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In my research I encountered the following double series involving reciprocals of Beta functions: \begin{equation} f(x,y) :=\sum_{j=0}^\infty \sum_{k=0}^\infty \frac{x^j}{j!}\frac{y^k}{k!} \frac{1}{\boldsymbol{B}(j+1,k+1)} = \ ? \end{equation} where $x,y$ are non-negative and $\boldsymbol{B}(j+1,k+1) := \frac{j!k!}{(j+k+1)!}. $

Is there a known closed form for $f(x,y)$?

Some ideas: 1.The summation over $k$ yields \begin{equation} f(x,y) = \sum_{j=0}^\infty \frac{x^j}{j!} (j+1) L_{-(j+2)}(y) = e^y\sum_{j=0}^\infty \frac{x^j}{j!} (j+1) L_{j+1}(-y), \end{equation} where $L_{n}(x)$ is the $n$-th Laguerre polynomial. In the same way, summing first over $j$ yields \begin{equation} f(x,y) = e^x\sum_{k=0}^\infty \frac{y^k}{k!} (k+1) L_{k+1}(-x). \end{equation}

  1. Alternatively, using the integral representation of the Gamma function it follows that \begin{equation} f(x,y) = \int_0^\infty I_0(2\sqrt{xt}) I_0(2\sqrt{yt}) t e^{-t} dt, \end{equation} where $I_0(x)$ is the zero-order modified Bessel function of the first kind.

  2. Thanks @Tyma Gaidash for pointing me towards hypergeometric functions in two variables. The series is a special case of a Humbert series: \begin{equation} f(x,y) = \Psi_2(2,1,1;x,y) \end{equation}

I am happy about ideas and suggestions. Thanks!

ARedder
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  • There definitely is a “closed form” in terms of hypergeometric functions. Not sure if these evaluate to anything more explicit for any $x,y$… – nejimban Feb 11 '23 at 09:46
  • Note: Mathematica finds $f(x,x)=\mathrm e^{2 x} ((2 x+1) I_0(2 x)+2 x I_1(2 x))$ where $I_0,I_1$ are modified Bessel functions of the first kind. – nejimban Feb 11 '23 at 09:50
  • Right I was already checking some properties of hypergeometric functions as $L_{-(k+2)}(x) = 1F1(2 + k, 1, x)$. – ARedder Feb 11 '23 at 09:56
  • Thanks for your suggestion. Appels series do not seem to work here (directly), due to the falling factorials in the expressions. By the way, it is fine to assume that |x|<1, |y|<1. – ARedder Feb 11 '23 at 14:45
  • You are right. Indeed the coefficients in the Appel series are rising factorials. I got confused since the notation on wiki is different. – ARedder Feb 11 '23 at 16:14
  • I derived the same integral representation earlier but with $t\operatorname e^{-t}$ instead of $\mathrm e^{-t}$. Are there particular values of $x$ and $y$ for which you seek a more explicit expression of $f(x,y)$? – nejimban Feb 11 '23 at 17:10
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    @TymaGaidash The Humbert series might help, there are some transformations in the literature. I will check for something useful. Thanks! – ARedder Feb 11 '23 at 17:34
  • @nejimban Thanks, the $t$ was missing. Indeed, the actual variables are $x=a z/(1+z) $ and $y=b /(1+z)$, where $a,b,z$ are positive real numbers and we can assume $|a|<1$ and $|b|<1$. Let me know if you are interested in the context. – ARedder Feb 11 '23 at 17:41
  • There is a simpler closed form for $x=-y$ too, but literature gives not much else. – Тyma Gaidash Feb 12 '23 at 00:34
  • Yes, I saw quite some closed forms for $x=-y$ here, but nothing useful for the general case with coefficients (2,1,1). – ARedder Feb 12 '23 at 09:08

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