Using Maple I am obtaining the numerical approximation


Please, let me know what is the closed form. Many thanks.

  • $\begingroup$ Do you know that there is one? Most of the time, with sums such as this, there are no closed-form expressions. Do you even know whether there's a closed-form expression for $J_0$? $\endgroup$ – Arthur Sep 20 '14 at 15:12
  • $\begingroup$ I am sure that a closed form exists. $\endgroup$ – Juan Ospina Sep 20 '14 at 15:25
  • $\begingroup$ Hint: Try to find a function which has the Fourier series $\sum_{n=1}^\infty \frac{J_0(n)}{n^2}$ and then use Percevals theorem. See this and this question. $\endgroup$ – Winther Sep 20 '14 at 15:43


Observe that

$$ J_0(n)=\frac 1\pi \int_0^\pi \cos (n \sin x)\:{\rm d}x \tag1 $$

and that $$\sum_{n=1}^\infty\frac{\cos nt}{n^4}=\frac{\pi^4}{90}-\frac{\pi^2 t^2}{12}+\frac{\pi t^3}{12}-\frac{t^4}{48},\quad 0\leq t\leq 2\pi. \tag2 $$ Then, due to normal convergence of the series on $[0,2\pi] $, we may write $$ \begin{align} \sum_{n=1}^\infty\frac{J_0^2(n)}{n^4} & =\frac{1}{\pi^2} \int_0^\pi \!\!\int_0^\pi \sum_{n=1}^\infty\frac{\cos (n \sin x)\cos (n \sin y)}{n^4}{\rm d}x\:{\rm d}y .\tag3 \end{align} $$ We may plug $$ 2\cos (n \sin x)\cos (n \sin y)=\cos (n(\sin x+\sin y))+\cos (n(\sin x-\sin y)) \tag4 $$ into $(3)$ and integrate as here.

Hence we obtain

$$ \sum_{n=1}^\infty\frac{J_0^2(n)}{n^4}=\frac{\pi ^4}{90}-\frac{\pi ^2}{12}-\frac{3}{64}+\frac{32}{27 \pi }. \tag5 $$

A numerical value is $$ \sum_{n=1}^\infty\frac{J_0^2(n)}{n^4} =0.5902373616900361395467798486 \ldots. $$

Some details.

Identity $(2)$ may be rewritten as $$\sum_{n=1}^\infty\frac{\cos nt}{n^4}=\frac{\pi^4}{90}-\frac{\pi^2 t^2}{12}+\frac{\pi |t|^3}{12}-\frac{t^4}{48},\quad -\pi \leq t\leq \pi. \tag6 $$

Then using $(3)$, $(4)$ and $(6)$, we get

$$ \begin{align} \sum_{n=1}^\infty\frac{J_0^2(n)}{n^4} =\frac{1}{2\pi^2}\int_0^\pi \!\!\int_0^\pi \left[\left(\frac{\pi^4}{90}-\frac{\pi^2}{12}(\sin x+\sin y)^2+\frac{\pi }{12}(\sin x+\sin y)^3-\frac{1}{48}(\sin x+\sin y)^4\right)+\left(\frac{\pi ^4}{90}-\frac{\pi ^2}{12}(\sin x-\sin y)^2+\frac{\pi }{12}\left|\sin x-\sin y\right|^3-\frac{1}{48}(\sin x-\sin y)^4\right)\right]{\rm d}x\:{\rm d}y .\tag7 \end{align} $$ We just have to be careful with the computation of the term ($|t|^3=\left({\rm Abs} (t)\right)^3$) $$ \begin{align} \frac{1}{2\pi^2}\times \frac{\pi }{12}\times \int_0^\pi \!\!\int_0^\pi \left|\sin x-\sin y\right|^3 {\rm d}x\:{\rm d}y & = \frac{1}{2\pi^2}\times\frac{\pi }{12} \times 4\int_0^{\pi/2} \!\!\int_0^{\pi/2} \left|\sin x-\sin y\right|^3 {\rm d}x\:{\rm d}y \\\\ &=-\frac{13}{36}+\frac{32}{27 \pi }. \end{align} $$

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    $\begingroup$ @JuanOspina My method is correct, you just have to be careful with the sign of the argument $(\sin x-\sin y)$ in the series, which is not always positive, leading to consider instead $$\int_0^\pi \!\!\int_0^\pi ({\rm Abs}(\sin x-\sin y))^3{\rm d}x\:{\rm d}y$$ probably you did not see this. Can you take it from there? Hoping to be clear. Thank you! $\endgroup$ – Olivier Oloa Sep 20 '14 at 21:49
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    $\begingroup$ @JuanOspina OK, I'm going to show the computations. Thanks. $\endgroup$ – Olivier Oloa Sep 20 '14 at 22:23
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    $\begingroup$ Hi Olivier, you are a STAR. Super very nice computation, many, many thanks. $\endgroup$ – Juan Ospina Sep 21 '14 at 0:01
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    $\begingroup$ @JuanOspina In general, we can use this identity $$\text{Li}_n(e^{2\pi ix})+(-1)^n\, \text{Li}_n(e^{-2\pi ix})=-\frac{(2\pi i)^n}{n!}B_n(x)$$ where $B_n(x)$ is the Bernoulli polynomials $\endgroup$ – Anastasiya-Romanova 秀 Sep 21 '14 at 13:27
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    $\begingroup$ Mr. @JuanOspina: Indeed, Andres Escobar is also from there. Say hi to James Rodriguez! ≧◠◡◠≦✌ $\endgroup$ – Anastasiya-Romanova 秀 Sep 21 '14 at 14:06

Using the wonderful method of Olivier implemented by Maple it is possible to prove that

$$\sum _{n=1}^{\infty }{\frac { {{\it J}_{0}\left(\,\alpha\,n\right)} ^{2}}{{n}^{4}}}=-{\frac {3 }{64}}\,{\alpha}^{4}+{\frac {1}{90}}\,{\pi }^{4}-{\frac {1}{12}}\,{\alpha}^{2}{ \pi }^{2}+{\frac {32}{27}}\,{\frac {{\alpha}^{3}}{\pi }} $$

$$\sum _{n=1}^{\infty }{\frac { {{\it J}_{0}\left(\,\alpha\,n\right)} ^{2}}{{n}^{2}}}=\frac{1}{4}\,{ \alpha}^{2}+\frac{1}{6}\,{\pi }^{2}-{\frac {4\alpha}{\pi }} $$

For this last sum, when $\alpha =2 $ we obtain

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

that is justly the result derived here.

Other results:

$$\sum _{n=1}^{\infty }{\frac { {{\it J}_{0}\left(\,\alpha\,n\right)} ^{2}}{{n}^{6}}}={\frac {1} {945}}\,{\pi }^{6}+{\frac {1}{64}}\,{\pi }^{2}{\alpha}^{4}-{\frac {1}{ 180}}\,{\pi }^{4}{\alpha}^{2}+{\frac {5}{1152}}\,{\alpha}^{6}-{\frac { 512}{3375}}\,{\frac {{\alpha}^{5}}{\pi }} $$

$$\sum _{n=1}^{\infty }{\frac { {{\it J}_{0}\left(\,\alpha\,n\right)} ^{2}}{{n}^{8}}}=-{\frac {{\pi }^{6}{\alpha}^{2} }{1890}}\,+{\frac {1}{960}}\,{\pi }^{4}{\alpha}^ {4}-{\frac {35}{147456}}\,{\alpha}^{8}-{\frac {5}{3456}}\,{\pi }^{2}{ \alpha}^{6}+{\frac {1}{9450}}\,{\pi }^{8}+{\frac {4096}{385875}}\,{ \frac {{\alpha}^{7}}{\pi }} $$

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    $\begingroup$ Nice results! Thank you. +1 $\endgroup$ – Olivier Oloa Sep 21 '14 at 10:09
  • $\begingroup$ Hi Olivier, thanks. These results were obtained using Maple with your equations (1), (3) and (7). Your equation (4) was not used. I am using the evaluation of (3) in terms of the polylogaritm function given by Maple. Anastasiya-Romanova is noting that point. Again, many thanks. All the best. $\endgroup$ – Juan Ospina Sep 21 '14 at 12:57

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