Well playing with integral and WA I have found something (maybe?) solid .
We begin with two examples and no proof is mentioned here and because I strongly believe it's not new I add the tag "reference-request".
First example :
Well we work with integral :
Let $n\geq 1$ a natural number then define : $$\int_{}^{}x^n\ln(e^x-1)dx$$
Put $n=5$ we get :
$$\int_{}^{}z^5\ln(e^z-1)dz= -x^5 \operatorname{Li_2}(e^x) + 5 x^4 \operatorname{Li_3}(e^x) - 20 x^3 \operatorname{Li_4}(e^x) + 60 x^2 \operatorname{Li_5}(e^x) - 120 x \operatorname{Li_6}(e^x) + 120 \operatorname{Li_7}(e^x) - \frac{1}{6} x^6 \log(1 - e^x) + \frac{1}{6} x^6 \log(e^x - 1) + \operatorname{constant} $$
In this example we see that if we multiply a coefficient by the exponent we get the next coefficient .Call this the "factorial law"
Second example :
Let $n\geq 1$ a natural number then define : $$\int_{}^{} y^n \operatorname{W(y)}dy $$
$$\int_{}^{} y^5 \operatorname{W(y)} dy = \frac{(x^6 (1944 \operatorname{W(x)}^7 - 324 \operatorname{W(x)}^6 + 324 \operatorname{W(x)}^5 - 270 \operatorname{W(x)}^4 + 180 \operatorname{W(x)}^3 - 90 \operatorname{W(x)}^2 + 30 \operatorname{W(x)} - 5))}{(11664 \operatorname{W(x)}^6)} + \operatorname{constant}$$
Here we have $x^6$ to find the "law" multiply the first coefficient (here $-5$) by $6$ and divide by the exponent of Lambert's function to get the next coefficient .
Well now my question :
As in my first example can we find all the function $f(x)$ continuous and integrable such that it follows the "factorial law" describe above ?
Be gentle because I think it's really hard .
Thanks in advance !!