Alternative Derivation of Recurrence Relation for Bessel Functions of the First Kind

Question

How can the recurrence relation

$J^{'}_n(x) = \frac{1}{2} [J_{n-1}(x) - J_{n+1}(x)]$

be derived directly from the following?

$J_n(x) = \frac{1}{\pi} \int_0^\pi \cos(n\theta - x \sin\theta) \text{d} \theta$

Answer 1

The easiest way to get the recurrence relation is to use the generating function for Bessel functions of the first kind:

$$\displaystyle \exp\left(\displaystyle z\displaystyle\frac{t-1}{2t}\right)=\displaystyle\sum\limits_{n=-\infty}^{\infty}t^nJ_n(z)$$

See, for example: http://mathworld.wolfram.com/BesselFunctionoftheFirstKind.html

If you really want to derive the relation directly from an integral definition, however, it might be easier to use the contour integral for the Bessel function of the first kind:

$$J_n(z)=\frac{1}{2\pi i} \oint \exp\left(\displaystyle z\displaystyle\frac{t-1}{2t}\right) t^{-n-1} \text{d}t$$

The relationship between this definition and the generating function is much clearer (use Cauchy's integral formula). Namely, we can derive the generating function from it as follows:

First, use the following version of Cauchy's integral formula:

$$\displaystyle \frac{f^{(n)}(0)}{n!} = \frac{1}{2\pi i} \oint f(t) t^{-n-1}$$

If we define $f(t)=\exp\left(z \frac{t-1}{2t}\right)$, then it follows immediately that:

$$J_n(z) =\displaystyle \frac{f^{(n)}(0)}{n!}$$

and then the generating function for the Bessel functions turns out to just be the Laurent series of $\exp \left( z \frac{t-1}{2t} \right)$ around the origin.

Answer 2

Hint: $J'_n(x)={1\over\pi}\int_0^{\pi}\sin(\theta)\sin(n\theta-x\sin\theta)d\theta$

Hint: use $\sin x\sin y ={1\over 2}(\cos(x-y)-\cos(x+y))$

$={1\over \pi}\int_0^{\pi}{1\over 2}(\cos(-\theta+n\theta-x\sin\theta)-\cos(\theta+n\theta-x\sin\theta))d\theta$