I think this is a reverse product rule but I could not figure out how to reverse this.
$$\int_0^\infty \frac{\beta^\alpha}{\Gamma(\alpha)}{\theta^{-(\alpha+1)}}*e^{-\frac{\beta}{\theta}}d\theta$$
I pulled out the constants but then got stuck here: $$\frac{\beta^\alpha}{\Gamma(\alpha)}\int_0^\infty {\theta^{-(\alpha+1)}}*e^{-\frac{\beta}{\theta}}d\theta$$
$\alpha$ and $\beta$ are constants while $\Gamma$ is a function.
The change of variable $\;u:=\dfrac {\beta}{\theta}\;$ should help much here giving $\;\theta:=\dfrac {\beta}u,\;d\theta=-\dfrac{\beta}{u^2}du\;$ and :
\begin{align} I(\alpha,\beta)&:=\dfrac{\beta^\alpha}{\Gamma(\alpha)}\int_0^{\infty} {\theta^{-(\alpha+1)}}\;e^{-\large{\frac{\beta}{\theta}}}d\theta\\ &=\dfrac{\beta^\alpha}{\Gamma(\alpha)}\int_{\infty}^0 \left(\frac {\beta}u\right)^{-(\alpha+1)}\;e^{-u}\frac {-\beta}{u^2}\,du\\ &=\dfrac{\beta^\alpha}{\Gamma(\alpha)}\int_0^{\infty} \beta\left(\frac{u}{\beta}\right)^{\alpha+1}\;e^{-u}\frac 1{u^2}\,du\\ &=\dfrac{1}{\Gamma(\alpha)}\int_0^{\infty} u^{\alpha-1}\;e^{-u}\,du\\ &=\dfrac{\Gamma(\alpha)}{\Gamma(\alpha)},\quad(*)\\ &=1 \end{align} $(*)$ from the definition of the $\Gamma$ function for $\,\alpha>0,\;\beta>0$.