# Deriving Bessel's Integrals

I'm trying to derive the integral form of the Bessel function by finding the $k$th coefficient of the Laurent series expansion of the function $f(z) =\exp [\lambda(z-\frac{1}{z})]$. I managed to get it down to the form

$J_k(\lambda) =\frac{1}{2\pi}\int_{0}^{2 \pi} e^{i(\lambda \sin\theta - k \theta)} d\theta =\frac{1}{2\pi}\int_{0}^{2 \pi} [\cos(\lambda \sin\theta - k \theta) +i\sin(\lambda \sin\theta - k \theta)]d\theta$

But, I need to show that this is equivalent to $\frac{1}{2\pi}\int_{0}^{2 \pi} \cos(\lambda \sin\theta - k \theta)d\theta$. In other words, I need to show that $\int_{0}^{2 \pi}\sin(\lambda \sin\theta - k \theta)d\theta =0$

But I can't figure out how to do this. I tried expanding using trig identities and then writing sin and cos as Taylor Series and integrating term by term, but no luck. What am I missing?

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Split the integral involving $\sin$ as $\int_0^{\pi}$ + $\int_{\pi}^{2\pi}$. Change variable in the second integral as $\phi = \theta - \pi$. Do some trigonometric manipulations like $\sin(\theta + n \pi) = (-1)^n \sin(\theta)$ and you will find that the second integral turns out to be negative of the first.
Another way to think of the same thing is that the function is antisymmetric about $\pi$ i.e. in the integral if you replace $\alpha = \theta - \pi$, you will find that the resulting integral has an odd integrand and the limits of the integral go from $-\pi$ to $\pi$. So an odd function integrated from $-a$ to $a$ is zero. – user17762 Nov 18 '10 at 5:05
Thanks. I found I didn't have to split the integral, and like you said, I got an odd function integrated from $-\pi$ to $\pi$. However, I'm pretty sure your sample manipulation of $\sin(\theta + n \pi) = \sin \theta$ is only true for even $n$. – jake Nov 18 '10 at 17:28
Use $\displaystyle \int_{a}^{b} f(x)\ dx = \int_{a}^{b} f(a+b-x) \ dx$