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 $(n_k-c_k)/c_k$

  • $\begingroup$ Perhaps a clever combination of these dlmf.nist.gov/15.8.ii work. $\endgroup$
    – Gary
    Jul 10, 2021 at 18:47
  • 2
    $\begingroup$ In terms of the Meijer G-function, $$f(x) = -\frac {\pi^{3/2}} {8 \omega} G_{1, 4}^{2, 0} {\left( \omega \middle| { \frac 3 2 \atop 0, 1, \frac 1 2, \frac 3 2 } \right)} - \frac 1 {8 \omega} \, {_0 F_2} {\left( ; \frac 1 2, 1; -\omega \right)} + \frac {\ln \omega} 4 \, {_0 F_2} {\left( ; \frac 3 2, 2; -\omega \right)},$$ where $\omega = x^2/8$ (the residues of $\Gamma(s) \Gamma(1 + s)$ at the double poles give the polygamma terms). $\endgroup$
    – Maxim
    Jul 11, 2021 at 13:13
  • $\begingroup$ I added in my post that a MeijerG function is not desired. $\endgroup$ Jul 11, 2021 at 13:18
  • $\begingroup$ Suppose I give the "normal" expression for the G() term? $\endgroup$
    – rrogers
    Jul 11, 2021 at 21:12
  • $\begingroup$ For $\Re(z)>0$, the digamma function maybe expressed in terms of the Hypergeometric PFQ function as follows: $\psi(z)=(z-1)\, _3F_2(1,1,2-z;2,2;1)-\gamma$ . So you end up with an infinite sum of Hypergeometric PFQ functions instead of digamma functions. $\endgroup$ Jul 11, 2021 at 22:47

2 Answers 2


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.


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

  • $\begingroup$ Could you give the reference of Watson? $\endgroup$ Jul 11, 2021 at 21:44
  • $\begingroup$ @granularbastard reference added. note, first (1922) or second edition (1966) has the same page reference for the series representation of Bessel function products I've used. Also note, since you seem to want hypergeometric functions, that instead of a product of Bessel function you could represent the series as a ${}_{p}{{F}_{q}}$ hypergeometric function under the integral. $\endgroup$ Jul 11, 2021 at 23:35
  • $\begingroup$ A Bessel function would be also ok. The sense is to convert the sum in eq.(1) to a known function, e.g. hypergeometric function, Bessel function, without additional integrals. A MeijerG function is not desired as it is not implemented in standard software and somehow obscure. $\endgroup$ Jul 11, 2021 at 23:48

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)
$\Gamma\left(K\right)\left(K\right)_{k}=\intop_{t=0}^{\infty}e^{-t}\cdot e^{ln(t)\left(K-1\right)}\cdot t^{k}dt$

$\frac{d\left(\Gamma\left(K\right)\left(K\right)_{k}\right)}{dK}=\intop_{t=0}^{\infty}e^{-t}\cdot ln\left(t\right)t^{K-1}\cdot t^{k}dt$
$Proof. \psi\left(k+K\right)=\frac{1}{\Gamma\left(k+K\right)}\intop_{t=0}^{\infty}e^{-t}\cdot ln\left(t\right)t^{K-1}\cdot t^{k}dt=\frac{1}{\Gamma\left(k+K\right)}\frac{d\left(\Gamma\left(K\right)\left(K\right)_{k}\right)}{dK}$
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 :)
Proof. A result that is amusing; if you do the internal reductions the result is:
The definition of $\psi\left(\right)$ :)

So the “proof” could have used the definition rewritten as pochammer functions directly :)


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