# What is the closed-form of $\int_{0}^{\pi}\text{sin}(n\tau-x\text{sin}\tau)d\tau$

I know the Bessel function can be formatted as follows. $$J_n\left(x\right)=\frac{1}{\pi}\int_{0}^{\pi}\text{cos}\left(n\tau-x\text{sin}\tau\right)d\tau$$ So can we get a Bessel-similar expression for the integration in the title? Many thanks!

• Have you looked on Wikipedia? – WA Don May 21 '20 at 13:39
• @WADon I have checked this Wiki page, but I think I didn't find direct answer to this question. I have also tried some manipulations, but all failed. I want to express this integration in the Bessel function form, so I tried different approaches to convert it to Bessel form. – Jason Zhou May 21 '20 at 13:52
• You are right - there is a similar integral similar for $Y_n(t)$ but not quite what you needed. Sorry for sending you on a needless search. – WA Don May 21 '20 at 14:15
• @WADon That's all right, thanks for spending your time! – Jason Zhou May 21 '20 at 14:22
• I am pretty sure you will get a solution, which is a sum of Struve functions $H_\alpha(x)$, since for $n=0$ your integral leads to $-\pi\,H_0(x)$, assuming $x\in\mathbb{R}$. For $n=1$ we get $-\pi\,H_{-1}(x)$. For higher $n$ we get more and more involved results. I guess that using $\sin(n \tau - x \sin(\tau))=\cos(x\, \sin(\tau)) \sin(n \tau) - \cos(n \tau) \sin(x \sin(\tau))$ and the general expansion formulas for $\sin(n \tau)$ and $\cos(n \tau)$ can do the trick. You will end up with a sum of Struve functions. Sorry I can not be of more help. – Michael_K May 21 '20 at 19:57

## 1 Answer

$$\int_0^\pi\sin(n\tau-x\sin\tau)~d\tau=\pi\mathbf{E}_n(x)$$