3
$\begingroup$

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?

$\endgroup$
4
  • $\begingroup$ 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

6
$\begingroup$

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 $$ S_{7,6}(z) = 3{\frac {24\,{z}^{10}+148\,{z}^{8}+398\,{z}^{6}+669\,{z}^{4}-280 \,{z}^{2}-35}{ 16\left( {z}^{2}+1 \right) ^{11/2}{z}^{4}}{E} \left( {\frac {z}{\sqrt {{z}^{2}+1}}} \right) }-{\frac {72\,{z}^{8}+ 381\,{z}^{6}+864\,{z}^{4}-1575\,{z}^{2}-210}{32\, \left( {z}^{2}+1 \right) ^{11/2}{z}^{4}}{K} \left( {\frac {z}{\sqrt {{z}^{ 2}+1}}} \right) } $$

$\endgroup$
1
$\begingroup$

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:

enter image description here

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.