Originally Posted by

**Alan0354** I worked on a few more steps:

$\displaystyle J_m(z)=\frac{1}{2\pi}\int_0^{2\pi}e^{j(z\sin\theta-m\theta)} d\theta=\frac{1}{2\pi}\int_0^{2\pi}e^{j(z\cos u-m\frac{\pi}{2}+mu)} du$

$\displaystyle =\frac{1}{2\pi}\int_0^{2\pi}e^{j(z\cos u+mu)}e^{-jm\frac{\pi}{2}} \;du=-\frac{j\sin(m\frac{\pi}{2})}{2\pi}\int_0^{2\pi}e^{ j(z\cos u+mu)}du$

With this, $\displaystyle J_m(z)=0$ for m=even. That is not going to get the answer. Also, how do you change $\displaystyle \;\int_0^{2\pi}e^{j(z\cos u+mu)}du\;$ to $\displaystyle \;2 \int_0^{\pi}e^{j(z\sin\theta-m\theta)}d\theta$?