I want prove two follow integral. Show that $$\int_{1}^{\infty} e^{-s} \log(s)\,  ds=\lim_{n\to \infty } \int_{1}^{n} \left( 1-\frac{s}{n}\right)^n \log(s) \, ds$$ 
and that $$\int_{0}^{1} e^{-s} \log(s)\, ds=\lim_{n\to \infty } \int_{\frac{1}{n}}^{1} \left( 1-\frac{s}{n}\right) ^n \log(s)\, ds$$
thanks for help.
 A: Let 
$$f_n(s)=\chi_{[1,n]}(s)\left(1-\frac{s}{n}\right)^n\log(s)$$
then 
$$\lim_{n\to\infty}f_n(s)=\chi_{[1,\infty)}(s)e^{-s}\log(s)$$
and we have
$$|f_n(s)|\le e^{-s}\log(s),\; \forall  s\ge1$$
and the function $s\mapsto e^{-s}\log(s)$ is integrable on $[1,\infty)$ so we can apply the dominated convergence theorem to find the desired result.
A: I thought it might be instructive to present a proof that does not rely on the Dominated Convergence Theorem.  To that end, we proceed.

We wish to show that $\displaystyle \int_1^\infty e^{-s}\log(s)\,ds=\lim_{n\to\infty}\int_1^n\log(s)\left(1-\frac sn\right)^n\,ds$. This is equivalent to showing that $\displaystyle \lim_{n\to\infty}\int_1^n \log(s)e^{-s}\left(1-e^s\left(1-\frac sn\right)^n\right)\,ds=0$.
To do so, we note the estimates
$$\begin{align}
1-e^s\left(1-\frac sn\right)^n&\le 1-\left(1+\frac sn\right)^n\left(1-\frac sn\right)^n\\\\
&=1-\left(1-\frac{s^2}{n^2}\right)^n\\\\
&\le 1-\left(1-\frac{s^2}{n}\right)\\\\
&=\frac{s^2}{n}
\end{align}$$
Similarly, we have
$$1-e^s\left(1-\frac sn\right)^n\ge 1-e^se^{-s}=0$$
Therefore, applying the squeeze theorem yields the coveted result
$$\lim_{n\to\infty}\int_1^n \log(s)\left(1-\frac sn\right)^n\,ds=\int_1^\infty \log(s)e^{-s}\,ds$$
