I am trying to find an expected value by integration. Not sure if it is hopeless but the statistics part is not that important in this case, rather I want to see if this integral does have a nice anti-derivative. I am sadly not too familiar with gamma function properties besides the very simple properties.
$$\int_{0}^{\infty}\frac{1}{x}\frac{(n\beta)^{n\alpha}}{\Gamma(n\alpha)}x^{-n\alpha-1}e^{-\frac{n\beta}{x}}dx $$
As usual pulling out constants and adding exponents for same base.
$$\frac{(n\beta)^{n\alpha}}{\Gamma(n\alpha)}\int_{0}^{\infty}x^{-n\alpha-2}e^{-\frac{n\beta}{x}}dx $$
But this is where I do not know how to continue, I tried following this https://math.stackexchange.com/a/3650717/633922 post, but I am unsure if it is correct or if it even applies in my case. Any help would be appreciated.
Take $n\beta/x=y$ getting$$I=\frac{(n\beta)^{-1}}{\Gamma(n\alpha)}\int_0^\infty y^{n\alpha} e^{-y}dy=\frac{\Gamma(n\alpha+1)}{n\beta\Gamma(n\alpha)}$$What property of the Gamma function can you use to simplify the numerator?
Edit: Expanding on my answer, $dx=-n\beta~dy/y^2$ so$$\begin{align*}I&=\frac{(n\beta)^{n\alpha}}{\Gamma(n\alpha)}\int_{0}^{\infty}x^{-n\alpha-2}e^{-\frac{n\beta}{x}}dx\\&=\frac{(n\beta)^{n\alpha}}{\Gamma(n\alpha)}\int^0_{\infty}\left(\frac{n\beta}y\right)^{-n\alpha-2}e^{-y}\left[\frac{-n\beta}{y^2}\right]dy\\&=\frac{(n\beta)^{-1}}{\Gamma(n\alpha)}\int_0^\infty y^{n\alpha} e^{-y}dy\end{align*}$$