# Fourier transform on $1/(x^2+a^2)$

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: $$i\omega R(\cos \theta+i\sin \theta) =i\omega R\cos \theta -\omega R \sin \theta$$
$$\int_{C_R}\frac{e^{i\omega R\cos \theta}iRe^{i\theta}}{(R^2e^{2i\theta}+a^2)}e^{-\omega R\sin \theta}d\theta$$
Hmm, if that was the case, I don't really see how he gets the bound $\frac{e^{i\omega R\cos \theta}iRe^{i\theta}}{(R^2e^{2i\theta}+a^2)}\leq \frac{R}{|R^2-a^2|}$ on the next line... – Jay C Mar 15 '12 at 20:18
Obviously, you need absolute values on the left side, too. The numerator on the left side has absolute value $R$, and th denominator on the left side has a minimum absolute value of $|R^2-a^2|$ – Thomas Andrews Mar 15 '12 at 20:21