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I obtain this expression in my calculations, and numerically I am sure that it is a real number.

exp=PolyLog[2, -(1/8) I (-I + Sqrt[15])] + PolyLog[2, 1/8 I (I + Sqrt[15])];

Question.

Is it possible to express this expression as an explicit real expression (not a numeric approximation)?

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    $\begingroup$ It can be expressed as infinite sum of rational numbers. Does it fulfill your requirements? I doubt it can be expressed as finite expression other than it is by polylogarithm. $\endgroup$ Commented Apr 14 at 16:32
  • $\begingroup$ What for? Isn't it Mathematica for Mathematica's sake? $\endgroup$
    – user64494
    Commented Apr 14 at 16:33
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    $\begingroup$ @azerbajdzan Thanks. If the argument of SUM will be a real function, it may help. $\endgroup$
    – MsMath
    Commented Apr 14 at 16:40
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    $\begingroup$ @user64494 Sorry, I did not understand your meaning. $\endgroup$
    – MsMath
    Commented Apr 14 at 16:41
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    $\begingroup$ 2 Re[PolyLog[2, I/8 (I-Sqrt[15])]] is a bit simpler; but I doubt you can find anything even simpler. $\endgroup$
    – Roman
    Commented Apr 14 at 17:22

1 Answer 1

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The expression can be expressed as infinite sum. All the sums are pretty much the same only in different form and all produce the same terms/summands.

Sum[((-1)^n 2^(1 - n) Cos[n ArcTan[Sqrt[15]]])/
   n^2, {n, 1, ∞}] == 
  Sum[((-1)^
    n Sum[((-4)^k n (-1 - k + n)!)/(k! (-2 k + n)!), {k, 0, n/2}])/(
   n^2 2^(2 n)), {n, 1, ∞}] == 
  Sum[((-1)^n ((1/2 - (I Sqrt[15])/2)^n + (1/2 + (I Sqrt[15])/2)^n))/(
   n^2 2^(2 n)), {n, 1, ∞}] == 
  Sum[((-1)^k 2 Binomial[n, 2 k] (-(1/8))^(n - 2 k) (Sqrt[15]/8)^(
    2 k))/n^2, {n, ∞}, {k, 0, n/2}] == 
  PolyLog[2, -(1/8) - (I Sqrt[15])/8] + 
   PolyLog[2, -(1/8) + (I Sqrt[15])/8] // FullSimplify

True

enter image description here

So it is infinite sum of rational numbers:

Table[((-1)^n 2^(1 - n) Cos[n ArcTan[Sqrt[15]]])/n^2, {n, 1, 10}] // RootReduce
N@Total@%

enter image description here

The only thing now remains is to have some Euler that can express this sum as finite sum of other known real constants ;-)

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2
  • $\begingroup$ Both PolyLog[2, -(1/8) I (-I + Sqrt[15])] + PolyLog[2, 1/8 I (I + Sqrt[15])] and Sum[((-1)^n ((1/2 - (I Sqrt[15])/2)^n + (1/2 + (I Sqrt[15])/2)^n))/( n^2 2^(2 n)), {n, 1, ∞}] == Sum[((-1)^k 2 Binomial[n, 2 k] (-(1/8))^(n - 2 k) (Sqrt[15]/8)^( 2 k))/n^2, {n, ∞}, {k, 0, n/2}] and so on are analytical expressions (see Wiki for the definitions). Why do you think the latter is better than the former? $\endgroup$
    – user64494
    Commented Apr 14 at 17:17
  • $\begingroup$ Er… How do you notice this? $\endgroup$
    – xzczd
    Commented Apr 15 at 2:57

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