29

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?

Tito Piezas III
  • 54,565
  • 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. – Tito Piezas III Nov 26 '16 at 15:42
  • 5
    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! – Sangchul Lee Nov 27 '16 at 02:14
  • @SangchulLee: Ahh, very nice observation. There might be a simple explanation yet. – Tito Piezas III Nov 27 '16 at 02:17
  • 2
    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 ${\displaystyle \int_{-\infty}^{\infty}}\dfrac{(x-1)^2}{(2x - 1)^2 + (x^2 - x)^2},dx=\pi$ – Mark S. Nov 29 '16 at 03:06
  • @MarkS.: Thanks for the spot. – Tito Piezas III Nov 29 '16 at 03:48
  • @MarkS: I had supposed that for $p=11$, the integral might be equal to $y\pi$ for some algebraic number $y$, but Mathematica couldn't recognize it. If so, it must be of some high degree. – Tito Piezas III Nov 29 '16 at 03:53
  • 3
    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 http://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. – Martin Nicholson Nov 29 '16 at 10:43
  • @TitoPiezasIII: I left Mathematica 11 churning away at the four different $ p = 11 $ integrals. After symbolically evaluating the integrals, it seems to think that the $ -+ $ (top, bottom), and $ ++ $ cases are $ 2\pi $ times some degree 100 algebraic numbers (those minimal polynomials have coefficients on the order of $ 10^{302} $. It also claims the $ +- $ and $ -- $ cases are $ 2\pi $ times some degree 25 algebraic numbers. A numerical check to ~1000 decimal places seems to confirm this, but it doesn't feel right... – Steven Charlton Nov 30 '16 at 12:56
  • @StevenCharlton: The deg 25 algebraic numbers is promising. One way to test is to look at their discriminants $d$. For example, for $p=9$, the nonic's $d$ is divisible by the $21$st power of the sextic's $d$. Is it feasible to check $d$ of the deg 25 poly? – Tito Piezas III Nov 30 '16 at 13:11
  • @StevenCharlton: By the way, if you are interested, here is another problem that seemingly stops at $p=11$. – Tito Piezas III Nov 30 '16 at 13:39
  • @TitoPiezasIII: The discriminant of each degree 100 polynomial is divisible by the 1065th power of the denominator's discriminant. And the discriminant of each degree 25 polynomial is divisible by the 65th power of the denominator's discriminant. Maybe Mathematica is correct after all. – Steven Charlton Nov 30 '16 at 23:04
  • @StevenCharlton: Thanks! A further check is the constant term. It should have the form $d^m,w^n$ for powers greater than 1 and where $d$ is the denominator's discriminant and $w$ some integer. Can you post the smallest 25-deg as a partial answer? – Tito Piezas III Dec 01 '16 at 02:41

1 Answers1

8

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.

Tito Piezas III
  • 54,565
Steven Charlton
  • 2,002
  • 2
    +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. – Tito Piezas III Dec 01 '16 at 11:03