Spence function

[Albert Chan]

(01-12-2021 05:28 PM)C.Ret Wrote:  

(01-11-2021 06:13 PM)Albert Chan Wrote:  I just learned how to proof this, using Spence's function (dilogarithm)



Break-up Integral to parts, and integrate each:

Wait a minute !

I just realize that I don't know how to integrate such an integral ; .

Just move complex z inside the function. Let \(u = z\; x\;,\;du = z\;dx\)

\(\displaystyle Li_2(z) = - \int _0^1 {\ln(1-z\;x) \over x} dx\)

This substitution of variable give neat result. Example, let \(u = x^k\;,\;du = k\;x^{k-1}\;dx\)

\(\displaystyle -k \int _0^1 {\ln(1-x^k) \over x} dx
= -k \int _0^1 {\ln(1-u) \over x} {du \over k\;x^{k-1}}
= - \int _0^1 {\ln(1-u) \over u} du
= Li_2(1) = {\pi^2 \over 6} \)

I thought above formula had something to do with Valentin's puzzle ... but no luck.