Finding Fourier transform of $\frac{x}{(x^2 + 4)^2}$

Question

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!

Answer 1

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 $$ \mathfrak{F}(\xi) = -i \xi \frac{\pi}{4} e^{-2 |\xi|} $$

Answer 2

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 $$\boxed{\displaystyle\mathfrak{F}(\xi)=-\frac{\pi i}{4}\xi\, e^{-2|\xi|}}$$

Answer 3

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 ${\displaystyle {\pi \over 2} e^{-2|\xi|} = {-{i \xi\pi \over 4} e^{-2|\xi|}}}$.