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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?

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I am not an expert but as I understand it the only hope is to prove this using the five term relation. In "Special Functions" by Andrews, Askey, Roy this is (2.6.16). – Bruce Westbury Jan 18 '12 at 12:56
up vote 13 down vote accepted

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).

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Thank you for the suggestion. Indeed $a_3(1−a_2)=1$, or $a2+a_3^{−1}=1$ so we can eliminate two dilogs. However for $a_1$ and $a_4$ we have $ma_1^{−1}+a_4=m^2$, not just $a_1^{−1}+a_4=1$ and m seems to remain in this relation. Let me know if I am missing something. Note $a_4=1/2(1+(−1+m)m+\sqrt{(1+(−3+m)m)(1+m+m2)})$, not $a_4=1/(2m^2)(1+(−1+m)m+\sqrt{(1+(−3+m)m)(1+m+m2)})$. – Masahito Yamazaki Jun 7 '12 at 6:26
I copy & pasted the arguments from your Mathematica expression, so I think I got them right. I just rechecked the expressions $(1-a_1) a_4 = 1$ and $a_2 a_3 = -1+a_3$ and I think they are correct as well. Modulo a weird [Pi] (which I think should be Pi, but maybe I'm wrong) in you initial expression, I think it is also correct, except for the placement of branch cuts which are a bit subtle. The useful identity among $a_1$ and $a_4$ is $(1-a_1) a_4 = 1$, not $a_1^{-1} + a_4 = 1$, which is the one you tried. – Sidious Lord Jun 9 '12 at 8:21
Thank you, you are right! – Masahito Yamazaki Jun 19 '12 at 1:42

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, $m=0$).

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Thank you for pointing this out. I agree that this should come from five-term relations, but I thought Wojtkowiak’s theorem is the case with rational functions as arguments whereas the above relation involves square roots. Also, what I really wanted to do is to show the above from known relations (five-term relations etc), rather than taking a derivative. The square root above comes from extremizing w.r.t. a certain variable, and there seems to be generalizations to the case with more variables where we cannot use eliminate variables. Let me see if I can make the question sharper. – Masahito Yamazaki Jan 18 '12 at 17:22

The last comment of Sidious Lord, above, is correct: in fact, $a_4=1/(1-a_1)$ and $a_3=1/(1-a_2)$. Therefore, the application of Euler's reflection formula ($\mathrm{Li}_2(z) \rightarrow \mathrm{Li}_2(1-z)$), followed by an inversion ($\mathrm{Li}_2(z) \rightarrow \mathrm{Li}_2(1/z)$), as suggested by Sidious, actually reduces the expression to a few logarithms, which then simplify to zero.

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