Integral representation of Euler's constant

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 $$ \gamma = \lim_{n \to \infty} \sum_{k=1}^{n} \frac{1}{k} - \log{n}. $$

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 $$ \int_0^1\log(\log(1/t))\,\mathrm{d}t=-\gamma\tag{5} $$

$$I = \int_0^1 \log (-\log x)\,dx = \int_0^\infty e^{-x} \log(x)\,dx$$

Noting that

$$\Gamma(s) = \int_0^\infty e^{-x} x^{s-1}\, dx$$

we find that

$$\Gamma'(1) = I = -\gamma$$