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I have noticed experimentally that: $$\int_0^1 \frac{\color{red}{x}}{x^4+2x^3+2x^2-2x+1} \,dx=\color{blue}{\frac{\pi}{8}},\tag{1}$$ $$\int_0^1 \frac{\color{red}{1-x^2}}{x^4+2x^3+2x^2-2x+1} \, dx=\color{blue}{\frac{\pi}{4}},\tag{2}$$ $$\int_0^1 \frac{\color{red}{1+x-x^2}}{x^4+2x^3+2x^2-2x+1} \, dx =\color{blue}{\frac{3\pi}{8}}.\tag{3}$$ So slight variations in the numerator always seem to produce something like $n\pi$, where $n$ is a rational number. At MathSE I have asked what the exact relationship is between $n$ and the numerator of the integrand, to which Quanto has responded with a general formula: $$\int_{0}^{1} \frac{ax^2 +b x + c}{x^4+2x^3+2x^2-2x+1} \, dx =\frac\pi8(c+b-a)+\frac\pi{3\sqrt3}(a+c).\tag{4}$$ However, is it necessary for the denominator to remain fixed? Not really. The following integral is formula (34) in this list of $\pi$ formulas: $$\int_0^1 \frac{\color{red}{16x-16}}{x^4-2x^3+4x-4}\,dx=\color{blue}{\pi}.\tag{5}$$ Notice that the denominator is different. But again, a slight variation in the numerator and it still produces something like $n\pi$: $$\int_0^1 \frac{\color{red}{x^2-x-1}}{x^4-2x^3+4x-4}\,dx=\color{blue}{\frac{3\pi}{16}}.\tag{6}$$ The fact that the denominator is not the same suggests that a further generalization is possible.

Here is my question: is it possible to characterize the integrand $\frac{P(x)}{Q(x)}$ in such a way that by simple inspection we can say $I=n\pi$? Or in other words, what should be the relationship between the coefficients of the numerator and the denominator for the integral to yield $n\pi$?

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    $\begingroup$ If Q is a real polynomial with no real roots, it is not unlikely that one finds as antiderivative a sum of arctan, by part.frac.dec. $\endgroup$
    – Pietro Majer
    Commented Jul 18, 2023 at 4:47
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    $\begingroup$ Yes, but very few rational combinations of arctan of rational numbers give rational multiples of $\pi$. $\endgroup$
    – Christophe Leuridan
    Commented Jul 18, 2023 at 12:14
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    $\begingroup$ The latter denominator ($x^4-2x^3+4x-4$) factors as $(x^2-2)(x^2-2x+2)$; the original denominator is irreducible over $\mathbb{Q}$, though, which I think makes it a much more interesting case. $\endgroup$
    – Steven Stadnicki
    Commented Jul 18, 2023 at 17:12
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    $\begingroup$ @StevenStadnicki Yes, although if you look at one of the answers on math.SE, you can see that the source of $\pi$ is still arguably the integral of $1/(1+y^2)$ "in disguise." $\endgroup$
    – Timothy Chow
    Commented Jul 18, 2023 at 23:18
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    $\begingroup$ @EmmanuelJoséGarcía I get the same thing Quanto got. It looks to me that you forgot that $dx = dt/(2\cos^2(t/2))$. $\endgroup$
    – Timothy Chow
    Commented Jul 20, 2023 at 3:26

2 Answers 2

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It is probably too optimistic to hope for a complete characterization of such integrands. For some context, see for example Some New Formulas for $\pi$, by Gert Almkvist, Christian Krattenthaler, and Joakim Petersson, Experimental Mathematics 12 (2003), 441–456. On the first page you can find the formula $$\int_0^1 {28x^6 − 56x^5 + 28x^4 − 97x^3 + 97x^2 − 6 \over (x^3 − x^2 + 2)^3}\,dx = {\pi\over 8},$$ but more importantly, they describe some general methods for searching for such formulas, and make it clear that their list is likely far from comprehensive.

A related paper is Integral Proofs that $355/113 > \pi$ by Stephen Lucas, Australian Mathematical Society Gazette 32 (2005), 263–266, which concerns itself with finding integrals such as $$\int_0^1 {x^8(1−x)^8(25+816x^2)\over 3164(1+x^2)\,dx}={355\over 113}-\pi.$$ Again, Lucas describes several empirical methods for searching for such formulas, but was not able to find a systematic characterization of all such formulas.

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  • $\begingroup$ $$\int_0^1{1\over x^3-x^2+2}={3\log2+\pi\over10}, \int_0^1{x\over x^3-x^2+2}={-6\log2+\pi\over20}$$ so $$\int_0^1{x+1\over x^3-x^2+2}={\pi\over4}$$ $\endgroup$
    – Gerry Myerson
    Commented Jul 18, 2023 at 23:41
  • $\begingroup$ @GerryMyerson: That suggests that by doing some arithmetic with the integrands, we can obtain more intimidating integrands that still yield $\pi$. The following one comes from adding integrands (2) and (5) with different denominators: $$\int_{0}^{1} \frac{2(1-x)(x^5-5x^4-10x^3-4x^2+8x-8)}{x^8-2x^6-2x^5+9x^4-2x^3-16x^2+12x-4}=\pi.$$ $\endgroup$
    – Emmanuel José García
    Commented Jul 19, 2023 at 15:30
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    $\begingroup$ @EmmanuelJoséGarcía The paper by Almkvist et al. describes how they used lattice-basis reduction methods to find suitable integer linear relations. This is a standard technique in experimental mathematics nowadays. $\endgroup$
    – Timothy Chow
    Commented Jul 19, 2023 at 20:47
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There are an infinite number of these. They reduce via partial fraction decomposition to inverse trig functions or logarithms in which the unusable parts cancel out except $\pi$ or some other constant depending on the choices of $P(x), Q(x)$.

Certain choices of denominator polynomials $Q(x)$ leads to an infinite series as done by Krattenthaler [3] via beta function. Any polynomial will not work even if it factors. It has to be in the form $z+x^a(1-x)^b$. Unfortunately, it does not generalize to multivariable polynomials. For example [2]:

$$\int_0^1 {x^4-5x^2+16 \over 48+x^2(1-x^2)^2}\,dx = \frac{\sqrt{3}\pi}{18}$$

Which can be put into the above form with substitution $x=y^{1/2}$

Attempts to generalize this for higher order constants are given by [1] requiring harmonic numbers.

[1]https://arxiv.org/abs/2307.03086

[2] https://papers.ssrn.com/sol3/papers.cfm?abstract_id=3919892

[3] https://www.plouffe.fr/simon/articles/Almkvist.pdf

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