Generalized formula for the polylogarithm

Question

Some time ago, I discovered the formula for repeated application of $z\frac{d}{dz}$ here. Recently, I thought about taking the function to which this would be applied to be the integral representation of the polylogarithm: $$ Li_{s}z = \frac{z}{(s-1)!} \int_{0}^{\infty}\frac{t^{s-1}}{e^t-z}dt.$$ Notice that $$(z\frac{d}{dz})^n Li_s z = Li_{s-n}z.$$ Take $s-n=p,$ multiply both sides by $x^n$, sum over all natural $n$, and apply $$ \sum_{n=j}^{\infty}{n \brace j}\frac{x^n}{n!}=\frac{(e^x-1)^j}{j!}. $$ In the end, it will yield $$ \frac{d^{p-1}}{dx^{p-1}}\frac{x^{p-1}}{1-x}Li_p z = \int_0^{\infty}\int_0 ^{\infty} (\frac{z}{e^t-z}+\frac{e^{tx}-z}{e^t-z})t^{p-1}e^{\frac{e^{tx} - 1}{e^t - z}zh-h}dtdh $$ for $|x|<1$.

I was unable to proceed further. Can one use a trick that would make this result yield an interesting outcome? Even though $x$ is embedded in the integrand, it is a rational function of $x$ on the left-hand side and independent of $Li$.

Notice that multiplying both sides by a power of $(1-x)$ and then taking $lim_{x\rightarrow 1^-}$ will result in an integral without an antiderivative, and that swapping the limits so that the limit of $x$ goes into the integral is disallowed for it does not coverge then.

Answer 1

This may not be a complete answer, but after interchanging the order of integration, I believe your double integral is equivalent to the Mellin transform

$$\int\limits_0^{\infty} \frac{e^{t x}+z-1}{e^t-z e^{t x}}\, t^{p-1} \, dt,\quad\Re\left(\frac{z \left(e^{t x}-1\right)}{z-e^t}\right)>-1\tag{1}.$$

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I haven't yet been able to make any further progress in a general sense, but for $p=1$ and $x=0$,

$$\int\limits_0^{\infty} \frac{e^{t x}+z-1}{e^t-z e^{t x}}\, t^{p-1} \, dt=\int_0^{\infty} \frac{z}{e^t-z} \, dt=-\log (1-z)=\text{Li}_1(z)\tag{2}.$$

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Also, with respect to your related question, setting $x=\frac{1}{2}$ and $z=\frac{1}{2}$ gives

$$\int\limits_0^{\infty} \frac{e^{t x}+z-1}{e^t-z e^{t x}}\, t^{p-1} \, dt=\int\limits_0^{\infty } e^{-\frac{t}{2}} t^{p-1} \, dt=2^p\, \Gamma(p),\quad\Re(p)>0\tag{3}.$$