I have stumbled upon the following sum over $x,y$ for non-negative integers $\kappa,\lambda$: $$ \sum_{x=0}^{\kappa}\sum_{y=0}^{\lambda}\left(-\right)^{x+y}{2\kappa \choose 2x}{2\lambda \choose 2y}\frac{\displaystyle{\kappa+\lambda \choose x+y}}{\displaystyle{2\left(\kappa+\lambda\right) \choose 2\left(x+y\right)}}=\begin{cases} \dfrac{2^{4\kappa-1+\delta_{\kappa0}}}{\displaystyle{4\kappa \choose 2\kappa}}, & \kappa=\lambda\\ 0, & \kappa\neq\lambda \end{cases} $$ To reiterate, this sum is zero if $\kappa\neq\lambda$, so it looks to me like some orthogonality relation. Does anyone else know how to prove this or can anyone suggest a resource where similar sums are listed?

Below is the confirmation of the above for different $\kappa,\lambda$:

  • $\begingroup$ Are $x,y$ fixed? $\endgroup$ – Ethan Alwaise Feb 19 '16 at 23:10
  • $\begingroup$ They are summed over; I've modified the description to make it clearer. $\endgroup$ – Victor V Albert Feb 19 '16 at 23:16
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    $\begingroup$ So if $\kappa + \lambda$ is odd then I think $$(x,y) \to (\kappa - x, \lambda - y)$$ gives a bijection between terms of equal magnitude but opposite sign, since $${{n}\choose{k}} = {{n}\choose{n - k}}.$$ Not sure about the case $\kappa + \lambda$ is even. $\endgroup$ – Ethan Alwaise Feb 19 '16 at 23:39
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    $\begingroup$ Try writing the right fraction as factorials and simplifying. $\endgroup$ – Simply Beautiful Art Feb 20 '16 at 0:20
  • $\begingroup$ Thanks for the helpful comments! @EthanAlwaise, I think you are right. I wasn't able to simplify significantly by writing in factorials, but I can tell you that the ratio of x+y binomials can also be expressed as a ratio of double factorials. $\endgroup$ – Victor V Albert Feb 23 '16 at 6:46

Consider the integral $$ I=\int_0^1 \left[\left(\sqrt{1-t^2}+it\right)^{2k}+\left(\sqrt{1-t^2}-it\right)^{2k}\right]\cdot \left[\left(\sqrt{1-t^2}+it\right)^{2\lambda}+\left(\sqrt{1-t^2}-it\right)^{2\lambda}\right]\frac{dt}{\sqrt{1-t^2}}. $$

One can easily check that it equals \begin{align} I=&\ 4\sum_{x=0}^{\kappa}\sum_{y=0}^{\lambda}\left(-1\right)^{x+y}{2\kappa \choose 2x}{2\lambda \choose 2y}\int_0^1 t^{2x+2y}\left(1-t^2\right)^{k+\lambda-x-y-\frac{1}{2}}dt=\\ &2\sum_{x=0}^{\kappa}\sum_{y=0}^{\lambda}\left(-1\right)^{x+y}{2\kappa \choose 2x}{2\lambda \choose 2y}\int_0^1 t^{x+y-\frac{1}{2}}\left(1-t\right)^{k+\lambda-x-y-\frac{1}{2}}dt=\\ &2\sum_{x=0}^{\kappa}\sum_{y=0}^{\lambda}\left(-1\right)^{x+y}{2\kappa \choose 2x}{2\lambda \choose 2y}\frac{\Gamma\left(x+y+\frac{1}{2}\right)\Gamma\left(k+\lambda-x-y+\frac{1}{2}\right)}{\Gamma\left(k+\lambda+1\right)}=\\ &\frac{ 2^{1-k-\lambda}\pi}{(k+\lambda)!}\sum_{x=0}^{\kappa}\sum_{y=0}^{\lambda}\left(-1\right)^{x+y}{2\kappa \choose 2x}{2\lambda \choose 2y}(2x+2y-1)!!(2k+2\lambda-2x-2y-1)!!=\\ &\frac{ \pi (2k+2\lambda)!}{2^{2k+2\lambda-1}(k+\lambda)!^2}\sum_{x=0}^{\kappa}\sum_{y=0}^{\lambda}\left(-1\right)^{x+y}{2\kappa \choose 2x}{2\lambda \choose 2y}\frac{{\kappa+\lambda \choose x+y}}{{2\left(\kappa+\lambda\right) \choose 2\left(x+y\right)}}. \end{align}

On the other hand, substitution $t=\sin\theta$ shows that (it is assumed that $k>0$, since the case $k=\lambda=0$ is trivial) \begin{align} I=4\int_0^{\frac{\pi}{2}}\cos{2k\theta}\cos{2\lambda\theta}\ d\theta=\pi\delta_{k,\lambda}. \end{align}

So we proved that $$ \frac{ \pi (2k+2\lambda)!}{2^{2k+2\lambda-1}(k+\lambda)!^2}\sum_{x=0}^{\kappa}\sum_{y=0}^{\lambda}\left(-1\right)^{x+y}{2\kappa \choose 2x}{2\lambda \choose 2y}\frac{{\kappa+\lambda \choose x+y}}{{2\left(\kappa+\lambda\right) \choose 2\left(x+y\right)}}=\pi\delta_{k,\lambda},\quad k\neq 0, $$ which is equivalent to $$ \sum_{x=0}^{\kappa}\sum_{y=0}^{\lambda}\left(-\right)^{x+y}{2\kappa \choose 2x}{2\lambda \choose 2y}\frac{{\kappa+\lambda \choose x+y}}{{2\left(\kappa+\lambda\right) \choose 2\left(x+y\right)}}=\frac{2^{4\kappa-1}}{{4\kappa \choose 2\kappa}}\delta_{k,\lambda},\quad k\neq 0. $$


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