I need to prove $$ \operatorname{J}_{n\,}\left(x\right) = \frac{1}{2\pi}\int_{-\pi}^{\pi} {\rm e}^{-{\rm i}x\sin\left(\theta + in\theta\right)} \hspace{5mm}{\rm d}\theta $$ using something I already proved previously:
$$ {\rm e}^{x\left(t - 1/t\right)/2}\,\,\, = \sum_{n = -\infty}^{\infty} \operatorname{J}_{n}\left(x\right)\,t^{n} $$
I can see that substituting $$ t = {\rm e}^{{\rm i}\theta} $$ will get me $99\,\%$ of the way there, but I can't figure out how to get rid of the infinite sum to have the integral, nor why the sign of the second term in the exponential is positive.
Thanks !.
There was a sign error in the integral representaiton of the first-kind Bessel function of order $n$ as stated in the OP.
The correct representation of the Bessel function is
$$J_{n}(x)=\frac1{2\pi}\int_{-\pi}^\pi e^{i\color{red}{n}\theta-ix\sin(\theta)}\,d\theta$$
where the red-colored $n$ shows the location where the OP had $-n$ instead.
Now, letting $t=e^{-i\theta}$ in the expression for the generating function reveals
$$e^{x(t-1/t)/2}=e^{-ix\sin(\theta)}=\sum_{n=-\infty}^\infty J_n(x)e^{-in\theta}$$
Expoiting the orthogonality of $e^{in\theta}$ on $[-\pi,\pi]$, we find that
$$\begin{align} \frac1{2\pi}\int_{-\pi}^\pi e^{-ix\sin(\theta)}e^{in\theta}&=\frac1{2\pi}\int_{-\pi}^\pi \sum_{m=-\infty}^\infty J_{m}(x)e^{-im\theta}e^{in\theta}\,d\theta\\\\ &\frac1{2\pi}\left(2\pi \sum_{m=-\infty}^\infty J_{m}(x)\delta_{m,n}\right)\\\\ &=J_n(x) \end{align}$$
as expected!