Spence function

[Albert Chan]

(01-14-2021 01:55 PM)Albert Chan Wrote:  >>> a, b, x = 1-j, 2+3j, 1.25
>>> conj(Li2(a) + F(a,x)) + F(x,b)                # == conj(Li2(x)) + F(x,b)
(-0.28098805537806 + 3.01725120636941j)

This is how WolframAlpha interpret F(a, b):

∫(-ln(1-x)/x, x = 1-I .. 2+3I)
= (Li2(2+3I) - conj(Li2(1.25))) + (Li2(1.25) - Li2(1-I))
= (Li2(2+3I) - Li2(1-I)) + (Li2(1.25) - conj(Li2(1.25)))
= (Li2(2+3I) - Li2(1-I)) + 2I * im(Li2(1.25))

Do not calculate F(a, b) directly !
Because of the discontinuity crossing the real axis, it it very hard to evaluate accurately.

Since we are interested with Li2(2+3I), interpolation of integral limits is not necessary.
Any real x will do. Preferably Li2(x) is also real, so no need to worry about conjugates.

>>> pi*pi/6 + F(1, 2+3j)       # x = 1
(-0.280988055378061 + 3.01725120636941j)

This simple shift doubled the speed of integration, compared against F(0, 2+3j)