# Why does this pattern of "nasty" integrals stop?

I. Integrals

We have (typo corrected),

\begin{align} I_1 &=\pi =\int_{-\infty}^{\infty}\frac{(x-1)^2}{\color{blue}{(2x - 1)}^2 + (x^2 - x)^2}\,dx,\quad\text{(by Mark S.)}\\[1.8mm] I_3 &=\pi =\int_{-\infty}^{\infty}\frac{(x+1)^2}{\color{blue}{(x + 1)}^2 + (x^2 + x)^2}\,dx\\[1.8mm] I_5 &=\pi =\int_{-\infty}^{\infty}\frac{(x+1)^2}{\color{blue}{(x^2 - x - 1) }^2 + (x^2 + x)^2}\,dx\\[1.8mm] \color{red}{I_7} &=\pi =\int_{-\infty}^{\infty}\frac{(x+1)^2}{\color{blue}{(x^3 + 2x^2 - x - 1)}^2 + (x^2 + x)^2}\,dx\\[1.8mm] I_9 &=\pi =\int_{-\infty}^{\infty}\frac{(x-1)^2}{\color{blue}{(x^3 - 3x^2 + 1)}^2 + (x^2 - x)^2}\,dx\\[1.8mm] I_{11} &=\, ?? =\int_{-\infty}^{\infty}\frac{(x\pm1)^2}{\color{blue}{(x^5 + 3x^4 - 3x^3 - 4x^2 + x + 1)}^2 + (x^2 \pm x)^2}\,dx \end{align}

where those in blue are the minimal polynomials of $$x=\frac{1}{2\cos(2\pi/p)}$$ for $$p=1,3,5,7,9,11$$. These integrals have the form,

$$I_p =\int_{-\infty}^{\infty}\frac{(x\pm 1)^2}{F_p(x)}$$

where $$F(x)=0$$ is an equation with a solvable Galois group excepting $$p=11$$. The red integral $$I_7$$ is the one in the post, A nasty integral of a rational function,

$$\int_0^{\infty} \frac{x^8 - 4x^6 + 9x^4 - 5x^2 + 1}{x^{12} - 10 x^{10} + 37x^8 - 42x^6 + 26x^4 - 8x^2 + 1} \, dx = \frac{\pi}{2}$$ as well as in this post after some manipulation.

II. Question 1

Q: Why did the "pattern" of using minimal polynomials work then stop at $$p=11$$, and how can we make it continue by adjusting other parameters?

III. Alternative forms

Based on an insight from an old post, for $$p=7$$ we use the "negative" case on both its numerator and denominator. The denominator is a sextic again with a solvable Galois group, discriminant factor $$12833,$$ and we find,

$$\int_{-\infty}^{\infty}\frac{(x\color{red}-1)^2}{\color{blue}{(x^3 + 2x^2 - x - 1)}^2 + (x^2 \color{red}- x)^2}\,dx=\pi\sqrt{\frac{u}{\color{green}{12833}}}$$

where $$u$$ is a root of a nonic also with a solvable Galois group,

$$\small -\color{green}{12833}^3*1782434241^2 - 41120374319577904376201744753 u - 354521093943488815427187669 u^2 - 550802363395052799639795 u^3 - 176617825075778391189 u^4 + 116970252692553921 u^5 - 20201478347596 u^6 + 1625465206 u^7 - 63997 u^8 + u^9=0$$

For $$p=9$$, if we use the positive case, the denominator still is solvable. However, for $$p=11$$, then $$F(x) = 0$$ is not solvable for either case.

IV. Question 2

Q: So was the pattern interrupted because the denominator of $$p=11$$ no longer has a solvable Galois group?

• Can somebody pls check $p=11$? I'm using a really old version of Mathematica and maybe it just didn't evaluate the integral correctly. Nov 26, 2016 at 15:42
• It mildly reminds me of the case of Borwein's integral. I am not sure if it can ever be related to this kind of integral, but it is possible that the same mechanism lies behind. Interesting! Nov 27, 2016 at 2:14
• @SangchulLee: Ahh, very nice observation. There might be a simple explanation yet. Nov 27, 2016 at 2:17
• For $p=1$, the minimal polynomial of $\dfrac{1}{2\cos(2\pi/p)}$ is $2x-1$, not $2x+1$. ${\displaystyle \int_{-\infty}^{\infty}}\dfrac{(x+1)^2}{(2x - 1)^2 + (x^2 + x)^2}\,dx=\pi\sqrt{\dfrac{109+17\sqrt{73}}{146}}$. However, we do have ${\displaystyle \int_{-\infty}^{\infty}}\dfrac{(x-1)^2}{(2x - 1)^2 + (x^2 - x)^2}\,dx=\pi$ Nov 29, 2016 at 3:06
• Note that for the first 5 cases the blue polynomials are of 3rd or lower degree. If I'm not mistaken the equality of corresponding integrals to $\pi$ can be checked using Boole's theorem mathworld.wolfram.com/GlassersMasterTheorem.html by simple algebraic transformations of the integrand using fractional linear substitutions. However, when the degree of the blue polynomial is higher than 3 this reduction is not possible. I think this might be the reason why this the pattern fails.
– Nemo
Nov 29, 2016 at 10:43

Partial answer. Mathematica 11.0.1.0 (64-bit Windows version) seems to symbolically evaluate $$\int_{-\infty}^{\infty} \frac{(x - 1)^2}{(1 + x - 4 x^2 - 3 x^3 + 3 x^4 + x^5)^2 + (x^2 - x)^2} \, \mathrm{d}x =2\pi y$$ where $y$ is a root (the 5th root in Mathematica's ordering), of \begin{align*} \small F(y)=55936138949897200689844509841956235222126377325 - 2082209926471466695895506312399091645554188710590 y + 49399208260228586110040380712822122163293326842296 y^2 - 904097593617672391563622547821611243428330356636656 y^3 + 14127632726315977701496334077804393041066245226028208 y^4 - 192534883415138070802102412843131348551007666040509024 y^5 + 2248032708977729589700543648210682328879792825038892288 y^6 - 22010013756272539692699272127690186099540607721493676800 y^7 + 177728824048935169179013735666882776433001119535910888192 y^8 - 1170270214760621202108304618484485542592211842152325435904 y^9 + 6226689208769791815298929222960276164825821302955689534464 y^{10} - 26437408929821178291367173439675032999610116594417230776320 y^{11} + 87275205150008062398776420803782617539547332212906935361536 y^{12} - 209632027731557385765045313738415590487122817707525011718144 y^{13} + 284829590179494874220555955086122649413365826411704845058048 y^{14} + 245738741392479529396402731465119601079938307163739661565952 y^{15} - 2744252632383133719563152613313766366008892259189754592296960 y^{16} + 9042239242455966498125021473251288480014205602523431668940800 y^{17} - 19642481348541153825949628077511851598796849639028033440972800 y^{18} + 31384454408136427453055038714389257858518560896664228069376000 y^{19} - 37847103175390150688294536889184184478935891337063789625344000 y^{20} + 34290036775233047407263179281808801381553538237009356390400000 y^{21} - 22732262960008031643227099738915285612779131750417374904320000 y^{22} + 10440433388762840105269721355193655567662001399784538112000000 y^{23} - 2974656530310569079556114222635017838466182586996359168000000 y^{24} + \color{blue}{831141777440}^5 y^{25}=0 \end{align*}

The discriminant $d$ of the integrand's denominator

$$G(x)=(1 + x - 4 x^2 - 3 x^3 + 3 x^4 + x^5)^2 + (x^2 - x)^2$$

is $d=-2^5\times\color{blue}{831141777440}$. The discriminant of the $25$-deg $F(y)$ is divisible by $d^{65}$. However, its constant term is not integrally divisible by $d$.

Moreover, $y' = 831141777440 y$ is an algebraic integer. The discriminant of the minimal polynomial of $y'$ is divisible by $d^{246}$. The constant term is divisible by $d^{10}$, but the quotient is not a perfect power of an integer.

The positive case ++ passes the first test, but also fails on the second. The constant term is divisible by $d^{990}$, but the quotient is not a perfect power.

• +1. I made some edits to smoothen the flow. I guess the second test was just a peculiarity of the solvable nonic. But I believe the primary test is, given the discriminant $d$ of the denominator $G(x)$, then the discriminant of $F(y)$ should be divisible by a very high power of $d$, in this case $d^{65}$. The odds of a non-valid polynomial having that property seem to be nil. Plus the fact that it just so happens $d^5$ appears in the leading coefficient is a good sign of $2F(y)$'s validity. Dec 1, 2016 at 11:03