Question

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?

Answer 1

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).

Answer 2

All functional (i.e., depending on a parameter) relations are known to be a consequence of the 5-term relation of the dilogarithmic function; see [D. Zagier, The dilogarithm function, in: Frontiers in Number Theory, Physics and Geometry II, P. Cartier, B. Julia, P. Moussa, P. Vanhove (eds.), Springer-Verlag, Berlin-Heidelberg-New York (2006), 3-65]. In your case, I would suggest to compare first the derivatives w.r.t. $m$ and then use the fact that your identity is true for a particular value of $m$ (for example, $m=0$).