# 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 $\kappa,\lambda$:

• Are $x,y$ fixed? – Ethan Alwaise Feb 19 '16 at 23:10
• They are summed over; I've modified the description to make it clearer. – Victor V Albert Feb 19 '16 at 23:16
• 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. – Ethan Alwaise Feb 19 '16 at 23:39
• Try writing the right fraction as factorials and simplifying. – Simply Beautiful Art Feb 20 '16 at 0:20
• 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. – Victor V Albert Feb 23 '16 at 6:46

Consider the integral $$I=\int_0^1 \left[\left(\sqrt{1-t^2}+it\right)^{2k}+\left(\sqrt{1-t^2}-it\right)^{2k}\right]\cdot \left[\left(\sqrt{1-t^2}+it\right)^{2\lambda}+\left(\sqrt{1-t^2}-it\right)^{2\lambda}\right]\frac{dt}{\sqrt{1-t^2}}.$$
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 $$\sum_{x=0}^{\kappa}\sum_{y=0}^{\lambda}\left(-\right)^{x+y}{2\kappa \choose 2x}{2\lambda \choose 2y}\frac{{\kappa+\lambda \choose x+y}}{{2\left(\kappa+\lambda\right) \choose 2\left(x+y\right)}}=\frac{2^{4\kappa-1}}{{4\kappa \choose 2\kappa}}\delta_{k,\lambda},\quad k\neq 0.$$