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Wikipedia states that

$$\operatorname{Li}_2\left(z\right)+\operatorname{Li}_2\left(\frac{1}{z}\right)=-\frac{\pi ^2}{6}-\frac{1}{2}\left[\ln\left(-z\right)\right]^2 \\z\notin \left[0;1\right[$$

I tried to verify this identity by plotting it on matlab. Source code: syms x; m=polylog(2,x)+polylog(2,1/x)+pi^2/6+1/2*(log(-x)^2); ezplot(m, [0,10]); Main problem, and you can test it yourself, the plot is not as i expected it to be. Which, to me, can lead to three conclusions:

  1. I'm wrong. There's something i'm missing.
  2. Matlab's polylog is not precise enough (i've actually experienced the opposite in previous works on polylog so this should not be the case)
  3. The identity is not true.
Ignore the fact that matlab's plot includes z for which the identity has no meaning because the error is actually not localized in such interval. So if anyone could actually give me further information about this or just prove the identity it would be great. Thank you in advance.

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    $\begingroup$ What is your exact problem? Your $m$ is imaginary for $0\le x \le 1$. What do you see if you plot $\Re(m)?$ or $\Im(m)?$ Plotting with Maple gives noise of order working precision for $x>1.$ $\endgroup$ Sep 12 '18 at 13:02
  • $\begingroup$ Basically given that the above quoted equation is an identity, if i subtract the second term to the first one i should expect as result the plot of the line $y=0$. $\endgroup$ Sep 12 '18 at 13:18
  • $\begingroup$ Honestly i don't know how to share pics with you so at the moment i can't show you the output of $\Re(m)$ nor $\Im(m)$, sorry. $\endgroup$ Sep 12 '18 at 13:21
  • $\begingroup$ Yes, theoretically it should be zero but only if $x\not \in (0,1)$. No need to change pics. What are the values for $x=-0.5, 0.5, 1.5?$ $\endgroup$ Sep 12 '18 at 13:25
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    $\begingroup$ Well bro i actually solved the arcane: it's all correct, basically when i was looking at the plot i didn't notice the scale of it. The whole distorsion is due to an approximation of the functions: meaning that there is an error, probably introduced by matlab of around $10^{-15}$. $\endgroup$ Sep 12 '18 at 13:28
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The question is to prove $$\operatorname{Li}_2\left(x\right)+\operatorname{Li}_2\left(\frac{1}{x}\right)=-\frac{\pi ^2}{6}-\frac{1}{2}\left[\ln\left(-x\right)\right]^2 $$ and since $\ln(-x)=\ln x+i\pi$

then the question can be written

$$\operatorname{Li}_2\left(x\right)+\operatorname{Li}_2\left(\frac{1}{x}\right)=i\pi-\frac12\ln^2x+2\zeta(2)$$


Solution:

Since $$\frac{d}{dy}\operatorname{Li}_2(1/y)=\frac{\ln(1-1/y)}{y}=\frac{\ln(y-1)}{y}-\frac{\ln y}{y}$$

Then $$\operatorname{Li}_2(1/y)|_1^x=\underbrace{\int_1^x\frac{\ln(y-1)}{y}\ dy}_{IBP}-\int_1^x\frac{\ln y}{y}\ dy$$ \begin{align}\operatorname{Li}_2(1/x)-\zeta(2)&=\ln y\ln(y-1)|_1^x+\int_1^x\frac{ln y}{1-y}\ dy-\frac12\ln^2y|_1^x\\ &=\ln x\ln(x-1)+\operatorname{Li}_2(1-y)|_1^x-\frac12\ln^2x\\ &=\ln x\ln(x-1)+\operatorname{Li}_2(1-x)-\frac12\ln^2x\\ &=\ln x(\ln(1-x)+i\pi)+\zeta(2)-\ln x\ln(1-x)-\operatorname{Li}_2(x)-\frac12\ln^2x\\ &=i\pi\ln x-\operatorname{Li}_2(x)-\frac12\ln^2x+\zeta(2) \end{align}

Reagganging the terms gives

$$\operatorname{Li}_2(1/x)+\operatorname{Li}_2(x)=i\pi\ln x-\frac12\ln^2x+2\zeta(2)$$


Note that we used the reflection formula $\ \operatorname{Li}_2(1-x)=\zeta(2)-\ln x\ln(1-x)+\operatorname{Li}_2(x)$ ( see here).

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