how can I integrate: $$ \int \frac{\log(\log(x))}{\log(x)}dx$$ I did in the following way but it seems to my kind sketchy.
$\text{li}(x)$ the logarithmic integral,$\gamma$ the Euler-Mascheroni constant and $\text{F}\left|\begin{array}{l}a_{1}...a_{p}\\b_{1}...b_{q}\end{array} ;z\right|$ the hypergeometric function.
$\frac{d}{dx} \text{li}(x)\log(\log(x)) =\frac{\log(\log(x))}{\log(x)}+\frac{\text{li}(x)}{x \log(x)}$
$\text{li}(x)=\gamma+\log(\log(x))+\sum_{k=1}^{\infty} \frac{\log(x)^{k}}{k! k}$
$\int \frac{\log(\log(x))}{\log(x)}dx=\text{li}(x)\log(\log(x)) - \int \frac{\text{li}(x)}{x \log(x)}dx+c=$
$=\log(\log(x))\Big(\text{li}(x)-\gamma -\frac{1}{2}\log(\log(x)) \Big) -\int (\sum_{k=1}^{\infty} \frac{\log(x)^{k}}{k! k})\frac{dx}{x \log(x)}+c=$
$=\log(\log(x))\Big(\text{li}(x)-\gamma -\frac{1}{2}\log(\log(x)) \Big) -\log(x)\sum_{k=0}^{\infty} \frac{\log(x)^{k}}{k! (k+1)^{3}}+c.$
$ \ $
$$ \int \frac{\log(\log(x))}{\log(x)}dx=\log(\log(x))\Big(\text{li}(x)-\gamma -\frac{1}{2}\log(\log(x)) \Big) -\log(x)\text{F}\left|\begin{array}{l}1,1,1\\2,2,2\end{array} ;\log(x)\right|+c.$$
is it ok? any suggestions?
I would have started using $x=e^t$ to make $$I=\int \frac{\log(\log(x))}{\log(x)}\,dx=\int \frac{e^t \log (t)}{t}\,dt$$ Now, using the expansion of $e^t$ we have $$I=\sum_{n=0}^\infty \int\frac{t^{n-1} \log (t)}{n!}\,dt=\int \frac{\log (t)}{t}\,dt+\sum_{n=1}^\infty \int\frac{t^{n-1} \log (t)}{n!}\,dt$$
For the first integral $$\int \frac{\log (t)}{t}\,dt=\frac{\log ^2(t)}{2}$$ and for the other $$J_n=\int\frac{t^{n-1} \log (t)}{n!}\,dt=\frac{t^n (n \log (t)-1)}{n^2 n!}$$ $$\sum_{n=1}^\infty J_n=\log (t) (\text{Ei}(t)-\log (t)-\gamma )-t \, _3F_3(1,1,1;2,2,2;t)$$