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I think that I discovered another (c.f. Series expansion of expressions with Log and PolyLog functions) issue related to Series and PolyLog.

Consider the following expression

exp = (-16*Pi^2*Log[1 + Sqrt[x]])/3 + (16*Pi^2*Log[1 + Sqrt[x]])/
   x^4 + (16*Pi^2*Log[1 + Sqrt[x]])/(3*x^2) + (32*Pi^2*
     Log[1 + Sqrt[x]])/(3*x) + (32*Pi^2*x*Log[1 + Sqrt[x]])/
   3 - (16*Log[1 + Sqrt[x]]^3)/3 + (16*Log[1 + Sqrt[x]]^3)/
   x^4 + (16*Log[1 + Sqrt[x]]^3)/(3*x^2) + (32*Log[1 + Sqrt[x]]^3)/(3*
     x) + (32*x*Log[1 + Sqrt[x]]^3)/3 + (16*Pi^2*Log[1 - x])/
   3 - (16*Pi^2*Log[1 - x])/
   x^4 - (16*Pi^2*Log[1 - x])/(3*x^2) - (32*Pi^2*Log[1 - x])/(3*
     x) - (32*Pi^2*x*Log[1 - x])/3 + 
  16*Log[1 + Sqrt[x]]^2*
   Log[1 - x] - (48*Log[1 + Sqrt[x]]^2*Log[1 - x])/
   x^4 - (16*Log[1 + Sqrt[x]]^2*Log[1 - x])/
   x^2 - (32*Log[1 + Sqrt[x]]^2*Log[1 - x])/x - 
  32*x*Log[1 + Sqrt[x]]^2*Log[1 - x] - 
  16*Log[1 + Sqrt[x]]*
   Log[1 - x]^2 + (48*Log[1 + Sqrt[x]]*Log[1 - x]^2)/
   x^4 + (16*Log[1 + Sqrt[x]]*Log[1 - x]^2)/
   x^2 + (32*Log[1 + Sqrt[x]]*Log[1 - x]^2)/x + 
  32*x*Log[1 + Sqrt[x]]*Log[1 - x]^2 + (16*Log[1 - x]^3)/
   3 - (16*Log[1 - x]^3)/
   x^4 - (16*Log[1 - x]^3)/(3*x^2) - (32*Log[1 - x]^3)/(3*x) - (32*x*
     Log[1 - x]^3)/3 - 
  8*Log[1 + Sqrt[x]]^2*Log[x] + (24*Log[1 + Sqrt[x]]^2*Log[x])/
   x^4 + (8*Log[1 + Sqrt[x]]^2*Log[x])/
   x^2 + (16*Log[1 + Sqrt[x]]^2*Log[x])/x + 
  16*x*Log[1 + Sqrt[x]]^2*Log[x] + 
  16*Log[1 + Sqrt[x]]*Log[1 - x]*
   Log[x] - (48*Log[1 + Sqrt[x]]*Log[1 - x]*Log[x])/
   x^4 - (16*Log[1 + Sqrt[x]]*Log[1 - x]*Log[x])/
   x^2 - (32*Log[1 + Sqrt[x]]*Log[1 - x]*Log[x])/x - 
  32*x*Log[1 + Sqrt[x]]*Log[1 - x]*Log[x] - 
  8*Log[1 - x]^2*Log[x] + (24*Log[1 - x]^2*Log[x])/
   x^4 + (8*Log[1 - x]^2*Log[x])/x^2 + (16*Log[1 - x]^2*Log[x])/x + 
  16*x*Log[1 - x]^2*Log[x] - 
  32*PolyLog[3, 1 - Sqrt[x]] + (96*PolyLog[3, 1 - Sqrt[x]])/
   x^4 + (32*PolyLog[3, 1 - Sqrt[x]])/
   x^2 + (64*PolyLog[3, 1 - Sqrt[x]])/x + 
  64*x*PolyLog[3, 1 - Sqrt[x]] - 
  32*PolyLog[3, Sqrt[x]] + (96*PolyLog[3, Sqrt[x]])/
   x^4 + (32*PolyLog[3, Sqrt[x]])/x^2 + (64*PolyLog[3, Sqrt[x]])/x + 
  64*x*PolyLog[3, Sqrt[x]] - 
  32*PolyLog[
    3, -(Sqrt[x]/(1 - Sqrt[x]))] + (96*
     PolyLog[3, -(Sqrt[x]/(1 - Sqrt[x]))])/
   x^4 + (32*PolyLog[3, -(Sqrt[x]/(1 - Sqrt[x]))])/
   x^2 + (64*PolyLog[3, -(Sqrt[x]/(1 - Sqrt[x]))])/x + 
  64*x*PolyLog[3, -(Sqrt[x]/(1 - Sqrt[x]))] + 
  32*Zeta[3] - (96*Zeta[3])/x^4 - (32*Zeta[3])/x^2 - (64*Zeta[3])/x - 
  64*x*Zeta[3]

exp can be made much shorter by applying Simplify to it, but in my case it is a part of a much larger expression and applying Simplify to the whole expression makes little sense.

In fact, for x between 0 and 1 exp vanishes identically. Therefore, if I expand this expression in x around 0 the expected result is zero. With Mathematica 8 and Mathematica 9 I get the desired result so that

Series[exp, {x, 0, 0}] // Normal

yields

0

However, starting with Mathematica 10 (I have version 10.3 here) something strange is going on. For some reason

Series[exp, {x, 0, 0}] // Normal

returns

-3201281/105840 + 2*Pi^2 + (84*(-3 + Log[x]))/(5*Sqrt[x]) + (1089*Log[x])/70

but

Series[Simplify[exp], {x, 0, 0}] // Normal

still gives me the correct result. The same behavior is observed with Mathematica 11.0 and 11.2. Using Assumptions -> x > 0 or Assumptions -> 0 < x < 1 does not change anything.

To me this very much looks like a bug. In any case I believe that the result returned by Series should not depend on whether one have applied Simplify to the initial expression or not. However, as this behavior might also be caused by a misunderstanding from my side, I'd like to hear the opinion of other users from this site first.

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  • $\begingroup$ These: Series[exp[[48]], {x, 0, 0}], Series[exp[[53]], {x, 0, 0}], Series[exp[[58]], {x, 0, 0}] return incorrect results unless sufficiently high order is calculated. $\endgroup$ – QuantumDot Dec 20 '17 at 3:28
  • $\begingroup$ @DanielLichtblau Is this issue related to mathematica.stackexchange.com/questions/159777/…? $\endgroup$ – vsht Jan 29 '18 at 9:46
  • $\begingroup$ Hard to be certain but it could be the same, or similar, underlying issues. $\endgroup$ – Daniel Lichtblau Feb 14 '18 at 17:17
  • $\begingroup$ @DanielLichtblau Seems like this is now fixed in Mathematica 11.3. Thanks a lot! $\endgroup$ – vsht May 30 '18 at 6:37

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