= Solution
Consider the counterclockwise <unit circle> and set
$$
I=\frac1{2\pi i}\int_{|z|=1}z^{n-1}e^{-xz/2}e^{x/(2z)}\,dz.
$$
The parametrization $z=e^{i\theta}$ gives
$$
I=\frac1{2\pi}\int_0^{2\pi}e^{i(n\theta-x\sin\theta)}\,d\theta.
$$
The <imaginary part> is zero: replacing $\theta$ by $2\pi-\theta$ conjugates the integrand, since $n$ is an integer and $x$ is real. The <real part> is precisely the defining Bessel integral, so $I=J_n(x)$.
Both exponential series converge absolutely and uniformly on the <circle>. Their product may therefore be integrated term by term:
$$
z^{n-1}e^{-xz/2}e^{x/(2z)}
=\sum_{j,k\geq0}\frac{(-1)^j(x/2)^{j+k}}{j!k!}z^{n-1+j-k}.
$$
Uniform <absolute convergence> can also be checked by summing the absolute coefficient bounds, which gives $e^{|x|}$. The <residue> is obtained exactly when $n-1+j-k=-1$, or $k=n+j$. The <Cauchy coefficient formula> consequently yields
$$
\boxed{J_n(x)=\sum_{j=0}^{\infty}\frac{(-1)^j}{j!(n+j)!}\left(\frac x2\right)^{n+2j},\qquad n\geq0.}
$$
This proves the <Bessel function of the first kind> series directly from its integral, including the sign and <factorial> normalization.
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