This question is in light of the question here https://stats.stackexchange.com/questions/445330/probability-distribution-function-expressed-in-terms-of-a-divergent-series.
Let the pdf of $x$ be: $$f_X(x)=\frac{2\gamma(k+2/v, R^vx/\theta)(R^vx/\theta)^{-1-2/v}}{v\Gamma(k)}$$
where $R$, $k$ and $\theta$ are positive real, and $v$ is positive integer. I want to find $E[x]$ and $E[x^2]$ where, $E$ is the expected value. I know to find them we need: $$\int_0^\infty xf_X(x)dx, $$ $$\int_0^\infty x^2f_X(x)dx$$
How can I integrate a gamma function?
A pair of answers from a CAS (Mathematica 11.3): (The first two moments do not exist if $v$ is too small, as indicated.) \begin{align*} \int_0^\infty \; x f_X(x) \,\mathrm{d}x &= \frac{2 R^{-2v} \theta^2 \Gamma(k+1)}{(v-2)\Gamma(k)} &&, v>2 \\ \int_0^\infty \; x^2 f_X(x) \,\mathrm{d}x &= \frac{R^{-3v}\theta^3 \Gamma(k+2)}{(v-1)\Gamma(k)} &&, v>1 \text{.} \end{align*} Given the fairly obvious pattern, does it extend?... \begin{align*} \int_0^\infty \; x^3 f_X(x) \,\mathrm{d}x &= \frac{2 R^{-4v}\theta^4 \Gamma(k+3)}{(3v-2)\Gamma(k)} \\ \int_0^\infty \; x^4 f_X(x) \,\mathrm{d}x &= \frac{-R^{-5v}\theta^5 \Gamma(k+4)}{(1-2v) \Gamma(k)} \end{align*} ... Not so much.