Evaluate $ \int_{0}^{1}\frac{\operatorname{li}(x)\ln^s(x)\ln(-\ln(x))}{x}dx$ $$G(s)=\large \int_{0}^{1}\frac{\operatorname{li}(x)\ln^s(x)\ln(-\ln(x))}{x}\mathrm dx$$
Where $\operatorname{li}(x)$;Logarithmic Integral
$$\large \operatorname{li(x)}=\gamma+\ln(\ln x)+\sum_{k=1}^{\infty}\frac{(\ln x)^k}{k!k}$$
$$G(s)=\large \gamma\int_{0}^{1}\frac{\ln^s(x)\ln(-\ln(x))}{x}\mathrm dx+\int_{0}^{1}\frac{\ln^s(x)\ln(\ln x)\ln(-\ln(x))}{x}\mathrm dx+\sum_{k=1}^{\infty}\frac{1}{k!k}\int_{0}^{1}\frac{(\ln(x))^{s+k}\ln(-\ln(x))}{x}\mathrm dx$$
Trying to work this one out first
$\large u=-\ln x$
$$\large \gamma\int_{0}^{1}\frac{\ln^s(x)\ln(-\ln(x))}{x}\mathrm dx=\gamma\int_{0}^{\infty}e^{-us}\ln u \mathrm du=-\frac{\gamma}{s}\left(\gamma+\ln s\right)$$
$$\int_{0}^{1}\frac{\ln^s(x)\ln(\ln x)\ln(-\ln(x))}{x}\mathrm dx=\int_{0}^{\infty}\ln u\ln(-u)e^{-us}\mathrm du$$
$$\sum_{k=1}^{\infty}\frac{1}{k!k}\int_{0}^{1}\frac{(\ln(x))^{s+k}\ln(-\ln(x))}{x}\mathrm dx=\sum_{k=1}^{\infty}\frac{1}{k!k}\int_{0}^{\infty}(-u)^{s+k}\ln u\mathrm du$$
Problem, the last two integrals does not converge.
How would we evaluate $G(s), s\ge0$?
 A: I will assume $s \in \mathbb{N}_0$ to avoid complex numbers. We can let $x = \mathrm{e}^{-t}$ and use $\operatorname{li}(\mathrm{e}^{-t}) = \operatorname{Ei}(-t)$, where $\operatorname{Ei}$ is the exponential integral defined by
$$ \operatorname{Ei}(z) = \int \limits_{-\infty}^z \frac{\mathrm{e}^{y}}{y} \, \mathrm{d} y \, , \, z \in \mathbb{R} \, .$$
The principal value is to be used for $z>0$ . Then we can integrate by parts (using $\frac{\mathrm{d}}{\mathrm{d} t} \frac{t^{s+1} [(s+1)\ln(t) - 1]}{(s+1)^2} = t^s \ln(t)$)
\begin{align}
 G(s) &= \int \limits_0^\infty (-1)^s \operatorname{Ei}(-t) t^s \ln(t) \, \mathrm{d} t = \frac{(-1)^s}{(s+1)^2} \int \limits_0^\infty \left[t^s \mathrm{e}^{-t} - (s+1) t^s \ln(t) \mathrm{e}^{-t}\right] \, \mathrm{d} t \\
&= \frac{(-1)^s}{(s+1)^2} [\Gamma(s+1) - (s+1) \Gamma'(s+1)] \, .
\end{align}
Using $\Gamma(s+1) = s!$ and $\Gamma'(s+1) = s! (H_s - \gamma)$ with the harmonic number $H_s$ now yields
$$G(s) = \frac{(-1)^s s!}{(s+1)^2} [ 1 + (s+1)(\gamma-H_s)] \, . $$
