$$\mathrm{Li}_2\left(\frac{1}{2}\right)= \frac{\pi^2}{12}-\frac{1}{2}\log^2 \left(\frac{1}{2}\right) $$

$$ proof $$

Using the duplication formula proved here

$$\mathrm{Li}_2\left(z\right)+\mathrm{Li}_2(1-z)\, = \frac{\pi^2}{6}-\log(z) \log(1-z) \,\, $$

We can easily deduce that for \( z=\frac{1}{2}\)

$$2\mathrm{Li}_2\left(\frac{1}{2}\right)= \frac{\pi^2}{6}-\log^2\left(\frac{1}{2}\right) \,\,$$

It follows by dividing by 2.

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