Convergence of $(u_n)_{n \in \mathbb{N}} := \sum\nolimits_{k=1}^n \frac{1}{k} - \log(n)$ was proven before
The egality $\Gamma'(1) = \lim_{n \to \infty}\int_0^n \log(t) (1-\frac{t}{n})^n dt$ was also proven before
Let $\gamma$ denote the limit of $(u_n)_{n \in \mathbb{Z}_{>0}}$. Show \begin{align*} \int_0^1 (1-u)^n \log(u)du = \frac{1}{n + 1} \int_0^1 \frac{(1-u)^{n+1} - 1}{u} du \end{align*} and conclude $\Gamma'(1) = -\gamma$.
I managed to prove the Integral identity but then I get lost !
Thanks in advance !
Calculation
\begin{align*} \Gamma'(1) &= \lim_{n \to \infty}\int_0^n \log(t)\left(1-\frac{t}{n}\right)^ndt \end{align*} Substitution: $u = \frac{t}{n}$, $du = \frac{1}{n} dt$ \begin{align*} \Gamma'(1) &= \lim_{n \to \infty}\int_0^1 [\log(u) + \log(n)](1-u)^n n du \\ &= \lim_{n \to \infty}\Big( n\int_0^1 \log(u)(1-u)^n du + n \log(n) \int_0^1 (1-u)^{n} du \Big) \\ &= \lim_{n \to \infty}\Big( \frac{n}{n+1} \int_0^1 \frac{(1-u)^{n+1} - 1}{u} du - \frac{n}{n+1} \log(n) \left.(1-u)^{n+1}\right|_{u=0}^1 \Big) \end{align*} Second substitution $x = 1-u$, $du = -dx$: \begin{align*} \Gamma'(1) &= \lim_{n \to \infty}\Big( \frac{-n}{n+1} \int_0^1 \frac{x^{n+1} - 1}{1-x} dx + \frac{n}{n+1} \log(n) \Big) \\ &= \lim_{n \to \infty} \frac{n}{n+1} \Big( \int_0^1 \frac{1-x^{n+1}}{1-x} dx + \log(n)\Big) \end{align*} We use that $H_n = \sum_{k=1}^n \frac{1}{k} = \int_0^1 \frac{1-x^n}{1-x}dx$: \begin{align*} \Gamma'(1) &= \lim_{n \to \infty}\Big( \frac{n}{n+1} \Big) \cdot \lim_{n \to \infty} \Big( \sum_{k=1}^{n+1}\frac{1}{k} + \log(n) \Big) \\ &= \lim_{n \to \infty} \Big( \sum_{k=1}^{n+1}\frac{1}{k} + \log(n) \Big) \\ \end{align*} But I should get !!! \begin{align*} -\lim_{n \to \infty} \Big( \sum_{k=1}^{n+1}\frac{1}{k} - \log(n) \Big) \\ \end{align*}
When you substitute $u = 1-x$ in
$$\int_0^1 \frac{(1-u)^{n+1}-1}{u}\,du,$$
you apparently forgot to adapt the integration limits (or maybe the sign from $du = -dx$),
$$\begin{align} \int_0^1 \frac{(1-u)^{n+1}-1}{u}\,du &= \int_{1-0}^{1-1} \frac{(1-(1-x))^{n+1}-1}{1-x}\,d(1-x)\\ &= \int_1^0\frac{1-x^{n+1}}{1-x}\,dx\\ &= -\int_0^1\frac{1-x^{n+1}}{1-x}\,dx. \end{align}$$
It turns out right then.