i have some problem with the integral representation of first kind bessel function $J_n(x)$
i have to proof this formula:
$$J{_n}(x)=\frac{i^{-n}}{2\pi}\int_{-\pi}^{\pi}e^{i(n\phi-x\cos\phi)}d\phi\tag{1}$$
To proof that equation, i have considered the generating function for first kind Bessel function $J_n$ with $n\in \mathbb{Z}$:
\begin{equation} e^{\frac{1}{2}x(t-\frac{1}{t})}=\sum_{k=-\infty}^\infty J_k(x) t^k \end{equation} Assuming $t=e^{-i\varphi}$ and multiplying both members of latter equation for $e^{in\varphi}$, we have: \begin{equation} e^{in\varphi}e^{\frac{1}{2}x\big(e^{-i\varphi}-\frac{1}{e^{-i\varphi}}\big)}=e^{in\varphi}e^{-ix\big(\frac{e^{i\varphi}-e^{-i\varphi}}{2i}\big)}=\sum_{k=-\infty}^\infty J_k(x)e^{-ik\varphi}e^{in\varphi} \end{equation} \begin{equation} e^{in\varphi}e^{-ix\sin\varphi}=\sum_{k=-\infty}^\infty J_k(x)e^{i(n-k)\varphi} \end{equation} integrating from $-\pi$ to $\pi$, follows: \begin{equation} \int_{-\pi}^\pi e^{in\varphi}e^{-ix\sin\varphi} d\varphi=\int_{-\pi}^\pi \sum_{k=-\infty}^\infty J_k(x)e^{i(n-k)\varphi} d\varphi=\sum_{k=-\infty}^\infty J_k(x)\int_{-\pi}^\pi e^{i(n-k)\varphi} d\varphi \end{equation} Since it results: \begin{equation}\tag{*} \int_{-\pi}^{\pi} e^{i (n-k)\theta}d\theta = \begin{cases} 2\pi&\text{per }n=k\\ 0&\text{per }n\neq k \end{cases} \end{equation} From previous exprension: \begin{equation} J_n(x)=\frac{1}{2\pi}\int_{-\pi}^\pi e^{i(n\varphi-x\sin\varphi)}d\varphi \end{equation}
Now we can substitutes $\varphi=\phi+\pi/2$, so the previous expression becames:
\begin{equation} J_n(x)=\frac{e^{in\frac{\pi}{2}}}{2\pi}\int_{-\pi}^{\pi} e^{i(n\phi-x\cos\phi)}d\phi=\frac{i^{n}}{2\pi}\int_{-\pi}^{\pi} e^{i(n\phi-x\cos\phi)}d\phi \tag{2} \end{equation}
But now i can proced to obtain the (1)? Is it wrong since there's $i^{-n}$ and instead of $i^{n}$? becouse i have seen the Jacobi-Anger expansion, and using it, i obtain the latter relation (2)