What is the Recurrence Relation for Bessel's Functions?

[Altabeh]

Homework Statement

The integral is

$$\int _{-\pi }^{\pi }\!{\frac {{{\rm e}^{x\cos \left( \theta \right) }} \cos \left( n\theta \right) }{\pi }}{d\theta}.$$

The Attempt at a Solution

I've tried a couple of alternative methods huddling in my mind to solve this integral, but none of them worked. Actually using De Moivre's formula and integrationa by parts this can be written as

$${\frac {{{\rm e}^{x+in\pi }}-{{\rm e}^{-x-in\pi }}}{\pi }}+\int _{-\pi }^{\pi }\!{\frac {x\sin \left( \theta \right) {{\rm e}^{x\cos\left( \theta \right) +in\theta}}}{in\pi }}{d\theta}.$$

where we hit the second integral which, in its indefinite form, cannot be described by the elementary functions.

Any help will be appreciated!

AB

 

[gabbagabbahey]

Is it safe to assume that $n$ is an integer and $x$ has no $\theta$ dependence?

If so, just begin by defining $I_n(x)\equiv\frac{1}{\pi}\int_{-\pi}^{\pi}{\rm e}^{x\cos\theta}\cos\left(n\theta\right)}d\theta$, then use integration by parts once, along with the trig identity $\sin(a)\sin(b)=\frac{1}{2}\left[\cos(a-b)-\cos(a+b)\right]$ to derive a recurrence relation. The resulting infinite series is very well known

 

[Altabeh]

Is it safe to assume that $n$ is an integer and $x$ has no $\theta$ dependence?

Yes and do you have any idea then?

AB

 

[gabbagabbahey]

See my edited post above

 

[Altabeh]

See my edited post above

Would you mind elaborating the integration by parts part a little more?

Thanks for your time.

AB

 

[gabbagabbahey]

Well, I assume you know how to differentiate $\rm{e}^{x\cos\theta}$ and integrate $\cos(n\theta)d\theta$, so...

 

[Altabeh]

Well, I assume you know how to differentiate $\rm{e}^{x\cos\theta}$ and integrate $\cos(n\theta)d\theta$, so...

Okay, I get this recurrence equation in the end:

$$\frac{x}{2n}(I_{n-1}-I_{n+1})=I_n$$

and this is..?!

AB

 

[gabbagabbahey]

Compare that to the recurrence relation you get for Bessel's functions.