Expectations of Indicator functions and Random Variables Let $ X_1, X_2, ... $ be i.i.d with $P(X_i>x)=\frac{e}{x\log(x)}$ for $x\geq e$.  If I were to try to find the expected value of  the
$$ EX_k1_{\{|X_k|\leq n\}}$$ for $e \leq x \le n$,   what exactly would that be? I know the answer is supposed to be $e\log\log n$ but I can't figure out how? Will we not take the integral of  $$\int _e^n \frac{ex}{x\log(x)}dx$$

You are given $P(X_k>x)$, so the CDF of $X_k$ is $P(X_k\le x)=1-P(X_k>x)=1-\frac e{x\log x}$.
The PDF of $X_k$ is$$\frac{dP(X_k\le x)}{dx}=\frac{e(1+\log x)}{x^2\log^2x}$$For $n>e$, the required expectation is $$\int_e^nx\frac{e(1+\log x)}{x^2\log^2x}dx=e\int_1^{\log n}\frac{(1+m)}{m^2}dm$$where $m=\log x$. This gives$$e\left[-\frac1m+\log|m|\right]_1^{\log n}=e\log\log n+e(1-1/\log n)$$
