Suppose $X$ has an exponential distribution with parameter $5$. How would I compute $E[1/x]$? If I use the definition of expected value, I get:
$\int_{0}^{\infty} \frac{5e^{-5 x}}{x} dx$
I've not the slightest clue as to how I would integrate this. I was told the answer is simply $\frac{1}{1/ \lambda} = \lambda$, but I don't know how this could be.
On
Taking $X \sim \text{Exp}(\lambda)$ and taking any $\epsilon > 0$ you get:
$$\begin{equation} \begin{aligned} \mathbb{E}(1/X) &= \int \limits_0^\infty \frac{1}{x} \cdot \frac{1}{\lambda} e^{-\lambda x} \ dx \\[6pt] &= \int \limits_0^\infty \frac{1}{y} e^{-y} \ dy \\[6pt] &\geqslant \int \limits_0^{\epsilon } \frac{1}{y} e^{-y} \ dy \\[6pt] &\geqslant e^{-\epsilon} \int \limits_0^{\epsilon } \frac{1}{y} \ dy \\[6pt] &= e^{-\epsilon} \Bigg[ \int \limits_1^{\epsilon } \frac{1}{y} \ dy - \int \limits_1^{0} \frac{1}{y} \ dy \Bigg] \\[6pt] &= e^{-\epsilon} \Big( \ln (\epsilon) - \ln(0) \Big) \\[6pt] &= e^{-\epsilon} \Big( \ln (\epsilon) - (-\infty) \Big) \\[6pt] &= \infty. \\[6pt] \end{aligned} \end{equation}$$
Thus, the expected value does not exist (i.e., it is infinite).
On
As the other answers have pointed out it is not possible to integrate this.
To get the expected value, I would consider the relationship between the exponential and gamma distributions. An exponential is a gamma distribution with an α of 1 and λ = λ.
Then we can find the expectation using the gamma distribution in a similar manner to top two answers on this post.
$$ E(X^m)= \frac{\Gamma(a+m)}{\lambda^{m}\Gamma(\alpha)} $$ In our case α=1, m=-1, so
$$ E(X^{-1})= \frac{\Gamma(1-1)}{\lambda^{-1}\Gamma(1)}= $$
$$ = \frac{1}{\frac{1}{\lambda}} = \lambda $$
$$\int_0^\infty \frac{5e^{-5 x}}{x} \, dx = 25 \int_0^\infty \frac{e^{-u}}{u} \, du$$ but the integral does not converge since $\frac{e^{-u}}{u} \ge \frac{1}{2u}$ as $u \to 0$. So $E[1/X]$ does not exist.