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 ($\gamma \approx 0.57721$).

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.
 A: $$I = \int_0^1 \log (-\log x)\,dx = \int_0^\infty e^{-x} \log(x)\,dx$$
Noting that
$$\Gamma(s) = \int_0^\infty e^{-x} x^{s-1}\, dx$$
we find that 
$$\Gamma'(1) = I = -\gamma$$
A: You can see a proof here where we use that $$\Gamma(z) = \frac{\exp{(-\gamma z)}}{z}\prod\limits_{n=1}^\infty\frac{\exp \left({\frac z n}\right)}{1+\dfrac z n }$$
There is another proof here where we use $$\gamma=\lim\limits_{n\to\infty}\left( H_n-\log n\right)$$
A: Another way, from another definition: by the Dominated Convergence Theorem,
$$ \int_0^{\infty} e^{-u} \log{u} \, du = \lim_{n \to \infty} \int_0^n \left( 1 - \frac{u}{n} \right)^{n-1} \log{u} \, du. $$
Then, changing variables to $v=1-u/n$,
$$ \begin{align}
\int_0^n \left( 1 - \frac{u}{n} \right)^{n-1} \log{u} \, du &= n \int_0^1 v^{n-1} \log{( n(1-v))} \, dv \\
&= n\log{n} \int_0^1 v^{n-1} \, dv + (n+1) \int_0^1 v^{n-1} \log{(1-v)} \, dv \\
&= \log{n} - n \int_0^1 \sum_{k=1}^{\infty} \frac{v^{k+n-1}}{k} \, dv \\
&= \log{n} - n \sum_{k=1}^{\infty} \int_0^1 \frac{v^{k+n-1}}{k} \, dv \\
&= \log{n} - n \sum_{k=1}^{\infty} \frac{1}{k(k+n)} \\
&= \log{n} -  \sum_{k=1}^{\infty} \left( \frac{1}{k} - \frac{1}{n+k} \right) \\
&= \log{n} -  \sum_{k=1}^{n} \frac{1}{k},
\end{align} $$
using uniform convergence and partial fractions. But this is precisely the definition
$$ \gamma = \lim_{n \to \infty} \sum_{k=1}^{n} \frac{1}{k} - \log{n}. $$
A: 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
$$
\int_0^1\log(\log(1/t))\,\mathrm{d}t=-\gamma\tag{5}
$$
