Sum with Digamma function expressed by hypergeometric function?

Question

The function

$$f(x)=\sum_ {k = 0}^{\infty} \frac {(-1)^ k x^{2 k}} {2^k k! (2 k + 2)! }\left(\psi (k + 2)+ \frac{1}{2}\psi\left(k + \frac{3}{2}\right)\right)\tag{1}$$

with $x>0, x\in\mathbb{R}$ is defined by an infinite sum that contains two Digamma functions $\psi$. It was not possible for me to convert $f(x)$ to a known function. If in $f(x)$ the Digamma functions would be replaced by simpler functions (e.g. $k^2$) then the whole expression could be expressed by hypergeometric functions. Is it possible to express this sum as an hypergeometric function or by any other known function (except by more complex functions like MeijerG)?

What I tried

The Digamma functions can be expressed by finite sums $$\psi(k+2)=-\gamma+\sum_{i=1}^{k+1}\frac{1}{i}\tag{2}$$ $$\frac{1}{2}\psi\left(k + \frac{3}{2}\right)=-\frac{\gamma}{2}-\textrm{ln}(2)+\sum_{i=1}^{k+1}\frac{1}{2i-1}\tag{3}$$ with $\gamma\approx0.577$ (Euler-Mascheroni constant). If eqs.(2,3) are inserted in eq.(1) one gets

$$f(x)=-\left(\frac{3}{2}\gamma+\textrm{ln}(2)\right)\sum_ {k = 0}^{\infty} \frac {(-1)^ k x^{2 k}} {2^k k! (2 k + 2)! }+ \sum_{k = 0}^{\infty} \frac {(-1)^ k x^{2 k}} {2^k k! (2 k + 2)! }\sum_{i=1}^{k+1}\left(\frac{1}{i}+ \frac{1}{2i-1}\right)\tag{4}$$ The first sum in eq.(4) can be expressed as generalized hypergeometric function $_0F_2$ $$f(x)=-\frac{1}{2}\left(\frac{3}{2}\gamma+\textrm{ln(2)}\right) {_0}F_2\left(;\frac{3}{2},\frac{1}{2};-\frac{x^2}{8}\right)+ \sum_{k = 0}^{\infty} \frac {(-1)^ k x^{2 k}} {2^k k! (2 k + 2)! }\sum_{i=1}^{k+1}\left(\frac{1}{i}+ \frac{1}{2i-1}\right)\tag{5}$$

Unfortunately, eq.(5) now contains a nested sum that does not simplify eq.(1). (A nested sum results also if the Digamma functions are replaced by asymptotic series.) The nested sum in eq.(5) can be written as $$\sum_{k = 0}^{\infty} \frac {(-1)^ k x^{2 k}} {2^k k! (2 k + 2)! }\sum_{i=1}^{k+1}\left(\frac{1}{i}+ \frac{1}{2i-1}\right)=\sum_{k=0}^\infty (-1)^k x^{2k} c_k\tag{6}$$

with $$c_k=1,\frac{17}{288},\frac{101}{172800},\frac{1579}{812851200}, \frac{5129}{1755758592000},\frac{59989}{25493614755840000},\ldots$$ The first terms for $c_k$ are shown in the plot below.

The asymptotic approximation of $c_k$ is

$$m_k=\lim_{k\to\infty}c_k=\left(\frac{\textrm{e}}{2k}\right)^{3k} \tag{7}$$ or with another correction term $$n_k=\lim_{k\to\infty}c_k=m_k \ \frac{3 \gamma + 2\ \textrm{ln}(2) + 3\ \textrm{ln}(k)}{k^3\ 2^{9/2}\ \pi} \tag{8}$$

$m_k$ converges to $c_k$ concerning the absolute error $m_k-c_k$

whereas $n_k$ is also convergent for the relative error $(n_k-c_k)/c_k$

Answer 1

Too long for a comment but perhaps this is of some use. Consider for example the series $${{S}_{1}}=\sum\limits_{k=0}^{\infty }{\frac{{{\left( -1 \right)}^{k}}{{x}^{2k}}\psi \left( k+2 \right)}{{{2}^{k}}k!\left( 2k+2 \right)!}}$$ Note $$\psi \left( k+2 \right)=\frac{1}{\Gamma \left( k+2 \right)}\int\limits_{0}^{\infty }{{{t}^{k+1}}\log \left( t \right){{e}^{-t}}dt}$$ And so $${{S}_{1}}=\sum\limits_{k=0}^{\infty }{\frac{{{\left( -1 \right)}^{k}}{{x}^{2k}}}{{{2}^{k}}k!\left( 2k+2 \right)!\left( k+1 \right)!}}\int\limits_{0}^{\infty }{{{t}^{k+1}}\log \left( t \right){{e}^{-t}}dt}$$ Now from Watson pg 148 $${{J}_{0}}\left( z \right){{I}_{0}}\left( z \right)=\sum\limits_{k=0}^{\infty }{\frac{{{\left( -1 \right)}^{k}}{{\left( \tfrac{1}{2}z \right)}^{4k}}}{\Gamma \left( k+1 \right)\Gamma \left( 2k+1 \right)k!}}$$ so $$\frac{d}{dz}{{J}_{0}}\left( z \right){{I}_{0}}\left( z \right)=-4{{\left( \tfrac{1}{2}z \right)}^{3}}\sum\limits_{k=0}^{\infty }{\frac{{{\left( -1 \right)}^{k}}{{\left( \tfrac{1}{2}z \right)}^{4k}}}{k!\left( 2k+2 \right)!\left( k+1 \right)!}}$$ Now $${{S}_{1}}=\int\limits_{0}^{\infty }{\log \left( t \right){{e}^{-t}}\sum\limits_{k=0}^{\infty }{\frac{{{\left( -1 \right)}^{k}}{{\left( \tfrac{1}{2}{{\left( 8{{x}^{2}}t \right)}^{1/4}} \right)}^{4k}}}{k!\left( 2k+2 \right)!\left( k+1 \right)!}}dt}$$ and therefore $${{S}_{1}}=-\frac{d}{dx}\int\limits_{0}^{\infty }{\log \left( t \right){{I}_{0}}\left( {{\left( 8t{{x}^{2}} \right)}^{1/4}} \right){{J}_{0}}\left( {{\left( 8t{{x}^{2}} \right)}^{1/4}} \right){{e}^{-t}}dt}$$ Or perhaps $${{S}_{1}}=-\frac{4}{{{x}^{2}}}\int\limits_{0}^{\infty }{\log \left( t \right){{e}^{-t}}\frac{d}{dt}\left\{ {{I}_{0}}\left( {{\left( 8t{{x}^{2}} \right)}^{1/4}} \right){{J}_{0}}\left( {{\left( 8t{{x}^{2}} \right)}^{1/4}} \right) \right\}dt}$$ depending on where you want to go with it.

References:

Watson, G. N., “A Treatise on the theory of Bessel functions”, Cambridge University Press 1922 (1st edition)

Answer 2

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 :)