Integral in Denominator

Relevant wiki: Gamma Function

$\begin{aligned} S & = \sum_{k=0}^\infty \frac 1{\int_0^1 \ln^k x\ dx} & \small \color{#3D99F6} \text{Let }u = \ln x \implies du = \frac {dx}x \\ & = \sum_{k=0}^\infty \frac 1{\int_{-\infty}^0 u^ke^u\ du} & \small \color{#3D99F6} \text{Let }t= - u \implies dt =- du \\ & = \sum_{k=0}^\infty \frac 1{(-1)^k\int_0^\infty t^ke^{-t}\ dt} & \small \color{#3D99F6} \text{As gamma function }\Gamma (z) = \int_0^\infty t^{z-1}e^{-t}\ dt \\ & = \sum_{k=0}^\infty \frac 1{(-1)^k\Gamma(k+1)} & \small \color{#3D99F6} \text{Note that }\Gamma (n+1) = n! \\ & = \sum_{k=0}^\infty \frac {(-1)^k}{k!} \\ & = e^{-1} \approx \boxed{0.368} \end{aligned}$

Note that even if you are not aware of the gamma function, you could deduce that by integrating by parts repeatedly (letting the lower bound be epsilon as epsilon approach 0+) the integral is equal to $(-1)^nn!$ , and from there solve the problem.