Originally Posted by

**halbard** Start with $\displaystyle I_{n-1}(z)-I_{n+1}(z)=\frac1\pi\int_0^\pi\mathrm e^{z\cos\theta}[\cos(n-1)\theta-\cos(n+1)\theta]\mathrm d\theta$ $\displaystyle =\frac2\pi\int_0^\pi\mathrm e^{z\cos\theta}\sin n\theta\,\sin\theta\,\mathrm d\theta=-\frac2{\pi z}\int_0^\pi\mathrm \sin\,n\theta\,\mathrm d(e^{z\cos\theta})$.

Integrate the last term by parts, so that

$\displaystyle I_{n-1}(z)-I_{n+1}(z)=\left[-\frac2{\pi z}\sin n\theta\,e^{z\cos\theta}\right]_0^\pi+\frac{2n}{\pi z}\int_0^\pi e^{z\cos\theta}\cos n\theta\,\mathrm d\theta=\frac{2n}z I_n(z)$.

For the second one, start with $\displaystyle I'_{n+1}(z)=\frac1\pi\int_0^\pi e^{z\cos\theta}\cos(n+1)\theta\,\cos\theta\,\mathr m d\theta$.

Thus $\displaystyle I_n(z)-I'_{n+1}(z)=\frac1\pi\int_0^\pi e^{z\cos\theta}[\cos n\theta-\cos(n+1)\theta\,\cos\theta]\mathrm d\theta$ $\displaystyle =\frac1\pi\int_0^\pi e^{z\cos\theta}\sin(n+1)\theta\,\sin\theta\,\mathr m d\theta=\frac{n+1}z I_{n+1}(z)$ (similar to the last part of the previous derivation).

Warning: the last identity is not quite the one given.