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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?

2 Answers 2

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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.

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    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.2010-11-18
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    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 $n$.2010-11-18
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    @jake: Right. Thanks for letting me know. I have edited the answer accordingly.2010-11-18
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Use $\displaystyle \int_{a}^{b} f(x)\ dx = \int_{a}^{b} f(a+b-x) \ dx$