Prove that \[ 60\sum_{\lambda\in\mathbb Z[i]\setminus\{0\}}\lambda^{-4} = 64\pi^2\frac{\Gamma(5/4)^4}{\Gamma(3/4)^4}. \]
Solution
Let \[ \Lambda=\mathbb Z[i], \qquad G_2(\Lambda)=\sum_{\lambda\in\Lambda\setminus\{0\}}\lambda^{-4}, \qquad g_2=60G_2(\Lambda). \] Write \[ E=\wp_\Lambda(1/2)>0. \]
1. For the square lattice,\[ g_3=0, \qquad (e_1,e_2,e_3)=(E,-E,0), \qquad g_2=4E^2. \]
Multiplication by \(i\) preserves \(\Lambda\). Hence \[ G_3(\Lambda) =\sum_{\lambda\ne0}\lambda^{-6} =\sum_{\lambda\ne0}(i\lambda)^{-6} =-G_3(\Lambda), \] so \(G_3(\Lambda)=0\) and therefore \(g_3=140G_3(\Lambda)=0\).
The square symmetry also gives \[ e_2=-e_1, \qquad e_3=0. \] With \(e_1=E>0\), the Weierstrass equation is therefore \[ \wp_\Lambda'(z)^2 =4(\wp_\Lambda(z)-E)\wp_\Lambda(z)(\wp_\Lambda(z)+E) =4\wp_\Lambda(z)^3-4E^2\wp_\Lambda(z). \] Comparing this with \[ \wp_\Lambda'(z)^2=4\wp_\Lambda(z)^3-g_2\wp_\Lambda(z)-g_3 \] gives \(g_2=4E^2\).
2. If\[ I=\int_1^\infty\frac{dt}{\sqrt{t^3-t}}, \]then\[ E=I^2 \qquad\text{and}\qquad g_2=4I^4. \]
The real half-period formula for the square lattice gives \[ \frac12 =\int_E^\infty\frac{dx}{\sqrt{4x^3-g_2x}}. \] By 1, \[ 4x^3-g_2x=4x(x^2-E^2), \] so \[ \frac12 =\frac12\int_E^\infty\frac{dx}{\sqrt{x(x^2-E^2)}}. \] Substituting \(x=Et\) yields \[ \frac12 =\frac1{2\sqrt E} \int_1^\infty\frac{dt}{\sqrt{t(t^2-1)}} =\frac I{2\sqrt E}. \] Thus \(E=I^2\), and 1 gives \(g_2=4I^4\).
3. We have\[ \boxed{ 60\sum_{\lambda\in\mathbb Z[i]\setminus\{0\}}\lambda^{-4} =64\pi^2\frac{\Gamma(5/4)^4}{\Gamma(3/4)^4}. } \]
In the integral from 2, put \(u=t^{-2}\). Since \[ dt=-\frac12u^{-3/2}\,du, \qquad t^3-t=u^{-3/2}(1-u), \] we obtain \[ \begin{aligned} I &=\frac12\int_0^1u^{-3/4}(1-u)^{-1/2}\,du\\ &=\frac12 B\left(\frac14,\frac12\right)\\ &=\frac{\sqrt\pi\,\Gamma(1/4)}{2\Gamma(3/4)}. \end{aligned} \] Therefore \[ \begin{aligned} 60\sum_{\lambda\in\mathbb Z[i]\setminus\{0\}}\lambda^{-4} =g_2 &=4I^4\\ &=\frac{\pi^2\Gamma(1/4)^4}{4\Gamma(3/4)^4}\\ &=64\pi^2\frac{\Gamma(5/4)^4}{\Gamma(3/4)^4}, \end{aligned} \] because \(\Gamma(5/4)=\frac14\Gamma(1/4)\).
4. Q.E.D.
Step 1 reduces the square-lattice invariant to its positive half-period, step 2 expresses that half-period by the beta integral, and step 3 evaluates the integral and obtains the required identity.