Prove $\int_{0}^{1} \frac{k^{\frac34}}{(1-k^2)^\frac38} K(k)\text{d}k=\frac{\pi^2}{12}\sqrt{5+\frac{1}{\sqrt{2} } }$

Question

The paper mentioned a proposition: $$ \int_{0}^{1} \frac{k^{\frac34}}{(1-k^2)^\frac38} K(k)\text{d}k=\frac{\pi^2}{12}\sqrt{5+\frac{1}{\sqrt{2} } }. $$ Its equivalent is $$ \int_{0}^{\infty}\vartheta_2(q)^3\vartheta_4(q)^2 \sqrt{\vartheta_2(q)\vartheta_4(q)}\text{d}x =\frac{1}{3} \sqrt{5+\frac{1}{\sqrt{2} } }. $$ I think that there is a $L$-series satisfying $$ \int_{0}^{\infty}x^{s-1}\vartheta_2(q)^3\vartheta_4(q)^2 \sqrt{\vartheta_2(q)\vartheta_4(q)}\text{d}x =L_f(s)\Gamma(s)\times\text{other components}. $$

Answer 1

Months later, I finally find a self-contained proof(which do lead to a general expression). We have for $\Re(s)\in(0,2)$, $$ \int_{0}^{1} x^{s-1}K(x)\text{d}x =\frac{\sin\left ( \frac{\pi s}{2} \right ) }{\pi} \frac{\Gamma\left ( \frac{s}2 \right )^2}{\Gamma(\frac{s+1}2)^2} \int_{0}^{1} x^{-s+1}(1-x^2)^{s/2-1/2}K(x)\text{d}x. $$ Substituting $s=1/4$ and as a matter of fact that $$ \int_{0}^{1} x^{-3/4}K(x)\text{d}x=\frac{\left ( 3+\sqrt{2} \right ) \Gamma\left ( \frac18 \right )^2\Gamma\left ( \frac38 \right )^2 }{ 48\pi\sqrt{2} }, $$ gives $$ \int_{0}^{1} \frac{x^{\frac34}}{(1-x^2)^{\frac38}}K(x)\text{d}x =\frac{\pi^2}{12}(2+3\sqrt{2})\sin\left(\frac\pi{8}\right), $$ with $$ (2+3\sqrt{2})\sin\left(\frac\pi{8}\right)=\sqrt{5+\frac{1}{\sqrt{2} } }. $$ A related one: $$ \int_{0}^{1} \frac{x^{\frac14}}{(1-x^2)^{\frac18}}K(x)\text{d}x=\frac{\pi^2}{12}\sqrt{5-\frac{1}{\sqrt{2} } }. $$