Evaluate $\int_{0}^1\ln(\ln(\frac{1}{x})) dx$

Question

I looking to evaluate $$\int_{0}^1\ln(\ln(\frac{1}{x})) dx$$

I know the answer is the Euler-Mascheroni constant, $\gamma$ but how do I get that result?

I've tried differentiating under the integral, but that didn't seem to work. Maybe series could work since $\gamma$ is connected to a lot of series and is defined by using harmonic series.

Answer 1

Enforcing $x= e^{-u}$ $$\int_{0}^1\ln\left(\ln\left(\frac{1}{x}\right)\right) dx =\int_{0}^\infty\ln(u)e^{-u} du =-\gamma $$

Where $\gamma$ is the Euler Mascheroni constant