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 $$ \int_{0}^{\infty}x^{s-1}\vartheta_2(q)^3\vartheta_4(q)^2 \sqrt{\vartheta_2(q)\vartheta_4(q)}\text{d}x =L_f(s)\Gamma(s)\times\text{other components}. $$
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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: $$ \int_{0}^{1} \frac{x^{\frac14}}{(1-x^2)^{\frac18}}K(x)\text{d}x=\frac{\pi^2}{12}\sqrt{5-\frac{1}{\sqrt{2} } }. $$
Setness Ramesory
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This question is trivial if you know $${}_3F_2\left(\frac12,\frac12,\frac18;1,\frac98;1\right)=\frac{\left ( 3+\sqrt{2} \right )\Gamma\left ( \frac18 \right )^2 \Gamma\left ( \frac38 \right )^2 }{96\pi^2\sqrt{2} }$$ i,e, $\int_0^1 x^{-3/4} K(x) dx$. Because it's obviously related to the one in question by standard $_3F_2$ transformation. – pisco Sep 29 '23 at 18:01
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@pisco does your coefficient-comparing method applicable to series with factor $(\frac12)_n^2/(1)_n^2$? – Setness Ramesory Sep 30 '23 at 07:22
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@pisco Just for curiosity. I still have obstacles dealing with extremely complicated computations(that should be done by computers). Could you share one? – Setness Ramesory Sep 30 '23 at 12:42
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then learn a bit about programing. I use Mathematica most of the time for complicated computations. – pisco Sep 30 '23 at 12:56
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@pisco frankly, I once used some programmes(such as your integrating), but my current conditions are not really good. If any inconvenience, I am quite sorry for that. – Setness Ramesory Sep 30 '23 at 13:19
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@pisco of course some of them aren't, like a obvious one $\sum_{n=1}^{\infty} \frac{\left ( \frac12 \right )n^3}{(1)_n^3} \frac{6n+1}{2^{2n}}\left ( H{2n}-H_n \right ) =\frac{16\ln(2)}{3\pi}-\frac{\sqrt{3},\Gamma\left ( \frac13 \right )^6 }{2^{8/3}\pi^4}$. – Setness Ramesory Oct 01 '23 at 07:07
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I believe if you uses certain software to aid those heavy computations, then you will definitely discover more amazing results. – pisco Oct 01 '23 at 08:05
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@pisco What is FL recurrence relation of $\frac{K(\sqrt{x})}{\sqrt{x}}$? Or are there any methods which can judge the complexity of recurrence relations when $f(x)$ is given(without explicit computations)? – Setness Ramesory Oct 01 '23 at 15:16
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If $f(n) = \int_0^1 K(\sqrt{x})/\sqrt{x} P_n(2x-1)dx$, then $$-(2 n+3) \left(2 n^2+6 n+5\right) f(n+1)-2 (n+1)^3 f(n)-2 (n+2)^3 f(n+2) + 6 + 4n = 0$$ this recurrence has no hypergeometric solution. Regarding your second question, the answer is no, one always needs to do some computations. – pisco Oct 04 '23 at 22:14
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@pisco Do you have any WZ identities which can be exploited to accelerate the sum $\sum_{n=0}^{\infty} \frac{\left ( \frac12 \right )n^3 }{(1)_n^3} \sum{j=0}^{n-1} \frac{1}{(2j+1)^2} =\frac{\left ( \pi^2-8G \right )\Gamma\left ( \frac14 \right )^4 }{32\pi^3}?$ – Setness Ramesory Oct 14 '23 at 14:09
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@pisco any idea with $,_3F_2\left ( \frac12,\frac12,1;\frac34,\frac34;-\frac18 \right ) =\frac23+\frac{\Gamma\left ( \frac34 \right )^2}{3\sqrt{\pi} }?$ – Setness Ramesory Oct 21 '23 at 10:30
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@pisco Do you have any insights into the connection between $K(k)$ integrals and MZVs? I could only found simple(and ugly) ones like$$\int_{0}^{1} \frac{(2k^2-1)K(k)^2\arctan\left ( \frac{K^\prime(k)}{K(k)} \right ) }{ \sqrt{1-k^2} }\text{d}k =-4\operatorname{Li}_4\left ( \frac12 \right ) -\frac72\zeta(3)\ln(2)+\frac{\pi^2}{6}\ln(2)^2-\frac{1}{6}\ln(2)^4+\frac{151}{2880}\pi^4.$$ – Setness Ramesory Jan 07 '24 at 06:41
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It could be seen as a generalization to $\int_{0}^{1} \frac{\arctan\left ( \frac{K^\prime(k)}{K(k)} \right ) }{ \sqrt{1-k^2} }\text{d}k =\frac{\pi^2}{8}$ – Setness Ramesory Jan 07 '24 at 06:51
EllipticK[x^2]instead – Setness Ramesory Apr 30 '23 at 01:29EllipticK[k^2]. – Setness Ramesory Apr 30 '23 at 02:50