Monotony of $x\psi(x)$ and $x^2\psi(x)$ on the positive reals. Are $f(x)=x\psi(x)$ and $g(x)=x^2\psi(x)$ monotonically increasing on $x\in(0,\infty)$?
I found $f^\prime(x)=\psi(x)+x\psi^{(1)}(x)$ and $g^\prime(x)=2x\psi(x)+x^2\psi^{(1)}(x)$ so if I could show these derivatives are positive then the answer would be yes.
 A: Neither of these functions are monotone increasing on $x\in(0,\infty)$.
Take for example $f(x)=x\psi(x)$. Evaluating the derivative gives $f^\prime(x)=\psi(x)+x\psi^{(1)}(x)$. Using the Laurent expansion for the polygamma function we find as $x\to0^+$
$$
f^\prime(0)=-\gamma=-0.577\dots,
$$
which is negative. However, using this relation we easily show
$$
f^\prime(1)=\frac{\pi^2}{6}-\gamma=1.067\dots,
$$
which is positive.
By the continuity of $f^\prime(x)$ on $x\in(0,\infty)$ and the intermediate value theorem we conclude $f^\prime$ must have at least one root on the interval $x\in(0,1)$; thus, $x\psi(x)$ cannot be monotonically increasing everywhere on $x\in(0,\infty)$.
Using this same line of reasoning we can choose $g(x)=x^2\psi(x)\implies g^\prime(x)=2x\psi(x)+x^2\psi^{(1)}(x)$ and show $g^\prime(0)=-1<0$, while $g^\prime(1)=\pi^2/6-2\gamma=0.49\dots >0$ and then again call on the intermediate value theorem to show $x^2\psi(x)$ cannot be monotonically increasing everywhere on $x\in (0,\infty)$.
