The polygamma function of order $2$ is defined as
$$\psi^{(2)}(z)= \frac{d^2}{dz^2} \psi(z) = \frac{d^{3}}{dz^{3}} \ln\Gamma(z)$$
where $\Gamma(z)$ is the usual gamma function: $\int_0^\infty x^{z-1}e^{-x}dx.$ Prove that this is strictly positive, or negative, or give the point(s) where the function switches sign.
I know that $\psi(z)$ and $\psi^{(1)}(z)$ are strictly decreasing/increasing and concave/convex respectively.
I'm trying to show that $\psi^{(1)}(z+h) -\psi^{(1)}(z)$ is either strictly negative or positive to prove a unique minimum exists.
I reformulated the difference as $\int_z^{z+h}\psi^{(2)}(y)dy$ and this means if $\psi^{(2)}(y)$ can be shown to be strictly positive for $y>2$ then I'm done but I don't think this is the case.
For $n \in \mathbb N$ and $x > 0$, we have an integral representation with a positive integrand: $$\psi^{(n)}(x) = (-1)^{n + 1} \int_0^\infty \frac {t^n e^{-x t}} {1 - e^{-t}} dt.$$