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V. Reshetnikov gave the interesting integral,

$$\int_0^1\frac{\mathrm dx}{\sqrt[4]x\ \sqrt{1-x}\ \sqrt[4]{2-x\,\sqrt3}}=\frac{2\,\sqrt2}{3\,\sqrt[8]3}\pi\tag1$$

After some experimentation, it turns out that more generally, given some integer/rational $N$, we are to find an algebraic number $\gamma$ that solves,

$$\int_0^1\frac{dx}{\sqrt[4]x\ \sqrt{1-x}\ \sqrt[4]{1-x\,\gamma^2}}=\frac{1}{N}\,\frac{2\pi}{\sqrt{2\gamma}}\tag2$$

(Compare to the similar integral in this post.) Equivalently, to find $\gamma$ such that,

$$\begin{aligned} \frac{1}{N} &=I\left(\gamma^2;\ \tfrac14,\tfrac14\right)\\[1.8mm] &= \frac{B\left(\gamma^2;\ \tfrac14,\tfrac14\right)}{B\left(\tfrac14,\tfrac14\right)} \end{aligned} \tag3$$

with beta function $B(a,b)$, incomplete beta $B(z;a,b)$ and regularized beta $I(z;a,b)$, and $B\left(\tfrac14,\tfrac14\right)=\frac{\sqrt\pi}{\Gamma^2\left(\frac14\right)}$. Reshetnikov's example, after tweaking, was just the case $N=\frac{3}{2}$ and $\gamma=\frac{3^{1/4}}{\sqrt{2}}$.

Solutions for prime $N=2,3,5,7$ are known. Let $v=\gamma$, then, $$-1 + 2 v^2 = 0\quad\quad N=2\\ - 1 + 2 v + 2 v^2 = 0\quad\quad N=3\\ - 1 + 8 v - 4 v^2 - 8 v^3 + 4 v^4 = 0\quad\quad N=5$$ etc, with $N=7$ using a $12$-deg equation. I found these using Mathematica's FindRoot command but, unlike the other post, I couldn't find a nice common form for $\gamma$. (The pattern of this family is also different. I had expected $N=7$ to also involve a sextic only.)

Q: Is it true one can find algebraic number $\gamma$ for all prime $N$? What is it for $N=11$?

Update, Aug 16, 2019

In this comment, Reshetnikov gave the explicit solution to,

$$I\left(\gamma^2;\ \tfrac14,\tfrac14\right) = \tfrac17$$ as, $$\small\gamma = \frac16\left(5\cos x-\sqrt3\sin x-1-\sqrt3\sqrt{7+4\sqrt7-(11+2\sqrt7)\cos x+\sqrt3(5+2\sqrt7)\sin x}\right)$$ where $x = \tfrac13\arccos\big(\tfrac{13}{14}\big)$.

P.S. I forgot I also found $\gamma$ in this 2016 post as,

$$\gamma = \tfrac12\left(2\cos\tfrac{2\pi}7-\sqrt{2\cos\tfrac{4\pi}7+\sqrt2\csc\tfrac{9\pi}{28}}\right)$$

pisco
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Tito Piezas III
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4 Answers4

11

For any rational $0<r<1$, there exists an algebraic $\gamma$ such that $$\int_0^1\frac{dx}{\sqrt[4]x\ \sqrt{1-x}\sqrt[4]{1-x\,\gamma^2}}=r\frac{2\pi}{\sqrt{2\gamma}}$$ Such $\gamma$ can always be expressed by radicals.

Following the first half of Achille hui's solution here, we have $$F(\mu ) = \int_0^1 {\frac{1}{{\sqrt[4]{x}\sqrt {1 - x} \sqrt[4]{{1 - \mu x}}}}dx} = \frac{{4\sqrt 2 {\pi ^{3/2}}}}{{{\Gamma ^2}(\frac{1}{4})}}\frac{1}{{{\mu ^{1/4}}}}\int_0^\omega {\frac{1}{{\sqrt {1 + {t^4}} }}dt} $$ with $\omega = {\left( {\frac{\mu }{{1 - \mu }}} \right)^{1/4}}$, let $\zeta = e^{i\pi/4}$, then $$F(\mu) = \frac{{4\sqrt 2 {\pi ^{3/2}}}}{{{\Gamma ^2}(\frac{1}{4})}}\frac{\zeta }{{{\mu ^{1/4}}}}\int_0^{\omega \zeta } {\frac{1}{{\sqrt {1 - {t^4}} }}dt} $$

Now let $\wp(z)$ be the Weierstrass elliptic function with $g_2 = 1,g_3 = 0$, its periods are $\{\lambda, \lambda i\}$ where $\lambda = \frac{1}{{2\sqrt \pi }}{\Gamma ^2}(\frac{1}{4})$, and $$x = \sqrt 2 \int_0^{1/\sqrt {2\wp (x)} } {\frac{1}{{\sqrt {1 - {t^4}} }}dt} $$ modulo an appropriate period lattice. Therefore, $$\omega \zeta = \frac{1}{{\sqrt {2\wp (\lambda r)} }}\implies F(\mu ) = \frac{{4\sqrt 2 {\pi ^{3/2}}}}{{{\Gamma ^2}(\frac{1}{4})}}\frac{\zeta }{{{\mu ^{1/4}}}}\frac{{\lambda r}}{{\sqrt 2 }} = 2\pi r\frac{\zeta }{{{\mu ^{1/4}}}}$$ Replace $\mu$ by $\gamma^2$ gives $${\gamma ^2} = \frac{1}{{1 - 4{\wp ^2}(\lambda r)}}\implies F({\gamma ^2}) = \frac{{2\pi }}{{\sqrt {2\gamma } }}r(1 + i)$$

Take $r = (1-i)/22$, which corresponds with your $N=11$, gives $${\gamma ^2} = \frac{1}{{1 - 4{\wp ^2}(\lambda \frac{{1 - i}}{{22}})}}$$ $\wp(\lambda r)$ here is a $22$-torsion of elliptic curve $E:y^2 = 4x^3-x$, it is then routine to verify that $\gamma$ indeed satisfies the desired equation.

Note that $E$ has complex multiplication by $i$, so $\mathbb{Q}(\gamma)/\mathbb{Q}$ is always an abelian extension for any $r$.

For $N=13$, the minimal polynomial for $\gamma$ is $$262144 x^{36}-524288 x^{35}-4456448 x^{34}-2621440 x^{33}+20381696 x^{32}+41418752 x^{31}-40894464 x^{30}-170655744 x^{29}+41484288 x^{28}+370671616 x^{27}-49610752 x^{26}-499843072 x^{25}+160481280 x^{24}+460292096 x^{23}-379060224 x^{22}-328843264 x^{21}+535066624 x^{20}+219197440 x^{19}-480704512 x^{18}-146997248 x^{17}+279806464 x^{16}+89368576 x^{15}-101117952 x^{14}-45646848 x^{13}+19158272 x^{12}+19104768 x^{11}-151808 x^{10}-5997824 x^9-834624 x^8+1179264 x^7+214528 x^6-105280 x^5-24924 x^4+2456 x^3+412 x^2+40 x-1 = 0$$

For $N=15$, the minimal polynomial for $\gamma$ is

1-64 x-96 x^2+128 x^3-5600 x^4-13504 x^5+86144 x^6+66176 x^7-697408 x^8-3584 x^9+3053056 x^10-1106944 x^11-7039488 x^12+5483520 x^13+7354368 x^14-15812608 x^15+1345024 x^16+32538624 x^17-9871360 x^18-47906816 x^19+778240 x^20+46448640 x^21+20676608 x^22-25067520 x^23-29114368 x^24+2752512 x^25+18481152 x^26+4980736 x^27-5636096 x^28-2883584 x^29+524288 x^30+524288 x^31+65536 x^32

For $N=17$, the minimal polynomial is 

1-80 x-288 x^2-7472 x^3+263520 x^4-1834944 x^5+7949120 x^6+9713984 x^7-133644672 x^8-24257792 x^9+562386944 x^10+2302738688 x^11+482554368 x^12-46943337472 x^13-7480857600 x^14+435187680256 x^15-25017156608 x^16-2451637374976 x^17+749950640128 x^18+9497241620480 x^19-7627550072832 x^20-26814443339776 x^21+50357313683456 x^22+56935237042176 x^23-233656552685568 x^24-96787065536512 x^25+794987687837696 x^26+168511477186560 x^27-2052161595637760 x^28-434303530696704 x^29+4129721034735616 x^30+1386031936176128 x^31-6608987067383808 x^32-3909549495943168 x^33+8496901307498496 x^34+8840840062435328 x^35-8705679109914624 x^36-15919713150304256 x^37+6741308655796224 x^38+22995053303562240 x^39-3118665184575488 x^40-26813854742740992 x^41-784242774638592 x^42+25326363753840640 x^43+3450707285901312 x^44-19377307395293184 x^45-4143992149114880 x^46+11962939040858112 x^47+3281065036742656 x^48-5905464873189376 x^49-1905340371173376 x^50+2291962858176512 x^51+830419578650624 x^52-678683215921152 x^53-271245438353408 x^54+144941187596288 x^55+65689377308672 x^56-19666655248384 x^57-11630771437568 x^58+1026497183744 x^59+1468878815232 x^60+120259084288 x^61-120259084288 x^62-17179869184 x^63+4294967296 x^64
pisco
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6

Taking $ \gamma $ to be the smallest positive root (the 4th root according to Mathematica) of the degree 30 polynomial \begin{align*} f(x) = -1 + 30 x + 170 x^2 + 672 x^3 - 6956 x^4 - 6808 x^5 + 42872 x^6 - 56576 x^7 - 241616 x^8 + 712800 x^9 + 1099296 x^{10} - 2718208 x^{11} - 3427264 x^{12} + 5028992 x^{13} + 8030592 x^{14} - 3956736 x^{15} - 14783232 x^{16} - 2065920 x^{17} + 20241920 x^{18} + 7954432 x^{19} - 19317760 x^{20} - 7817216 x^{21} + 12445696 x^{22} + 3342336 x^{23} - 5435392 x^{24} - 122880 x^{25} + 1662976 x^{26} - 393216 x^{27} - 344064 x^{28} + 98304 x^{29} + 32768 x^{30} \end{align*} seems to give a solution for $ N = 11 $.

Steven Charlton
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    +1. I trust this result. Its discriminant is a neat power $d=2^{870}\times11^{29}$, the leading coefficient is $2^{15}$, and its reciprocal $f(1/x)=f(y)=0$ is the minpoly of the related $I(y^2;1/2,1/4)=\frac{1}{11}$ as discussed in this post. – Tito Piezas III Dec 02 '16 at 11:22
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    Furthermore, if $u=(x_4+x_{10}+x_{17}+x_{18}+x_{19}+x_{20})/2$ using the root ordering of Mathematica, then $u$ is a root of $-23 - 343u + 390 u^2 - 42u^3 - 15u^4 + u^5=0$ which is a solvable quintic with a cyclic Galois group. A random $f(x)$ will not have all these properties. – Tito Piezas III Dec 02 '16 at 11:51
5

Remarkably, the root of the 30-degree equation in question $\small(\gamma \approx 0.028402981583...)$ has an elementary closed form. Let $$ \small \begin{align} \alpha &= \frac15\,\arccos\left(\frac{255300649+251245525 \sqrt{5}}{4\cdot11^8}\right),\\ \beta &= \sqrt{25085-6802 \sqrt{5}},\\[1mm] \kappa &= 11 \cdot \beta\,\sin\alpha,\\[2mm] \lambda &= \beta \, \sin 2\alpha,\\[2mm] \mu &= 11\cdot 811 \cos\alpha,\\[2mm] \nu &= 811 \cos 2\alpha,\\[2mm] \eta &= 11^3\cdot811 \left(55-14 \sqrt{11}\right)\\ &-11\,\kappa\left(2474+1479 \sqrt{5}-763 \sqrt{11}-385 \sqrt{55}\right)+ \lambda\left(22127+140925 \sqrt{5}+3421 \sqrt{11}-35431 \sqrt{55}\right)\\ &-11\, \mu\left(365 + 38 \sqrt{5} - 107 \sqrt{11} - 9 \sqrt{55}\right)- \nu\left(39510-14118 \sqrt{5} - 12837 \sqrt{11} + 3179 \sqrt{55}\right)\!,\\[2mm] \rho &=11^3\cdot811 \left(\sqrt{11} - 3\right)\\ &+4 \, \kappa \left(1397+132 \sqrt{5}-466 \sqrt{11}+122 \sqrt{55}\right) + \lambda \left(46291+43551 \sqrt{5}-9923 \sqrt{11}-9501 \sqrt{55}\right) \\ &-2 \, \mu \left(638 + 22 \sqrt{5} - \sqrt{11} - 63 \sqrt{55}\right) + \nu\,\left(5479-3061 \sqrt{5}-4723 \sqrt{11}-1341 \sqrt{55}\right)\!,\\[2mm] \sigma &= 2\cdot 11^3 \cdot 811 \left(11-2 \sqrt{11}\right)\\ &- \kappa\left(13673 - 2431 \sqrt{5} - 4514 \sqrt{11} - 120 \sqrt{55}\right) -\lambda \left(172406+140252 \sqrt{5}-23673 \sqrt{11}-31171 \sqrt{55}\right) \\ &+ \mu\left(3179+143 \sqrt{5}-798 \sqrt{11}-180 \sqrt{55}\right) - \nu\left(15896 - 1642 \sqrt{5} - 6967 \sqrt{11} + 1035 \sqrt{55}\right)\!,\\[2mm] \tau &= 2\cdot11^3\cdot811 \left(20 \sqrt{11} - 33\right) + 6\cdot11 \sqrt{3\cdot5\cdot11\cdot811 \,\eta }\\ &+ \kappa\left(166573+3025 \sqrt{5} - 49167 \sqrt{11} - 2979 \sqrt{55}\right) \\ &+2\,\lambda \left(576143+229934 \sqrt{5}-195022 \sqrt{11}-64418 \sqrt{55}\right) \\ &+ \mu\left(2453-2035 \sqrt{5}-505 \sqrt{11}-113 \sqrt{55}\right)\\ &+\nu\left(292348+26650 \sqrt{5}-75990 \sqrt{11}-2062 \sqrt{55}\right)\!. \end{align} $$ Then the root can be represented as $$\small\gamma = \frac1{2\cdot3\cdot5\cdot11^3\cdot811}\left(\rho +11^2 \cdot 2^{1/3} \cdot 5^{2/3} \cdot 811^{2/3} \cdot \tau^{1/3} + 11 \cdot 2^{2/3} \cdot 5^{1/3} \cdot 811^{1/3} \cdot \tau^{-1/3} \cdot \sigma \right)\!.$$

Mathematica expression is here.

I obtained this form by factoring Steven's 30-degree polynomial over rationals extended with $\sqrt{11}$ and a root of Tito's solvable quintic $-23 - 343u + 390 u^2 - 42u^3 - 15u^4 + u^5$ (I solved it as shown here). It factored into a product of cubics over that extension, which were easy to solve. The most difficult part was simplification of the result to a form that fits on one page. $811$ is a magical prime number that for some reason appears many times throughout the result.

Vladimir Reshetnikov
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  • It is no surprise that the $30$-deg equation is solvable by radical. As I proved, the value of $\gamma$ lies in field $K(E[22])$, which is generated by $22$-torsion for $E: y^2 = 4x^3-x$, but $E$ has CM by $K=\mathbb{Q}(i)$, so $K(E[22])/K$ is abelian. – pisco Sep 08 '19 at 11:37
  • Also because $j(E)\in \mathbb{Q}$, it should not be difficult to show $\gamma \in K(E[11])$, which is also the ray class field modulo $11$, whose degree over $\mathbb{Q}$ can be computed, it predicts accurately the degree of $\gamma$ over $\mathbb{Q}$, which is $30$. – pisco Sep 08 '19 at 11:48
  • Thanks for your comments. Do you expect polynomials of higher degrees arising from the inverse Beta regularized to be also solvable in radicals? – Vladimir Reshetnikov Sep 08 '19 at 18:19
  • Any idea why $811$ appears all over the result? Or it just happens so? – Vladimir Reshetnikov Sep 08 '19 at 18:20
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    For any rational $0<r<1$, the algebraic $\gamma$ in $$\int_0^1\frac{dx}{\sqrt[4]x\ \sqrt{1-x}\sqrt[4]{1-x,\gamma^2}}=r\frac{2\pi}{\sqrt{2\gamma}}$$ is solvable by radical. I need sometime to contemplate the genesis of $811$ though. – pisco Sep 10 '19 at 06:48
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    Sometimes posts on the Mathematics Stack exchange just deserve to be in papers. – Alexander Conrad Aug 02 '21 at 20:50
3

It seems solving $I(\beta^2;\tfrac{1}{2},\tfrac{1}{4})=\frac1N$ also solves $I(\gamma^2;\tfrac{1}{4},\tfrac{1}{4})=\frac1N$ as they involve reciprocal polynomials. So again following Nemo's lead in this answer, we employ the duplication formula, $$\frac{1}{2}I(p^2;\tfrac{1}{2},\tfrac{1}{4})=I(1-q^2;\tfrac{1}{2},\tfrac{1}{4})$$ where $p,q$ are related by the $8$-deg, $$p^2 (-1 + 2 q + q^2)^4 = 16 q (-1 + q^2) (1 + q^2)^2\tag1$$ So given a solution to $\displaystyle I(\beta^2;\tfrac{1}{2},\tfrac{1}{4})= \frac{1}{2^n N}$ for $n=0$, the duplication formula enables us to find infinitely many for $n\geq1$.

For example, the solution to $I(p^2;\tfrac{1}{2},\tfrac{1}{4})=\frac{1}{3}$ is $p = 1-\sqrt3$. One can then solve for $I(\beta^2;\tfrac{1}{2},\tfrac{1}{4})=\frac{1}{6}$ by plugging the known $p$ into $(1)$.

Tito Piezas III
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