TITLE: Integral representation of Euler's constant
QUESTION [19 upvotes]: 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.

REPLY [15 votes]: 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
$$ \gamma = \lim_{n \to \infty} \sum_{k=1}^{n} \frac{1}{k} - \log{n}. $$