I've tried to use an integral representation of the Bessel Function of the First Kind $J_n(x)$ to derive a series representation of the function. My end result is pretty close to the answer that it should be, but it still doesn't quite match up and I'd be grateful for anybody that could point out where my work is faulty.
\begin{align} J_n(x) &= \frac{1}{2\pi}\int_0^{2\pi}\cos(n\phi-x\sin\phi)d\phi\\ &= \frac{1}{2\pi}\int_0^{2\pi}e^{in\phi}e^{-ix\sin\phi}d\phi \tag1\\ &=\frac{1}{2\pi}\int_0^{2\pi}\left(e^{i\phi}\right)^ne^{-\frac{x}{2}\left(e^{i\phi}-e^{-i\phi}\right)}d\phi \tag2 \end{align}
In $(1)$, I used the fact that the integral of the imaginary part of the integrand over a complete period is equal to $0$.
A parametrisation of the unit circle in the complex plane is given by $z=e^{i\phi}, \phi\in(0,2\pi)$, thus $e^{-i\phi}=\left(e^{i\phi}\right)^{-1}=\frac{1}{z}$.
Letting $\Gamma$ be the unit circle in the complex plane traversed in the counterclockwise direction, $(2)$ becomes
\begin{align} J_n(x) &= \frac{1}{2\pi}\oint_\Gamma z^n e^{-\frac{x}{2}z}e^{\frac{x}{2z}}dz\\ &= \frac{1}{2\pi}\oint_\Gamma z^n e^{-\frac{x}{2}z}\sum_{k=0}^{\infty}\frac{x^k}{2^kk!z^k}dz \tag3\\ &=\frac{1}{2\pi}\sum_{k=0}^{\infty}\frac{x^k}{2^kk!}\oint_\Gamma z^{n-k}e^{-\frac{x}{2}z}dz\\ &=i\sum_{k=0}^{\infty}\frac{x^k}{2^kk!}\text{Res}\left(z^{n-k}e^{-\frac{x}{2}z},0\right) \tag4 \end{align}
$(3)$ uses the Laurent series expansion of $e^{\frac{1}{z}}$ and $(4)$ uses the Residue Theorem.
Let $m:=k-n\rightarrow k=m+n$. Then the Residue of $z^{-m}e^{-\frac{x}{2}z}$ is only non-zero for strictly positive values of $m$ and $(4)$ becomes
\begin{align} J_n(x) &= i\sum_{m=1}^{\infty}\frac{x^{m+n}}{2^{m+n}(m+n)!}\text{Res}\left(z^{-m}e^{-\frac{x}{2}z},0\right)\\ &= i\sum_{m=1}^{\infty}\frac{x^{m+n}}{2^{m+n}(m+n)!}\frac{1}{(m-1)!}\lim\limits_{z\rightarrow 0}\frac{d^{m-1}}{dz^{m-1}}e^{-\frac{x}{2}z}\\ &=i\sum_{m=1}^{\infty}\frac{x^{m+n}}{2^{m+n}(m+n)!}\frac{(-1)^{m-1}}{(m-1)!}\frac{x^{m-1}}{2^{m-1}} \tag5 \end{align}
Letting $r=m-1\rightarrow m=r+1$, $(5)$ becomes
\begin{align} J_n(x)=i\sum_{r=0}^{\infty}(-1)^r\frac{x^{2r+n+1}}{2^{2r+n+1}r!(r+n+1)!}\tag6 \end{align}
However, comparing this with the known series representation, the $i$ shouldn't be there (obviously, since $J_n(x)$ is a real function) and the sum (disregarding the $i$ for a moment) is actually a series representation of $J_{n+1}(x)$, not $J_n(x)$.
I've been over this calculation a few times, and I simply cannot find where I made a mistake that either allowed the $i$ to creep in there or that raised the order of the Bessel Function that I am trying to find a series representation for.
Any ideas?