Random variable does not have an expectation

So, I was transforming a random variable $f_X$, to get the density of $Y = \exp X$

$$f_X =\begin{cases} 2x^{-3} & \text{if } x \geqslant 1\\ 0 & \text{otherwise} \end{cases}$$

I went trough this process and go that $f_Y=\frac{2}{y(\ln y)^{3}}$ for $y\geqslant e$. The obtained pdf integrates to one, but when I try to get the expectation of Y, it does not exist. $E( Y) =\int _{e}^{\infty }\frac{2}{(\ln y)^{3}} dy$. Could you find my problem?

$$\begin{gather*} f( x) =\begin{cases} 2x^{-3} & \text{if } x\geqslant 1\\ 0 & \text{otherwise} \end{cases}\\ \text{We first estimate} \ F_{X}\\ F_{X} =\int _{1}^{x} 2t^{-3} dt\\ F_{X} =1-x^{-2}\text{ for } x\geqslant 1\\ \text{By definition we know}\\ F_{Y} =P( Y\leqslant y)\\ F_{Y} =P\left( e^{X} \leqslant y\right)\\ F_{Y} =P\left(\ln e^{X} \leqslant \ln y\right)\\ F_{Y} =P( X\leqslant \ln y)\\ \\ P( X\leqslant \ln y) =1-\frac{1}{(\ln y)^{2}}\\ \text{We can now get} \ F_{Y}\\ F_{Y} =1-\frac{1}{(\ln y)^{2}}\text{ for } y\geqslant e\\ \text{We need to take the derivative to get } f_{Y}\\ f_{Y} =\frac{d}{dy}\left( 1-\frac{1}{(\ln y)^{2}}\right)\\ f_{Y} =\begin{cases} \frac{2}{y(\ln y)^{3}} & \text{ for } y\geqslant e\\ 0 & \text{otherwise} \end{cases} \end{gather*}$$

However, $$E( Y) =\int _{e}^{\infty }\frac{2}{(\ln y)^{3}} dy$$ is not defined. I don't know why.

Yes, you are right. Observe that there are a lot of random variables which do no admit an expectation, i.e. Cauchy, Student's t distribution for particular d.o.f.'s but also, simply, the reciprocal of a uniform over $(0;1)$

In fact, if $X\sim U(0;1)$, and $Y=1/X$ it is easy to verify that

$$f_Y(y)=\frac{1}{y^2}\cdot\mathbb{1}_{(0;+\infty)}(y)$$

with mean

$$\mathbb{E}[Y]=\int_0^{\infty}\frac{1}{y}dy=\infty$$

Not every r.v. has finite expectation. In this case $EY=\infty$. If you are looking for a proof of the fact that $\int_e^{\infty} \frac 2 {(\ln y)^{2}} dy=\infty$ make the change of variable $t=\ln y$. You get $\int_1^{\infty} \frac 2 {t^{3}} e^{t}dt$. Now use the fact that $e^{t} >\frac {t^{3}} {3!}$.