chisigma said:
Some years ago I arrived to the following Taylor expansion of the function $z\ \ln z$ around z=1...
$\displaystyle z\ \ln z = (z-1) + \sum_{n=2}^{\infty} (-1)^{n}\ \frac{(z-1)^{n}}{n\ (n-1)},\ |z - 1| \le 1\ (1)$
Kind regards
$\chi$ $\sigma$
Very nice indeed, Chisigma! Thanks for sharing... (Yes)It strikes me that your series above could be used to find a series acceleration formula for the Dilogarithm. For example, replacing the $$z$$ in your series with the dummy variable $$x$$, dividing both sides of the expansion by $$(1-x)$$, and then integrating from $$1$$ to $$z$$ gives$$\int_1^z \frac{x\log x}{1-x}\,dx= - \int_1^z \,dx + \sum_{k=2}^{\infty} \frac{ (-1)^{k-1} }{k(k-1)} \, \int_1^z (x-1)^{k-1} \,dx=$$$$1-z + \sum_{k=2}^{\infty} \frac{1}{k(k-1)} \, \int_1^z (1-x)^{k-1} \,dx$$Substitute $$y=1-x$$ in that last integral to get:$$ 1-z - \sum_{k=2}^{\infty} \frac{1}{k(k-1)} \, \int_0^{1-z} y^{k-1} \,dy=$$$$(01) \quad 1-z - \sum_{k=2}^{\infty} \frac{(1-z)^k}{k^2(k-1)} $$On the other hand,$$\int_1^z \frac{x\log x}{1-x}\,dx= \int_1^z \frac{ [1-(1-x)] }{1-x} \log x \,dx=$$$$\int_1^z\frac{ \log x}{(1-x)}\,dx - \int_1^z \log x \,dx=$$$$-\log x\log(1-x)\, \Bigg|_1^z+\int_1^z\frac{\log(1-x)}{x}\, dx - \Bigg[ z\log z - z +1 \Bigg]=$$$$-\log z \log(1-z) - z\log z +z - 1 + \Bigg[ \int_1^z\frac{\log(1-x)}{x}\, dx \Bigg]=$$$$-\log z \log(1-z) - z\log z +z - 1 + \Bigg[ \text{Li}_2(1) - \text{Li}_2(z) \Bigg]=$$$$ (02) \quad -\log z \log(1-z) - z\log z +z - 1 + \Bigg[ \zeta(2) - \text{Li}_2(z) \Bigg] $$
Finally, equating
(01) and
(02) then gives the final result:
$$\text{Li}_2(z)= -\log z \log(1-z) - z\log z +2z - 2 + \zeta(2) + \sum_{k=2}^{\infty} \frac{(1-z)^k}{k^2(k-1)}$$
A few observations:Firstly, although faster series acceleration formulae exist for the Dilogarithm - search for
BBP-type series online - and some modest restrictions would need to be put on $$z$$, this series is self-evidently a good deal faster than the standard Dilogarithmic series:$$\text{Li}_2(z) = \sum_{k=1}^{\infty} \frac{z^k}{k^2}$$Secondly, in many years of exploring Polylogarithms, and reading many related papers, I'm pretty certain I've not seen this series before.
And finally, repeating the same steps as above on this series will give an acceleration formula for the Trilogarithm. Similarly, repeated iterated integration in this way will lead to series acceleration formulae for additional higher order Polylogarithms.As I said,
Chisigma, very nice work indeed... :D
