Expected value of $1/x$ when $x$ follows a Beta distribution Let $x$ have the probability density:
$$f(x) = \frac{x^{\alpha - 1}(1-x)^{1 - \beta}}{\mathrm{B}(\alpha,\beta)}$$
where $\alpha,\beta$ are two positive parameters and $0 \le x \le 1$ is the domain of $x$. What is the expected value of $1/x$? That is, what is the value of:
$$\langle1/x\rangle = \int_0^1 \frac{f(x)}{x} \mathrm d x$$
 A: I want to point out another interesting solution method, which also generalizes the result to the expectation of $X^{-m}$ for integer $m=1,2,3,\dots$. I will use moment generating functions (mgf) and the results from the paper by N Cressie et.al http://amstat.tandfonline.com/doi/abs/10.1080/00031305.1981.10479334?journalCode=utas20  "The Moment-Generating Function and Negative Integer Moments".
They give the result that when $X$ is a positive random variable with mgf $M_X(t)$ which is defined in an open neighbourhood of the origin, then we have
$$
\DeclareMathOperator{\E}{\mathbb{E}}
\E X^{-m} = \Gamma(m)^{-1} \int_0^\infty t^{m-1} M_X(-t) \; dt
$$
for positive integers $m$.  
It is known that for the beta distribution, the mgf is given by a confluent hypergeometric function as
$$
  M_X(t) = {}_1F_1(\alpha;\alpha+\beta;t)
$$
so using the result above gives that
$$
  \E X^{-m} = \Gamma(m)^{-1} \int_0^\infty t^{m-1} {}_1F_1(\alpha;\alpha+\beta;-t)\; dt
$$
I evaluated that integral with the help of maple:
assume( a>0, b>0 );assume(m-1,posint)
GAMMA(m)^(-1) * int( t^(m-1)*hypergeom([a],[a+b],-t), t=0..infinity )
                   GAMMA(a + b) GAMMA(a - m)
                   -------------------------
                   GAMMA(a) GAMMA(a + b - m)

so finally we can write the result as 
$$
\E X^{-m} = \frac{\Gamma(\alpha+\beta)\Gamma(\alpha-m)}{\Gamma(\alpha)\Gamma(\alpha+\beta-m)}
$$
which coinsides with the other answer for $m=1$. Then some human mathematics is needed to conclude that we need the assumption $\alpha > m$ for this to be valid.
A: First note that the pdf of a Beta$(\alpha, \beta)$ distribution is only defined for $\alpha, \beta > 0$. Which means that for when $\alpha \leq 0$ or $\beta \leq 0$
$$\int_0^1  \dfrac{x^{\alpha - 1} (1 - x)^{1 - \beta}}{B(\alpha, \beta)} $$
is not finite. Now,
\begin{align*}
E\left(\dfrac{1}{X} \right) & = \int_0^1 \dfrac{1}{x} \dfrac{x^{\alpha - 1} (1 - x)^{1 - \beta}}{B(\alpha, \beta)}dx\\
& = \int_0^1 \underbrace{\dfrac{x^{(\alpha - 1) - 1} (1 - x)^{1 - \beta}}{B(\alpha-1, \beta)}}_{\text{Beta}(\alpha-1, \beta)} \dfrac{B(\alpha-1, \beta)}{B(\alpha, \beta)} dx\\
& = \dfrac{B(\alpha-1, \beta)}{B(\alpha, \beta)} \text{ if $\alpha - 1$ > 0} \,.
\end{align*}
Thus when $\alpha > 1$, the expectation $E(1/X)$ is finite and is as above (this can be further simplified as per Nate Pope's comment). Otherwise, it is undefined.
