Any ideas how to compute or to approximate integral $$\int_{0}^{1}\frac{(x+a)^{2q}(x+b)^{2q}}{(x1)^{4q}+x^{4q}}\exp({2\pi i x y})dx$$ where $q \in \mathbb{N}$ and $a,b =2,1,0,1$, $y \in (0,1)$
Let $r_j, j=1\ldots,4q$ be the roots of the polynomial $(x1)^{4q} + x^{4q}$. I suspect that these are distinct for all $q$ (I checked up to $q=100$). Then you have a partial fraction expansion $$ \dfrac{(xa)^{2q}(xb)^{2q}}{(x1)^{4q} + x^{4q}} = \frac{1}{2} + \sum_{j=1}^{4q} \dfrac{a_j}{x  r_j}$$ where $a_j = \lim_{z \to r_j} \dfrac{(z  r_j)(z+a)^{2q}(z+b)^{2q}}{(z1)^{4q} + z^{4q}}$ Each term can be evaluated in terms of the exponential integral function: in Maple's notation $$\int_0^1 \dfrac{1}{x  r_j} \exp(2\pi i x y)\ dx ={{\rm e}^{2 \pi i yr_j}} \left( {\it Ei} \left( 1,2\pi i yr_j \right) {\it Ei} \left( 1,2\pi i y \left( r_j1 \right) \right) \right) $$ 

