Finding Fourier transform of $\frac{x}{(x^2 + 4)^2}$ So I have this function
$$
f(x) = \frac{x}{(x^2 + 4)^2}
$$
and I have to find its Fourier transform.
This is however much harder than what I have done before so I don't have a clue where to start. I have stared at it for 2 hours and come up with nothing.
I know that
$$
\mathfrak{F}(\xi) = \int_{-\infty}^{\infty}e^{-i \xi x} f(x) dx
$$
but this is as far as I get. Thankful for any help!
 A: We can use residue calculus here to find the integral value:
$$
\int_{-\infty}^{\infty}e^{-i \xi x}f(x)dx = \int_\gamma e^{-i \xi z} \cdot \frac{z}{(z^2 + 4)^2}dz
$$
which have double poles at $z = \pm 2i \Rightarrow z_1 = (z-2i)^{-2}, z_2 = (z + 2i)^{-2}$. 
We need to find the residues for 
$$
e^{-i \xi z}f(z) = (z - 2i)^{-2}ze^{-i \xi z}(z + 2i)^{-2} = (z - 2i)^{-2}g(z)
$$
where we let
$$
g(z) = e^{-i \xi z}(z+2i)^{-2}z
$$
differentiation yields
$$
g'(z) = e^{-i \xi z}((z+2i)^{-2} - i \xi z(z + 2i)^{-2} - 2z(z + 2i)^{-3})
$$
insertion of $z = 2i$ yields
$$
e^{2 \xi}((4i)^{-2} + 2 \xi(4i)^{-2} - 2\cdot 2i(4i)^{-3}) = -\frac{1}{8} \xi e^{2 \xi}
$$
thus
$$
Res_{z = 2i}(e^{-i \xi z}f(z)) = \frac{g'(2i)}{1!} = -\frac{1}{8} \xi e^{2 \xi}
$$
and equally
$$
Res_{z = -2i}(e^{-i \xi z}f(z)) = \frac{g'(2i)}{1!} = \frac{1}{8} \xi e^{-2 \xi}
$$
and finally the Fourier transform is given by
$$
\mathfrak{F}(\xi) = \left\{
  \begin{array}{l l}
    2 \pi i (- \frac{1}{8} \xi e^{2 \xi}) = - i \xi \frac{\pi}{4} e^{2 \xi}, & \xi \lt 0\\
    -2 \pi i (\frac{1}{8} \xi e^{-2 \xi}) = - i \xi \frac{\pi}{4} e^{-2 \xi}, & \xi \gt 0
  \end{array} \right.
$$
and to conclude
$$
\mathfrak{F}(\xi) = -i \xi \frac{\pi}{4} e^{-2 |\xi|}
$$
A: One can rather easily compute the integral by residues. For example, for $\xi\geq0$ we can write
\begin{align*}
\mathfrak{F}(\xi)=-2\pi i\cdot \operatorname{res}_{z=-2i}\frac{ze^{-i\xi z}}{(z-2i)^2(z+2i)^2}
=-2\pi i \cdot\left(\frac{ze^{-i\xi z}}{(z-2i)^2}\right)'_{z=-2i}=-\frac{\pi i}{4}\xi\, e^{-2\xi}.
\end{align*}
But since the answer should be an odd function of $\xi$ (why?), the general result for $\xi\in\mathbb{R}$ is given by
$$\boxed{\displaystyle\mathfrak{F}(\xi)=-\frac{\pi i}{4}\xi\, e^{-2|\xi|}}$$
A: The function ${\displaystyle {x \over (x^2 + 4)^2}}$ is ${\displaystyle -{1 \over 2}}$ times the derivative of ${\displaystyle {1 \over x^2 + 4}}$, so its Fourier transform will be ${\displaystyle -{i \xi \over 2}}$ times the fourier transform of ${\displaystyle {1 \over x^2 + 4}}$. This one is standard and is given by ${\displaystyle {\pi \over 2} e^{-2|\xi|}}$. (To verify this, you can just compute the inverse Fourier transform directly using basic calculus).
Thus the desired Fourier transform is ${\displaystyle -{i \xi \over 2}}$ times ${\displaystyle {\pi \over 2} e^{-2|\xi|} = {-{i \xi\pi \over 4} e^{-2|\xi|}}}$. 
