$\int_0^1\int_0^1\binom{\text{something}}{\text{something}}\binom{\text{something}}{\text{something}}\cdot \text{something }dxdy$ with closed-form

Question

I've calculated an approximation of integrals like than $$\int_0^1\int_0^1\binom{f(x)}{f(y)}\binom{f(y)}{f(x)}dxdy\tag{1}$$ for simple functions $f(x)$. I don't know if some of these were in the literature or have a nice closed-form.

Question . I would like to know how to create, if it is feasible, nice examples of double integrals of binomials similar than $(1)$. Do you know how to calculate a nice example using different functions $$\int_0^1\int_0^1\binom{\text{something}}{\text{something}}\binom{\text{something}}{\text{something}}\cdot \text{something }dxdy\,?\tag{2}$$ If you know an example from the literature with a nice closed-form, please answer this question as a reference request, then I am going to try search such literature and read the example. Many thanks.

Your closed-form can be expressed as a series of special functions ( I'm especially interested in how to create such an example).

Answer 1

Well, first of all you can realise that:

$$\binom{\text{f}\left(x\right)}{\text{f}\left(y\right)}\cdot\binom{\text{f}\left(y\right)}{\text{f}\left(x\right)}=\frac{\sin\left(\pi\cdot\left(\text{f}\left(x\right)-\text{f}\left(y\right)\right)\right)}{\pi\cdot\left(\text{f}\left(x\right)-\text{f}\left(y\right)\right)}=$$ $$\sum_{\text{k}=0}^\infty\frac{\left(-1\right)^\text{k}\cdot\left(\text{f}\left(x\right)-\text{f}\left(y\right)\right)^{1+2\text{k}}\cdot\pi^{1+2\text{k}}}{\pi\cdot\left(\text{f}\left(x\right)-\text{f}\left(y\right)\right)\cdot\left(1+2\text{k}\right)!}\tag1$$

So taking the integral gives:

$$\int_0^1\int_0^1\binom{\text{f}\left(x\right)}{\text{f}\left(y\right)}\cdot\binom{\text{f}\left(y\right)}{\text{f}\left(x\right)}\space\text{d}x\space\text{d}y=$$ $$\int_0^1\int_0^1\frac{\sin\left(\pi\cdot\left(\text{f}\left(x\right)-\text{f}\left(y\right)\right)\right)}{\pi\cdot\left(\text{f}\left(x\right)-\text{f}\left(y\right)\right)}\space\text{d}x\space\text{d}y=$$ $$\int_0^1\int_0^1\sum_{\text{k}=0}^\infty\frac{\left(-1\right)^\text{k}\cdot\left(\text{f}\left(x\right)-\text{f}\left(y\right)\right)^{1+2\text{k}}\cdot\pi^{1+2\text{k}}}{\pi\cdot\left(\text{f}\left(x\right)-\text{f}\left(y\right)\right)\cdot\left(1+2\text{k}\right)!}\space\text{d}x\space\text{d}y=$$ $$\sum_{\text{k}=0}^\infty\frac{\left(-1\right)^\text{k}\cdot\pi^{1+2\text{k}}}{\pi\cdot\left(1+2\text{k}\right)!}\int_0^1\int_0^1\frac{\left(\text{f}\left(x\right)-\text{f}\left(y\right)\right)^{1+2\text{k}}}{\text{f}\left(x\right)-\text{f}\left(y\right)}\space\text{d}x\space\text{d}y\tag2$$

It was too big for a comment, just an idea.