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Is there a closed form for the fallowing series, $$\sum_{n=1}^\infty \Gamma\left(n+\frac12+\frac12k\right)\Gamma\left(n+\frac12-\frac12k\right)\frac{x^{n}}{(2n+1)!}$$ where $k\notin\mathbb{Z}$.

I tried this using some properties of gamma function, but I didn't get any closed form. Also it can be proved that this series converges when $|x|<2$.

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    $\begingroup$ Mathematica seems to think so: Sum[Gamma[n + 1/2 + k/2] Gamma[n + 1/2 - k/2] x^n/(2 n + 1)!, {n, 1, \[Infinity]}] // FullSimplify gives the result: $$\frac{\pi \sec \left(\frac{\pi k}{2}\right) \left(2 \sin \left(k \csc ^{-1}\left(\frac{2}{\sqrt{x}}\right)\right)-k \sqrt{x}\right)}{k \sqrt{x}}$$ $\endgroup$
    – flinty
    Sep 6, 2021 at 15:33
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    $\begingroup$ We can write $\sin(kx)$ as a power series of $\sin(x)$ and it's related to this. $\endgroup$ Sep 6, 2021 at 15:53
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    $\begingroup$ Consider the product $\prod_{j=0}^{n-1}((2j+1)^2-k^2)$. $\endgroup$ Sep 7, 2021 at 18:01

2 Answers 2

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Consider the Chebyshev differential equation $(1-x^2)\frac{d^2y}{dx^2}-x\frac{dy}{dx}+k^2y=0$ which has the solution $y=a\sin(k\sin^{-1}(x))$ when $y(0)=0$ for abitrary $a\in\mathbb{C}$. Assuming $\sum_{k=0}^\infty a_kx^k$ to be a solution, we get $$a_{k+2}=\frac{k^2-\alpha^2}{(k+1)(k+2)}a_k\text{ for all }k\geq 0$$ Taking $a_0=0$ and $a_1\in\mathbb{R}$ to be arbitrary we get $$a_{2n}=0\text{ for all }n\geq 0$$ $$\text{and}$$ $$a_{2n+1}=\frac{a_1}{(2n+1)!}\prod_{j=0}^{n-1}((2j+1)^2-k^2)\text{ for all }n\geq 1$$ $$\implies \Gamma\left(1+\frac k2\right)\Gamma\left(1-\frac k2\right)a_{2n+1}=\frac{4^na_1}{(2n+1)!}\Gamma\left(n-\frac{1+k}2\right)\Gamma\left(n+\frac{k-1}2\right)$$ Using $\Gamma(z)\Gamma(1-z)\sin(\pi z)=\pi$ for $z\in\mathbb{C-Z}$ we get, $$a_{2n+1}=\frac{4^n k a_1}{(2n+1)!\pi}\Gamma\left(n-\frac{1+k}2\right)\Gamma\left(n+\frac{k-1}2\right)\cos\left(\frac{\pi k}2\right)$$ Finally we get, $$\sin(k\sin^{-1}(x))=kx\left(1+\frac{\cos\left(\frac{\pi k}2\right)}{\pi}\sum_{n=1}^\infty 4^n\Gamma\left(n-\frac{1+k}2\right)\Gamma\left(n+\frac{k-1}2\right)\frac{x^{2n}}{(2n+1)!}\right)$$ By the recurrence relation it is easy to check that the radius of convergence of this power series is $1$. Now taking $x\to \sqrt{x}/2$ we have $$\sum_{n=1}^\infty \Gamma\left(n+\frac12+\frac12k\right)\Gamma\left(n+\frac12-\frac12k\right)\frac{x^{n}}{(2n+1)!}=\frac{2\pi}{k\sqrt{x}\cos\left(\frac{\pi k}2\right)}\left(\sin\left(k\sin^{-1}\left(\frac{\sqrt{x}}2\right)\right)-k\frac{\sqrt{x}}2\right)$$ for all $0<|x|<2$ and, $0$ when $x=0$.

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    $\begingroup$ Thank you ! Nice one $\endgroup$
    – user965146
    Sep 8, 2021 at 11:08
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Your series equals $$ \sum_{n\geq 1}\frac{x^n}{2n+1}B\left(n+\frac{1}{2}+\frac{k}{2},n+\frac{1}{2}-\frac{k}{2}\right) $$ which, assuming $\text{Re}(k)<3$, equals $$ \sum_{n\geq 1}\frac{x^n}{2n+1}\int_{0}^{1} z^{n-1/2+k/2}(1-z)^{n-1/2-k/2}\,dz = 2 \int_{0}^{\pi/2}\sum_{n\geq 1}\frac{x^n}{2n+1}\sin(\theta)^{2n+k}\cos(\theta)^{2n-k}\,d\theta. $$ In order to ensure the validity of the next identities, we make the further assumptions $0<x<2$ and $\text{Re}(k)\in(-1,1)$. The RHS of the previous equation equals $$ 2\int_{0}^{\pi/2}\tan(\theta)^k \sum_{n\geq 1}\frac{(\sqrt{x} \sin\theta\cos\theta)^{2n}}{2n+1}\,d\theta = 2 \int_{0}^{+\infty}\frac{u^k}{u^2+1}\sum_{n\geq 1}\frac{1}{2n+1}\left(\frac{u\sqrt{x}}{u^2+1}\right)^{2n}\,du $$ or $$ 2\int_{0}^{+\infty}\frac{u^{k-1}}{\sqrt{x}}\left(-\frac{u\sqrt{x}}{u^2+1}+\operatorname{arctanh}\frac{u\sqrt{x}}{u^2+1}\right)\,du $$ or $$ -\frac{\pi}{\cos\frac{\pi k}{2}}+\frac{2}{\sqrt{x}}\int_{0}^{+\infty}u^{k-1}\operatorname{arctanh}\left(\frac{u\sqrt{x}}{u^2+1}\right)\,du $$ where the last integral can be computed through Feynman's trick. We have

$$\frac{\partial}{\partial \alpha}\int_{0}^{+\infty}u^{k-1}\operatorname{arctanh}\left(\frac{\alpha u}{u^2+1}\right)\,du = \int_{0}^{+\infty}\frac{u^k(1+u^2)}{(1-\alpha u+u^2)(1+\alpha u+u^2)}\,du$$ which equals $$ \frac{1}{2}\int_{0}^{+\infty}\frac{u^k}{1-\alpha u+u^2}\,du +\frac{1}{2}\int_{0}^{+\infty}\frac{u^k}{1+\alpha u+u^2}\,du. $$ It is not difficult to compute a closed form for this sum of moments through a partial fraction decomposition and the Beta function, then integrate the outcome with respect to $\alpha$, leading to the closed form mentioned in the comments.

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    $\begingroup$ Thank you. I didn't got it at once, I will try this $\endgroup$
    – user965146
    Sep 8, 2021 at 11:09

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