The trouble (as was already explained to you) lies in the starting point $t=0$ of the integral in the exponential. Fortunately, W+W are only interested in steady solutions of equation (15) of their arXiv preprint and these can be written as the function in their equation (16) provided one replaces the starting point $t=0$ of the integral by any starting point $t=x_0$ with $x_0>0$. Changing $x_0$ only modifies $f$ by a multiplicative factor so the normalisation condition saves the day.

Assuming for example that $x_0=1$, one gets
$$f(x)=\frac{c}{x^2}\exp\left(\int^x_1 \left(\frac{\mu}{t^2}+\frac{1-2\alpha}{t}\right) \mathrm{d}t\right)=\frac{c}{x^2}\exp\left(\mu-\frac{\mu}{x}+(1-2\alpha)\log(x)\right),
$$
hence
$$
f(x)=c\mathrm{e}^{\mu}x^{-1-2\alpha}\mathrm{e}^{-\mu/x}.
$$
This is not W+W's formula (18) so either I made a mistake in this post or there is a misprint in W+W's preprint. Note that the function $f_{0}$ written in (18) of W+W's preprint and in your question here is not integrable if $\alpha\le2$ because $f_{0}(x)$ behaves like a multiple of $x^{1-\alpha}$ when $x\to\infty$, hence for such values of $\alpha$, $f_{0}$ cannot be normalized to get a probability density function (even assuming that one got rid of the problem of the starting point $t=0$ as I explained). The function I obtained above is integrable for every positive $\alpha$ and $\mu$.

If $f$ is the density of the distribution of a random variable $X$, the distribution of $Y=\mu/X$ has density
$$
g(y)=c\mathrm{e}^{\mu}\mu^{-2\alpha}y^{2\alpha-1}\mathrm{e}^{-y},
$$
hence $Y$ is a standard Gamma random variable of exponent $2\alpha$ and
$$
c=\mathrm{e}^{-\mu}\mu^{2\alpha}/\Gamma(2\alpha).
$$
(As I said before, this question belongs to math.SE.)

any(!) point? – Mariano Suárez-Alvarez♦ Feb 17 '11 at 17:33