I am trying to re-scale this integrand to prove that it is the density function of a Gamma distribution:
$$\int^∞_0 \frac{(αθ^α)\cdot λe^{-αλ}}{x^{α+1}} \, d\alpha$$
I pulled out $λ/x$ to get:
$$\frac λ x \int^∞_0 α \left(\frac\theta x\right)^\alpha e^{-\lambda\alpha} \, d\alpha$$
From here I am unsure how to re-scale this to become a Gamma. Any help would be great!
\begin{align} \frac \lambda x \int^\infty_0 \alpha \left(\frac\theta x\right)^\alpha e^{-\lambda\alpha} \, d\alpha = {} & \frac \lambda x \int_0^\infty \alpha \left( \frac \theta x e^{-\lambda} \right)^\alpha \, d\alpha \\[10pt] = {} & \frac \lambda x \int_0^\infty \alpha \left( e^{-\lambda\log(\theta/x)} \right)^\alpha \, d\alpha \\[10pt] = {} & \frac \lambda x \int_0^\infty \alpha e^{-\beta\alpha} \,d\alpha \\ & \text{where } \beta = \lambda\log(\theta/x) \\[10pt] = {} & \frac \lambda {x\beta^2} \int_0^\infty (\beta\alpha) e^{-\beta\alpha} (\beta\, d\alpha) \\[10pt] = {} & \frac \lambda {x\beta^2} \int_0^\infty u^{2-1} e^{-u} \, du \\[10pt] = {} & \frac \lambda {x\beta^2} \Gamma(2). \end{align} Then change $\beta$ back to what it is.