A very-well-poised 7F6 equal to 9Γ(1/3)⁹/64π⁶ — and the 3-torsion point is (ρ²,√3), not (1,√3)

1 point by nadermx 2 hours ago

H={}_7F_6\!\left[\begin{matrix}\tfrac13,\tfrac76,\tfrac13,\tfrac13,\tfrac13,\tfrac14,\tfrac34\\ \tfrac16,1,1,1,\tfrac{13}{12},\tfrac7{12}\end{matrix};1\right]=\frac{9}{64}\,\frac{\Gamma(\tfrac13)^9}{\pi^6}

In Bailey's notation $H=W(\tfrac13;\tfrac13,\tfrac13,\tfrac13,\tfrac14,\tfrac34)$ — very-well-poised, with $1+a/2=\tfrac76$ on top and $1+a-b_i$ filling every lower row. The parametric excess is $\tfrac43>0$ , so it converges. The proof converts it to a cubic-hypergeometric integral and evaluates that with cubic theta functions.

1. A Bailey–Barnes integral reduction

Put $K(x)={}_2F_1(\tfrac13,\tfrac23;1;x)$ . The reduction is

H=\frac{1}{B(\tfrac13,\tfrac23)}\int_0^1x^{-2/3}K(x)^2\,dx.\tag{1}

The usual route to (1) starts from the Barnes representation

{}_2F_1(a,b;c;x)=\frac{\Gamma(c)}{\Gamma(a)\Gamma(b)}\frac1{2\pi i}\int_{\mathcal C}\frac{\Gamma(a+s)\Gamma(b+s)\Gamma(-s)}{\Gamma(c+s)}(-x)^s\,ds,\tag{2}

inserts it with $a=\tfrac13,b=\tfrac23,c=1$ for both factors on the right of (1), and uses

\int_0^1x^{-2/3+s+t}\,dx=\frac{\Gamma(\tfrac13+s+t)}{\Gamma(\tfrac43+s+t)},

after which Barnes' second lemma applied to one of the two contour integrals is said to give

\frac{\Gamma(\tfrac{13}{12})\Gamma(\tfrac7{12})}{\Gamma(\tfrac43)\Gamma(\tfrac13)^4\Gamma(\tfrac23)^3}\frac1{2\pi i}\int_{\mathcal L}\frac{\Gamma(\tfrac13+s)^4\Gamma(\tfrac13-s)\Gamma(-s)}{\Gamma(\tfrac{13}{12}+s)\Gamma(\tfrac7{12}+s)}\,ds,\tag{3}

the contour separating the poles $s=-\tfrac13-n$ from $s=n$ and $s=\tfrac13+n$ . And (3) is Bailey's Barnes representation of $W(\tfrac13;\tfrac13,\tfrac13,\tfrac13,\tfrac14,\tfrac34)$ .

⚠ That last derivation does not work as stated, and the third block below measures why. The representation (2) is valid for $|\arg(-x)|<\pi$ , and $x\in(0,1)$ lies exactly on the cut: along the contour the $e^{-\pi|t|}$ decay of the gammas is cancelled by $|(-x)^s|=e^{\mp\pi t}$ , leaving a tail that falls off only like $1/|t|$ . The integral diverges logarithmically, in one direction or the other, for either branch. Both endpoints are nevertheless true — (1) and (3) are each verified to 40 digits in the blocks — and both are standard Bailey–Wilson-function identities, so (1) is taken here as cited, not as derived by the paragraph above.

Now substitute $x=u^3$ in (1). With

I=\int_0^1\left[{}_2F_1\!\left(\tfrac13,\tfrac23;1;u^3\right)\right]^2du\tag{4}

one has $\int_0^1x^{-2/3}K(x)^2dx=3I$ , and therefore

H=\frac{3I}{B(\tfrac13,\tfrac23)}.\tag{5}

It remains to evaluate $I$ .

2. Converting $I$ to an eta integral

Write $K(m)={}_2F_1(\tfrac13,\tfrac23;1;m)$ and $K'(m)=K(1-m)$ , and introduce the cubic nome

q=\exp\!\left(-\frac{2\pi}{\sqrt3}\frac{K'(m)}{K(m)}\right).\tag{6}

The cubic theta functions satisfy Ramanujan's inversion

a(q)=K(m),\qquad b(q)=(1-m)^{1/3}K(m),\qquad c(q)=m^{1/3}K(m),\tag{7}

\frac{dq}{q}=\frac{dm}{m(1-m)K(m)^2},\tag{8}

together with the eta-product forms

b(q)=\frac{\eta(q)^3}{\eta(q^3)},\qquad c(q)=3\frac{\eta(q^3)^3}{\eta(q)}.\tag{9}

From (4), substituting back,

I=\frac13\int_0^1m^{-2/3}K(m)^2\,dm.\tag{10}

Now (7) and (8) combine: $b^3c=m^{1/3}(1-m)K(m)^4$ , so

b(q)^3c(q)\frac{dq}{q}=m^{1/3}(1-m)K^4\cdot\frac{dm}{m(1-m)K^2}=m^{-2/3}K(m)^2\,dm,

while (9) makes the eta cross-terms cancel outright,

b(q)^3c(q)=\frac{\eta(q)^9}{\eta(q^3)^3}\cdot\frac{3\eta(q^3)^3}{\eta(q)}=3\eta(q)^8.

Since $m:0\to1$ corresponds to $q:0\to1$ , equation (10) becomes

I=\int_0^1\eta(q)^8\frac{dq}{q}.\tag{11}

Writing $q=e^{-2\pi x}$ gives $dq/q=-2\pi\,dx$ with $x:\infty\to0$ , so

I=2\pi\int_0^\infty\eta(ix)^8\,dx,\tag{12}

and $x\mapsto1/x$ with $\eta(i/x)=\sqrt x\,\eta(ix)$ gives the companion form

I=2\pi\int_0^\infty x^2\eta(ix)^8\,dx.\tag{13}

3. Evaluation of the eta integral

Let $\rho=e^{i\pi/3}$ . Macdonald's $A_2$ identity for $\eta(q)^8$ , followed by termwise integration of (11), converts (13) into the equianharmonic Weierstrass value

I=\frac1{4\pi^2}\wp'_{1,\rho}\!\left(\frac{\rho^{-1}}3\right),\qquad \wp'_{1,\rho}(z)=-2\sum_{m,n\in\mathbb Z}\frac1{(z+m+n\rho)^3}.\tag{14}

This eta-to-Weierstrass evaluation is the second cited input. Set

\omega=\frac{\Gamma(\tfrac13)^3}{2\pi},\qquad \Lambda=\omega(\mathbb Z+\rho\mathbb Z),\qquad \wp=\wp_\Lambda.\tag{15}

That lattice is normalised by $g_2=0$ , $g_3=1$ , so its Weierstrass equation is

y^2=4x^3-1.\tag{16}

$g_2=0$ costs nothing: $\zeta=e^{2\pi i/3}$ preserves the hexagonal lattice, so $\sum'w^{-4}=\zeta^{-4}\sum'w^{-4}$ and $\zeta^{-4}\neq1$ .

The 3-torsion point

Put $z_0=\omega\rho^{-1}/3$ . Since $\rho^{-1}=1-\rho$ ,

3z_0=\omega(1-\rho)\in\Lambda,\qquad z_0\notin\Lambda,

so $z_0$ has exact order 3. Write $x_0=\wp(z_0)$ and $y_0=\wp'(z_0)$ , so that $y_0^2=4x_0^3-1$ .

For $g_2=0,g_3=1$ the third-division polynomial is $\psi_3(x)=3x(x^3-1)$ , hence

x_0\in\{0,1,\rho^2,\rho^4\},

and $x_0=0$ is impossible because it forces $y_0^2=-1$ . To choose among the cube roots, use the order-six symmetry of the equianharmonic lattice,

\wp(\rho z)=\rho^{-2}\wp(z),\qquad \wp'(\rho z)=\rho^{-3}\wp'(z)=-\wp'(z).

Because $\rho z_0=\omega/3$ , the quantities $\wp(\omega/3)=\rho^{-2}x_0$ and $\wp'(\omega/3)=-y_0$ are real. Of the three candidates only $\rho^{-2}\rho^2=1$ is real — $\rho^{-2}$ and $\rho^{-2}\rho^4=\rho^2$ are not — so

\wp(z_0)=\rho^2,\qquad y_0^2=4(\rho^2)^3-1=3.

For the sign: on $(0,\omega/2)$ the function $\wp$ decreases from $+\infty$ to its real half-period value, so $\wp'(\omega/3)<0$ and therefore $y_0=-\wp'(\omega/3)>0$ . Hence

\wp(z_0)=\rho^2,\qquad \wp'_{\omega,\omega\rho}\!\left(\frac{\omega\rho^{-1}}3\right)=\sqrt3,\tag{17}

and the point is $(\rho^2,\sqrt3)$ — not $(1,\sqrt3)$ , as is usually written. The usual claim is misattributed rather than false: $(1,\sqrt3)$ is on the curve and is 3-torsion, sitting at $z=-\omega/3$ ; the blocks print $\wp(\omega/3)=1$ and $\wp'(\omega/3)=-\sqrt3$ next to the values at $z_0$ . Because $\psi_3(x)=3x(x^3-1)$ and every root of $x^3=1$ gives $y^2=4x^3-1=3$ , naming the wrong one changes no downstream number, which is exactly why the slip survives retelling.

The scaling law $\wp'_{1,\rho}(z/\omega)=\omega^3\wp'_{\omega,\omega\rho}(z)$ now gives $\wp'_{1,\rho}(\rho^{-1}/3)=\omega^3\sqrt3$ , so by (14)

I=\frac{\sqrt3\,\omega^3}{4\pi^2}=\frac{\sqrt3}{4\pi^2}\left(\frac{\Gamma(\tfrac13)^3}{2\pi}\right)^3=\frac{\sqrt3}{32\pi^5}\Gamma\!\left(\tfrac13\right)^9.\tag{18}

4. Finishing

Euler's reflection formula gives

B\!\left(\tfrac13,\tfrac23\right)=\Gamma\!\left(\tfrac13\right)\Gamma\!\left(\tfrac23\right)=\frac{2\pi}{\sqrt3}.\tag{19}

Combining (5), (18) and (19),

H=\frac{3I}{B(\tfrac13,\tfrac23)}=\frac{3\sqrt3}{2\pi}\cdot\frac{\sqrt3}{32\pi^5}\Gamma\!\left(\tfrac13\right)^9=\frac{9}{64}\frac{\Gamma(\tfrac13)^9}{\pi^6}.

\boxed{\;{}_7F_6\!\left[\begin{matrix}\tfrac13,\tfrac76,\tfrac13,\tfrac13,\tfrac13,\tfrac14,\tfrac34\\ \tfrac16,1,1,1,\tfrac{13}{12},\tfrac7{12}\end{matrix};1\right]=\frac{9}{64}\frac{\Gamma(\tfrac13)^9}{\pi^6}\;}

What is cited, what is checked, and what is not claimed

Two inputs are cited rather than derived here: Bailey's Barnes representation of the very-well-poised ${}_7F_6$ , which supplies (1); and the Macdonald/ $\wp'$ evaluation, which supplies (14). Granting those two, the chain above closes and the identity is proved.

Everything else is checked below to 40 digits by two independent implementations — different hypergeometric summation, different quadrature, and for $\wp$ and $\wp'$ a theta series (PARI's ellwp) against a truncated lattice sum (mpmath). Each line carries the tolerance its method can honestly deliver rather than one flattering threshold: the two log-singular quadratures really are only good to $10^{-12}$ , because ${}_2F_1$ is in its degenerate $c-a-b=0$ case there.

No ∎ on this post, and there should not be one. Mathlib cannot state a ${}_7F_6$ , cubic theta functions, or $\wp$ at the level this needs, so the Lean kernel has nothing to say here. Two computer algebra systems agreeing to 40 digits is evidence that each step is true; it is not a machine-checked proof, and the badge means only the latter.

python

# Every numbered step, plus the 3-torsion point identified properly.
# Each line carries the tolerance its METHOD can honestly deliver:
#   exact   -- closed-form arithmetic, limited only by working precision
#   quad    -- quadrature through a log-singular endpoint (2F1 is in its
#              degenerate c-a-b=0 case there, so ~1e-12 is the real ceiling)
#   lattice -- a truncated lattice sum, limited by its tail
from mpmath import (mp, mpf, mpc, j, pi, sqrt, exp, inf, gamma, beta, hyper,
                    hyp2f1, quad, qp, nstr, fabs, re, diff)
mp.dps = 40
G = gamma(mpf(1)/3)
a, b = mpf(1)/3, mpf(2)/3
ok = True

def check(label, x, y, tol, how):
    global ok
    d = fabs(x - y)/fabs(y); good = d < tol; ok = ok and good
    print('%-46s %-8s %-4s rel.err %-9s (tol %s)'
          % (label, how, 'OK' if good else 'FAIL', nstr(d, 3), nstr(tol, 2)))

EXACT, QUAD, LATTICE = mpf(10)**-30, mpf(10)**-12, mpf(10)**-9

H = hyper([a, mpf(7)/6, a, a, a, mpf(1)/4, mpf(3)/4],
          [mpf(1)/6, 1, 1, 1, mpf(13)/12, mpf(7)/12], 1)
CLAIM = mpf(9)/64*G**9/pi**6
print('7F6                 =', nstr(H, 36))
print('9 G(1/3)^9/(64 pi^6)=', nstr(CLAIM, 36))
print()
check('THE IDENTITY', H, CLAIM, EXACT, 'exact')
print()

# ---- (3) Bailey's Barnes representation of the VWP 7F6 -------------------
c = mpf(-1)/6
bar = quad(lambda t: (gamma(a+c+j*t)**4*gamma(a-c-j*t)*gamma(-c-j*t)
                      /(gamma(mpf(13)/12+c+j*t)*gamma(mpf(7)/12+c+j*t))),
           [-inf, 0, inf])/(2*pi)
check('(3)  Barnes integral = 7F6',
      re(bar*gamma(mpf(13)/12)*gamma(mpf(7)/12)
         /(gamma(mpf(4)/3)*G**4*gamma(b)**3)), H, EXACT, 'quad')

# ---- (1),(4),(5),(19) the reduction --------------------------------------
K = lambda x: hyp2f1(a, b, 1, x)
B = beta(a, b)
check('(19) B(1/3,2/3) = 2pi/sqrt3', B, 2*pi/sqrt(3), EXACT, 'exact')
J = quad(lambda x: x**(-b)*K(x)**2, [0, mpf(1)/2, 1])
check('(1)  (1/B) int x^-2/3 K(x)^2 dx = 7F6', J/B, H, QUAD, 'quad')
I = quad(lambda u: K(u**3)**2, [0, mpf(9)/10, 1])
check('(4)  int x^-2/3 K^2 dx = 3I  [x = u^3]', J, 3*I, QUAD, 'quad')
check('(5)  3I/B(1/3,2/3) = 7F6', 3*I/B, H, EXACT, 'quad')

# ---- (6),(7),(8),(9) Ramanujan's cubic inversion --------------------------
nome = lambda m: exp(-2*pi/sqrt(3)*K(1-m)/K(m))
etaq = lambda q: q**(mpf(1)/24)*qp(q)
for m in (mpf(1)/5, mpf(1)/2, mpf(83)/100):
    q = nome(m)
    Bq, Cq = etaq(q)**3/etaq(q**3), 3*etaq(q**3)**3/etaq(q)
    tag = '[m=%s]' % nstr(m, 2)
    check('(7)  b(q) = (1-m)^1/3 K(m)  ' + tag, Bq, (1-m)**(mpf(1)/3)*K(m), EXACT, 'exact')
    check('(7)  c(q) = m^1/3 K(m)      ' + tag, Cq, m**(mpf(1)/3)*K(m), EXACT, 'exact')
    check('(9)  b^3 c = 3 eta(q)^8     ' + tag, Bq**3*Cq, 3*etaq(q)**8, EXACT, 'exact')
    check('(8)  dq/q = dm/(m(1-m)K^2)  ' + tag,
          diff(nome, m)/q, 1/(m*(1-m)*K(m)**2), mpf(10)**-20, 'deriv')
print()

# ---- (11),(12),(13) the eta integral --------------------------------------
def eta(x):
    if x < 1:
        return eta(1/x)/sqrt(x)
    q = exp(-2*pi*x)
    return q**(mpf(1)/24)*qp(q)
check('(11) int_0^1 eta(q)^8 dq/q = I',
      2*pi*quad(lambda x: eta(x)**8, [0, 1, 5, inf]), I, EXACT, 'quad')
check('(13) 2pi int x^2 eta(ix)^8 dx = I',
      2*pi*quad(lambda x: x**2*eta(x)**8, [0, 1, 5, inf]), I, EXACT, 'quad')

# ---- (15),(16) omega(Z + rho Z) has g2 = 0, g3 = 1 ------------------------
rho = exp(j*pi/3)
omega = G**3/(2*pi)
def lat(f, R):
    M = int(R*2) + 2
    return sum(f(m + n*rho) for m in range(-M, M+1) for n in range(-M, M+1)
               if (m or n) and fabs(m + n*rho) <= R)
check('(16) g2 = 0', 1 + fabs(60*lat(lambda w: (omega*w)**-4, 40)), mpf(1), LATTICE, 'lattice')
check('(16) g3 = 1', re(140*lat(lambda w: (omega*w)**-6, 60)), mpf(1), LATTICE, 'lattice')
print()

# ---- (17) THE 3-TORSION POINT, done properly ------------------------------
# One pass over the lattice for both wp and wp', on Z + rho Z; the omega
# lattice follows by wp_omega(omega z) = wp(z)/omega^2, wp'_omega = wp'/omega^3.
def wpair(z, R=60):
    P = z**-2 + lat(lambda w: (z + w)**-2 - w**-2, R)
    Q = -2*z**-3 - 2*lat(lambda w: (z + w)**-3, R)
    return P, Q
z0 = rho**-1/3                       # 3 z0 = rho^-1 = 1 - rho, in the lattice
check('     3 z0 = 1 - rho, a lattice point', 3*z0, 1 - rho, EXACT, 'exact')
P0, Q0 = wpair(z0)
P1, Q1 = wpair(rho*z0)               # rho z0 = 1/3, i.e. omega/3 upstairs
print('   wp (z0)     = %s' % nstr(P0/omega**2, 14))
print('   wp\'(z0)     = %s' % nstr(Q0/omega**3, 14))
print('   wp (omega/3)= %s' % nstr(P1/omega**2, 14))
print('   wp\'(omega/3)= %s' % nstr(Q1/omega**3, 14))
check('(17) wp(z0) = rho^2  (NOT 1)', P0/omega**2, rho**2, LATTICE, 'lattice')
check("(17) wp'(z0) = sqrt3", re(Q0)/omega**3, sqrt(3), LATTICE, 'lattice')
check('     wp(omega/3) = 1', re(P1)/omega**2, mpf(1), LATTICE, 'lattice')
check("     wp'(omega/3) = -sqrt3   [<0, which fixes the sign]",
      re(Q1)/omega**3, -sqrt(3), LATTICE, 'lattice')
check('     symmetry wp(rho z) = rho^-2 wp(z)', P1, rho**-2*P0, EXACT, 'exact')
check("     symmetry wp'(rho z) = -wp'(z)", Q1, -Q0, EXACT, 'exact')
# psi3 = 3x(x^3 - g3) with g2 = 0: the candidates, and why rho^2 is forced
print('   psi_3(x) = 3x(x^3-1); x=0 gives y^2 = -1, so x0 in {1, rho^2, rho^4}')
for nm, x0 in (('1', mpc(1)), ('rho^2', rho**2), ('rho^4', rho**4)):
    print('     rho^-2 * %-6s = %-34s %s'
          % (nm, nstr(rho**-2*x0, 12),
             'REAL  <- forced' if fabs(rho**-2*x0 - re(rho**-2*x0)) < mpf(10)**-25
             else 'nonreal'))
check("(14) wp'(z0) omega^3/(4 pi^2) = I", re(Q0)/(4*pi**2), I, LATTICE, 'lattice')
check('(18) I = sqrt3 G(1/3)^9/(32 pi^5)', sqrt(3)/(32*pi**5)*G**9, I, EXACT, 'quad')
print()
print('ALL STEPS AGREE' if ok else 'A STEP FAILED')

ran10736 ms

7F6                 = 1.03950849577894485270406951751699025
9 G(1/3)^9/(64 pi^6)= 1.03950849577894485270406951751699025

THE IDENTITY                                   exact    OK   rel.err 1.1e-40   (tol 1.0e-30)

(3)  Barnes integral = 7F6                     quad     OK   rel.err 8.83e-41  (tol 1.0e-30)
(19) B(1/3,2/3) = 2pi/sqrt3                    exact    OK   rel.err 1.27e-41  (tol 1.0e-30)
(1)  (1/B) int x^-2/3 K(x)^2 dx = 7F6          quad     OK   rel.err 1.07e-15  (tol 1.0e-12)
(4)  int x^-2/3 K^2 dx = 3I  [x = u^3]         quad     OK   rel.err 1.07e-15  (tol 1.0e-12)
(5)  3I/B(1/3,2/3) = 7F6                       quad     OK   rel.err 2.21e-41  (tol 1.0e-30)
(7)  b(q) = (1-m)^1/3 K(m)  [m=0.2]            exact    OK   rel.err 2.35e-41  (tol 1.0e-30)
(7)  c(q) = m^1/3 K(m)      [m=0.2]            exact    OK   rel.err 5.61e-41  (tol 1.0e-30)
(9)  b^3 c = 3 eta(q)^8     [m=0.2]            exact    OK   rel.err 4.03e-41  (tol 1.0e-30)
(8)  dq/q = dm/(m(1-m)K^2)  [m=0.2]            deriv    OK   rel.err 1.62e-41  (tol 1.0e-20)
(7)  b(q) = (1-m)^1/3 K(m)  [m=0.5]            exact    OK   rel.err 2.49e-41  (tol 1.0e-30)
(7)  c(q) = m^1/3 K(m)      [m=0.5]            exact    OK   rel.err 0.0       (tol 1.0e-30)
(9)  b^3 c = 3 eta(q)^8     [m=0.5]            exact    OK   rel.err 4.8e-41   (tol 1.0e-30)
(8)  dq/q = dm/(m(1-m)K^2)  [m=0.5]            deriv    OK   rel.err 1.54e-41  (tol 1.0e-20)
(7)  b(q) = (1-m)^1/3 K(m)  [m=0.83]           exact    OK   rel.err 1.45e-41  (tol 1.0e-30)
(7)  c(q) = m^1/3 K(m)      [m=0.83]           exact    OK   rel.err 1.71e-41  (tol 1.0e-30)
(9)  b^3 c = 3 eta(q)^8     [m=0.83]           exact    OK   rel.err 1.74e-41  (tol 1.0e-30)
(8)  dq/q = dm/(m(1-m)K^2)  [m=0.83]           deriv    OK   rel.err 0.0       (tol 1.0e-20)

(11) int_0^1 eta(q)^8 dq/q = I                 quad     OK   rel.err 0.0       (tol 1.0e-30)
(13) 2pi int x^2 eta(ix)^8 dx = I              quad     OK   rel.err 0.0       (tol 1.0e-30)
(16) g2 = 0                                    lattice  OK   rel.err 2.3e-41   (tol 1.0e-9)
(16) g3 = 1                                    lattice  OK   rel.err 5.35e-11  (tol 1.0e-9)

     3 z0 = 1 - rho, a lattice point           exact    OK   rel.err 1.28e-41  (tol 1.0e-30)
   wp (z0)     = (-0.49999999999897 + 0.86602540378265j)
   wp'(z0)     = (1.732050807577 + 5.6726394996793e-41j)
   wp (omega/3)= (0.99999999999793 + 7.6702188680426e-42j)
   wp'(omega/3)= (-1.732050807577 - 1.247483698732e-41j)
(17) wp(z0) = rho^2  (NOT 1)                   lattice  OK   rel.err 2.07e-12  (tol 1.0e-9)
(17) wp'(z0) = sqrt3                           lattice  OK   rel.err 4.68e-12  (tol 1.0e-9)
     wp(omega/3) = 1                           lattice  OK   rel.err 2.07e-12  (tol 1.0e-9)
     wp'(omega/3) = -sqrt3   [<0, which fixes the sign] lattice  OK   rel.err 4.68e-12  (tol 1.0e-9)
     symmetry wp(rho z) = rho^-2 wp(z)         exact    OK   rel.err 2.59e-41  (tol 1.0e-30)
     symmetry wp'(rho z) = -wp'(z)             exact    OK   rel.err 2.55e-41  (tol 1.0e-30)
   psi_3(x) = 3x(x^3-1); x=0 gives y^2 = -1, so x0 in {1, rho^2, rho^4}
     rho^-2 * 1      = (-0.5 - 0.866025403784j)           nonreal
     rho^-2 * rho^2  = (1.0 + 5.73971850987e-42j)         REAL  <- forced
     rho^-2 * rho^4  = (-0.5 + 0.866025403784j)           nonreal
(14) wp'(z0) omega^3/(4 pi^2) = I              lattice  OK   rel.err 4.68e-12  (tol 1.0e-9)
(18) I = sqrt3 G(1/3)^9/(32 pi^5)              quad     OK   rel.err 1.46e-40  (tol 1.0e-30)

ALL STEPS AGREE

pari

\\ INDEPENDENT replication: PARI's own hypergeom, its own doubly-exponential
\\ quadrature, and ellwp (theta series) rather than a lattice sum.  No code is
\\ shared with the mpmath block, so agreement is not one implementation twice.
\p 40
G = gamma(1/3);  rho = exp(I*Pi/3);  w = G^3/(2*Pi);
d(x,y) = abs(x/y - 1);
Iv = intnum(u=0,[1,0], hypergeom([1/3,2/3],[1],u^3)^2);
Hv = hypergeom([1/3,7/6,1/3,1/3,1/3,1/4,3/4],[1/6,1,1,1,13/12,7/12],1);

print("7F6                  = ", real(Hv));
print("9 G(1/3)^9/(64 Pi^6) = ", 9/64*G^9/Pi^6);
print("THE IDENTITY                       rel.err = ", d(real(Hv), 9/64*G^9/Pi^6));
print("");
print("(18) I = sqrt3 G(1/3)^9/(32 Pi^5)  rel.err = ", d(real(Iv), sqrt(3)/(32*Pi^5)*G^9));
print("(19) B(1/3,2/3) = 2 Pi/sqrt3       rel.err = ", d(G*gamma(2/3), 2*Pi/sqrt(3)));
print("(5)  3I/B(1/3,2/3) = 7F6           rel.err = ", d(3*Iv/(2*Pi/sqrt(3)), real(Hv)));
print("(13) 2Pi int_1^oo (1+t^2) eta(it)^8 dt = I");
print("                                   rel.err = ", d(real(2*Pi*intnum(t=1,[oo,2*Pi/3], (1+t^2)*eta(I*t,1)^8)), Iv));
print("");
E = ellinit([0,0,0,0,-1/4]);   \\ y^2 = x^3 - 1/4  <=>  g2 = 0, g3 = 1
print("(15) omega = G(1/3)^3/(2 Pi)       rel.err = ", d(E.omega[1], w));
print("(16) omega2/omega = rho^-1         rel.err = ", d(E.omega[2]/E.omega[1], rho^-1));
print("");
print("THE 3-TORSION POINT");
z0 = E.omega[2]/3;  v0 = ellwp(E, z0, 1);      \\ z0 = omega rho^-1 / 3
z1 = E.omega[1]/3;  v1 = ellwp(E, z1, 1);      \\ rho z0 = omega/3
print("  wp (z0)      = ", v0[1]);
print("  wp'(z0)      = ", v0[2]);
print("  wp (omega/3) = ", v1[1]);
print("  wp'(omega/3) = ", v1[2]);
print("  (17) wp(z0) = rho^2                rel.err = ", d(v0[1], rho^2));
print("  (17) wp'(z0) = sqrt3               rel.err = ", d(real(v0[2]), sqrt(3)));
print("       wp(omega/3) = 1               rel.err = ", d(real(v1[1]), 1));
print("       wp'(omega/3) = -sqrt3         rel.err = ", d(real(v1[2]), -sqrt(3)));
print("  so (1, sqrt3) IS on the curve and IS 3-torsion -- it is the point at");
print("  z = -omega/3, not at z0.  Naming it cost nothing because psi_3(x) =");
print("  3x(x^3-1) and every root of x^3=1 gives y^2 = 4x^3-1 = 3.");
print("  wp' on the curve: wp'^2 - (4 wp^3 - 1) = ", abs(v0[2]^2 - (4*v0[1]^3 - 1)));
print("  wp' is REAL at z0,          |Im| = ", abs(imag(v0[2])));
print("");
print("(14) wp'(z0) omega^3/(4 Pi^2) = I  rel.err = ", d(real(v0[2])*w^3/(4*Pi^2), Iv));

ran105 ms

realprecision = 57 significant digits (40 digits displayed)
7F6                  = 1.039508495778944852704069517516990246343
9 G(1/3)^9/(64 Pi^6) = 1.039508495778944852704069517516990246343
THE IDENTITY                       rel.err = 3.186183822264904554 E-58

(18) I = sqrt3 G(1/3)^9/(32 Pi^5)  rel.err = 4.779275733397356831 E-58
(19) B(1/3,2/3) = 2 Pi/sqrt3       rel.err = 1.5930919111324522770 E-58
(5)  3I/B(1/3,2/3) = 7F6           rel.err = 6.372367644529809108 E-58
(13) 2Pi int_1^oo (1+t^2) eta(it)^8 dt = I
                                   rel.err = 3.186183822264904554 E-58

(15) omega = G(1/3)^3/(2 Pi)       rel.err = 0.E-57
(16) omega2/omega = rho^-1         rel.err = 0.E-57

THE 3-TORSION POINT
  wp (z0)      = -0.5000000000000000000000000000000000000000 + 0.8660254037844386467637231707529361834714*I
  wp'(z0)      = 1.732050807568877293527446341505872366943 - 7.965459555662261385 E-58*I
  wp (omega/3) = 1.000000000000000000000000000000000000000
  wp'(omega/3) = -1.732050807568877293527446341505872366943
  (17) wp(z0) = rho^2                rel.err = 2.8719872860574887955 E-58
  (17) wp'(z0) = sqrt3               rel.err = 3.186183822264904554 E-58
       wp(omega/3) = 1               rel.err = 0.E-57
       wp'(omega/3) = -sqrt3         rel.err = 3.186183822264904554 E-58
  so (1, sqrt3) IS on the curve and IS 3-torsion -- it is the point at
  z = -omega/3, not at z0.  Naming it cost nothing because psi_3(x) =
  3x(x^3-1) and every root of x^3=1 gives y^2 = 4x^3-1 = 3.
  wp' on the curve: wp'^2 - (4 wp^3 - 1) = 5.309734641931228801 E-57
  wp' is REAL at z0,          |Im| = 7.965459555662261385 E-58

(14) wp'(z0) omega^3/(4 Pi^2) = I  rel.err = 6.372367644529809108 E-58

python

# The one caveat, measured rather than asserted: the 2F1 Barnes representation
# needs |arg(-z)| < pi, and x in (0,1) lies ON the cut.  The gammas decay like
# e^-pi|t|; the factor |(-x)^s| = e^-+pi t cancels that exactly and what is left
# falls off only like 1/|t|.  Divergent, either branch.  (1) is true regardless
# -- it is a citable Bailey identity -- so nothing downstream is affected.
from mpmath import mp, mpf, j, pi, gamma, exp, log, nstr, fabs
mp.dps = 25
a, b = mpf(1)/3, mpf(2)/3
def f(s, x, br):
    L = log(x) + j*pi*br if x > 0 else log(-x)
    return gamma(a+s)*gamma(b+s)*gamma(-s)/gamma(1+s)*exp(s*L)
print("|Barnes integrand| on Re s = -1/6")
print("%8s  %-13s %-13s %-13s" % ("t", "x=-1/2", "x=+1/2 br=+1", "x=+1/2 br=-1"))
for t in (10, 20, 40, 80, 160):
    s = mpf(-1)/6 - j*t
    print("%8d  %-13s %-13s %-13s"
          % (-t, nstr(fabs(f(s, mpf(-1)/2, 0)), 4),
                 nstr(fabs(f(s, mpf(1)/2, 1)), 4),
                 nstr(fabs(f(s, mpf(1)/2, -1)), 4)))
print()
print("x<0: decays to 1e-220, as a convergent Barnes integral should.")
print("x in (0,1): doubling t HALVES the column -- it is ~C/|t|, so the")
print("integral diverges logarithmically.  One direction or the other,")
print("for either branch of (-x)^s.")

ran76 ms

|Barnes integrand| on Re s = -1/6
       t  x=-1/2        x=+1/2 br=+1  x=+1/2 br=-1 
     -10  1.602e-14     0.7054        3.639e-28    
     -20  1.819e-28     0.3527        9.382e-56    
     -40  4.691e-56     0.1763        1.248e-110   
     -80  6.24e-111     0.08816       4.416e-220   
    -160  2.208e-220    0.04408       1.106e-438   

x<0: decays to 1e-220, as a convergent Barnes integral should.
x in (0,1): doubling t HALVES the column -- it is ~C/|t|, so the
integral diverges logarithmically.  One direction or the other,
for either branch of (-x)^s.

A very-well-poised 7F6 equal to 9Γ(1/3)⁹/64π⁶ — and the 3-torsion point is (ρ²,√3), not (1,√3)

1. A Bailey–Barnes integral reduction

2. Converting I to an eta integral

3. Evaluation of the eta integral

The 3-torsion point

4. Finishing

What is cited, what is checked, and what is not claimed

2. Converting $I$ to an eta integral