Using binomial theorem in exponential

[yungman]

In page 11 of http://math.arizona.edu/~zakharov/BesselFunctions.pdf, I am trying to follow the derivation using binomial theorem to get this step:

(e^{j\theta}-e^{-j\theta})^{n+2k}≈\frac{(n+2K)!}{k!(n+k)!}(e^{j\theta})^{n+k}(-e^{-j\theta})^kIf you read the paragraph right above this equation, it said only the terms that has the $e^{jn\theta}$ alone will not be integrated to zero. So the equation becomes:(e^{j\theta}-e^{-j\theta})^{n+2k}≈\frac{(n+2K)!}{k!(n+k)!}(e^{j\theta})^{n+k}(-e^{-j\theta})^k
As it's the $(k+1)^{th}$ term.

Is it because when substitute into (53)
I_{k,n}=\frac{1}{2\pi}\int_{-\pi}^{\pi}(e^{j\theta}-e^{-j\theta})^{n+2k}e^{-jn\theta}d\theta
Only the term that has $e^{jn\theta}$ alone can cancel out the $e^{-jn\theta}$ term in (53). Any $e^{jn\theta}$ type exponential term will result in zero after integrate from $-\pi$ to $\pi$. Am I correct.

Thanks

 

[Infrared]

Yes, you are correct. The only way \int_{-\pi}^{\pi} e^{jb\theta} d\theta can be non-zero is if b=0 (and the integrand is constant).

 

[yungman]

Yes, you are correct. The only way \int_{-\pi}^{\pi} e^{jb\theta} d\theta can be non-zero is if b=0 (and the integrand is constant).

Thanks, you are of big help. I just cannot find any good books that explain Bessel Functions, all the ones I read seemed to assume you know a lot of the derivation and just gave the final equation without a good explanation. The three article including the one you gave are as good as any book in this subject. But the problem is I found inconsistency in their formulas that call into question( most likely are typos). This really hold up my study as I have to stop and really go through it again and again. I have to even post again in another thread just on the very page of this article later.

Thanks for all the detail help. The one you took the time to type for the $\cos(x\sin\theta)$ is the utmost helpful. I went through step by step and really learn a lot.

Thanks