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Using the method showed here proposed by Olivier Oloa with simplifications proposed by Anastasiya-Romanova, it is possible to prove that

$$\sum _{n=1}^{\infty }{\frac {{{\it J}_{0}\left(\,2\,n\right)} {{\it J}_{0}\left(\,n\right)}}{{n}^{2}}}={\frac {5}{8}}+{\frac {1}{6}}\,{\pi }^{2}-4\,{\frac { {\it EllipticE} \left( {\frac {1}{2}} \right) }{\pi }}-\frac{2}{{\pi }}\,\int _{0}^{1}\!{\frac { v\arcsin \left( {\frac {1}{2}}\,v \right) }{\sqrt {1-{v}^{2}}}}{dv} $$

but for the last integral it is possible to have a closed form given by Jack D'Aurizio (Please note that Jack is using the notation for elliptic functions in Mathematica and I am using the more standard notation used by Maple). Then we have

$$\sum _{n=1}^{\infty }{\frac {{{\it J}_{0}\left(\,2\,n\right)} {{\it J}_{0}\left(\,n\right)}}{{n}^{2}}}=-8\,{\frac {{\it EllipticE} \left( \frac {1}{2} \right) }{\pi }}+3\,{\frac {{\it EllipticK} \left( \frac {1}{2} \right) }{\pi }}+\frac {5}{8}+\frac {1}{6}\,{\pi }^{2} $$

According with this result the answer to my question maybe yes. Do you agree?

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    $\begingroup$ The last integral equals $$\frac{2 \left(2 \text{EllipticE}\left[\frac{1}{4}\right]-\frac{3}{2} \text{EllipticK}\left[\frac{1}{4}\right]\right)}{\pi }$$. $\endgroup$ – Jack D'Aurizio Sep 21 '14 at 16:50
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    $\begingroup$ You are very right, many thanks. I am using your result but with the standard notation. $\endgroup$ – Juan Ospina Sep 21 '14 at 17:14
  • $\begingroup$ @JuanOspina Are there any assumptions on the values of $\alpha$ and $\beta$? Can we assume $\alpha,\beta>0$? It would also be convenient if we could assume $\alpha+\beta<2\pi$. $\endgroup$ – David H Sep 21 '14 at 21:17
  • $\begingroup$ Hi @DavidH, given that $J_{0}$ is an even function it is sufficient to consider only the case when $\alpha>0$ and $\beta>0$. The assumption that you are mentioning could be used if you think that is possible to obtain a closed form. Please let me know. Many thanks. $\endgroup$ – Juan Ospina Sep 21 '14 at 21:21
  • $\begingroup$ @JuanOspina See my edit below for the advantages of assuming $\alpha+\beta<2\pi$. It allows us to simplify one of the trigonometric sums to a quadratic function. $\endgroup$ – David H Sep 21 '14 at 22:04
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Using the method showed here proposed by Olivier Oloa with simplifications proposed by Anastasiya-Romanova and which is essentially the method that David H is showing, I am obtaining

$$\sum _{n=1}^{\infty }{\frac {{{\it J}_{0}\left(\,\alpha\,n\right)} {{\it J}_{0}\left(\,\beta\,n\right)}}{{n}^{2}}}=\frac {{\beta}^{2}}{8}\,+\frac{1}{6}\,{ \pi }^{2}+\frac{1}{8}\,{\alpha}^{2}-2\,{\alpha}^{2}{\it EllipticF} \left( { \frac {\beta}{\alpha}},{\frac {\alpha}{\beta}} \right) {\beta}^{-1}{ \pi }^{-1}+2\,\beta\,{\it EllipticF} \left( {\frac {\beta}{\alpha}},{ \frac {\alpha}{\beta}} \right) {\pi }^{-1}-2\,\beta\,{\it EllipticE} \left( {\frac {\beta}{\alpha}},{\frac {\alpha}{\beta}} \right) {\pi } ^{-1}-2\,\beta{\pi }^{-1}\,\int _{0}^{1}\!v\arcsin \left( {\frac {\beta\,v}{ \alpha}} \right) {\frac {1}{\sqrt {1-{v}^{2}}}}{dv} $$

Then there is a closed form if the last integral has a closed form. But for the last integral it is possible to have a closed form given by Daniel H (Please note that Daniel is using the notation for elliptic functions in Mathematica and I am using the more standard notation used by Maple). Then we have the following closed form:

$$\sum _{n=1}^{\infty }{\frac {{{\it J}_{0}\left(\,\alpha\,n\right)} {{\it J}_{0}\left(\,\beta\,n\right)}}{{n}^{2}}}=\frac{1}{8}\,{\beta}^{2}+\frac{1}{6}\,{ \pi }^{2}+\frac{1}{8}\,{\alpha}^{2}-2\,{\alpha}^{2}{\it EllipticF} \left( { \frac {\beta}{\alpha}},{\frac {\alpha}{\beta}} \right) {\beta}^{-1}{ \pi }^{-1}+2\,\beta\,{\it EllipticF} \left( {\frac {\beta}{\alpha}},{ \frac {\alpha}{\beta}} \right) {\pi }^{-1}-2\,\beta\,{\it EllipticE} \left( {\frac {\beta}{\alpha}},{\frac {\alpha}{\beta}} \right) {\pi } ^{-1}-2\,{\beta}^{2}{\it EllipticK} \left( {\frac {\beta}{\alpha}} \right) {\pi }^{-1}{\alpha}^{-1}+2\,\alpha\,{\it EllipticK} \left( { \frac {\beta}{\alpha}} \right) {\pi }^{-1}-2\,\alpha\,{\it EllipticE} \left( {\frac {\beta}{\alpha}} \right) {\pi }^{-1} $$

It is worthwhile to note the following particular case: $$\sum _{n=1}^{\infty }{\frac {{{\it J}_{0}\left(\,\alpha\,n\right)} {{\it J}_{0}\left(\,\frac{1}{2}\,\alpha\,n\right)}}{{n}^{2}}}={\frac {5}{32}}\,{ \alpha}^{2}+\frac{1}{6}\,{\pi }^{2}-3\,{\frac {\alpha\,{\it EllipticF} \left( \frac{1}{2},2 \right) }{\pi }}-{\frac {\alpha\,{\it EllipticE} \left( \frac{1}{2},2 \right) }{\pi }}-2\,{\frac {\alpha\,{\it EllipticE} \left( \frac{1}{2} \right) }{\pi }}+\frac{3}{2}\,{\frac {\alpha\,{\it EllipticK} \left( \frac{1}{2} \right) }{\pi }} $$

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  • $\begingroup$ The last integral does in fact have a closed form: $\int _{0}^{1}\!v\arcsin \left( {\frac {\beta\,v}{ \alpha}} \right) {\frac {1}{\sqrt {1-{v}^{2}}}}{dv}=\frac{\beta^2-\alpha^2}{\alpha\beta}K{\left(\frac{\beta^2}{ \alpha^2}\right)}+\frac{\alpha}{\beta}E{\left(\frac{\beta^2}{\alpha^2}\right)}$. Assuming everything else is correct, you have your closed form. $\endgroup$ – David H Sep 21 '14 at 23:06
  • $\begingroup$ Hi @DavidH, you are great. Now we have a correct closed form. I am using your result with the standard notation used by Maple. Many thanks. $\endgroup$ – Juan Ospina Sep 21 '14 at 23:29
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(Too long for comment)

The Bessel function of the first kind of order zero, $J_{0}{\left(z\right)}$, has the following integral representation:

$$J_{0}{\left(z\right)}=\frac{2}{\pi}\int_{0}^{1}\mathrm{d}t\,\frac{\cos{\left(zt\right)}}{\sqrt{1-t^2}}.$$

Represent the Bessel functions in the series by the integrals,

$$J_{0}{\left(\alpha n\right)}=\frac{2}{\pi}\int_{0}^{1}\mathrm{d}x\,\frac{\cos{\left(\alpha nx\right)}}{\sqrt{1-x^2}},$$

$$J_{0}{\left(\beta n\right)}=\frac{2}{\pi}\int_{0}^{1}\mathrm{d}y\,\frac{\cos{\left(\beta ny\right)}}{\sqrt{1-y^2}}.$$

Assume that $0<\alpha,\beta$ and $\alpha+\beta<2\pi$. Then,

$$\sum_{n=1}^{\infty}\frac{\cos{\left[\left(\alpha x+\beta y\right)n\right]}}{n^2}=\frac{\pi^2}{6}-\frac{\pi \left(\alpha x+\beta y\right)}{2}+\frac{\left(\alpha x+\beta y\right)^2}{4}.$$

We can then write the series as a double integral:

$$\begin{align} S{\left(\alpha,\beta\right)} &=\sum_{n=1}^{\infty}\frac{J_{0}{\left(\alpha n\right)} J_{0}{\left(\beta n\right)}}{n^2}\\ &=\frac{4}{\pi^2}\sum_{n=1}^{\infty}\frac{1}{n^2} \int_{0}^{1}\mathrm{d}x\,\frac{\cos{\left(\alpha nx\right)}}{\sqrt{1-x^2}} \int_{0}^{1}\mathrm{d}y\,\frac{\cos{\left(\beta ny\right)}}{\sqrt{1-y^2}}\\ &=\frac{4}{\pi^2}\int_{0}^{1}\mathrm{d}x\int_{0}^{1}\mathrm{d}y\sum_{n=1}^{\infty}\frac{\cos{\left(\alpha nx\right)}\,\cos{\left(\beta ny\right)}}{n^2\,\sqrt{1-x^2}\,\sqrt{1-y^2}}\\ &=\frac{4}{\pi^2}\int_{0}^{1}\mathrm{d}x\int_{0}^{1}\mathrm{d}y\,\frac{1}{\sqrt{1-x^2}\,\sqrt{1-y^2}}\sum_{n=1}^{\infty}\frac{\cos{\left(\alpha nx\right)}\,\cos{\left(\beta ny\right)}}{n^2}\\ &=\frac{2}{\pi^2}\int_{0}^{1}\mathrm{d}x\int_{0}^{1}\mathrm{d}y\,\frac{1}{\sqrt{1-x^2}\,\sqrt{1-y^2}}\sum_{n=1}^{\infty}\frac{\cos{\left[(\alpha x-\beta y)n\right]}+\cos{\left[(\alpha x+\beta y)n\right]}}{n^2}\\ &=\frac{2}{\pi^2}\int_{0}^{1}\mathrm{d}x\int_{0}^{1}\mathrm{d}y\,\frac{1}{\sqrt{1-x^2}\,\sqrt{1-y^2}}\sum_{n=1}^{\infty}\frac{\cos{\left[(\alpha x-\beta y)n\right]}}{n^2}\\ &~~~~~ + \frac{2}{\pi^2}\int_{0}^{1}\mathrm{d}x\int_{0}^{1}\mathrm{d}y\,\frac{1}{\sqrt{1-x^2}\,\sqrt{1-y^2}}\sum_{n=1}^{\infty}\frac{\cos{\left[(\alpha x+\beta y)n\right]}}{n^2}\\ &=\frac{2}{\pi^2}\int_{0}^{1}\mathrm{d}x\int_{0}^{1}\mathrm{d}y\,\frac{1}{\sqrt{1-x^2}\,\sqrt{1-y^2}}\sum_{n=1}^{\infty}\frac{\cos{\left[(\alpha x-\beta y)n\right]}}{n^2}\\ &~~~~~ + \frac{2}{\pi^2}\int_{0}^{1}\mathrm{d}x\int_{0}^{1}\mathrm{d}y\,\frac{1}{\sqrt{1-x^2}\,\sqrt{1-y^2}}\left[\frac{\pi^2}{6}-\frac{\pi \left(\alpha x+\beta y\right)}{2}+\frac{\left(\alpha x+\beta y\right)^2}{4}\right]\\ &=\frac{2}{\pi^2}\int_{0}^{1}\mathrm{d}x\int_{0}^{1}\mathrm{d}y\,\frac{1}{\sqrt{1-x^2}\,\sqrt{1-y^2}}\sum_{n=1}^{\infty}\frac{\cos{\left[(\alpha x-\beta y)n\right]}}{n^2}\\ &~~~~~ + \frac{2}{\pi^2}\left[\frac{\pi^4}{24}-\frac{\pi^2}{4}\left(\alpha+\beta\right)+\frac12\alpha\beta+\frac{\pi^2}{32}\left(\alpha^2+\beta^2\right)\right]\\ &=\frac{2}{\pi^2}\int_{0}^{1}\mathrm{d}x\int_{0}^{1}\mathrm{d}y\,\frac{1}{\sqrt{1-x^2}\,\sqrt{1-y^2}}\sum_{n=1}^{\infty}\frac{\cos{\left[(\alpha x-\beta y)n\right]}}{n^2}\\ &~~~~~ + \left[\frac{\pi^2}{12}-\frac{1}{2}\left(\alpha+\beta\right)+\frac{\alpha\beta}{\pi^2}+\frac{1}{16}\left(\alpha^2+\beta^2\right)\right]\\ \end{align}$$

...

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  • $\begingroup$ Your method is essentially the method that I am using. Many thanks. $\endgroup$ – Juan Ospina Sep 21 '14 at 19:42
  • $\begingroup$ Hi @DavidH, very nice simplification. Do you thinks that it is possible to obtain a closed form for the remaining integrals. Many thanks. $\endgroup$ – Juan Ospina Sep 21 '14 at 22:11
  • $\begingroup$ @JuanOspina My gut says yes, it should be possible to find a closed form for the general case. Have I found it yet? Nope. :) $\endgroup$ – David H Sep 21 '14 at 22:20

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