Orthogonality relation as double sum of products of binomial coefficients 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 $\kappa,\lambda$:

 A: 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
$$
\sum_{x=0}^{\kappa}\sum_{y=0}^{\lambda}\left(-\right)^{x+y}{2\kappa \choose 2x}{2\lambda \choose 2y}\frac{{\kappa+\lambda \choose x+y}}{{2\left(\kappa+\lambda\right) \choose 2\left(x+y\right)}}=\frac{2^{4\kappa-1}}{{4\kappa \choose 2\kappa}}\delta_{k,\lambda},\quad k\neq 0.
$$
