Expected value of negative and positive part of a random variable We have $M_n=\max\left\{X_1,\ldots,X_n\right\}$ where $X_i\sim \exp(1)$ iid , where the cdf of $M_n$ is $(1-e^{-x})^n$.And I want to prove that 
$$\lim_{n\rightarrow \infty}\mathbb{E}[(M_n-\ln n)^{+}]=\int_0^\infty 1-e^{-e^{-t}} \, dt$$
$$\lim_{n\rightarrow \infty}\mathbb{E}[(M_n-\ln n)^{-}]=\int_0^\infty e^{-e^{-t}} \, dt$$
So first I try to find the dsitribution of $(M_n-\ln n)^{+}$
$$\mathbb{P}[(M_n- \ln n)^{+}\leq x] = \begin{cases}
0 &  M_n-\ln n<0\\ 
\mathbb{P}[M_n-\ln n] & M_n-\ln n>0
\end{cases}=\begin{cases}
0 &  M_n-\ln n<0\\
\left(1-\frac{e^{-x}}n\right)^n & M_n-\ln n>0
\end{cases}$$
Then $$\lim_{n\rightarrow \infty}\mathbb{E}[(M_n-\ln n)^{+}]=\lim_{n\rightarrow \infty}\int_0^\infty xe^{-x}\left(1-\frac{e^{-x}} n \right)^{n-1}\,dx=-\int_0^\infty e^{-e^{-x}} \, dx$$
Same calculation for the negative part , but I'm not getting the correct result .There has to be a mistake , but I dont know what I did wrong.
 A: So, first of all, your derivation of the distribution of $(M_n - \ln n)^+$ is confusing; in particular, nothing depends on $M_n - \ln n$, it can only depend on $x$. The correct way of arguing is the following:
$P(M_n - \ln n)^+ \le x) = 0$ if $x \le 0$. Assuming $x > 0$, we can write that 
$$\{(M_n - \ln n)^+ \le x\} = \{M_n \le \ln n\} \cap \{\ln n \le M_n \le \ln n + x\}$$
Which implies, for $x > 0$, 
$$P(M_n - \ln n)^+ \le x) = P(M_n \le \ln n) + P(\ln n \le M_n \le \ln n + x) $$$$= F_{M_n}(\ln n) + F_{M_n}(\ln n + x) - F_{M_n} (\ln n) = F_{M_n}(\ln n + x) $$$$= \left(1 - \frac{e^{-x}}n\right)^n $$
At this point, we note that $(M_n - \ln n)^+$ is a r.v. with positive support. We can then use the formula
$$E(X) = \int_0^\infty 1 - F_X(x) dx$$
and bringing the limit inside we arrive at the desired result.

The error in your calculations is probably here:
The density is
$$f_{M_n}(x) = e^{-x}\left(1 - \frac{e^{-x}}n\right)^{n-1}$$
We want the limit
$$\lim_{n\rightarrow \infty}\mathbb{E}[(M_n-\ln n)^{+}]=\lim_{n\rightarrow \infty}\int_0^\infty xe^{-x}\left(1-\frac{e^{-x}} n \right)^{n-1}\,dx$$
Bringing the limit inside we find we get 
$$\int_0^\infty xe^{-x}e^{-e^{-x}}\,dx$$
Now, I guess that to get to your result you used integration by parts as this point, since $(e^{-e^{-x}})' = e^{-x}e^{-e^{-x}}$. The issue is that the term outside the integral would be $\displaystyle (xe^{-e^{-x}})\mid_0^\infty$ which does not converge, so integration by parts is invalid. 
