I am trying to show that:
$$\frac{2}{n}\log\Gamma\left(\frac{x}{2}\right) - \log\Gamma\left(\frac{x+n-1}{n}\right)$$
is an increasing function for $x \ge 5$ and $n > 2$
One way to do this would be to show that $\frac{x}{2} > \frac{x+n-1}{n}$ since:
$n(x-2) > 2x-2$
and then using $\frac{\Gamma'}{\Gamma}(s) = \int_0^{\infty}\left(\frac{e^{-t}}{t} - \frac{e^{-st}}{1 - e^{-1}}\right)dt$ to show that for $x \ge 5$, $n > 2$:
$$\frac{\Gamma'}{\Gamma}\left(\frac{x}{2}\right) - \frac{\Gamma}{\Gamma}\left(\frac{x+n-1}{n}\right)=\int_0^{\infty}\frac{1}{1-e^{-t}}(e^{-\frac{x+n-1}{n}t} - e^{-\frac{x}{2}})dt > 0$$
Is this the best way to show this? Is there a better way?
Thanks,
-Larry