Let $X$ be a random variable with probability function $f(x)=\lambda e^{-\lambda x} \text{ if } x>0$ ($\lambda >1$ constant), and $0$ otherwise. 
*

*Let $X$ be a random variable with probability function $\operatorname{f}\left(x\right)=\lambda{\rm e}^{-\lambda x}$ if $x>0$ and $\lambda >1$ Is a $constant$. $0$ otherwise.

*Calculate the expected value of the variable $Y=e^X$ by first finding
the density function of $Y$ and applying the elementary definition of
expectation.

*As a second method use the unconscious statistician's
theorem.

Attempt
I have already tried, and according to my calculations, the density function of $Y$ is
$$g(y)=\frac{1}{y}\lambda  e^{-\lambda (\frac{1}{y})} \text{ if } y>1 (\lambda >1 \text{ constant}),$$
and $0$ otherwise. Are my calculations correct?
Therefore, the expected value is
$$E(Y)=\int_{1}^{\infty}\, y\cdot g(y)\, dy=\int_{1}^{\infty}\, \lambda  e^{-\lambda (\frac{1}{y})}\, dy$$
How do I calculate this integral?
 A: The correct Y-density is
$$f_Y(y)=\lambda y^{-(\lambda+1)}$$
$y,\lambda>1$
thus
$$\mathbb{E}[Y]=\int_1^{\infty}\lambda y^{-\lambda}dy=\frac{\lambda}{\lambda-1}$$
LOTUS
$$\mathbb{E}[Y]=\int_0^{\infty}\lambda e^{-\lambda x}e^x dx=\int_0^{\infty}\lambda e^{-(\lambda-1)x}dx=\frac{\lambda}{\lambda-1}\int_0^{\infty}(\lambda-1)e^{-(\lambda-1)x}dx=$$
$$\frac{\lambda}{\lambda-1}\underbrace{\int_0^{\infty}\theta e^{-\theta x}dx}_{=1}=\frac{\lambda}{\lambda-1}$$

Calculation's detail for $f_Y$
$$f_Y(y)=\frac{\lambda}{y}\exp\left\{ -\lambda\log y  \right\}=\frac{\lambda}{y}\exp\left\{ \log y^{-\lambda}  \right\}=\lambda y^{-\lambda-1}$$
A: No, that is not the density of $Y$. In fact your function does not integrate to $1$. The correct one is $\frac {\lambda} {y^{1+\lambda }}$ for $y>1$ . Hence, $EY=\int_0^{\infty} \frac {\lambda} {y^{\lambda }}dy$ and the value of this integral is $\frac {\lambda} {\lambda -1}$.
A: First,
$$
\begin{align}
P(X\ge t)
&=\int_t^\infty\lambda e^{-\lambda x}\,\mathrm{d}x\tag{1a}\\
&=e^{-\lambda t}\tag{1b}
\end{align}
$$
Therefore, for $Y=e^X$,
$$
\begin{align}
P(Y\ge t)
&=P\!\left(e^X\ge t\right)\tag{2a}\\
&=P(X\ge\log(t))\tag{2b}\\
&=e^{-\lambda\log(t)}\tag{2c}\\
&=t^{-\lambda}\tag{2d}
\end{align}
$$
Thus, the density function of $Y=e^X$ is
$$
\lambda t^{-\lambda-1}\tag3
$$
The expected value of $Y=e^X$ is
$$
\begin{align}
\int_1^\infty t\lambda t^{-\lambda-1}\,\mathrm{d}t
&=\left.\frac{\lambda}{1-\lambda}t^{1-\lambda}\right|_1^\infty\tag{4a}\\
&=\frac{\lambda}{\lambda-1}\tag{4b}
\end{align}
$$
