I try to calculate the following series \begin{align*} S_{n,m}(z)=\sum_{k=0}^{\infty} {\frac{(-1)^{k} \Gamma(\frac{2k+n+m}{2})\,\Gamma(\frac{2k+m-1}{2})}{k!\Gamma(k+\frac{m}{2})}} \, z^{2k}, \end{align*} where $\Gamma(z)$ is the Gamma function and where $n,m\in \mathbb Z^+$ (positive integers).

More precisely I would like to show that the series $S_{n,m}(z)$ is an elementary function.

I used the Legendre's duplication formula: \begin{align*} \Gamma (z)\;\Gamma \left(z+{\frac {1}{2}}\right)=2^{1-2z}\;{\sqrt {\pi }}\;\Gamma (2z), \end{align*} to simplify the expression of the series $S_{n,m}(z)$, but without success.

Otherwise, I thought of the hypergeometric functions: $$\displaystyle {}_{2}F_{1}(a,b;c;z)=\frac{\Gamma(c)}{\Gamma(a)\Gamma(b)}\sum_{n=0}^{\infty} \frac{\Gamma(a+n)\Gamma(b+n)}{\Gamma(c+n)} \frac{z^{n}}{n!}.$$ Then, returning back to $S_{n,m}(z)$, keeping in mind the expression of the hypergeometric functions $\displaystyle {}_{2}F_{1}(a,b;c;z)$, we get \begin{align*} S_{n,m}(z)&=\sum_{k=0}^{\infty} {\frac{(-1)^{k} \Gamma(\frac{2k+n+m}{2})\,\Gamma(\frac{2k+m-1}{2})}{k!\Gamma(k+\frac{m}{2})}} \, z^{2k}\\ &=\sum_{k=0}^{\infty} {(-1)^{k} \frac{\Gamma(k+\frac{n+m}{2})\,\Gamma(k+\frac{m-1}{2})}{\Gamma(k+\frac{m}{2})}} \, \frac{(z^{2})^k}{k!}\\ &=\frac{\Gamma(\frac{n+m}{2})\Gamma(\frac{m-1}{2})}{\Gamma(\frac{m}{2})}\, {}_{2}F_{1}\left(\frac{n+m}{2},\frac{m-1}{2};\frac{m}{2};-z^2\right)\\ &=\frac{\Gamma(\frac{n+m}{2})\Gamma(\frac{m-1}{2})}{\Gamma(\frac{m}{2})}\, {}_{2}F_{1}\left(\frac{m}{2}+\frac{n}{2},\frac{m}{2}-\frac{1}{2};\frac{m}{2};-z^2\right)? \end{align*} If that's right, how can I show that $S_{n,m}(z)$ is an elementary function?

  • $\begingroup$ What is the definition of an elementary function? And why should this sum in particular be one? $\endgroup$
    – LSpice
    Commented Nov 1, 2021 at 12:46
  • $\begingroup$ From Wikipedia, in mathematics, an elementary function is a function of a single variable (typically real or complex) that is defined as taking sums, products, and compositions of finitely many polynomial, rational, trigonometric, hyperbolic, and exponential functions, $\endgroup$
    – Z. Alfata
    Commented Nov 1, 2021 at 12:50
  • 1
    $\begingroup$ for $n$ an even integer, the sum $S_{n,m}(z)$ equals $(1+z^2)^{-(m+n-1)/2}$ times a polynomial in $z$ of degree $n$, so that is an "elementary" function; for odd $n$ there appear arcsinh functions, which you may or may not consider "elementary". $\endgroup$ Commented Nov 1, 2021 at 13:11
  • 1
    $\begingroup$ @CarloBeenakker ... of course $\operatorname{arcsinh}(z) = \log(z+\sqrt{1+z^2})$ is elementary. You should make your comment into an answer. $\endgroup$ Commented Nov 1, 2021 at 15:34

2 Answers 2


As Carlo noted, for $n$ an even integer, $S_{n,m}(z)$ is an elementary function of $z$.
What about $n$ odd?
When $n,m$ are both odd, I get something in terms of arcsinh, also elementary.

But for $n$ odd and $m$ even, Maple gets complete elliptic integrals, which are not elementary... Examples $$ S_{1,2}(z) = \frac{1}{\sqrt{z^2+1}}\;E\left(\frac{z}{\sqrt{z^2+1}\;}\right) $$ and $$ S_{7,6}(z) = 3{\frac {24\,{z}^{10}+148\,{z}^{8}+398\,{z}^{6}+669\,{z}^{4}-280 \,{z}^{2}-35}{ 16\left( {z}^{2}+1 \right) ^{11/2}{z}^{4}}{E} \left( {\frac {z}{\sqrt {{z}^{2}+1}}} \right) }-{\frac {72\,{z}^{8}+ 381\,{z}^{6}+864\,{z}^{4}-1575\,{z}^{2}-210}{32\, \left( {z}^{2}+1 \right) ^{11/2}{z}^{4}}{K} \left( {\frac {z}{\sqrt {{z}^{ 2}+1}}} \right) } $$


The arguments of the term ${}_{2}F_{1}\left(\frac{m}{2}+\frac{n}{2},\frac{m}{2}-\frac{1}{2};\frac{m}{2};-z^2\right)$ are not special. So, it is very unlikely that this term can be expressed anyhow other than tautologically. Mathematica cannot do anything with it:

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