As always, watch for typo's with me.
I am not sure that “this really simplificates the problem”
We can use @mathstackuser12 's technique
$\psi\left(z\right)={\displaystyle \frac{1}{\Gamma\left(z\right)}\int_{t=0}^{\infty}}\left(t^{z-1}\cdot ln\left(t\right)\cdot e^{-t}\right)dt$
Incidentally for 16.5.3 we have
$\psi\left(z\right)={\displaystyle \frac{1}{\Gamma\left(z\right)}\int_{t=0}^{\infty}}\left(\frac{d\left(t^{\left(z-1\right)}\right)}{dz}\cdot e^{-t}\right)dt$
Setting $z=k+K$
$\psi\left(k+K\right)={\displaystyle \frac{1}{\Gamma\left(k+K\right)}\int_{t=0}^{\infty}}\left(t^{k+K-1}\cdot ln\left(t\right)\cdot e^{-t}\right)dt$
and from Sagemath
$f(x)=\sum_{k=0}^{\infty}\frac{(-1)^{k}x^{2k}}{2^{k}k!(2k+2)!}=\frac{1}{2}\,\,_{0}F_{2}\left(\begin{matrix}\\
2,\frac{3}{2}
\end{matrix};-\frac{1}{8}\,x^{2}\right)$
Applying the composition
$\int_{t=0}^{\infty}e^{-t}\cdot ln\left(t\right)\cdot t^{\left(K-1\right)}\cdot\sum_{k=0}^{\infty}\frac{(-1)^{k}x^{2k}}{2^{k}k!(2k+2)!}\cdot\frac{t^{k}}{\Gamma\left(k+K\right)}$
$\int_{t=0}^{\infty}e^{-t}\cdot ln\left(t\right)\cdot t^{\left(K-1\right)}\cdot\frac{1}{2\cdot\Gamma\left(K\right)}\cdot\,_{0}F_{3}\left(\begin{matrix}\\
2,\frac{3}{2},K
\end{matrix};-\frac{1}{8}\,tx^{2}\right)$
$\frac{1}{2\cdot\Gamma\left(K\right)}\cdot\int_{t=0}^{\infty}e^{-t}\cdot ln\left(t\right)\cdot t^{\left(K-1\right)}\cdot\frac{1}{2\cdot\Gamma\left(K\right)}\cdot\,_{0}F_{3}\left(\begin{matrix}\\
2,\frac{3}{2},K
\end{matrix};-\frac{1}{8}\,tx^{2}\right)$
Now we examine DLMF 16.5.3
$_{p}F_{q}\left(\begin{array}{c}
a_{0,}a_{1},\cdots,a_{p}\\
b_{1},\cdots,b_{q}
\end{array};y\right)={\displaystyle \frac{1}{\Gamma\left(a_{0}\right)}\int_{t=0}^{\infty}}\left(t^{a_{0}-1}\cdot e^{-t}\right)\cdot{}_{p}F_{q}\left(\begin{array}{c}
a_{1},\cdots,a_{p}\\
b_{1},\cdots,b_{q}
\end{array};t\cdot y\right)dt$
$\Gamma\left(a_{0}\right)\cdot_{p}F_{q}\left(\begin{array}{c}
a_{0,}a_{1},\cdots,a_{p}\\
b_{1},\cdots,b_{q}
\end{array};y\right)={\displaystyle \int_{t=0}^{\infty}}\left(e^{ln(t)\cdot(a_{0}-1)}\cdot e^{-t}\right){}_{p}F_{q}\left(\begin{array}{c}
a_{1},\cdots,a_{p}\\
b_{1},\cdots,b_{q}
\end{array};t\cdot y\right)dt$
$\frac{d}{da_{0}}\left(\Gamma\left(a_{0}\right)\cdot_{p}F_{q}\left(\begin{array}{c}
a_{0,}a_{1},\cdots,a_{p}\\
b_{1},\cdots,b_{q}
\end{array};y\right)\right)={\displaystyle \int_{t=0}^{\infty}}\frac{d}{da_{0}}\left(e^{ln(t)\cdot(a_{0}-1)}\right)\cdot e^{-t}{}_{p}F_{q}\left(\begin{array}{c}
a_{1},\cdots,a_{p}\\
b_{1},\cdots,b_{q}
\end{array};t\cdot y\right)dt$
$={\displaystyle \int_{t=0}^{\infty}}\left(t^{\left(a_{0}-1\right)}\cdot ln\left(t\right)\cdot e^{-t}\right){}_{p}F_{q}\left(\begin{array}{c}
a_{1},\cdots,a_{p}\\
b_{1},\cdots,b_{q}
\end{array};t\cdot y\right)dt$
Notice that the assignment $a_{0}=K $ has to be done after the differentiation. This is more obvious using the Mellin transform route. Which I wrote up, until I noticed 16.5.3 and decided this was more obvious; not involving outside transforms. Also the extra factor of 2 is constant and basically attached to the summation $\frac{1}{2}\,\,_{0}F_{2}\left(\begin{matrix}\\
2,\frac{3}{2}
\end{matrix};-\frac{1}{8}\,x^{2}\right)$
I would like to also like to mention that this process is term-wise and doesn't depend on the summation or HyperGeometric functions/series.
I would like to also like to mention that this process is term-wise and doesn't depend on the summation or HyperGeometric functions/series.
The conversion of $\psi(k+K)$ to semi-hypergeometric terms can be done via:
(You might want to skip to Lemma 2 :) but redundancy in alternate proofs is not always bad)
Lemma 1. $\psi\left(k+K\right)=\frac{d\left(\Gamma\left(K'\right)\left(K'\right)_{k}\right)}{dK'}\cdot\frac{1}{\Gamma\left(K\right)\cdot\left(K\right)_{k}}$ and $K'->K$ afterwards.
Proof. Let k be summation index and K be a parameter
i.e. $f(K,x)={\displaystyle \sum_{k=0}^{\infty}f_{k}(K)\cdot x^{k}}$
By definition (sometimes) $\Gamma\left(k+K\right)=\intop_{t=0}^{\infty}e^{-t}\cdot t^{k+K-1}dt$
$\left(K\right)_{k}=\frac{\Gamma\left(K+k\right)}{\Gamma\left(K\right)}=\frac{1}{\Gamma\left(K\right)}\intop_{t=0}^{\infty}e^{-t}\cdot t^{K-1}\cdot t^{k}dt $
(Which is just the core of DLMF 16.5.3)
Rewriting
$\Gamma\left(K\right)\left(K\right)_{k}=\intop_{t=0}^{\infty}e^{-t}\cdot e^{ln(t)\left(K-1\right)}\cdot t^{k}dt$
$\frac{d\left(\Gamma\left(K\right)\left(K\right)_{k}\right)}{dK}=\intop_{t=0}^{\infty}e^{-t}\cdot ln\left(t\right)t^{K-1}\cdot t^{k}dt$
$Proof. \psi\left(k+K\right)=\frac{1}{\Gamma\left(k+K\right)}\intop_{t=0}^{\infty}e^{-t}\cdot ln\left(t\right)t^{K-1}\cdot t^{k}dt=\frac{1}{\Gamma\left(k+K\right)}\frac{d\left(\Gamma\left(K\right)\left(K\right)_{k}\right)}{dK}$
$\psi\left(k+K\right)=\frac{1}{\Gamma\left(K\right)\cdot\left(K\right)_{k}}\frac{d\left(\Gamma\left(K\right)\left(K\right)_{k}\right)}{dK}$
QED
Which is amenable to semi-hypergeometrc and other summation.
If $f(K,x)={\displaystyle \sum_{k=0}^{\infty}f_{k}(K)\cdot x^{k}}$ then
$\sum_{k=0}^{\infty}f_{k}(K)\cdot x^{k}\cdot\psi\left(k+K\right)=\frac{1}{\Gamma\left(K\right)}\cdot\left[\frac{d}{dK'}\left(\sum_{k=0}^{\infty}f_{k}(K)\cdot\left[\begin{array}{c}
\Gamma\left(K^{'}\right)\left(K'\right)_{k}\\
\left(K\right)_{k}
\end{array}\right]\cdot x^{k}\right)\right]_{K'->K}$
Where the $K'->K$ reduction is done after the differentiation.
Lemma 2. Alternate proof of lemma 1 :)
$\psi\left(k+K\right)=\frac{1}{\Gamma\left(K\right)\cdot\left(K\right)_{k}}\frac{d\left(\Gamma\left(K\right)\left(K\right)_{k}\right)}{dK}$
Proof. A result that is amusing; if you do the internal reductions the result is:
$\psi\left(k+K\right)=\frac{1}{\Gamma(k+K)}\frac{d\left(\Gamma\left(k+K\right)\right)}{d\left(k+K\right)}$
The definition of $\psi\left(\right)$ :)
QED
So the “proof” could have used the definition rewritten as pochammer functions directly :)