Spence function

[C.Ret]

(01-12-2021 07:49 PM)Albert Chan Wrote:  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.

Thank you , for this quick response.

You exactly respond to my question. No more worries now about complex values in integration's limits.

Effectively, integrating from \(0\) to \(z\) on variable \(du\) give the same integration as from \(0\) to \(1\) on variable \(dx\) where \(x=\frac{u}{z}\)

You make my day. I learn something today. (I also discover that Latex is enable on this forum. Two nice discoveries in one post ).

As usual, a lot of nice people to meet here.
Thank you again.