closed form of $\sum_{n=1}^\infty \frac{J_{2n-1}(z)}{2n-1}$

Question

Does the following infinite series have a closed form: \begin{equation} \sum_{n=1}^\infty \frac{J_{2n-1}(z)}{2n-1}? \end{equation} Here, $J_n$ is the Bessel function. (If the denominator does not depend on $n$, then there's a nice expression given here.)

As an aside: I'm interested in this problem because I encountered the following integral: \begin{equation} \int_0^\pi \cos(\cos(x))\sin(z\sin(x)) dx, \end{equation} which, by using Jacobi-Anger expansion, can be expressed as \begin{equation} 2J_0(1)\sum_{n=1}^\infty J_{2n-1}(z) \int_0^\pi \sin((2n-1)x) dx \propto \sum_{n=1}^\infty \frac{J_{2n-1}(z)}{2n-1}. \end{equation}

Answer 1

Using the generating function for those functions: $$\sum_{n\in \mathbb Z}J_n(x)t^n = e^{\frac x 2 \left( t - \frac 1 t\right) }$$ $$e^{\frac x 2 \left( t - \frac 1 t\right)}-e^{-\frac x 2 \left( t - \frac 1 t\right)}=\sum_{n\in \mathbb Z}J_n(x)t^n - \sum_{n\in \mathbb Z}J_n(x)(-t)^n= 2\sum_{n\in \mathbb Z}J_{2n-1}(x)t^{2n-1}$$ Furthermore, $J_{-(2n+1)}(x)=-J_{2n+1}(x)$, thus $$e^{\frac x 2 \left( t - \frac 1 t\right)}-e^{-\frac x 2 \left( t - \frac 1 t\right)}= 2\sum_{n\geq 1}J_{2n-1}(x)\left(t^{2n-1}-\frac 1 {t^{2n-1}}\right)$$ Plugging $t=e^{i\theta}$: $$e^{\frac x 2 \left( e^{i\theta} - e^{-i\theta}\right)}-e^{-\frac x 2 \left( e^{i\theta} - e^{-i\theta}\right)}= 2\sum_{n\geq 1}J_{2n-1}(x)\left(e^{(2n-1)i\theta}-e^{-(2n-1)i\theta}\right)$$ $$e^{ix\sin\theta }-e^{-ix\sin\theta}= 4i\sum_{n\geq 1}J_{2n-1}(x)\sin\left((2n-1)\theta\right)$$ $$\sin(x\sin \theta) = 2\sum_{n\geq 1}J_{2n-1}(x)\sin\left((2n-1)\theta\right)$$ Integrating w.r.t. $\theta$: $$\int_0^\theta \sin(x\sin u)du = -2\sum_{n\geq 1}\frac{J_{2n-1}(x)}{2n-1}\left[\cos\left((2n-1)\theta\right)-1\right]$$ Evaluating at $\theta=\frac \pi 2$: $$\int_0^{\frac \pi 2} \sin(x\sin u)du = 2\sum_{n\geq 1}\frac{J_{2n-1}(x)}{2n-1}$$ The left-hand-side is given by the Struve function: $$\sum_{n\geq 1}\frac{J_{2n-1}(z)}{2n-1} = \frac \pi 4 H_0(z)$$