Sum of odd Bessel Functions 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? 
$$\sum_{k=0}^\infty (-1)^k J_{2k+1}(x)\ =\ ?$$
$$\sum_{k=0}^\infty  J_{2k+1}(x)\ =\ ?$$
 A: 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:

$$\sum_{m=0}^{+\infty}(-1)^m J_{2m+1}(z) = \frac{1}{2}\sin z.$$

A: 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.
