Spence function

[Albert Chan]

(01-12-2021 05:28 PM)C.Ret Wrote:  How are mathematicians dealing with the computation of, for example :

(01-12-2021 02:17 PM)Albert Chan Wrote:  Li2(-1) = -pi^2/12
Li2(2) = (pi^2/6 - Li2(-1)) - ln(2)*pi*i = pi^2/4 - ln(2)*pi*i

I was playing around, and tried integrating using Li2(2) as base.
0 to 2+3i → 0 to 2, then 2 to 2+3i

Turns out, we have to use the conjugate of the base.
Anyone knows how and when we apply the conjugate ?

XCas> b := pi^2/4 - ln(2)*pi*i                                   // Li2(2)
XCas> float(b)
2.46740110027 - 2.1775860903*i

XCas> conj(b) - i*int(ln(-1-x*i)/(2+x*i), x = 0. .. 3)    // let u = 2+x*i, du = i*dx
-0.280988055379 + 3.01725120637*i

XCas> -int(ln(1-(2+3i)*x)/x, x = 0. .. 1)                     // confirmed Li2(2+3i)
-0.280988055378 + 3.01725120637*i

FYI, picking a closer base speed up integration. (about 5X for above example)
The problem is how to make it right ...