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I have obtained the Dilog function in some output and was just wondering if anyone knew how to interpret its meaning. The mathematica notes tell me that

$\text{DiLog}(x,\alpha)$ gives the dilogarithm function, which for real $x>1$ evaluates on the side of the branch prescribed by $$\text{lim}_{\epsilon \rightarrow 0^+} \text{Li}_2(x+i\alpha\epsilon)$$

I'm just wondering how to proceed with this prescription - should I first check whether my $\alpha$ is positive or negative in which case the expression becomes $$\text{lim}_{\epsilon \rightarrow 0^+} \text{Li}_2(x\pm i\epsilon),$$ with $\pm$ for positive/negative $\alpha$ respectively. I believe it's ok to absorb $\alpha$ into $\epsilon$ here as the latter is infinitesimal, but please correct if I'm mistaken in any of my thoughts, I don't have much familiarity with polylogs. I believe I can then write $$\text{lim}_{\epsilon \rightarrow 0^+} \text{Li}_2(x \pm i\epsilon) = -\left(\text{Li}_2(x) \mp i\pi \ln x \right)$$ I've seen this last display in some literature but I have not yet understood it's derivation. Mathematica tells me (from the above) that it holds only for $x>1$ which might not be the case for me, so I'm not sure either how to deal with the case $x>0,x<1$.

Thanks for any comments!

J. M.'s missing motivation
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  • The dilogarithm in Mathematica is continuous from below (corresponding to the $-$ sign); to get something continuous from above, use the solutions here. – J. M.'s missing motivation Oct 25 '17 at 10:36
  • @J.M. Thanks for the comment. For $x<1$ is $\text{DiLog}(x,\alpha)$ just equal to $\text{PolyLog}(2,x)$ as there is no imaginary part for x<1? – CAF Oct 25 '17 at 10:45
  • Why not do a few experiments yourself to convince yourself further? With[{x = -3}, {Limit[PolyLog[2, x + I ε], ε -> 0, Direction -> -1], Limit[PolyLog[2, x - I ε], ε -> 0, Direction -> -1]}] – J. M.'s missing motivation Oct 25 '17 at 10:53
  • Thanks! So I concluded that for x<1, $\text{DiLog}(x,\alpha) = \text{PolyLog}(2,x)$ and that for x>1, I use the equation $$\text{lim}_{\epsilon \rightarrow 0^+} \text{Li}_2(x\pm i \epsilon) = -(\text{Li}_2(x) \mp i \pi \ln x), $$ assuming you agree with what I did with the alpha as described in my post was ok. – CAF Oct 25 '17 at 11:02

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