How can I calculate the Fourier Transform of $f(x) =\frac{ax^2}{x^2 +a^2}$

Question

We have been taught in University, the Fourier Transform of a function $f(x)$ is $$\mathcal F{f(x)}=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty{e}^{-2{i}kx}f(x)\,\mathrm{d}x,$$ but I'm having difficulty in calculating this, because of the $x^2$ in numerator, so kindly help me in this, if you do know how to do it...

Answer 1

First about Fourier transform conventions. The usual $(a, b)$ parameter conventions for

$ \sqrt{\frac{|b|}{(2\pi)^(1-a)}} \int_{-\infty}^{\infty} f(t) e^{i b \omega t} dt $

are:

• either $(0,1)$ (default)
• or $(1, -1)$
• or $(-1, 1)$
• or $(0, -2 \pi)$ (signal processing convention)

I suspect you forgot your exponential exponent is $k$ not $-2k$.
I'll take the default convention, as it is closest to the question. Just change $k$ into $-2k$ to have a direct answer to the question in the following results.

You just have to follow Thomas Andrews's comment, which simplifies the question, as the Fourier transform of a constant is a known distribution, a multiple of the delta distribution. Now for a a real, you get the following transform for $(x - i\, a)^{-1}$ at s:

$i \, e^{-a s} \sqrt{2 \pi} \, \theta(s \, \text{sign}(a))\, \text{sign(a)}$

where $\theta$ is the Heaviside step function. This directly results from the transform of $\frac{1}{x}$, with a prior translation. The transformation for $\frac{1}{x}$ can be found in tables everywhere (check in the Wikipedia article) and translation formulas are standard.
Now separate the quadratic denominator into simple poles of the form $(x \pm i a)$ and decompose the Fourier transform of the ratio into the sum of simple transforms of $(x \pm i a)^{-1}$. This yields:

$\mathscr{F}(\frac{1}{x^2 + a^2}, s) = \frac{\sqrt{\frac{\pi }{2}} e^{-a\, s} \text{sgn}(a) \left(e^{2 a s} \, \theta (-s \, \, \text{sgn}(a))\, + \, \theta (s \, \, \text{sgn}(a))\right)}{a}$

Now the end of the reasoning is obvious : multiply by $-a^2$, add the delta distribution, multiply the whole result by $a$, which results in (general form for a complex $a$, I leave it to you to simplify for reals):

$a \left(\sqrt{2 \pi } \delta (s)-\sqrt{\frac{\pi }{2}} a e^{-a s}\,\, \text{sgn}(\Re(a)) \left(e^{2 a s} \theta (-s \,\,\text{sgn}(\Re(a)))+\theta (s \,\,\text{sgn}(\Re(a)))\right)\right)$

In particular for $a > 0, \,s > 0$:

$\sqrt{2 \pi } a \delta (s)-\sqrt{\frac{\pi }{2}} a^2 e^{-a s}$