The following integral appears to be true.
Consider
$$S = \int_1^{\infty} \dfrac{50 \zeta(3) \operatorname{li}(x)^6 (x - 1)^2 }{\ln(27) x^{10}} dx = 1$$
where $\operatorname{li}(x)$ is the logarithmic integral.
Is this correct and if so how to prove it ?
In similar style :
Integral $y = \int_1^{\infty} \frac{\operatorname{li}(x)^2 (x - 1)}{x^4} dx$
where the comment also mentions
$$\int_0^{1}\frac {\operatorname{li}(x)^3 (x-1)}{x^3} = \frac{\zeta(3)}{4}$$
and this one :
Is there an integral for $\frac{1}{\zeta(3)} $?
what is relevant because if the conjecture is true then
$$ \int_1^{\infty} \dfrac{50 \operatorname{li}(x)^6 (x - 1)^2 }{\ln(27) x^{10}} dx = \frac{1}{\zeta(3)}$$
