Is $\frac{\pi}{4}L_0(z) = \sum\limits_{n=1}^{+\infty} (-1)^{n+1} \frac{I_{2n-1}(z)}{2n-1}$ between Bessel and Struve known?

Question

Based of the detailed attempt to solve the integral $\int e^{\sin(x)} dx$ I stumbled upon a connection between modified Struve and modified Bessel function of the first kind. But, I cannot find a confirmation in the literature, and the connection looks too nice to be easily missed.

The integral can be shifted to $e^{\cos(u)}$

$\displaystyle \int e^{\sin(x)} dx, u=x-\frac{\pi}{2}$

This is making

$\displaystyle \int e^{\sin(x)} dx = \int e^{\sin(u+\frac{\pi}{2})} du = \int e^{\cos(u)} du$

Use Fourier transform

$\displaystyle e^{\cos(u)}=\frac{1}{2}a_0 +\sum_{n=1}^{+\infty} a_n \cos \left ( nu \right )+\sum_{n=1}^{+\infty} b_n \sin \left ( nu \right )$

$\displaystyle a_n=\frac{1}{\pi}\int_{-\pi}^{\pi} e^{\cos(t)}\cos(nt) dt$

$\displaystyle b_n=\frac{1}{\pi}\int_{-\pi}^{\pi} e^{\cos(t)}\sin(nt) dt$

All $b_n=0$

$a_n$ is a multiple of modified Bessel function of the first kind

$\displaystyle I_n(z)=\frac{1}{\pi} \int_{0}^{\pi} e^{z\cos(t)} \cos(nt) dt$

So $a_n = 2I_n(1)$

making

$\displaystyle e^{\cos(u)}= I_0(1) + 2\sum_{n=1}^{+\infty} I_n(1) \cos \left ( nu \right )$

or

$\displaystyle \int e^{\cos(u)} du=I_0(1)u + 2\sum_{n=1}^{+\infty} \frac{I_n(1)}{n} \sin \left ( nu \right )$

Knowing that boundaries have to be shifted $(\int_{\frac{\pi}{2}}^{x+\frac{\pi}{2}} \to \int_{0}^{x})$ we need first

$\displaystyle \int_{0}^{\frac{\pi}{2}} e^{\sin(x)} dx=\frac{\pi}{2}(L_0(1)+I_0(1))$

where $L_n(z)$ is modified Struve function.

Finally:

$\displaystyle \int_{0}^{x} e^{\sin(t)} dt=I_0(1)x + \frac{\pi}{2}L_0(1) + 2\sum_{n=1}^{+\infty} \frac{I_n(1)}{n} \sin \left ( nx - \frac{n\pi}{2} \right ) $

So generally we can write:

$\displaystyle \int e^{\sin(x)} dx=I_0(1)x + 2\sum_{n=1}^{+\infty} \frac{I_n(1)}{n} \sin \left ( nx - \frac{n\pi}{2} \right )+c $

Plainly extending beyond just $z=1$ for $x=0$ the definite integral gives

$\displaystyle \frac{\pi}{4}L_0(z) = \sum_{n=0}^{+\infty} (-1)^{n} \frac{I_{2n+1}(z)}{2n+1}$

But, there is no mentioning this anywhere I looked. And it is too nice to be missed. (Notice that the extension is related to the integral $\int e^{z\cos(x)} dx$)

Answer 1

This relation is a special case of a more general one: $$L_\nu(z)=\frac{4}{\sqrt{\pi}\,\Gamma\left(\nu+\frac{1}{2}\right)}\sum_{n=0}^\infty\frac{(-1)^n\,(2n+\nu+1)\,\Gamma(n+\nu+1)}{n!\,(2n+1)(2n+2\nu+1)}\,I_{2n+\nu+1}(z),$$ which can be found in https://arxiv.org/abs/1301.5432 (Integral representations and summations of modified Struve function, byÁrpád Baricz, Tibor K. Pogány, formula (2.1) on page 4).