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Let $${ a \choose b } = \frac{\Gamma(a+1)}{\Gamma(b+1)\Gamma(a-b+1)}$$ be the continuous extension of the binomial coefficient to non-integer arguments. I noticed this morning that $$\int_{\mathbb{R}}{t \choose x}^2{x \choose t}~dx = 1$$ For all real $t \geq0$. I tried to simplify the integrand according to the propeties on the MathWorld page for the gamma function, though I don't see where to apply even the reflection formula to:

$$\int_{\mathbb{R}}\frac{\Gamma(t+1)}{\Gamma(x+1)\,\Gamma(t-x+1)^2\,\Gamma(x-t+1)}~dx$$ To show that the above equals $1$. I couldn't find many similar questions on MSE about these sorts of integrals and am unfamiliar with how to approach them, so I bring this one here.