I was playing around with dilogarithms and numerically found the following dilogarithm identity:
$$\text{Li}_2\left(\frac{2 m}{m^2+m-\sqrt{((m-3) m+1)
\left(m^2+m+1\right)}-1}\right)$$
$$-\text{Li}_2\left(\frac{m^2+m-\sqrt{((m-3) m+1)
\left(m^2+m+1\right)}-1}{2 m^2}\right)$$
$$+\text{Li}_2\left(\frac{2 m^2}{(m-1)
m+\sqrt{((m-3) m+1) \left(m^2+m+1\right)}+1}\right)$$
$$-\text{Li}_2\left(\frac{1}{2}
\left((m-1) m+\sqrt{((m-3) m+1) \left(m^2+m+1\right)}+1\right)\right)$$
$$-\log(m)\log \left(-m^2-\sqrt{((m-3) m+1) \left(m^2+m+1\right)}+m-1\right)$$
$$+2\log(m)\log
\left(\frac{m^2-\sqrt{((m-3) m+1) \left(m^2+m+1\right)}+m-1}{m^{3/2}}\right)$$
$$+\log(m)\left(\log(m)+i\pi -\log(2)\right)=0$$
where m is a real number in a neighborhood of 1 (such that the square root is real).
For those who use mathematica, please copy below
expr = Log[
m] (I [Pi] – Log[2] + Log[m] –
Log[-1 + m – m^2 – Sqrt[(1 + (-3 + m) m) (1 + m + m^2)]] +
2 Log[(-1 + m + m^2 – Sqrt[(1 + (-3 + m) m) (1 + m + m^2)])/m^(
3/2)]) +
PolyLog[2, (
2 m)/(-1 + m + m^2 – Sqrt[(1 + (-3 + m) m) (1 + m + m^2)])] –
PolyLog[2, (-1 + m + m^2 – Sqrt[(1 + (-3 + m) m) (1 + m + m^2)])/(
2 m^2)] +
PolyLog[2, (2 m^2)/(
1 + (-1 + m) m + Sqrt[(1 + (-3 + m) m) (1 + m + m^2)])] –
PolyLog[2,
1/2 (1 + (-1 + m) m + Sqrt[(1 + (-3 + m) m) (1 + m + m^2)])]
Does anybody have any idea how to prove this?
Best Answer
It looks to me like you don't even need the five-term identities. If I denote the arguments of your dilogarithms by $a_1$, $a_2$, $a_3$ and $a_4$ I find that $(1-a_1) a_4 = 1$ and $a_2 a_3 = -1 + a3$. Then I think using the simple formulas $$ \text{Li}_2(z)=-\text{Li}_2(1-z)-\log (1-z) \log (z)+\frac{\pi ^2}{6}, $$ and $$ \text{Li}_2(z)=-\text{Li}_2\left(\frac{1}{z}\right)-\frac{1}{2} \log ^2(z)-\frac{\pi ^2}{6} $$ should be enough (but some care is needed when placing the branch cuts).