# Calculating $\lim_{x\to 1} \, \int -\frac{i \text{Li}_2\left(x-x^2\right)}{\sqrt{3} \left(x-\frac{1}{2}-\frac{i \sqrt{3}}{2}\right)} \, dx$

How can we force Mathematica to compute this limit? $$\lim_{x\to 1} \, \int -\frac{i \text{Li}_2\left(x-x^2\right)}{\sqrt{3} \left(x-\frac{1}{2}-\frac{i \sqrt{3}}{2}\right)} \, dx$$

It seems it works when trying $x\to 0$, but not in the previous case. What possibilities might I have?

This is the command I used

Limit[
Integrate[-((I PolyLog[2, x - x^2])/(Sqrt[3] (-(1/2) - (I Sqrt[3])/2 + x))), x],
x -> 1]

• Does this question make sense mathematically? Since you have an indefinite integral, there's the arbitrary constant to consider. $\lim_{x\to1}\int_1^x f(t)dt = 0$, but $\lim_{x\to1}\int_0^x f(t)dt$ might not. Commented Nov 11, 2016 at 20:41
• @ChipHurst I think you didn't see the picture behind because it's about calculating $\int_0^1$. Commented Nov 11, 2016 at 21:53
• Ah I see, thanks. Commented Nov 11, 2016 at 23:22
• Actually I also (in line with @ChipHurst) do not see what do you mean by taking the limit of indefinite integral. It is actually an arbitrary number. Commented Apr 18, 2018 at 21:08

Adding the assumption that x is real (which it is in this case) and then simplifying allows Mathematica to compute a symbolic answer:

\$Assumptions = Element[x, Reals];
Integrate[-((I PolyLog[2, x - x^2])/(Sqrt[3] (-(1/2) - (I Sqrt[3])/2 + x))), x];
Simplify @ %;
Limit[%, x -> 1]


On 10.1 these commands yield the following (after some computation time):

1/108 ((8 π^3+9 π Log[(I-Sqrt[3])/(I+Sqrt[3])]^2-72 π Log[(I-Sqrt[3])/(I+Sqrt[3])] Log[(I+Sqrt[3])/(-I+Sqrt[3])]+36 π Log[(I+Sqrt[3])/(-I+Sqrt[3])]^2-36 π PolyLog[2,(-1)^(1/3)]-36 π PolyLog[2,-(-1)^(2/3)]-36 π PolyLog[2,1-(-1)^(1/3)]-72 π PolyLog[2,1+(-1)^(2/3)]+36 π PolyLog[2,(I+Sqrt[3])/(-I+Sqrt[3])])/Sqrt[3]+(1/Sqrt[3])I (-54 Log[1+I Sqrt[3]] Log[(I-Sqrt[3])/(I+Sqrt[3])]^2+54 Log[-I+Sqrt[3]] Log[(I-Sqrt[3])/(I+Sqrt[3])]^2+12 π^2 Log[I+Sqrt[3]]-24 π^2 Log[(I+Sqrt[3])/(-I+Sqrt[3])]+108 Log[I+Sqrt[3]] PolyLog[2,(-1)^(1/3)]+108 Log[I+Sqrt[3]] PolyLog[2,-(-1)^(2/3)]+108 Log[(I-Sqrt[3])/(I+Sqrt[3])] PolyLog[2,1-(-1)^(1/3)]-108 Log[(I-Sqrt[3])/(I+Sqrt[3])] PolyLog[2,(1-I Sqrt[3])/(2 (1+(-1)^(2/3)))]+108 Log[(I-Sqrt[3])/(I+Sqrt[3])] PolyLog[2,(I-Sqrt[3])/(I+Sqrt[3])]-108 Log[(I-Sqrt[3])/(I+Sqrt[3])] PolyLog[2,(I+Sqrt[3])/(-I+Sqrt[3])]+108 PolyLog[3,1-(-1)^(1/3)]+216 PolyLog[3,1+(-1)^(2/3)]+216 PolyLog[3,(1-I Sqrt[3])/(2 (1+(-1)^(2/3)))]-108 PolyLog[3,(I-Sqrt[3])/(I+Sqrt[3])]+108 PolyLog[3,(I+Sqrt[3])/(-I+Sqrt[3])]+324 Zeta[3]))


With FullSimplify this can be reduced to:

(π (26 π^2+36 I π Log[2]+(-5+4 I Sqrt[3]) PolyGamma[1,1/6]+(5-12 I Sqrt[3]) PolyGamma[1,1/3]+(5+12 I Sqrt[3]) PolyGamma[1,2/3]+(-5-4 I Sqrt[3]) PolyGamma[1,5/6])+840 I Zeta[3])/(216 Sqrt[3])


Both results have a numerical value of 1.04978 + 4.17538 I

This value can be independently confirmed with NLimit and without using any assumptions:

Needs["NumericalCalculus"]
Integrate[-((I PolyLog[2, x - x^2])/(Sqrt[3] (-(1/2) - (I Sqrt[3])/2 + x))), x];
NLimit[%, x -> 1]

1.04978 + 4.17538 I


Note that with NLimit, setting the Direction option to I or -I yields a different limit, the same one that bbgodfrey found:

Needs["NumericalCalculus"]
Integrate[-((I PolyLog[2, x - x^2])/(Sqrt[3] (-(1/2) - (I Sqrt[3])/2 + x))), x];
NLimit[%, x -> 1, Direction -> I]

3.03883 + 1.54225 I

• Many thanks (+1) Commented Jun 29, 2015 at 16:16
• And you are Ok with the fact that adding a constant to the integral is still the same indefinite integral? Commented Apr 18, 2018 at 21:10

The idea is to let x = 1 + eps, expand the integral (antiderivative) into a series with respect to eps, and then let eps -> 0.

Mathematica 10.1 quickly returns the result to which some "beautifying" was done afterwards by hand.

Here's the code

FullSimplify[
Limit[Series[
Integrate[-((I PolyLog[2, x - x^2])/(Sqrt[
3] (-(1/2) - (I Sqrt[3])/2 + x))), x] /.
x -> 1 + \[Epsilon], {\[Epsilon], 0, 0},
Assumptions -> \[Epsilon] > 0] // Normal //
Simplify, \[Epsilon] -> 0]]

(*
(\[Pi] (26 \[Pi]^2 +
36 I \[Pi] Log[2] + (-5 + 4 I Sqrt[3]) PolyGamma[1, 1/
6] + (5 - 12 I Sqrt[3]) PolyGamma[1, 1/
3] + (5 + 12 I Sqrt[3]) PolyGamma[1, 2/
3] + (-5 - 4 I Sqrt[3]) PolyGamma[1, 5/6]) +
840 I Zeta[3])/(216 Sqrt[3])
*)

%//N

(* 1.04978 + 4.17538 I *)


This conincides with Nick's solution.

• Many thanks (+1) Commented Jun 29, 2015 at 16:16

Edit: material reordered for clarity

The formal definition of the PolyLogorithm is Sum[z^k/k^n, {n, 1, Infinity}], which converges for Abs[z] < 1. Thus, the integrand can be integrated term-wise.

summand = Integrate[-I (x (1 - x))^k/(Sqrt[3] (-(1/2) - (I Sqrt[3])/2 + x)), x]
// FullSimplify
(* ((-1)^(1/6) x^(1 + k) AppellF1[1 + k, -k, 1, 2 + k, x, -(-1)^(2/3) x])/
(Sqrt[3] (1 + k)) *)


The value of summand at x = 1 is

summand /. x -> 1
(* ((-1)^(1/6) 2^(-1 - 2 k) Sqrt[\[Pi]/3] Gamma[2 + k]
Hypergeometric2F1[1, 1 + k, 2 + 2 k, -(-1)^(2/3)])/((1 + k) Gamma[3/2 + k]) *)


Of course, summand /. x -> 0 is 0. The first ten terms of the series are adequate to give a good answer.

Sum[(summand /. x -> 1) k^-2, {k, 1, 10}] // N // Chop
(* 0.110503 *)


This same result can be obtained by numerically integrating the integrand from x = 0 to x = 1.

NIntegrate[-((I PolyLog[2, x - x^2])/(Sqrt[3] (-(1/2) - (I Sqrt[3])/2 + x))), {x, 0, 1}]
// Chop
(* 0.110503 *)


For completeness, note that the integrand is well behaved throughout the range of integration.

Plot[Evaluate[
ReIm[-((I PolyLog[2, x - x^2])/(Sqrt[3] (-(1/2) - (I Sqrt[3])/2 + x)))]], {x, 0, 1}]


(The Real and Imaginary parts are blue and orange, respectively.)

However, it is important to remember that the anti-derivative of any function is determined only up to an arbitrary constant. Thus,

int = Integrate[-((I PolyLog[2, x - x^2])/(Sqrt[3] (-(1/2) - (I Sqrt[3])/2 + x))), x];
int /. x -> .9999
(* 3.03883 + 1.54225 I*)


gives a different answer, because Integrate has thrown in a different constant in this case. Varying x shows that this numerical solution has converged.

(int /. x -> .99999) - (int /. x -> .9999)
(* 2.47505*10^-9 - 1.42876*10^-9 I *)


Which result is preferable depends on how the OP plans to use it.

• Many thanks (+1) Commented Jun 29, 2015 at 16:16
• @Chris'ssistheartist Please note the cautionary statement at the end of my Answer. I recommend that you add to your question how you plan to use the result, which may impact how it should be derived. Commented Jun 29, 2015 at 16:42