Hopefully this is my last question of this kind... I've tried this in several different ways, but I always end up stuck in a loop of iterative integration by parts and proving 1 = 1. As important a proof this is for fundamental mathematics, it isn't what I am after unfortunately... Here's the integral: $$ \int_0^\infty t^{3/\xi-1} e^{-t} \Gamma(\tfrac{2}{\xi},t) \ dt. $$
For information: $\xi$ is real and positive.
Thanks, everyone.
The only closed form I see is via the incomplete beta function: $$\int_0^\infty t^{\alpha-1}e^{-t}\,\Gamma(\beta,t)\,dt=\Gamma(\alpha+\beta)\,\mathrm{B}(1/2;\alpha,\beta),$$ obtained using the definition $\Gamma(\beta,t)=\int_t^\infty x^{\beta-1}e^{-x}\,dx$ and interchanging integrations: \begin{align} \int_0^\infty t^{\alpha-1}e^{-t}\,\Gamma(\beta,t)\,dt &=\int_0^\infty x^{\beta-1}e^{-x}\int_0^x t^{\alpha-1}e^{-t}\,dt\,dx \\\color{gray}{[\text{substitute }t=xy]} &=\int_0^\infty x^{\alpha+\beta-1}e^{-x}\int_0^1 y^{\alpha-1}e^{-xy}\,dy\,dx \\\color{gray}{[\text{integrate over }x]} &=\int_0^1 y^{\alpha-1}\frac{\Gamma(\alpha+\beta)}{(1+y)^{\alpha+\beta}}\,dy \\\color{gray}{[\text{substitute }y/(1+y)=t]} &=\Gamma(\alpha+\beta)\int_0^{1/2}t^{\alpha-1}(1-t)^{\beta-1}\,dt. \end{align}