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 $a,b =-2,-1,0,1$, $y \in (0,1)$
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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
$$\int_0^1 \dfrac{1}{x - r_j} \exp(-2\pi i x y)\ dx ={{\rm e}^{-2 \pi i yr_j}} \left( {\it Ei} \left( 1,-2\pi i yr_j \right) -{\it Ei} \left( 1,-2\pi i y \left( r_j-1 \right) \right) \right) $$
answered Jun 28, 2012 at 17:38
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2$\begingroup$ You can easily compute the roots (which are indeed simple) writing $\left(\frac{x-1}{x}\right)^{4q}=-1$. So $\frac{x-1}{x}$ is a root $\omega_j$ of $-1$ of order $4q$ ($j=1\dots 4q$), and $r_j=\frac{1}{1-\omega_j}$, which are $4q$ distinct numbers for $j=1\dots 4q$. $\endgroup$ Commented Jun 28, 2012 at 18:24