There's no Typo, right??

These type of integrals are called Multiplicative Integrals or Product Integral. The above integral is represented as $\prod_1^0 f(x)^{dx}$

Where $f(x)=\ln\left(\frac 1x\right)$

Using a property of Multiplicative integrals which is stated as $\displaystyle \prod_a^b f(x)^{dx}=\exp {\left(\int_a^b \ln f(x) dx\right)}$

Thus in this case $\displaystyle I=\int_1^0 \left(\ln\left(\frac 1x\right)\right)^{dx}=\prod_1^0 \left(\ln\left(\frac 1x\right)\right)^{dx}$ $=\displaystyle \exp {\left(\int_1^0 \ln\left(\ln\left(\frac 1x\right)\right)dx\right)}$

Now using that $\displaystyle \int_0^1 \ln\left(\ln\left(\frac 1x\right)\right)dx=-\gamma$

Where $\gamma$ is the Euler Mascheroni Constant, we get $\displaystyle I=\exp {\left(\int_1^0 \ln\left(\ln\left(\frac 1x\right)\right)dx\right)}=e^{\gamma}=\alpha$

Hence $\ln \alpha=\gamma\approx \boxed {0.5772}$

$\pmb{\text{Edit:}}$

For the Evaluation of $\displaystyle \int_0^1 \ln\left(\ln\left(\frac 1x\right)\right)dx$

Just substitute $-\ln x=t$ , You would get $\displaystyle\int_0^{\infty} (\ln t )e^{-t} dt$ . Now if $\displaystyle I(a)=\int_0^{\infty} t^{a-1}e^{-t}dt=\Gamma(a)$ then what we need is $I'(1)=\Gamma(1)\psi(1)=-\gamma$