I try to calculate the following series
\begin{align*}
\sum_{k=0}^{\infty} {\frac{(-1)^{k}\, \Gamma(\frac{4k+n+2m-1}{2})\Gamma(\frac{8k+3n+4m-1}{4})\,\Gamma (\frac{8k+n+4m-3}{4})}{2^{-2k}k!\, \Gamma(k+\frac{m}{2})\Gamma(4k+n+2m-1)}} \, x^{2k} \qquad (*)
\end{align*}
where $\Gamma(z)$ is the Gamma function and where $n,m\in \mathbb Z^+$ (non-negative integers).
For this, this is what I tried to do. By using Legendre's duplication formula
\begin{align*}
\Gamma (z)\;\Gamma \left(z+{\frac {1}{2}}\right)=2^{1-2z}\;{\sqrt {\pi }}\;\Gamma (2z),
\end{align*}
we have $$ \Gamma(4k+n+2m-1)=\pi^{-1/2} 2^{4k+n+2m-1} \, \Gamma\left(\frac{4k+n+2m-1}{2}\right) \Gamma\left(\frac{4k+n+2m}{2}+\frac{1}{2}\right),$$
then, we get
\begin{align*}
(*) &= \sum_{k=0}^{\infty} {\frac{(-1)^{k}\Gamma(\frac{4k+n+2m-1}{2})\Gamma(\frac{8k+3n+4m-1}{4})\,\Gamma (\frac{8k+n+4m-3}{4})}{2^{-2k}k!\Gamma(k+\frac{m}{2})\Gamma(4k+n+2m-1)}} \, x^{2k} \\
&= \frac{\sqrt{\pi}}{2^{n+2m-1}} \sum_{k=0}^{\infty} {\frac{(-1)^{k}\, \Gamma(\frac{8k+3n+4m-1}{4})\,\Gamma (\frac{8k+n+4m-3}{4})}{2^{2k}k!\, \Gamma(k+\frac{m}{2})\Gamma(\frac{4k+n+2m}{2})}} \, x^{2k}\\
&= \frac{\sqrt{\pi}}{2^{n+2m-1}} \sum_{k=0}^{\infty} {\frac{(-1)^{k}\, \Gamma(\frac{8k+3n+4m-1}{4})\,\Gamma (\frac{8k+n+4m-3}{4})}{k!\, \Gamma(k+\frac{m}{2})\Gamma(\frac{4k+n+2m}{2})}} \, \left(\frac{x}{2}\right)^{2k}\\
&=\, …
\end{align*}
Can someone help me to continue the calculation!! Thank you in advance
Best Answer
If your last expression is correct,as you can expect, you face a very nasty hypergeometric function $$ \sum_{k=0}^{\infty} (-1)^{k}\frac{\Gamma \left(\frac{1}{4} (8 k+4 m+n-3)\right) \,\Gamma \left(\frac{1}{4} (8 k+4 m+3 n-1)\right)}{k!\, \Gamma \left(k+\frac{m}{2}\right)\, \Gamma \left(\frac{1}{2} (4 k+2 m+n)\right)} \, \left(\frac{x}{2}\right)^{2k}$$ is "just" (given by a CAS) $$\frac{\Gamma \left(\frac{4m+3 n-1}{4}\right) \Gamma \left(\frac{4 m+n-3}{4} \right)}{\Gamma \left(\frac{m}{2}\right) \Gamma \left(\frac{2m+n}{2}\right)}\, \, _4F_3\left(a,b,c,d;e,f,g;-x^2\right)$$
with $$a=\frac{4m+n-3}{8}\qquad b=\frac{4m+n+1}{8}\qquad c=\frac{4m+3n-1}{8}\qquad d=\frac{4m+3n+3}{8}$$ $$e=\frac{m}{2}\qquad f=\frac{2m+n}{4}\qquad g=\frac{2m+n+2}{4}$$