I have been struggling calculating the Fourier transform of $f(x)=\frac{x}{(x^2+1)^2}$.
I tried to calculate $f(t)=\int\frac{x}{(x^2+1)^2}e^{-ixt}\,dx$ directly by integration by parts, but it is not working. I have been thinking about using residue theorem, sine it has a singularity at $x=i$, but then I got stuck.
$2\pi i Res(\frac{x}{(x^2+1)^2}e^{-ixt})=\int_{C_M}\frac{x}{(x^2+1)^2}e^{-ixt}\,dx=\int_{real}\frac{x}{(x^2+1)^2}e^{-ixt}\,dx+\int_{complex}\frac{z}{(z^2+1)^2}e^{-izt}\,dz$. $\int_{complex}\frac{z}{(z^2+1)^2}e^{-izt}\,dz$ goes to zero.
Does anyone know how to do this? Please help me. Thanks a lot.:)
The residue theorem works here. Note that we have poles at $z=\pm i$.
Consider the case $k \gt 0$; then we use a contour in the upper half plane, and the integral is $i 2 \pi$ times the residue at the pole $z=i$. Since we have a double pole here, the residue takes the value
$$i 2 \pi \left [\frac{d}{dz} \frac{z \, e^{i k z}}{(z+i)^2} \right ]_{z=i} = i \frac{\pi}{2} k \, e^{-k}$$
For the case $k \lt 0$, however, we must use a contour in the lower half plane (why? Think about Jordan's lemma.) In this case, then, the integral is
$$-i 2 \pi \left [\frac{d}{dz} \frac{z \, e^{i k z}}{(z-i)^2} \right ]_{z=-i} = i \frac{\pi}{2} k \, e^{k}$$
Thus,
$$\int_{-\infty}^{\infty} dx \frac{x}{(1+x^2)^2} e^{i k x} = i \frac{\pi}{2} k \, e^{-|k|} $$