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CAF123
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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!
##\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!