The polygamma function is generally given by
$$\psi^{(n)}(x)=\frac{d^{n+1}}{dx^{n+1}}\ln(\Gamma(x)),~n\in\mathbb N_{\ge0}$$
where $\Gamma$ is the gamma function. This can be extended to negative integers by letting
$$\psi^{(n-1)}(x)=\int_a^x\psi^{(n)}(x)$$
Unfortunately, there is no fixed $a$ such that the above holds true for any $n$, but it does capture the general idea.
Using fractional calculus, one can extend the polygamma function to complex values. Here, I could have
$$\psi^{(z)}_u(x)=\frac1{\Gamma(1-\{a\}-bi)}\frac{d^{\lceil a\rceil+2}}{dx^{\lceil a\rceil+2}}\int_u^x\frac{\ln(\Gamma(t))}{(x-t)^{\{a\}+bi}}~\mathrm dt$$
where $z=a+bi$, $a>-1$, $\lceil a\rceil$ is the ceiling function, and $\{a\}$ is the fractional part of $a$. $u$ could be any constant such that the integral converges. For $z\in\mathbb R^+$, the above expression reduces down to the normal polygamma function regardless of $u$ (at least for $x>0$).
For $a<-1$,
$$\psi^{(z)}_u(x)=\frac1{\Gamma(z)}\int_u^x(x-t)^{z-1}\ln(\Gamma(t))~\mathrm dt$$
Particularly of interest, I'm trying to find the closed form of $\psi^{(1/2)}_u(x)$.
$$\psi^{(1/2)}_u(x)=\frac1{\sqrt\pi}\frac{d^3}{dx^3}\int_u^x\frac{\ln(\Gamma(t))}{\sqrt{x-t}}~\mathrm dt$$
By applying Leibniz rule, this reduces down to
$$\frac{d}{dx}\int_u^x\frac{\ln(\Gamma(t))}{\sqrt{x-t}}~\mathrm dt=\lim_{y\to x}\left[\frac{\ln(\Gamma(y))}{\sqrt{x-y}}-\frac12\int_u^y\frac{\ln(\Gamma(t))}{(x-t)^{3/2}}~\mathrm dt\right]$$
However, I can't figure out how to simplify this, if possible. Trying to attack the original integral, then differentiating the result, I'm tempted to use this series expansion, but integrating term by term of that expression is quite nasty and the resulting series looks quite doubtful from anywhere near a closed form.
Also unsure if this is any help, but if we appropriately define $\sqrt z=|z|^{1/2}e^{i\theta/2}$, where $\theta\in[0,2\pi)$, one could attempt a keyhole contour of radius $u$ centered at $x$ to deduce that (hopefully this is correct)
$$\int_u^x\frac{\ln(\Gamma(t))}{\sqrt{x-t}}~\mathrm dt=\frac12\int_\pi^{3\pi}\frac{\Gamma(x+ue^{i\theta})}{\sqrt{-ue^{i\theta}}}~d\theta$$
But again, doesn't seem much helpful
What about applying a fractional derivative to each term of the Kummer-Malmsten Fourier series for $\log\Gamma$? It is not difficult to apply $D^{\alpha}$ (fractional derivative with respect to $x$) to the integral representation
$$ \psi(x)=\int_0^\infty \left(\frac{e^{-t}}{t}-\frac{e^{-xt}}{1-e^{-t}}\right)\, dt,$$ too.