I would like to show that,for $m \geq 2$,
$$I_m := \int_0^1 \frac{\operatorname{Li}_{1 – 2m}(1 – 1/x)}{x} dx = 0$$ where $\operatorname{Li}_{1 – 2m}$ is the $1-2m$ polylogarithm (https://en.wikipedia.org/wiki/Polylogarithm#Particular_values). According to wolfram-alpha $I_m = 0$ until $m = 7$, but I suspect that $I_m = 0$ for all $m \geq 2$.
Do you have any suggestion ? I am not used to polylogarithm, and I don't know how to deal with is integral.
Context : the integral comes from the sum $\sum_{n = 1}^m S(m, n) (n-1)! (-1)^{n} \frac1{n}$ wich is equal to $I_m$ according to the equation (17) of https://mathworld.wolfram.com/StirlingNumberoftheSecondKind.html ($S(m, n)$ are the stirling number of second kind)
Edit : I tried to use the idea of ChewyPro's answer, but I wasn't able to make them rigorous. It seems to have a link between this function and the $\zeta$ function evaluated at $-2m$. Does someone have an idea to prove it ?
Best Answer
$\operatorname{Li}_{-n}$ for positive integer $n$ has the following identities: \begin{align} \operatorname{Li}_{-n}(x) &= (-1)^{n+1}\operatorname{Li}_{-n}\left(\frac{1}{x}\right) \label{inv} \tag{1} \\ \operatorname{Li}_{-n}(x) &= \frac{1}{(1-x)^{n+1}}\sum_{k=0}^{n-1}\left\langle {n \atop k}\right\rangle x^{n-k} \label{expl} \tag{2} \\ \frac{\mathrm{d}}{\mathrm{d}x}\operatorname{Li}_{-n}(x) &= \frac{1}{x}\operatorname{Li}_{-n-1}(x), \label{diff} \tag{3} \end{align} where the second sum uses Eulerian numbers. Equation \eqref{expl} implies \begin{align} \operatorname{Li}_{-n}(0) = 0 \text{ and } \lim\limits_{x \to \pm \infty} \operatorname{Li}_{-n}(x) = 0. \tag{4} \end{align}
With help of \eqref{diff} we obtain \begin{align} \frac{\mathrm{d}}{\mathrm{d}x}\operatorname{Li}_{-n}\left(1 - \frac{1}{x}\right) &\overset{(3)}{=} \frac{1}{x(x - 1)} \operatorname{Li}_{-n-1}\left(1 - \frac{1}{x}\right). \label{diff_s} \tag{5} \end{align}
Then your integral is equal to \begin{align} \mathfrak{I}_{m} &= \int\limits_{0}^{1} \frac{1}{x}\operatorname{Li}_{1-2m}\left(-\frac{1-x}{x}\right)\,\mathrm{d}x \tag{6} \\ &= \int\limits_{0}^{1} \frac{1}{1-x}\operatorname{Li}_{1-2m}\left(-\frac{x}{1-x}\right)\,\mathrm{d}x \\ &\overset{(1)}{=} \int\limits_{0}^{1} \frac{1}{1-x}\operatorname{Li}_{1-2m}\left(-\frac{1-x}{x}\right)\,\mathrm{d}x \tag{7} \\ &= \frac{(6) + (7)}{2} \\ &= \frac{1}{2}\int\limits_{0}^{1} \frac{1}{x(1-x)}\operatorname{Li}_{1-2m}\left(1-\frac{1}{x}\right)\,\mathrm{d}x \\ &\overset{(5)}{=} -\frac{1}{2}\left.\operatorname{Li}_{2-2m}\left(1-\frac{1}{x}\right)\right|_{0}^{1} \\ &\overset{(4)}{=} 0. \end{align}