real analysis - Integral representation of Euler's constant

Prove that : $$ \gamma=-\int_0^{1}\ln \ln \left ( \frac{1}{x} \right) \ \mathrm{d}x.$$

where $\gamma$ is Euler's constant ($\gamma \approx 0.57721$).

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This integral was mentioned in Wikipedia as in Mathworld , but the solutions I've got uses corollaries from this theorem. Can you give me a simple solution (not using much advanced theorems) or at least some hints.

Answer

In this answer, it is shown that since $\Gamma$ is log-convex, $$ \frac{\Gamma'(x)}{\Gamma(x)}=-\gamma+\sum_{k=1}^\infty\left(\frac1k-\frac1{k+x-1}\right)\tag{1} $$ Setting $x=1$ yields $$ \Gamma'(1)=-\gamma\tag{2} $$ The integral definition of $\Gamma$ says $$ \begin{align} \Gamma(x)&=\int_0^\infty t^{x-1}\,e^{-t}\,\mathrm{d}t\\ \Gamma'(x)&=\int_0^\infty\log(t)\,t^{x-1}\,e^{-t}\,\mathrm{d}t\\ \Gamma'(1)&=\int_0^\infty\log(t)\,e^{-t}\,\mathrm{d}t\tag{3} \end{align} $$ Putting together $(2)$ and $(3)$ gives $$ \int_0^\infty\log(t)\,e^{-t}\,\mathrm{d}t=-\gamma\tag{4} $$ Substituting $t\mapsto\log(1/t)$ transforms $(4)$ to $$ \int_0^1\log(\log(1/t))\,\mathrm{d}t=-\gamma\tag{5} $$