As you can see, here I performed a derivation of a quite simple formula, not much differing from the standard integral representation of the Polylogarithm. Seeking to make it fancier, I arrived at generalized Stirling numbers of the second kind and once again applied the result.
However, the resulting integral is still not simple. This kept me experimenting and in order to simplify the integral, I discovered the following.
Suppose $f(g(z))=g'(z)$. Then $$ (f(z)\frac{d}{dz})^n u(z) = \sum_{j=0}^{n} <n, j> u^{(j)}(z) $$ where $$ \forall j \in \mathbb{N} | \sum_{n=j}^{\infty}<n, j>\frac{x^n}{n!} = \frac{(g(x)-g(0))^j}{j!} $$ and $z$ is evaluated at $g(0)$. When $f(z)=z$, this yields the already known exponential generating series with Stirling numbers of the second kind, whereas taking $f=exp$ yields a sum with Stirling numbers of the first kind.
Replacing the Stirling numbers with $<n, j>$ in the original derivation will lead one to conclude that $$\frac{d^{s-1}}{dy^{s-1}}\sum_{n=0}^{\infty} y^{n+s-1} (f(z)\frac{d}{dz})^n Li_{s+n}(z) = z\int_{0}^{\infty}t^{s-1}\frac{1}{\frac{e^t}{g(t)}-1}dt, $$ where $z=g(0)$ and $y=1$. You will notice that in the original derivation I use $x$ in $g$, but the $x$ cancels out on both sides eventually. Also, the $z^j$ factor from the derivation is absent.
Taking $g(z)=\frac{e^z}{e^z+1}$ gives $$ \frac{d^{s-1}}{dy^{s-1}}\sum_{n=0}^{\infty} y^{n+s-1} ((z-z^2)\frac{d}{dz})^n Li_{s+n}(z) = \frac{\Gamma(s)}{2}.$$
Do you think this qualifies as advanced enough to be considered research? I am an undergraduate student of computing science. Do you have any ideas on what Riemann Zeta function identities could be derived from this?
