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 $\int e^{z\cos(x)} dx$)



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?
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). 


