Let $P(Y \ge y) = {1 \over y \log(y)}$ and $y \ge e$, then $F(y) = 1- {1 \over y \log(y)}$ and $F' = {\log(y) + 1 \over (y \log (y))^2}$. Assuming that $Y$ is non-negative, you can get the expected value by :
$$EY = \int_e^{\infty} P(Y \ge y) dy \quad \text{or} \quad EY = \int_e^\infty y dF = \int_e^\infty y F'(y) dy$$
The first integral will get me $\int_e^\infty {1 \over y \log(y)} dy$ but the second integral gives me $\int_e^\infty {1 \over y \log(y)}dy + \int_e^\infty {1 \over y(\log(y))^2}dy$.
What am I missing here? Why am I getting different results?
Best Answer
The random variable does not have density. $P(Y=e)=1-\frac 1 e$. So when you compute $EY$ you have to add $eP(Y=e)=e-1$ to $\int_{(e,\infty)} yF'(y)dy$.
Note that $\int_e^{\infty} \frac 1 {y \ln y} dy=\infty$. So $EY=\infty$.