Integrate all powers of log

By Integration by Parts: $\large F(a)=x \ln^a x|_{\small 0}^{\small 1} - a\int_0^1 \ln^{a-1} (x) \, dx$ $\large =-aF(a-1)~~\small{(\color{#3D99F6}{\because x \ln^a x|_{\small 0}^{\small 1}=0})}$ $\large =a(a-1)F(a-2)$ $\large =-a(a-1)(a-2)F(a-3)$ $\large \cdots\cdots$ $\large =(-1)^a a! F(0)=(-1)^a a! ~~\small{(\color{#3D99F6}{\because F(0)=1})}$ $\large \therefore \ln\left(\sum_{a=0}^\infty\dfrac{1}{F(a)}\right)=\ln(\dfrac{1}{e})=\huge \boxed{\color{#007fff}{-1}}$

$\mathcal{NOTE:}\\ \text{Consider expansion of }e^x :\\ \color{#D61F06}{e^x=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}\cdots}$ Putting x=-1, we get $\dfrac{1}{e}= \sum_{a=0}^\infty\dfrac{1}{(-1)^a a!}=\sum_{a=0}^\infty\dfrac{1}{F(a)}$

consider $\int_0^1 x^n \text{dx} = \dfrac{1}{n+1}$ d.w.r.n a time and putting n = 0 to get $\int_0^1 \ln^a(x) \text{dx}=(-1)^a a!$ now put this in the summation $\sum_{a=0}^\infty \dfrac{(-1)^a}{a!}=e^{\boxed{-1}}$ In the last step the taylor series of $e^x$ was used.