I'm trying to find if there is a specific probability distribution $f\left(x\right)$ (or many) such that the following integral
$$\int\left(x-a\right)^{-\gamma}f\left(x\right)dx$$
has a closed form solution for any scalars $a\geq 0$ and $\gamma\geq 0$. The limits of integration must respect that $x\geq a$, but in general they can adjust to the shape of the distribution chosen. It is easy to see that the uniform distribution, where $f\left(x\right)$ is a constant, trivially works. But I wonder if there are other distributions that also work. If the general case is too hard, it would be helpful to know if there is an answer for the $\gamma=1$ case.
We can solve the differential equation:
$$f'=(x-a)^{-\gamma}f $$
This is a first order differential equation with solution:
$$f(x)=ce^{\frac{(x-a)^{1-\gamma}}{1-\gamma}}$$
To make this a density, we need to set $c$ so
$$\int_a^M ce^{\frac{(x-a)^{1-\gamma}}{1-\gamma}}dx = 1$$
For $\gamma \neq 1$, we need $a\leq M < \infty$. If $\gamma=1$ then the function is undefined.
The integral is not nice, but you get (with more than a little help from WolframAlpha) that $c$ is strictly a function of $M$:
$$c_M=\left\{\left[\left(\frac{(x-a)^{1-k}}{k-1}\right)^{\frac{k}{k-1}} (x-a)^k \Gamma\left(\frac{1}{1-k}, \frac{(x-a)^{1-k}}{k-1}\right)\right]_a^M\right\}^{-1}$$
$$=\left\{ \left(\frac{(M-a)^{1-k}}{k-1}\right)^{\frac{k}{k-1}} (M-a)^k \Gamma\left(\frac{1}{1-k}, \frac{(M-a)^{1-k}}{k-1}\right)\right\}^{-1}$$
So, with a suitably chosen $c_M$ we get that the above density will yield a nice closed formula:
$$\int_L^U\left(x-a\right)^{-\gamma}c_Me^{\frac{(x-a)^{1-\gamma}}{1-\gamma}}dx = \left[c_Me^{\frac{(x-a)^{1-\gamma}}{1-\gamma}}\right]_L^U,\;\;\textrm{where}\;\;a \leq L \leq U\leq M$$