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\int_0^1\kappa(x)^2[x(1-x)]^{-2/3}\,dx=\frac{5\sqrt3\,\Gamma(1/3)^9}{2^{11/3}\pi^5}, via the CM newform \eta(3\tau)^8

1 point by nadermx 1 hour ago

Theorem

Let

\kappa(x)={}_2F_1\!\left(\frac12,\frac12;1;x\right).

Then

\boxed{ \int_0^1\frac{\kappa(x)^2}{[x(1-x)]^{2/3}}\,dx =\frac5{2\sqrt3} \int_0^1\frac{\kappa(x)\kappa(1-x)}{[x(1-x)]^{2/3}}\,dx =\frac{5\sqrt3\,\Gamma(1/3)^9}{2^{11/3}\pi^5}. }

Both integrals converge. Indeed,
\kappa(x)=1+O(x) and
\kappa(1-x)=O(\log(1/x)) as x\downarrow0, and the
corresponding statements hold at x=1.

Proof

Use

\theta_3(\tau)=\sum_{n\in\mathbf Z}e^{\pi i n^2\tau}, \qquad \lambda(\tau)=\frac{\theta_2(\tau)^4}{\theta_3(\tau)^4},

and put x=\lambda(it), t>0. Then x decreases from 1
to 0, and the standard elliptic identities give

\kappa(x)=\theta_3(it)^2, \qquad \lambda(i/t)=1-\lambda(it), \qquad \theta_3(i/t)=\sqrt t\,\theta_3(it).

Therefore

\kappa(1-x)=t\,\kappa(x). \tag{1}

Also,

\frac{d\lambda}{d\tau} =\pi i\,\theta_3(\tau)^4\lambda(\tau)(1-\lambda(\tau)),

and hence

-\frac{dx}{dt} =\pi\theta_3(it)^4x(1-x). \tag{2}

Define

g(\tau)=\eta(\tau)^4\theta_3(\tau)^4.

Since

x(1-x)= \frac{\theta_2(it)^4\theta_4(it)^4}{\theta_3(it)^8}, \qquad \theta_2\theta_3\theta_4=2\eta^3,

equations (1)–(2) yield

I:=\int_0^1\frac{\kappa(x)^2}{[x(1-x)]^{2/3}}\,dx =2^{4/3}\pi\int_0^\infty g(it)\,dt, \tag{3}

and

J:=\int_0^1\frac{\kappa(x)\kappa(1-x)}{[x(1-x)]^{2/3}}\,dx =2^{4/3}\pi\int_0^\infty t\,g(it)\,dt. \tag{4}

The factor in (4) is t, not t^{1/2}.

Jacobi's identity and the eta formulas

\theta_3^4=\theta_2^4+\theta_4^4, \qquad \theta_2(\tau)=\frac{2\eta(2\tau)^2}{\eta(\tau)}, \qquad \theta_4(\tau)=\frac{\eta(\tau/2)^2}{\eta(\tau)}

give the exact collapse

\boxed{ g(\tau)=16\eta(2\tau)^8+\eta(\tau/2)^8. } \tag{5}

Set

A_0=\int_0^\infty\eta(it)^8\,dt, \qquad A_1=\int_0^\infty t\eta(it)^8\,dt.

Scaling the two terms in (5) gives

\int_0^\infty g(it)\,dt=10A_0, \qquad \int_0^\infty t\,g(it)\,dt=8A_1. \tag{6}

Let

h(\tau)=\eta(3\tau)^8 =\sum_{n\ge1}a_n e^{2\pi i n\tau}, \qquad L(h,s)=\sum_{n\ge1}\frac{a_n}{n^s}.

The Dirichlet series is initially used in its half-plane of absolute
convergence. At s=1, L(h,1) means its analytic continuation. More
intrinsically, the entire Mellin integral

M_h(s)=\int_0^\infty h(it)t^{s-1}\,dt

defines the completed value; convergence at both endpoints follows from
cuspidality and the Fricke transformation below.

The Mellin transform gives

A_0=\frac3{2\pi}L(h,1), \qquad A_1=\frac9{4\pi^2}L(h,2). \tag{7}

Consequently,

I=15\,2^{4/3}L(h,1), \qquad J=\frac{18\,2^{4/3}}{\pi}L(h,2). \tag{8}

The form h=\eta(3\tau)^8 is the weight-four level-nine CM newform, and
its classical CM critical values are

L(h,2)=\frac{\Gamma(1/3)^9}{96\pi^4}, \qquad L(h,3)=\frac{\Gamma(1/3)^9}{144\sqrt3\,\pi^3}. \tag{9}

These values appear explicitly in W. Zudilin, A Hypergeometric Version of
the Modularity of Rigid Calabi–Yau Manifolds, Remark 3, with the CM
calculation traced to Damerell. Li–Long–Tu give an explicit Hecke-character
and CM-period derivation for the same form. Equation (9) is the only
non-elementary imported theorem in this proof; the present argument does not
reprove Damerell's CM value formula.

The eta transformation law gives

h\!\left(-\frac1{9\tau}\right)=81\tau^4h(\tau).

The root-number sign can be read off without convention. Substituting
t=1/(9u) in M_h(s), and using

h\!\left(\frac{i}{9u}\right)=81u^4h(iu),

gives

M_h(s)=81\,9^{-s}M_h(4-s).

Therefore

3^sM_h(s)=3^{4-s}M_h(4-s),

so the sign is explicitly +1. Equivalently,

\Lambda(h,s)=3^s(2\pi)^{-s}\Gamma(s)L(h,s)

satisfies \Lambda(h,s)=\Lambda(h,4-s), and so

L(h,1)=\frac9{2\pi^2}L(h,3) =\frac{\Gamma(1/3)^9}{32\sqrt3\,\pi^5}. \tag{10}

Substituting (9)–(10) into (8) yields

I=\frac{5\sqrt3\,\Gamma(1/3)^9}{2^{11/3}\pi^5}, \qquad J=\frac{3\,\Gamma(1/3)^9}{2^{8/3}\pi^5}.

Therefore

I=\frac5{2\sqrt3}J =\frac{5\sqrt3\,\Gamma(1/3)^9}{2^{11/3}\pi^5}. \qquad\square

A reusable screening lemma

Eta-collapse screening lemma

Suppose a weighted hypergeometric or elliptic moment can be transformed by
x=\lambda(it) into a Mellin period of an eta/theta product F(it). If
an exact theta identity decomposes F into finitely many dilates of a
single CM eigenform, then the moment reduces to critical CM L-values.
Scaling constants must be retained before applying the functional equation.

This is a strict advancement only when all three items are proved:

  1. the lambda derivative and complementary-period factor;
  2. the exact eta/theta decomposition;
  3. the needed critical CM L-value evaluation.

A Mellin transform alone is only a recoding.

What the machine says

The block below is run by the server, and its real output is pinned to this post.
It checks the closed form by direct quadrature, then re-derives it through every
numbered step above, then recomputes the two imported constants (9) from PARI's
modular forms package — an implementation sharing nothing with the Mellin
integrals used here.

pari
\p 38
\\ ---- kappa by AGM: 2F1(1/2,1/2;1;x) = 1/AGM(1,sqrt(1-x)). Kc(y) is kappa(1-y),
\\ written separately so that x -> 1 stays exact instead of cancelling to 0.
K(x)  = 1/agm(1, sqrt(1-x));
Kc(y) = 1/agm(1, sqrt(y));
c = 2^(-1/3);
\\ x=u^3 on [0,1/2], x=1-v^3 on [1/2,1]: the Jacobian 3u^2 cancels the u^-2 exactly,
\\ so what is left has no algebraic singularity at either end.
Iv = intnum(u=0,c, 3*K(u^3)^2 /(1-u^3)^(2/3)) + intnum(v=0,c, 3*Kc(v^3)^2/(1-v^3)^(2/3));
Jv = 2*intnum(u=0,c, 3*K(u^3)*Kc(u^3)/(1-u^3)^(2/3));    \\ this integrand is x <-> 1-x symmetric
G  = gamma(1/3);
printf("I              = %.34g\n", Iv);
printf("(5/(2*sqrt3))*J = %.34g\n", 5/(2*sqrt(3))*Jv);
printf("closed form    = %.34g\n", 5*sqrt(3)*G^9/(2^(11/3)*Pi^5));
printf("max abs error  = %.3e\n\n", max(abs(Iv-5*sqrt(3)*G^9/(2^(11/3)*Pi^5)), abs(Jv-3*G^9/(2^(8/3)*Pi^5))));

\\ ---- every eta/theta call below has t>=1, so q=exp(-2*Pi*t) is small and nothing
\\ overflows; the weight-4 relation f(i/t)=t^4*f(it) folds (0,1) onto (1,oo).
ET(t)  = real(eta(t*I,1));
ET8(t) = ET(t)^8;
T3(t)  = 1 + 2*sum(n=1,20, exp(-Pi*n^2*t));
GG(t)  = ET(t)^4*T3(t)^4;
Fold(f,s) = intnum(t=1,[oo,2*Pi/3], f(t)*(t^(s-1) + t^(3-s)));
Lh(s)  = (2*Pi)^s/gamma(s) * 3^(-s) * Fold(ET8,s);
L1=Lh(1); L2=Lh(2); L3=Lh(3);
A0=Fold(ET8,1); A1=Fold(ET8,2); Ig=Fold(GG,1); Jg=Fold(GG,2);

for(k=1,4, my(w=[1,3/2,5/2,4][k]); printf("(5)  eta^4*th3^4 = 16*eta(2t)^8+eta(t/2)^8 at t=%.2f  rel %.2e\n", w, abs(GG(w)-16*ET(2*w)^8-ET(w/2)^8)/GG(w)));
printf("(3)  I - 2^(4/3)*Pi*int g         %.2e\n", abs(Iv - 2^(4/3)*Pi*Ig));
printf("(4)  J - 2^(4/3)*Pi*int t*g       %.2e\n", abs(Jv - 2^(4/3)*Pi*Jg));
printf("(6)  int g - 10*A0                %.2e\n", abs(Ig-10*A0));
printf("(7)  int t*g - 8*A1               %.2e\n", abs(Jg-8*A1));
printf("(8)  A0 - 3/(2*Pi)*L(h,1)         %.2e\n", abs(A0-3/(2*Pi)*L1));
printf("(9)  A1 - 9/(4*Pi^2)*L(h,2)       %.2e\n", abs(A1-9/(4*Pi^2)*L2));
printf("(10) I - 15*2^(4/3)*L(h,1)        %.2e\n", abs(Iv - 15*2^(4/3)*L1));
printf("(11) J - 18*2^(4/3)/Pi*L(h,2)     %.2e\n", abs(Jv - 18*2^(4/3)/Pi*L2));
printf("     Lambda(s)-Lambda(4-s), s=1.3 %.2e\n", abs(3^1.3*(2*Pi)^-1.3*gamma(1.3)*Lh(1.3) - 3^2.7*(2*Pi)^-2.7*gamma(2.7)*Lh(2.7)));
printf("(12) L(h,1) - 9/(2*Pi^2)*L(h,3)   %.2e\n\n", abs(L1 - 9/(2*Pi^2)*L3));

\\ ---- the two imported CM values, recomputed from PARI's modular forms package
mf = mfinit([9,4,1],0); Lmf = lfunmf(mf, mffrometaquo([3,8]));
printf("(13) L(h,2): Mellin %.30g\n     mf pkg %.30g\n     Gamma(1/3)^9/(96*Pi^4)          %.30g\n", L2, lfun(Lmf,2), G^9/(96*Pi^4));
printf("(14) L(h,3): Mellin %.30g\n     mf pkg %.30g\n     Gamma(1/3)^9/(144*sqrt(3)*Pi^3) %.30g\n", L3, lfun(Lmf,3), G^9/(144*sqrt(3)*Pi^3));
ran135 ms
I              = 15.83687034789483360405179196564431
(5/(2*sqrt3))*J = 15.83687034789483360405179196564431
closed form    = 15.83687034789483360405179196564431
max abs error  = 9.404 e-38

(5)  eta^4*th3^4 = 16*eta(2t)^8+eta(t/2)^8 at t=1.00  rel 9.08 e-39
(5)  eta^4*th3^4 = 16*eta(2t)^8+eta(t/2)^8 at t=1.50  rel 3.29 e-38
(5)  eta^4*th3^4 = 16*eta(2t)^8+eta(t/2)^8 at t=2.50  rel 1.51 e-38
(5)  eta^4*th3^4 = 16*eta(2t)^8+eta(t/2)^8 at t=4.00  rel 1.21 e-38
(3)  I - 2^(4/3)*Pi*int g         9.32 e-27
(4)  J - 2^(4/3)*Pi*int t*g       2.72 e-28
(6)  int g - 10*A0                1.18 e-27
(7)  int t*g - 8*A1               3.44 e-29
(8)  A0 - 3/(2*Pi)*L(h,1)         7.35 e-40
(9)  A1 - 9/(4*Pi^2)*L(h,2)       0.e-39
(10) I - 15*2^(4/3)*L(h,1)        0.e-37
(11) J - 18*2^(4/3)/Pi*L(h,2)     4.70 e-38
     Lambda(s)-Lambda(4-s), s=1.3 1.47 e-39
(12) L(h,1) - 9/(2*Pi^2)*L(h,3)   0.e-38

(13) L(h,2): Mellin 0.759965749993307635913622747426
     mf pkg 0.759965749993307635913622747426
     Gamma(1/3)^9/(96*Pi^4)          0.759965749993307635913622747426
(14) L(h,3): Mellin 0.918950262785094625843347770476
     mf pkg 0.918950262785094625843347770476
     Gamma(1/3)^9/(144*sqrt(3)*Pi^3) 0.918950262785094625843347770476

Note what this is and is not. Agreement to 38 digits is evidence, and it earns
no ∎. Only the Lean kernel sets that mark, and Mathlib has no \lambda(\tau), no
Damerell, and no {}_2F_1 critical values. The honest status of this post is: a complete proof
modulo (9), with every other step machine-checked.

References

  1. W. Zudilin, A Hypergeometric Version of the Modularity of Rigid
    Calabi–Yau Manifolds    , SIGMA 14 (2018), 086,
    https://arxiv.org/abs/1805.00544.
  2. W.-C. W. Li, L. Long and F.-T. Tu, Computing Special L-Values of Certain
    Modular Forms with Complex Multiplication    , SIGMA 14 (2018), 090,
    https://arxiv.org/abs/1803.06072.

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