1}{2}(\ln(z))^2 = C$ come from? I do not see that at all. 
I don't see how $-(\displaystyle\frac{1}{2})Li_{2}(1)$ is the alternating series. Theres no way to generate $-1$ from it which is the first term of the alternating series.

Extending the hint in the comment section, one has
$$\frac{d}{dx} Li_2\left(-\frac{1}{x}\right)=\frac{\ln\left(1+\frac{1}{x}\right)}{x}=\frac{\ln(1+x)-\ln(x)}{x},$$
and
$$\frac{d}{dx} Li_2\left(-x\right)= -\frac{\ln\left(1+x\right)}{x}. $$
This implies
$$\frac{d}{dx} Li_2\left(-\frac{1}{x}\right)+ \frac{d}{dx} Li_2\left(-x\right) = 
-\frac{\ln(x)}{x}. $$
Integrating the above expression one arrives at
$$ Li_2\left(-\frac{1}{x}\right)+  Li_2\left(-x\right) = 
-\frac{1}{2}\ln^2 x + K.  $$
To determine the constant of integration $K$ we select $x=1$ in the above equality: 
$$2Li_2(-1)= 0+K= K $$
In summary,
$$Li_2\left(-\frac{1}{x}\right)+Li_2(-x)=2 Li_2(-1)-\frac{1}{2}\ln^2(x), $$
as claimed.
