I start from the following obvious observation, which is declared to be($q=e^{-\pi x}$): \begin{aligned} \int_{1}^{\infty}x\vartheta_2(q)^4\vartheta_4(q)^4 \text{d}x&=\int_{0}^{1}x\vartheta_2(q)^4\vartheta_4(q)^4 \text{d}x\\ &=\frac12\int_{0}^{\infty}x\vartheta_2(q)^4\vartheta_4(q)^4 \text{d}x=\frac14, \end{aligned} in which the functional equation $\vartheta_2(e^{-\pi/x})=\sqrt{x}\vartheta_4(e^{-\pi x})$ is used. The last equality is derived from $$ \int_{0}^{\infty}x^{s-1}\vartheta_2(q)^4\vartheta_4(q)^4\text{d}x =16\frac{\Gamma(s)}{\pi^s}\lambda(s)\eta(s-3). $$ Then, I am confronted with the new one $$ \int_{1}^{\infty}\theta_2(q)^4\theta_4(q)^4 \text{d}x=\frac{2G}{\pi^2}. $$ These two integrals have cousins widely as well. For example, some experiments suggest \begin{aligned} &\int_{1}^{\infty}(x^2-1)\vartheta_2 (q)^4\vartheta_3(q)^4 =\frac{8G}{\pi^2}-\frac{14}{\pi^3}\zeta(3),\\ &\int_{1}^{\infty}(1+x)^2 \vartheta_2(q)^2\vartheta_3(q)^4\vartheta_4(q)^4\text{d}x =\frac{\Gamma\left ( \frac14 \right)^8}{8\pi^6}, \end{aligned} the integral $$ \int_{1}^{\infty} \left [ (x-1)\vartheta_2(q)^4\vartheta_3(q)^2 +2\sqrt{2}x\vartheta_2(q)^3\vartheta_4(q)^3 \right ]\text{d}x =2-\frac{8G}{\pi^2} $$ included. By substituting $x=K^\prime(k)/K(k)$, we obtain \begin{aligned} &\int_{0}^{\frac{1}{\sqrt{2} } } K(k)\left ( K(k)+K^\prime(k) \right )^2\text{d}k =\frac{\Gamma\left ( \frac14 \right )^8 }{128\pi^2},\\ &\int_{\frac{1}{\sqrt{2} } }^{1} \frac{K(k)^2-K^\prime(k)^2}{k} \text{d}k =\pi G-\frac{7}{4}\zeta(3),\\ &\int_{\frac{1}{\sqrt{2} } }^{1} \left(\frac{K(k)-K^\prime(k)}{k}+\frac{2\sqrt{2}\sqrt{k}K(k) }{(1-k^2)^{1/4}} \right)\text{d}k =\frac{\pi^2}{2}-2G. \end{aligned}
Question. Is there any ways being able to prove the integrals listed above?
The 2nd Question. I wonder if it exists similar ones with the interval $[\sqrt{n},+\infty)$, where $n\in\mathbb{Q}_{>0}$.
The methodology to solve those integrals is somewhat complex. Some fair results could be shown first. \begin{align} &\int_{0}^{\frac{1}{\sqrt{2}} }K(k)\text{d}k =\frac{\pi}{2\sqrt{2} }\,_3F_2 \left ( \frac12,\frac12,\frac12;1,\frac32;\frac12 \right ),\\ &\int_{0}^{\frac{1}{\sqrt{2}} }K^\prime(k)\text{d}k =\frac{\pi^2}{8}+ \frac{\pi}{4 }\,_3F_2 \left ( \frac12,\frac12,\frac12;1,\frac32;-1 \right ),\\ &\int_{0}^{\frac{1}{\sqrt{2}} }\frac{K(k)}{1-k^2}\text{d}k =\frac{\pi}{2}\,_3F_2 \left ( \frac12,\frac12,\frac12;1,\frac32;-1 \right ),\\ &\int_{0}^{\frac{1}{\sqrt{2}} }\frac{K^\prime(k)}{1-k^2}\text{d}k =\frac{\pi}{\sqrt{2}}\,_3F_2 \left ( \frac12,\frac12,\frac12;1,\frac32;\frac12 \right ),\\ &\begin{aligned}\int_{\frac{1}{\sqrt{2} } }^1 \frac{K(k)}{k} \text{d}k &=\frac{5\pi}{4}\ln(2)-\frac{\pi}{32}\,_4F_3\left ( 1,1,\frac32,\frac32;2,2,2;\frac12 \right )-2G,\\ &=\frac{\pi}{2}\,_3F_2\left ( \frac14,\frac14,\frac14;1,\frac54;1 \right ) +\frac{1}{2}\sqrt{\frac{\pi}{2} } \Gamma\left ( \frac34 \right )^2 \,_3F_2\left ( \frac14,\frac14,\frac34;1,\frac54;1 \right ) -2G,\end{aligned}\\ &\int_{\frac{1}{\sqrt{2} } }^1 \frac{K^\prime(k)}{k} \text{d}k =\frac{\pi}{2}\,_3F_2\left ( \frac14,\frac14,\frac14;1,\frac54;1 \right ) -\frac{1}{2}\sqrt{\frac{\pi}{2} } \Gamma\left ( \frac34 \right )^2 \,_3F_2\left ( \frac14,\frac14,\frac34;1,\frac54;1 \right ),\\ &\int_{0}^{\frac{1}{\sqrt{2} } } \frac{K(k)}{\sqrt{1-k^2} } \text{d}k =\frac{\pi^2}{8}\,_3F_2\left ( \frac14,\frac14,\frac12;1,1;1 \right ),\\ &\begin{aligned} \int_{0}^{\frac{1}{\sqrt{2} } } \frac{K(k)}{\sqrt{k}\left ( 1-k^2 \right )^{3/4} } \text{d}k & = \frac{\Gamma\left ( \frac14 \right )^4 }{8\pi\sqrt{2} }- \frac{\pi^2}{4\sqrt{2} }\,_3F_2\left ( \frac14,\frac14,\frac12;1,1;1 \right ),\\ &=\pi\,_3F_2\left ( \frac14,\frac12,\frac12;1,\frac54;-1 \right ),\\ &=\frac{\sqrt{\pi} }{8}\Gamma\left ( \frac14 \right )^2\,_3F_2\left ( \frac14,\frac14,\frac14;\frac34,1;1 \right ), \end{aligned}\\ &\int_{\frac{1}{\sqrt{2} } }^{1} \frac{K(k)^2}{k}\text{d}k =\frac{3\pi^2}{8}\ln(2)+\frac{\pi G}{2} -\frac{7}{4}\zeta(3)-\frac{\pi^2}{128} \,_5F_4\left ( 1,1,\frac32,\frac32,\frac32;2,2,2,2;1 \right ),\\ &\int_{\frac{1}{\sqrt{2} } }^{1} \frac{K^\prime(k)^2}{k}\text{d}k =\frac{3\pi^2}{8}\ln(2)-\frac{\pi G}{2} -\frac{\pi^2}{128} \,_5F_4\left ( 1,1,\frac32,\frac32,\frac32;2,2,2,2;1 \right ),\\ &\int_{0}^{\frac{1}{\sqrt{2} } } kK(k)K^\prime(k)\left ( K^\prime(k)^2-K(k)^2 \right ) \text{d}k=\frac{1}{2} \int_{0}^{\frac{1}{\sqrt{2}}} \frac{K(k)^4}{\sqrt{1-k^2} }\text{d}k,\\ &\int_0^1K(k)^2\text{d}k+\int_{\frac{1}{\sqrt{2} }}^{1} \left ( K(k)^2-K^\prime(k)^2 \right )\text{d}k =\frac{1}{\sqrt{2} } \int_{0}^{\frac{1}{\sqrt{2} }} \left ( \frac{1}{\sqrt{k}\left ( 1-k^2 \right )^{1/4} } +\frac{\sqrt{k} }{\left ( 1-k^2 \right )^{3/4} } \right ) K(k)^2 \text{d}k,\\ &\int_{\frac{1}{\sqrt{2} }}^{1} K(k)K^\prime(k)\text{d}k =\frac{1}{2\sqrt{2} } \int_{0}^{\frac{1}{\sqrt{2} }} \left ( \frac{1}{\sqrt{k}\left ( 1-k^2 \right )^{1/4} } -\frac{\sqrt{k} }{\left ( 1-k^2 \right )^{3/4} } \right ) K(k)^2 \text{d}k,\\ &\int_{0}^{\frac{1}{\sqrt{2} }} \frac{K(k)^2}{\sqrt{1-k^2} } \text{d}k =\frac{\pi^3}{16}\,_4F_3\left ( \frac12,\frac12,\frac12,\frac12;1,1,1;1 \right ),\\ &\frac{1}{2\sqrt{2} } \int_{0}^{\frac{1}{\sqrt{2} }} \frac{K(k)^3}{\sqrt{k} \left ( 1-k^2 \right )^{3/4} } \text{d}k-\int_{0}^{\frac{1}{\sqrt{2} }} kK(k)^3\text{d}k =\frac{\Gamma\left ( \frac14 \right )^8 }{512\pi^2} -\frac{3\pi^2}{8}\,_3F_2\left ( \frac12,\frac12,\frac12;\frac54,\frac54;1 \right ). \end{align}