The Closed-Form of $\displaystyle \int_{0}^{1}\mathrm{li}(x)\ln\Big(\ln\Big(\frac{1}{x}\Big)\Big) \mathrm{d}x$ $$\mathrm{Prove \;that } \int_{0}^{1}\mathrm{li}(x)\ln\Big(\ln\Big(\frac{1}{x}\Big)\Big) \mathrm{d}x\;\;=\;\; \frac{1}{2}\zeta(2)+\frac{1}{2}\ln^2(2)+\gamma \ln(2)$$
I found this problem on a group in facebook, I would like to share my solution with you and hope you like it.
 A: To solve this problem I will use a very nice result 
        $$\int_{0}^{1}\mathrm{li}(x)\sin\Big(u\ln\Big(\frac{1}{x}\Big)\Big) \mathrm{d}x \;\;=\;\; \frac{1}{u^2+1}\Bigg(\tan^{-1}\Big(\frac{u}{2}\Big)-\frac{u\ln(u^2+4)}{2}\Bigg)$$
        Multiply both sides by $\displaystyle \frac{\ln(u)}{u} $ and integrate both sides from $\displaystyle  u \;=\;0\;\; \mathrm{to} \;\;\infty$ , we get 
            $$\int_{0}^{1}\int_{0}^{\infty}\mathrm{li}(x)\sin\Big(u\ln\Big(\frac{1}{x}\Big)\Big)\frac{\ln(u)}{u}\mathrm{d}u \mathrm{d}t \;\;=\;\; \int_{0}^{\infty}\frac{\ln(u)}{u(u^2+1)}\Bigg(\tan^{-1}\Big(\frac{u}{2}\Big)-\frac{u\ln(u^2+4)}{2}\Bigg)\mathrm{d}u$$
and it's very easy to prove that $\displaystyle \int_{0}^{\infty}\frac{\sin(ax)\ln(u)}{u}\mathrm{d}u \;\;=\;\;- \frac{\pi}{2}\Big(\ln(a)+\gamma\Big)$, then we get that 
    $$\frac{\pi}{2}\;\;\int_{0}^{1}\mathrm{li}(x)\ln\Big(\ln\Big(\frac{1}{x}\Big)\Big) +\gamma\;\mathrm{li}(x)\mathrm{d}x \;\;=\;\; \int_{0}^{\infty}\frac{\ln(u)}{u(u^2+1)}\Bigg(\frac{u\ln(u^2+4)}{2}-\tan^{-1}\Big(\frac{u}{2}\Big)\Bigg)\mathrm{d}u$$
$$\mathrm{Let \;\; I \;\;=\;\; }\int_{0}^{1}\mathrm{li}(x)\ln\Big(\ln\Big(\frac{1}{x}\Big)\Big) \mathrm{d}x \;\;=\;\;\frac{2}{\pi}\underbrace{\int_{0}^{\infty}\frac{\ln(u)}{u(u^2+1)}\Bigg(\frac{u\ln(u^2+4)}{2}-\tan^{-1}\Big(\frac{u}{2}\Big)\Bigg)\mathrm{d}u}_\text{ $\mathrm{I}_1$}  -\gamma\;\underbrace{\int_{0}^{1}\;\mathrm{li}(x)\mathrm{d}x}_\text{ $\mathrm{I}_2$}     $$
$$\mathrm{I}_1 \;\;=\;\;\int_{0}^{1}\int_{0}^{\infty} \frac{\partial}{\partial a }\frac{\ln(u)}{u(u^2+1)}\Bigg(\frac{u\ln(a^2 u^2+4)}{2}-\tan^{-1}\Big(\frac{au}{2}\Big)\Bigg)\mathrm{d}u\mathrm{d}a  $$
$$=\;\;\int_{0}^{1}\int_{0}^{\infty} \frac{\ln(u)}{(u^2+1)}.\frac{a u^2-1}{(a^2u^2+4)}\mathrm{d}u \mathrm{d}a\;\;=\;\; \int_{0}^{1}\frac{1}{a-2}\int_{0}^{\infty} \frac{\ln(u)}{(u^2+1)}-\frac{2a \ln(u)}{(a^2u^2+4)}\mathrm{d}u\mathrm{d}a $$
    but $\displaystyle \int_{0}^{\infty} \frac{\ln(x)}{(x^2+a^2)}\mathrm{d}x \;\;=\;\;\frac{1}{a}\int_{0}^{\infty} \frac{\ln(a)+\ln(x)}{(x^2+1)}\mathrm{d}x \;\;=\;\; \frac{\pi}{2} \ln(a)+\int_{0}^{\infty} \frac{\ln(x)-\ln(x)}{(x^2+1)}\mathrm{d}x \;\;=\;\;\frac{\pi}{2} \ln(a)  $
$$\mathrm{Then \;\; I_1} \;\;=\;\;\frac{\pi}{2}\int_{0}^{1}\frac{\ln(a)}{a-2}\mathrm{d}a \;\;=\;\;\frac{\pi}{4}\int_{0}^{1}\frac{\ln(a)}{\frac{a}{2}-1}\mathrm{d}a \;\;=\;\;-\frac{\pi}{4}\;\sum_{n=0}^{\infty}\frac{1}{2^n}\int_{0}^{1}\ln(a) a^n\mathrm{d}a   $$
$$ =\;\;\frac{\pi}{4}\;\sum_{n=0}^{\infty}\frac{1}{2^n(n+1)^2} \;\;= \;\;\frac{\pi}{2} \mathrm{Li}_2\Big(\frac{1}{2}\Big) \;\;=\;\;\frac{\pi}{4} \Big(\zeta(2)-\ln^2(2)\Big)$$
    and $\mathrm{ I_2}$ is know in the mathematical literature and has the value : $-\ln(2)$
By combining these results we get that 
$$\qquad\int_{0}^{1}\mathrm{li}(x)\ln\Big(\ln\Big(\frac{1}{x}\Big)\Big) \mathrm{d}x \;\;=\;\;\frac{1}{2}\zeta(2)+\frac{1}{2}\ln^2(2)+\gamma \ln(2) $$
                [by Ahmad Albow November 8,2019 ]

