Trying to compute the following expression by hand and mathematica, namely $\int_{\mathbb{R}}dx \frac{x^2}{x^2+n^2}\frac{e^{ixk}}{x^2+a^2}$, where $a^2\in \mathbb{R}$ and $n\in\mathbb{Z}-\{0\}$ gives me different results.
Clearly, it has four purely imaginary poles and I would construct $2$ half-circles eincirling either the $2$ with positive imaginary part (for $k>0$) or the $2$ with negative imaginary part (for $k<0$). Interestingly, mathematica gives different results if one feeds it with the above expression (FourierTransform vs. Integrate). How is this possible?
Edit: The calculation per hand gives: $I=\frac{-\pi|n|e^{-|n||k|}}{-|n|^2+a^2}+\frac{\pi|a|e^{-|a||k|}}{-|n|^2+a^2}$
It's probably easier to just use a little partial fractions and write $${x^2 \over (x^2 + a^2)(x^2 + n^2)} = -{a^2 \over n^2 - a^2}{1 \over x^2 + a^2} + {n^2 \over n^2 - a^2}{1 \over x^2 + n^2}$$ So you need to find $$-{a^2 \over n^2 - a^2}\int_{\mathbb R} {e^{ikx} \over x^2 + a^2} + {n^2 \over n^2 - a^2}\int_{\mathbb R} {e^{ikx} \over x^2 + n^2}$$ A simple application of residues in the upper half plane then gives for $k \geq 0$ that this is equal to $$-{a^2 \over n^2 - a^2}{\pi e^{-|a|k} \over |a|} + {n^2 \over n^2 - a^2}{\pi e^{-|n|k} \over |n|}$$ $$=-{\pi|a|\over n^2 - a^2} e^{-|a|k} + {\pi |n| \over n^2 - a^2}e^{-|n|k}$$ If $k < 0$ we get the same result as for $|k| > 0$ since it just changes the imaginary part of the integral which is $0$. So for any $k$ the integral is $$=-{\pi|a|\over n^2 - a^2} e^{-|ak|} + {\pi |n| \over n^2 - a^2}e^{-|nk|}$$ This agrees with what you got.