Evaluation for some integer $k$ using wolfram alpha show that this integral
$\int_{0}^{n}\prod_{k=0}^{n}{\cot^{-1}(x^k+\frac{1}{x^k})}dx$ converges fast and deacreasing to $0$ , Now my question here is :
Question: What is the exact value of that integral if it is always converges for any integer $k$ ?
Note: I have edited the question because i meant the partial sum product not infinit product because infinit product here is :$0$
I guess you mean
$$\int_0^\infty\prod_{k=0}^n \cot^{-1}(x^k+x^{-k})$$
If $n=0$, it is the integral of a constant $\cot^{-1}2$, so it's infinite.
If $n=1$, the arccot is close to the reciprocal. So $$\int\cot^{-1}(x+1/x)dx\approx\int\frac1{x+1/x}dx\\ =\frac12\ln(x^2+1)|_0^\infty$$ which is infinite again.
If $n>1$, I doubt it has an 'exact' value. Again approximating the arccot by the reciprocal; and then approximating that reciprocal by $x^n$ when $x<1$ and $x^{-1}$ when $x>1$, $$\int_0^1\prod x^kdx+\int_1^\infty\prod x^{-k}dx\\=\int_0^1x^{n(n+1)/2}dx+\int_1^\infty x^{-n(n+1)/2}dx=\frac4{n(n+1)}$$ However, a more important effect is that $\cot^{-1}(x^k+x^{-k})$ never gets above $\cot^{-1}2\approx 0.464$, so each new factor reduces the final answer by that factor.
My back-of-the-envelope calculation is that it looks like $$\frac{0.464^n}{n^2}$$