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Physgeek64
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Homework Statement
calculate ## \int_{-\infty}^{\infty}{\frac{2}{1+x^2} e^{-ikx}dx}##, where k is any positive number
Homework Equations
The Attempt at a Solution
So first consider the closed contour ##I= \int{\frac{2}{1+z^2} e^{-ikz}dx}##
We can choose the contour to be along the real axis ## [-R,R]## then along the semi circle for which ##z=Re^{i \theta}, [0, \pi] ##
then ##I= \int_{-\infty}^{\infty}{\frac{2}{1+x^2} e^{-ikx}dx} +\int_{0}^{\pi} {\frac{2}{1+(Re^{i \theta})^2} e^{-ikRe^{i\theta}}dx}##
I calculated the residue to be ## \pi e^{k} ##, with the poles being at ## z=i## and ##z=-i##
I think the integral around the semi-circle is supposed to go to zero in the limit of R going to infinity, but I am struggling to see why. I can't use Jordan's lemma (?) since it is not in the usual formal, and if i expand
##e^{-ikRe^{i\theta}} = e^{-ikR(cos(\theta)+isin(\theta))}##
##= e^{-ikRcos(\theta)}e^{kRsin(\theta)}##
but since ##\theta ## is between ##[0, \pi], sin(\theta)## is greater than, or equal to zero, meaning that the integrand does not drop off as R goes to infinity..
Any help would be greatly appreciated- thanks in advance :)
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