<s>[INST] Local maximum uniqueness of the logarithm of the Student-t distribution The negative logarithm of the Student-t distribution partial density function is
$$f(\nu,x) := -\ln\Gamma\left(\frac{\nu+1}{2}\right)
            +\ln\Gamma\left(\frac{\nu}{2}\right)
            +\frac{1}{2}\ln(\pi\nu)
            +\frac{\nu+1}{2}\ln\left(1+\frac{x^2}\nu\right)$$
How would one prove or disprove there is only one local minimum with respect to $\nu>0$ for any given $x$?
Numerical computation seems to suggest $f(\nu,x)$ strictly decreases with $\nu\in(0,\infty)$ for $x\in[0,1.5]$, and $f(\nu,x)$ is convex in $(0,a)$ and concave in $(a,\infty)$ for some $a>0$ for $x\in (b,\infty)$ for some $b\ge 1.5$.

To facilitate the solution, I post the first and second partial derivative of $f$ as follows.
\begin{align}2\frac{\partial f}{\partial \nu}=\frac{1-x^2}{\nu+x^2}+\ln\Big(1+\frac{x^2}\nu\Big)-\int_0^\infty \frac{e^{-\frac \nu2t}}{1+e^{-\frac t2}}\,dt,
\end{align}
$$4\frac{\partial^2 f}{\partial \nu^2}=-2\frac{\nu+x^4}{\nu(\nu+x^2)^2}+\int_0^\infty \frac{te^{-\frac \nu2t}}{1+e^{-\frac t2}}\,dt$$
 [/INST] Just some thoughts (I will add proofs in the future, if possible.)
For convenience, we replace $x^2$ with $y$.
We have
$$
 \frac{\partial f}{\partial v} 
 = \frac{1 - y}{2v + 2y} + \frac12 \ln \left(1 + \frac{y}{v}\right) - \frac12 \psi\left(\frac{v + 1}{2}\right) + \frac12\psi\left(\frac{v}{2}\right)
$$
where $\psi(\cdot)$ is the digamma function defined by $\psi(u) = \frac{\mathrm{d} \ln \Gamma(u)}{\mathrm{d} u} = \frac{\Gamma'(u)}{\Gamma(u)}$.
(i) If $0 < y \le 1$, then
\begin{align*}
 \frac{\partial f}{\partial v}
 &\le \frac{1 - y}{2v} + \frac12 \cdot \frac{y}{v}
 - \frac12 \left(\ln \frac{v + 1}{2} - \frac{2}{v + 1}\right) + \frac12\left(\ln \frac{v}{2} - \frac{1}{v}\right)\\
 &= \frac{1}{v + 1} - \frac12\ln\left(1 + \frac{1}{v}\right)\\
 &< 0
\end{align*}
where we have used $\ln z - \frac{1}{z} \le \psi(z) \le \ln z - \frac{1}{2z}$ for all $z > 0$, and $\ln(1 + u) \le u$ for all $u \ge 0$.
Note: The last inequality is easy to prove by taking derivative.
(ii) If $1 < y \le 1 + \sqrt{2}$, then
\begin{align*}
 \frac{\partial f}{\partial v}
 &\le \frac{1 - (1 + \sqrt2)}{2v + 2(1 + \sqrt2)} + \frac12 \ln \left(1 + \frac{1 + \sqrt2}{v}\right)\\
 &\quad - \frac12\left(\ln \frac{v + 1}{2} - \frac{1}{v + 1} - \frac{1}{3(v + 1)^2}\right)\\
 &\quad + \frac12\left(\ln \frac{v}{2} - \frac{1}{v} - \frac{1}{12(v/2 + 1/14)^2}\right)\\
 &< 0
\end{align*}
where we have used $\ln u - \frac{1}{2u} - \frac{1}{12u^2} < \psi(u) < \ln u - \frac{1}{2u} - \frac{1}{12(u + 1/14)^2}$ for all $u > 0$ (Theorem 5, [1]), and $y \mapsto \frac{1 - y}{2v + 2y} + \frac12 \ln \left(1 + \frac{y}{v}\right)$ is strictly increasing on $(1, \infty)$.
Note: The last inequality is easy to prove by taking derivative.
Remark: Where does $1 + \sqrt2$ come from? We rewrite $\frac{\partial f}{\partial v}$ as
$$\frac{\partial f}{\partial v}
= \frac{v \mathrm{e}^{A}}{2(v + y)}
\left\{\left(1 + \frac{y}{v}\right)\mathrm{e}^{-A}
\ln \left[\left(1 + \frac{y}{v}\right)\mathrm{e}^{-A}\right]
+ \left(1 + \frac{1}{v}\right)\mathrm{e}^{-A}\right\}$$
where
$A = 1 + \psi\left(\frac{v + 1}{2}\right) - \psi\left(\frac{v}{2}\right)$.
From $\frac{\partial f}{\partial v} = 0$, we solve
$y = {{\mathrm e}^{{\mathrm{LambertW}}\left(
  - (1 + 1/v)\mathrm{e}^{-A}
  \right) +A}}\,v - v$
where $W(\cdot)$ is the Lambert W function.
Maple tells us $\lim_{v\to \infty} \left( {{\mathrm e}^{{\mathrm{LambertW}}\left(
  - (1 + 1/v)\mathrm{e}^{-A}
  \right) +A}}\,v - v\right) = 1 + \sqrt{2}$.
By the way, @heropup also pointed out in comment that the point $x = \sqrt{1 + \sqrt{2}}$ is a demarcation point.
(iii) If $y > 1 + \sqrt{2}$ is fixed, we claim that $f$ has exactly one global minimizer $v^\ast$ on $(0, \infty)$, furthermore, $f$ is strictly decreasing on $(0, v^\ast)$ and strictly increasing on $(v^\ast, \infty)$.
We can prove the claim if the following conjecture is true:
Conjecture 1:  $Q < 0$ for all $v > 0$, where
$$Q = - 4 - v(v + 2)^2 \int_0^\infty \frac{t^2\mathrm{e}^{-vt/2}}{1 + \mathrm{e}^{-t/2}}\mathrm{d} t
+ (6v^2 + 16v + 8)\int_0^\infty \frac{t\mathrm{e}^{-vt/2}}{1 + \mathrm{e}^{-t/2}}\mathrm{d} t.$$
Numerical evidence shows its truth. However, we have not yet proved it.
References
[1] L. Gordon, “A stochastic approach to the gamma function”, Amer. Math. Monthly, 9(101), 1994, 858-865.
</s>