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I'm reading a book to review Fourier transforms, and I came across the following example, which is here: http://books.google.com/books?id=JokQD5nK4LMC&pg=PR36&lpg=PR36#v=onepage&q&f=false

I'm following the argument up until the author writes let $z=Re^{i\theta}$, and then he has the equality $\frac{1}{\sqrt{2\pi}}\int_{C_R}\frac{e^{i\omega z}}{(z^2+a^2)}dz=\frac{1}{2\pi}\int_{C_R}\frac{e^{i\omega R(\cos \theta+i\sin \theta)}iRe^{i\theta}}{(R^2e^{2i\theta}+a^2)}e^{-\omega R\sin \theta}d\theta.$ I'm a bit confused as to where the extra $e^{-\omega R\sin \theta}$ comes from in the RHS above. Would anybody care to explain? Sorry this may seem like a dumb question.

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Looks like an editing error in the book. Eventually, he'll want to use: $i\omega R(\cos \theta+i\sin \theta) =i\omega R\cos \theta -\omega R \sin \theta$

Alternatively, the author meant:

$\int_{C_R}\frac{e^{i\omega R\cos \theta}iRe^{i\theta}}{(R^2e^{2i\theta}+a^2)}e^{-\omega R\sin \theta}d\theta$

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    Oh, I must need more sleep-- sorry $f$or the stupid question, and thanks a lot!!2012-03-15