Proving $\int_0^n \left(1-\frac{t}{n}\right)^n\ln(1/t)\,dt \to \gamma$ I have to prove that $\displaystyle \lim_{n\to\infty}\int_0^n \left(1-\frac{t}{n}\right)^n\ln(1/t)\,dt= \gamma$
I tried to expand $\left(1-\frac{t}{n}\right)^n$ and swap sum and integral, which yields $$\int_0^n \left(1-\frac{t}{n}\right)^n\ln(1/t)\,dt=n\sum_{k=0}^n \binom n k\frac{(-1)^{k+1}}{k+1} \left(\ln(n)-\frac{1}{k+1}\right)$$
I can't manage to show that the last quantity goes to $\displaystyle \gamma=\lim_{n\to\infty} (H_n- \ln(n))$.
Any help is appreciated.
 A: $$\begin{eqnarray*}\int_{0}^{n}\left(1-\frac{t}{n}\right)^n (-\log t)\,dt &=& n \int_{0}^{1}(1-u)^n(-\log(nu))\,du\\&=&-\frac{n}{n+1}\log n-n\int_{0}^{1}(1-u)^n\log u\,du\end{eqnarray*}$$
but, through a change of variable and integration by parts:
$$\begin{eqnarray*} -(n+1)\int_{0}^{1}(1-u)^n\log u\,du &=& \int_{0}^{1}(n+1)u^n\left(-\log(1-u)\right)\,du\\&=&\int_{0}^{1}\frac{u^{n+1}-1}{u-1}\,du=H_{n+1}\end{eqnarray*} $$
so:

$$\int_{0}^{n}\left(1-\frac{t}{n}\right)^n(-\log t)\,dt = \frac{n}{n+1}\left(H_{n+1}-\log n\right).$$

A: As $n \to \infty$, $(1-t/n)^n \to \exp(-t)$, so (applying appropriate estimates
and limit theorems) the limit should be
$$\int_0^\infty \exp(-t) \ln(1/t)\; dt$$
which is indeed $\gamma$.  
A: We can consider your sum $$n\sum_{k=0}^{n}\dbinom{n}{k}\frac{\left(-1\right)^{k+1}}{k+1}\left(\log\left(n\right)-\frac{1}{k+1}\right)=n\log\left(n\right)\sum_{k=0}^{n}\dbinom{n}{k}\frac{\left(-1\right)^{k+1}}{k+1}-n\sum_{k=0}^{n}\dbinom{n}{k}\frac{\left(-1\right)^{k+1}}{\left(k+1\right)^{2}}.
 $$We have $$\sum_{k=0}^{n}\dbinom{n}{k}x^{k}=\left(1+x\right)^{n}\,\,\,\,\,\,(1)
 $$ then if we integrate $$ \sum_{k=0}^{n}\dbinom{n}{k}\frac{\left(-1\right)^{k+1}}{k+1}=-\int_{-1}^{0}\left(1+x\right)^{n}dx=-\frac{1}{n+1}.
 $$ Again from $(1)
 $ we have $$\sum_{k=0}^{n}\dbinom{n}{k}\int_{0}^{t}x^{k}=\sum_{k=0}^{n}\dbinom{n}{k}\frac{t^{k+1}}{k+1}=\frac{\left(1+t\right)^{n+1}-1}{n+1}
  $$ hence $$-\sum_{k=0}^{n}\dbinom{n}{k}\frac{\left(-1\right)^{k+1}}{\left(k+1\right)^{2}}=\frac{1}{n+1}\int_{-1}^{0}\frac{\left(1+t\right)^{n+1}-1}{t}dt
 $$ and the last integral is $$\int_{-1}^{0}\frac{\left(1+t\right)^{n+1}-1}{x}dx=\int_{0}^{1}\frac{u^{n+1}-1}{u-1}du=H_{n+1}
 $$ then your integral, as Jack wrote, is $$\frac{n}{n+1}\left(H_{n+1}-\log\left(n\right)\right).
 $$
