How many integrals can give multiples of $\pi$?

Question

This question notes a few families of rational functions whose integrals (from $0$ to $1$) give rational multiples of $\pi$. A fairly straightforward explanation is given there and in the related Math.SE question. The results in the Math.SE post imply that the set of polynomials $p\in\mathbb{Q}[x]$ for which $\int_0^1\frac{p(x)}{x^4+2x^3+2x^2-2x+1}\ dx$ is a rational multiple of pi is a space of dimension two. In general, it's easy to construct polynomials $q(x)$ of degree $2n$ for which the set of polynomials $p$ with $\int_0^1\frac{p(x)}{q(x)}\ dx\in\mathbb{Q}\pi$ has dimension $n$: just sum enough inverse quadratics. Can we do any better than this? Specifically:

Is there a polynomial $q(x)$ of degree $n$ for which the space of polynomials $p(x)$ with $\int_0^1\frac{p(x)}{q(x)}\ dx\in\mathbb{Q}\pi$ has dimension greater than $n/2$?

Answer 1

Try $q(x) = (x^2+1)^3$ with $n = 6$. The space of polynomials $p(x)$ such that $\int_0^1 \frac{p(x)}{q(x)} \; dx \in \mathbb Q \pi$ is infinite-dimensional. If you meant polynomials of degree $< n$, that has dimension $4$, as the
following $4$ linearly independent polynomials $p(x)$ qualify: $$ 4 x - 3, x^2, 4 x^3 - 1, x^4 + 1 $$ Note that, if $J_n = \int_0^1 x^n/q(x)\; dx$, we have the recurrence $$J_{n+6} + 3 J_{n+2} + 3 J_{n+1} + J_n - \frac{16}{2(n+1)} J_1 = 0$$ Thus there are polynomials of all degrees $\ge 6$ that satisfy the condition.

Similarly, for your $q(x) = x^4+2x^3+2x^2-2x+1$, there are $p(x)$ with all degrees $\ge 5$ that satisfy the condition.