4 Rewrote equation to avoid column overflow

I was playing around with dilogarithms and numerically found the following dilogarithm identity:

$$\text{Li}_2\left(\frac{2 m}{m^2+m-\sqrt{((m-3) m+1) \left(m^2+m+1\right)}-1}\right)-\text{Li}_2\left(\frac{m^2+m-\sqrt{((m-3) left(m^2+m+1\right)}-1}\right)$$ $$-\text{Li}_2\left(\frac{m^2+m-\sqrt{((m-3) m+1) \left(m^2+m+1\right)}-1}{2 m^2}\right)$$ $$+\text{Li}_2\left(\frac{2 m^2}{(m-1) m+\sqrt{((m-3) m+1) \left(m^2+m+1\right)}+1}\right)-\text{Li}_2\left(\frac{1}{2} left(m^2+m+1\right)}+1}\right)$$ $$-\text{Li}_2\left(\frac{1}{2} \left((m-1) m+\sqrt{((m-3) m+1) \left(m^2+m+1\right)}+1\right)\right)$$ left(m^2+m+1\right)}+1\right)\right)+\log (m) \left(-\log \$-\log(m)\log \left(-m^2-\sqrt{((m-3) m+1) \left(m^2+m+1\right)}+m-1\right)+2 \log 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)+i \pi left(m^2+m+1\right)}+m-1}{m^{3/2}}\right)+\log(m)\left(\log(m)+i\pi -\log (2)\right)=0$$\log(2)\right)=0$$

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?

3 breaking long eqn into 3 lines

I was playing around with dilogarithms and numerically found the following dilogarithm identity:

$$\text{Li}_2\left(\frac{2 m}{m^2+m-\sqrt{((m-3) m+1) \left(m^2+m+1\right)}-1}\right)-\text{Li}_2\left(\frac{m^2+m-\sqrt{((m-3) m+1) \left(m^2+m+1\right)}-1}{2 m^2}\right) \ +\text{Li}_2\left(\frac{2 m^2}\right)$$ $$+\text{Li}_2\left(\frac{2 m^2}{(m-1) m+\sqrt{((m-3) m+1) \left(m^2+m+1\right)}+1}\right)-\text{Li}_2\left(\frac{1}{2} \left((m-1) m+\sqrt{((m-3) m+1) \left(m^2+m+1\right)}+1\right)\right)+\log left(m^2+m+1\right)}+1\right)\right)$$ $$+\log (m) \left(-\log \left(-m^2-\sqrt{((m-3) m+1) \left(m^2+m+1\right)}+m-1\right)+2 \log \left(\frac{m^2-\sqrt{((m-3) m+1) \left(m^2+m+1\right)}+m-1}{m^{3/2}}\right)+\log (m)+i \pi -\log (2)\right)=0$$

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?

2 edited title

1