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.

  • $\begingroup$ For clarification: You are asking for the proof concerning especially the values for the Trigamma Function $\psi^{(2)}(z)$ or am I mistaken? Hence it is not clear at all. Further you should consider to add some details of what you have tried so far since these kinds of question - "Here is the task. Do it for me!" - are poorly received on MSE. $\endgroup$ – mrtaurho Oct 7 '18 at 22:58
  • $\begingroup$ @mrtaurho, yes I'm interested in m=2 case, I've reformulated the question somewhat to clarify the context and what I'm trying to show. Please let me know if this still doesn't meet the standards for MSE. I've tried some bounds on $\psi^1(z)$ but I don't really know anything about $\psi^2(z)$. $\endgroup$ – Lucas Roberts Oct 7 '18 at 23:33

For $n \in \mathbb N$ and $x > 0$, we have an integral representation with a positive integrand: $$\psi^{(n)}(x) = (-1)^{n + 1} \int_0^\infty \frac {t^n e^{-x t}} {1 - e^{-t}} dt.$$

  • $\begingroup$ if I understand correctly this integral representation indicates that for odd $n$ the function is positive, yet for even $n$ the function is negative. Implying that it is not possible for $\psi^{(2)}(x) >0$. If you could either provide a demonstration of this integral derivation (or a link in your answer) I'll gladly mark as correct. $\endgroup$ – Lucas Roberts Oct 12 '18 at 15:17
  • $\begingroup$ Correct, $\psi^{(2)}(x)$ is negative for all positive $x$. Whittaker and Watson (A Course of Modern Analysis, Chapter XII) derive Gauss's integral expression for the digamma function by starting with the Weierstrass's definition of the gamma function, differentiating the product and writing the terms in the resulting sum $S$ as integrals. The formula above can be obtained by repeatedly differentiating Gauss's integral expression. Alternatively, repeatedly differentiating $S$ gives $\psi^{(n)}(x) = (-1)^{n+1} n! \sum_{k=0}^\infty (k+x)^{-n-1}$, where the summand on the rhs is positive. $\endgroup$ – Maxim Oct 12 '18 at 20:58
  • $\begingroup$ ah, I figured it was somewhere in that one :) here is a link to electronic archive for those interested in that masterpiece archive.org/details/courseofmodernan00whit/page/n7 $\endgroup$ – Lucas Roberts Oct 12 '18 at 22:02

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