In an answer, I showed that: $$\sin(1)=2\sum_{k=0}^\infty(-1)^k J_{2k+1}(1)$$ Where $J_n(x)$ is the Bessel function of the first kind. Is there a more general result for the infinite sum of odd Bessel function? $$\sum_{k=0}^\infty (-1)^k J_{2k+1}(x)\ =\ ?$$ $$\sum_{k=0}^\infty J_{2k+1}(x)\ =\ ?$$
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1You might find my incomplete answer to another question interesting in this context. (I can't make anymore headway on that approach myself.) – Semiclassical Jul 25 '14 at 18:22
2 Answers
From the Jacobi-Anger expansion: $$e^{i z\cos\theta} = J_0(z) + 2\sum_{n=1}^{+\infty}i^n J_n(z) \cos(n\theta)$$ we have, by considering the imaginary part: $$\sin(z\cos\theta) = 2\sum_{m=0}^{+\infty}(-1)^m J_{2m+1}(z)\cos((2m+1)\theta)\tag{1}$$ and we can remove the cosine-dependent term by exploiting the identities: $$\int_{0}^{\pi/2}\cos((2n+1)x)\cos((2m+1)x)dx = \frac{\pi}{4}\delta_{m,n},$$ $$\sum_{m=0}^{+\infty}(-1)^m\cos((2m+1)\theta)=\frac{1}{2\cos\theta},$$ that give (integrating against the proper kernel): $$\frac{4}{\pi}\int_{0}^{\pi/2}\frac{\sin(z\cos\theta)}{2\cos\theta}d\theta=2\sum_{m=0}^{+\infty}J_{2m+1}(z),$$ hence:
$$\begin{eqnarray*}\sum_{m=0}^{+\infty}J_{2m+1}(z) &=& \frac{1}{\pi}\int_{0}^{\pi/2}\frac{\sin(z\cos\theta)}{\cos\theta}d\theta = \frac{1}{\pi}\int_{0}^{1}\frac{\sin(z t)}{t\sqrt{1-t^2}}dt\\&=&\frac{1}{2}\sum_{r=0}^{+\infty}\frac{(-1)^r}{(2r+1)4^r (r!)^2}z^{2r+1}=\frac{1}{2}\int_{0}^{z}J_0(u)du.\end{eqnarray*}$$
For the alternating sum, it is sufficient to take $\theta=0$ in $(1)$ in order to have:
$$\sum_{m=0}^{+\infty}(-1)^m J_{2m+1}(z) = \frac{1}{2}\sin z.$$

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I wonder if that kind of analysis could provide anything useful for this incompete answer of mine. (I fear that'll just return the original integral, though...) – Semiclassical Jul 25 '14 at 18:16
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@Semiclassical: that problem is a tough nut to crack. Continued fractions are a powerful tool for irrationality proofs, but after computing a few convergents for $f^{-1}(z)$ I lost hope in finding a simple pattern. I do not really believe we can prove the irrationality of a root of an analytic function by simply writing down some Taylor or Fourier coefficients, so I think the identites stated here are quite useless in the other context. But, obviously, I may be wrong, and I would be quite happy to be wrong. – Jack D'Aurizio Jul 25 '14 at 19:42
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I see your point: it's likely to be simply "equivalently hopeless." – Semiclassical Jul 25 '14 at 21:27
There is probably a typo in : $\sin(1)=2\sum_{k=1}^\infty J_{2k+1}(1)$
because $\sin(1)=0.841471...$ and $2\sum_{k=1}^\infty J_{2k+1}(1)=0.0396292...$
An exact similar relationship is : $$\sin(1)=2\sum_{k=\infty0}^\infty (-1)^kJ_{2k+1}(1)$$ More general results can be found in : http://functions.wolfram.com/Bessel-TypeFunctions/BesselJ/23/01/
From J.Spanier, K.B.Oldham, "An atlas of functions", 1st Edit. 1987, page 518, Eqs. 52:10:4 and 52:10:6 $$2\sum_{k=0}^\infty J_{2k+1}(x)=\int_{0}^{x}J_0(t)dt=xJ_0(x)+\frac{\pi x}{2}(J_1(x)h_0(x)-J_0(x)h_1(x))$$ $h_0(x)$ and $h_1(x)$ are the Struve functions.

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