# Showing a finite sum involving Gamma functions adds to zero

In the process of proving the Wrongskian identity for the Bessel function $J_\nu(x)J_{-\nu}'(x)-J_\nu'(x)J_{-\nu}(x)=-\frac{2\sin(\pi \nu)}{\pi x}$ (where the primes are differentiation by $x$), I came across the following sum: $$\sum_{n=0}^j\frac{j-2n+\nu}{n!(j-n)!\Gamma(n-\nu+1)\Gamma(j-n+\nu+1)}$$ where $\nu$ is not necessarily an integer. When $j=0$, this is equal to $\sin(\pi\nu)/\pi$ from the reflection relation for the gamma function. I am trying to show that for $j\geq 1$, this sum is equal to $0$. However, none of the typical methods for tackling this have succeeded, since $\nu$ is not an integer, and the terms do not cancel in an obvious manner. I have reduced it to the sum $$\frac{\sin(\pi \nu)}{\pi} \sum_{n=0}^j\frac{(-1)^n(j-2n+\nu)}{n!(j-n)!(\nu-n)_{j+1}}$$ where $(a)_n=a(a+1)...(a+n-1)$, which seems easier to work with, but I still do not get anywhere. Combining denominators gets $$\frac{\sin(\pi \nu)}{\pi\ j!(\nu-j)_{2j+1}} \sum_{n=0}^j (-1)^n(j-2n+\nu){{j}\choose{n}}(\nu-j)_{j-n}(\nu+j+1-n)_n$$ One can see that each power of $\nu$ cancels for specific $j$'s, but the general case is mess of Stirling numbers, and I feel like there is a better way of proving that this is $0$.

Edit: I have found my own answer, provided below, but other responses are still welcome.

I found a way to evaluate this sum using residues. In the following, let $b!$ denote $\Gamma(b+1)$ for non-integer $b$.

Rewrite the original sum as $$\sum_{n=0}^j \frac{2n-j-\nu}{n!(j-n)!(n-\nu)!(j-n+\nu)!} =\frac{1}{j!^2}\sum_{n=0}^j (2n-j-\nu)\binom{j}{n}\binom{j}{n-\nu}$$ Now we use $$\binom{a}{b}=\text{Res}_{x=0}\frac{(1+x)^a}{x^{b+1}},\ a\in\mathbb{Z}_{\geq 0}$$ Our sum then equals $$\text{Res}_{x=0}\frac{1}{j!^2}\sum_{n=0}^j (2n-j-\nu)\binom{j}{n}\frac{(1+x)^j}{x^{n-\nu+1}}=\text{Res}_{x=0}\frac{(1+x)^j}{j!^2 x^{1-\nu}}\sum_{n=0}^j (2n-(j+\nu))\binom{j}{n}\frac{1}{x^n}$$ Then we use $$\sum_{n=0}^j\binom{j}{n}x^n=(1+x)^j,\ \sum_{n=0}^j n\binom{j}{n}x^n=jx(1+x)^{j-1}$$ to see our sum is $$\text{Res}_{x=0}\frac{(1+x)^j}{j!^2 x^{1-\nu}}\left(2j\frac{1}{x}(1+1/x)^{j-1} -(j+\nu)(1+1/x)^j\right) =\text{Res}_{x=0}\frac{(1+x)^{2j-1}}{j!^2 x^{j+1-\nu}}(2j-(j+\nu)(1+x)) =\text{Res}_{x=0}\frac{(1+x)^{2j-1}}{j!^2 x^{j+1-\nu}}(j-\nu-(j+\nu)x)$$ $$=\frac{1}{j!^2}\left((j-\nu)\binom{2j-1}{j-\nu}-(j+\nu)\binom{2j-1}{j-\nu-1}\right)=\frac{1}{j!^2}\left((j-\nu)\binom{2j-1}{j-\nu}-(j-\nu)\binom{2j-1}{j-\nu}\right)=0$$ In evaluating the residue we assume $j>0$.

I am still interested in seeing if there are simpler or alternate techniques for showing this equality.

$\newcommand{\bbx}{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}{\displaystyle{#1}} \newcommand{\expo}{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}{\mathcal{#1}} \newcommand{\mrm}{\mathrm{#1}} \newcommand{\pars}{\left(\,{#1}\,\right)} \newcommand{\partiald}[]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}{\left\vert\,{#1}\,\right\vert}$ \begin{align} \mrm{f}\pars{\nu} & \equiv \sum_{n = 0}^{j}\pars{2n - j - \nu}{j \choose n}{j \choose n - \nu} \\[1cm] \mrm{f}\pars{\nu} & = \sum_{n = 0}^{j}\bracks{2\pars{j - n} - j - \nu} {j \choose j - n}{j \choose \bracks{j - n} - \nu} = -\sum_{n = 0}^{j}\bracks{2n -j + \nu}{j \choose n}{j \choose n + \nu} \\[5mm] & = -\,\mrm{f}\pars{-\nu} \\[1cm] \mrm{f}\pars{\nu} & = \sum_{n = -\infty}^{\infty}\pars{2n - j - \nu}{j \choose n} {j \choose n - \nu} = \sum_{n = -\infty}^{\infty}\pars{2n - j + \nu}{j \choose n + \nu} {j \choose n} = \,\mrm{f}\pars{-\nu} \end{align}

$$\bbox[15px,#ffe,border:1px dotted navy]{\ds{% \mrm{f}\pars{\nu} = -\,\mrm{f}\pars{-\nu}\quad\mbox{and}\quad \,\mrm{f}\pars{\nu} = \,\mrm{f}\pars{-\nu} \implies \bbox[15px,#eef,border:1px dotted navy]{\ds{\,\mrm{f}\pars{\nu} = 0}}}}$$

• Interesting proof; however, it seems you assume that $\nu$ is an integer in showing that $f(\nu)$ is even, since you shift the index $n$ by $\nu$. (Correct me if I am wrong.) – Grant B. Mar 20 '17 at 6:21
• @GrantB. Yes. You're right. It would be interesting to check the general result. I'll think about it. – Felix Marin Mar 20 '17 at 18:21