If $$\frac{\Gamma \bigg(\frac{\nu+1}{2}\bigg) }{\Gamma (\frac{\nu}{2})\sqrt{\pi}}= \frac{\Gamma \bigg(\frac{\nu+1}{2}\bigg) }{\Gamma (\frac{\nu}{2})\Gamma (\frac{1}{2})}= \frac{1}{B \left(\frac{1}{2}, \frac{\nu}{2} \right)},$$ by way of the property $\Gamma \left( \frac{1}{2} \right) = \int_0^\infty x^{-1/2}e^{-x} \, dx = 2\int_0^\infty e^{-u^2} \, du = \sqrt{\pi},$
What is $$\frac{\Gamma \bigg(\frac{\nu+k}{2}\bigg) }{\Gamma (\frac{\nu}{2})(\nu\pi)^{\frac{k}{2}}}$$ equal to in terms of the beta function?