# How to prove negativity of a $3\times3$ determinant whose elements involve trigamma, tetragamma, and pentagamma functions?

The classical Euler gamma function can be defined by the integral $$\begin{equation*} \Gamma(z)=\int_0^{\infty}t^{z-1}\operatorname{e}^{-t}\operatorname{d}t, \quad \Re(z)>0. \end{equation*}$$ Its logarithmic derivative $$\psi(z)=[\ln\Gamma(z)]'=\frac{\Gamma'(z)}{\Gamma(z)}$$ is called the digamma function. In sequence, the functions $$\psi'(z)$$, $$\psi''(z)$$, $$\psi'''(z)$$, and $$\psi^{(4)}(z)$$ are called the trigamma, tetragamma, pentagamma, and hexagamma functions respectively. In a word, the derivatives $$\psi^{(n)}(z)$$ for $$n\ge1$$ are called polygamma functions.

How to prove that the determinant $$\begin{gather*} \begin{vmatrix} \psi'(x) & \psi'\bigl(x+\frac{1}{2}\bigr) & 0\\ \psi''(x) & \psi''\bigl(x+\frac{1}{2}\bigr) & \psi'(x)-\psi'\bigl(x+\frac{1}{2}\bigr)\\ \psi'''(x) & \psi'''\bigl(x+\frac{1}{2}\bigr) & 2\bigl[\psi''(x)-\psi''\bigl(x+\frac{1}{2}\bigr)\bigr] \end{vmatrix}\\ =\begin{vmatrix} \int_{0}^{\infty}p(t,x)\operatorname{d}t & \int_{0}^{\infty} p(t,x)\operatorname{e}^{-t/2}\operatorname{d}t & 0\\ \int_{0}^{\infty}p(t,x) t\operatorname{d}t & \int_{0}^{\infty} p(t,x)\operatorname{e}^{-t/2}t\operatorname{d}t & \int_{0}^{\infty} p(t,x)(1-\operatorname{e}^{-t/2})\operatorname{d}t\\ \int_{0}^{\infty} p(t,x) t^2\operatorname{d}t & \int_{0}^{\infty} p(t,x)\operatorname{e}^{-t/2}t^2\operatorname{d}t & 2\int_{0}^{\infty} p(t,x)(1-\operatorname{e}^{-t/2})t\operatorname{d}t \end{vmatrix} \end{gather*}$$ is negative for $$x>0$$? where $$\begin{equation*} p(t,x)=\frac{t\operatorname{e}^{-xt}}{1-\operatorname{e}^{-t}}>0, \quad (t,x)\in(0,\infty)\times(0,\infty) \end{equation*}$$ and $$$$\psi^{(n)}(z)=(-1)^{n+1}\int_{0}^{\infty}\frac{t^{n}}{1-\operatorname{e}^{-t}}\operatorname{e}^{-zt}\operatorname{d}t, \quad \Re(z)>0, \quad n\in\mathbb{N}.$$$$ This problem is related to the problems at the sites How to prove the convexity of a simple function involving a ratio of two polygamma functions? and How to prove convexity of a simple function involving a ratio of two polygamma functions?.

In order to answer this question, it is sufficient to show $$\begin{equation*} \begin{vmatrix} \frac{\psi'(x)}{\psi''(x)} & \frac{\psi'(x+\frac{1}{2})}{\psi''(x+\frac{1}{2})}\\ 1-\frac{1}{2}\frac{\psi'''(x)}{\psi''(x)}\frac{\psi'(x)-\psi'(x+\frac{1}{2})}{\psi''(x)-\psi''(x+\frac{1}{2})} & 1-\frac{1}{2} \frac{\psi'''(x+\frac{1}{2})}{\psi''(x+\frac{1}{2})}\frac{\psi'(x)-\psi'(x+\frac{1}{2})}{\psi''(x)-\psi''(x+\frac{1}{2})} \end{vmatrix} >0, \quad x>0. \end{equation*}$$

• TeX note: $\operatorname{d}t$ \operatorname{d}t probably doesn't space as most people would want. See Should I \mathrm the d in my integrals? and questions linked there for best practices in TeXing integrals. (One common approach is to use $\mathrm dt$ \mathrm dt instead. Personally, I agree that the $\operatorname d$ is an operator, and so am stuck using the unsatisfactory ${\operatorname d}t$ {\operatorname d}t.) I didn't edit, in case this is really what you wanted. Commented Sep 23, 2023 at 19:14
• @LSpice In my opinion, the letter "d" should be regarded as an operator when it is used to express the differential in mathematics. So the simplest way to realize this is to define a command via \DeclareMathOperator{\td}{d} in the whole document or to employ \operatorname{d} locally and occationally. Moreover, the command "\td" comes from "\textup{d}" in AmS-LaTeX. Commented Sep 24, 2023 at 0:59
• Re, certainly you can do that, even in MathJax, but it will produce the (usually) undesireable result $\operatorname dt$. If you want that, then of course it's your choice, but notice the result is different from \textup{d}t, which doesn't include the extra spacing. Commented Sep 24, 2023 at 1:01
• @LSpice Many years ago, I used the command "\newcommand{\td}{\textup{d}}" and "\DeclareMathOperator{\td}{d\!}". Currently I am using the standard "\DeclareMathOperator{\td}{d}". Commented Sep 24, 2023 at 1:08
• Please read texts at the sites tex.stackexchange.com/questions/60545/… and tex.stackexchange.com/questions/60545/… Commented Sep 24, 2023 at 13:48