Valid gamma distribution pdf For a gamma probability distribution function of the form: 
$\frac{\beta ^{\alpha }}{\Gamma\left(\alpha \right)}x^{\alpha -1}e^{-\beta x}$,
is it sill a valid pdf if instead of $x$ we have $1/x$? 
More specifically, does the density function $\frac{\beta ^{\alpha }}{\Gamma\left(\alpha \right)}(1/x)^{\alpha -1}e^{-\beta /x}$ integrate to 1 over the interval (0, $\infty$) and can we still say that E(x) = $\frac{\alpha}{\beta}$? 
 A: No, it's not.  When you are working with probability density functions, you are really working with the derivative of the associated cumulative density function.  If we have a 1-1 function, let us say $y = 1/x$, we can make the straightforward substitution  $1/y$ for $x$ everywhere in the CDF, and that works, but not so with the PDF - you need to account for the change of variable by multiplying by the Jacobian of the transform, otherwise, although the variable has been transformed, the derivative (i.e., the PDF) hasn't been.  The first slide of this presentation explains why very well.
Let's work through this as an example.  We'll define $y = g(x) = 1/x$.  The  transform from $y$ back to $x$ is $g^{-1}(y)$ and $= 1/y$.  The Jacobian of the transform $g^{-1}(y)$ is $|dx/dy| = 1/y^2$.  Substituting $1/y$ for $x$ in the expression for the Gamma density function and multiplying by the Jacobian gives us:
$$f(y) = y^{-2}{\beta^{\alpha} \over \Gamma(\alpha)}y^{1-\alpha}e^{-\beta/y} ={\beta^{\alpha} \over \Gamma(\alpha)}y^{-1-\alpha}e^{-\beta/y} $$
which is the density function of the inverted Gamma distribution.  As the link shows, the mean of $y$ is $\beta / (\alpha - 1)$ but exists only when $\alpha > 0$.
