Any ideas how to compute or to approximate integral
$$\int\_{0}^{1}\frac{(x+a)^{2q}(x+b)^{2q}}{(x-1)^{4q}+x^{4q}}\exp({-2\pi i x y})dx$$
where $q \in \mathbb{N}$ and $a,b =-2,-1,0,1$, $y \in (0,1)$



Fourier coefficients of a rational function
Let $r\_j, j=1\ldots,4q$ be the roots of the polynomial $(x-1)^{4q} + x^{4q}$. I suspect that these are distinct for all $q$ (I checked up to $q=100$). Then you have a partial fraction expansion
$$ \dfrac{(x-a)^{2q}(x-b)^{2q}}{(x-1)^{4q} + x^{4q}} = \frac{1}{2} + \sum\_{j=1}^{4q} \dfrac{a\_j}{x - r\_j}$$ where
 $a\_j = \lim\_{z \to r\_j} \dfrac{(z - r\_j)(z+a)^{2q}(z+b)^{2q}}{(z-1)^{4q} + z^{4q}}$
Each term
can be evaluated in terms of the exponential integral function: in Maple's notation


$$\int\_0^1 \dfrac{1}{x - r\_j} \exp(-2\pi i x y)\ dx ={{\rm e}^{-2 \pi i yr\_j}} \left( {\it Ei} \left( 1,-2\pi i yr\_j
 \right) -{\it Ei} \left( 1,-2\pi i y \left( r\_j-1 \right) \right) 
 \right)
$$ 


